## Page 1

1Unit I

1 MATRICES

Unit Structure

1.0 Objectives

1.1 Introduction

1.2 Types of Matrices

1.3 Operations and Properties of Matrices

1.4 Elementary Transformation

1.5 Inverse of Matrix

1.5.1 Inverse of matrix by Elementary Row Operations

1.5.2 Inverse of matrix by using formula

1.6 Rank of Matrix

1.6.1 Echelon or Normal Matrix

1.7 Linear equations

1.8 Linear dependence and linear independence of vectors

1.9 Linear transformation

1.9.1 Matrix representation of Linear Transformation

1.10 Characteristics roots and characteristics vectors

1.11 Properties of characteristic vectors

1.12 Caley Hamilton Theorem

1.13 Similarity of matrices

1.14 Reduction of matrix to a diagonal matrix which has elements as characteristics values

1.15 Summary

1.16 References

1.17 Exercise

munotes.in

## Page 2

2 APPLIED MATHEMATICS

1.0 Objectives

After going through this chapter , students will able to learn

• Concept of adjoint of a matrix.

• Perform the matrix operations of addition, multiplication and express a

system of simultaneous linear equations in matrix form.

• Determine whether or not a given matrix is invertible and if it is, find its

inverse

• Rank of a matrix and methods finding these

• Solve a system of linear equations by row -reducing its augmented form

• Characteristics roots and characteristics vectors

• Reduction of matrix to a diagonal matrix

1.1 Introduction

A matrix is a rectangular arrangement of numbers into rows and columns .

Matrices provide a method of organizing, storing, and working with mathemati cal

information. We shall mostly be concerned with matrices having real numbers as

entries. The horizontal arrays of a matrix are called its rows and the vertical arrays

are called its columns. A matrix having m rows and n columns is said to have the

order m × n.

The numbers in a matrix can represent data, and they can also represent mathematical equations. Matrices have an abundance of applications and use in the

real world. Matrices have wide applications in engineering, physics, economics,

and statistics as well as in various branches of mathematics. In computer science,

matrix mathematics lies behind animation of images in movies and video games.

Matrices provide a useful t ool for working with models based on systems of linear

equations .

Definitions: A system of m x n numbers arranged in the form of an ordered set of

m horizontal lines called rows & n vertical lines called columns is called an m x n

matrix.

munotes.in

## Page 3

3Chapter 1: Matrices

A matrix A of order m × n can be represented in the following form

[ 𝑎𝑎11𝑎𝑎12………….𝑎𝑎1𝑛𝑛

𝑎𝑎21𝑎𝑎22………….𝑎𝑎2𝑛𝑛

…………………..……

………………………….

𝑎𝑎𝑎𝑎1𝑎𝑎𝑎𝑎2………..𝑎𝑎𝑎𝑎𝑛𝑛]

where a ij is the entry at the intersection of the ith row and jth column

Matrices are generally denoted by capital letters and the elements are generally

denoted by corresponding small letters.

1.2 Types of Matrices

1. Transpose of Matrix: Let A be an (m x n) matrix. Then, the matrix obtained

by interchanging the rows and columns of A is called the transpose of A,

denoted by A’ or AT. Thus, if A = [ a ij ]m x n then A’ = [a ij]n x m

eg. If A = [2−48

−359] then A’ = [2−3

−45

89]

Note: 1. If A is any matrix, then (A’)’ = A

2. If A is any matrix and k is scalar, then (kA)’ =k A’

3. If A and B are two matrices of same order then (A + B)’ = A’ + B’

2. Determinant of a square matrix: Corresponding to each square matrix

A = [𝑎𝑎11𝑎𝑎12𝑎𝑎13………𝑎𝑎1𝑛𝑛

𝑎𝑎21𝑎𝑎22𝑎𝑎23………𝑎𝑎2𝑛𝑛

……………………………

𝑎𝑎𝑛𝑛1𝑎𝑎𝑛𝑛2𝑎𝑎3𝑛𝑛………𝑎𝑎𝑛𝑛𝑛𝑛]

There is associated an expression, called the determinant of A, denoted by

det A or |𝐴𝐴|, written as

det A = |𝐴𝐴|= |𝑎𝑎11𝑎𝑎12𝑎𝑎13………𝑎𝑎14

𝑎𝑎21𝑎𝑎22𝑎𝑎23………𝑎𝑎24

……………………………

𝑎𝑎𝑛𝑛1𝑎𝑎𝑛𝑛2𝑎𝑎𝑛𝑛3………𝑎𝑎𝑛𝑛𝑛𝑛|

A matrix is an arrangement of numbers and so it has no fixed value, while

each determinant has a fixed value. A determinant having n rows and n

columns is known as a determinant of order n. The determinants of non -

square matrices are not defined. munotes.in

## Page 4

4 APPLIED MATHEMATICS

Value of a determinant of order 1: The value of a deter minant of a (1 x 1)

matrix [a] is defined as |𝑎𝑎|=a.

Value of a determinant of order 2: ቚ𝑎𝑎11𝑎𝑎12

𝑎𝑎21𝑎𝑎22ቚ = (a11.a22 x a21.a12)

Value of a determinant of order 3 or more: For Finding the value of an

order 3 or more, we need following definitions.

Minor of a ij in | |ǣ Minor of a ij in |𝐴𝐴 |defined as the value of the

determinant obtained by deleting the ith row and jth column of |𝐴𝐴 | is denoted

by M ij.

Cofactor of a ij in | |ǣ The cofactor Cij of an element a ij is defined as

Cij = (-1)i+ j . M ij

Eg. 1 Find the minor and cofactor of each element of A = อ1−32

4−12

352อ

Sol: The minors of the elements of A are given by,

M11 = ቚ−12

52ቚ= -2-10= -12 M12 = ቚ42

32ቚ= 8-6= 2

M13 = ቚ4−1

35ቚ= 20 +3= 23 M21 = ቚ−32

52ቚ= -6-10= -16

M22 = ቚ12

32ቚ= 2- 6= -4 M23 = ቚ1−3

35ቚ= 5+9= 14

M31 = ቚ−32

−12ቚ= -6+2= -4 M32 = ቚ12

42ቚ= 2-8= -6

M33 = ቚ1−3

4−1ቚ= -1+12= 11

SO, the cofactors of the corresponding elements of A are,

C11 = (-1)1+1 .M11 = M 11 = -12; C12 = (-1)1+2 .M12 = -M12 = -2;

C13 = (-1)1+3 .M13 = M 13 = 23; C21 = (-1)2+1 .M21 = -M21 = 16;

C22 = (-1)2+2 .M22 = M 22 = -4; C23 = (-1)2+3 .M23 = -M23 = -1;

C31 = (-1)3+1 .M31 = M 31 = 4; C32 = (-1)3+2 .M32 = -M32 = 6;

C33 = (-1)3+3 .M33 = M 33 = 11;

Value of Determinant: The value of determinant is the sum of the products

of elements of a row (or a column) with their corresponding cofactors.

We may expand a determinant by any arbitrarily chosen row or column. munotes.in

## Page 5

5Chapter 1: Matrices

Expansion of a Determinant: Expanding the given determinant by 1st row, we

have

อ𝑎𝑎11𝑎𝑎12𝑎𝑎13

𝑎𝑎21𝑎𝑎22𝑎𝑎23

𝑎𝑎31𝑎𝑎32𝑎𝑎33อ = a11. (its cofactor) + a12. (its cofactor) + a 13. (its cofactor)

= a 11. C11 + a 12. C12 + a 13. C13

= a11. M11 - a12. M12 + a 13. M13 [ C12 = - M12]

Eg. Evaluate A = อ345

−2−3

81อ

Sol: Expanding the given determinant by 1st row, we get

A = 3. ቚ2−3

1ቚ – 4 ቚ−−3

8ቚ + 5 ቚ−2

81ቚ

= 3 (14 +3) -4. (-42+24) + 5 (-6-16)

= 3 (1 7) + 4(18) - 5(22) = 51 +72 -110 = 13

3. Adjoint of Matrix: Let A = [aij] be a square matrix of order n and let A ij

denote the cofactor of a ij in |𝐴𝐴|. Then, the adjoint of A, denoted by adj A, is

defined as adj A=[a ji]n x n

Thus, adj A is the transpose of the matrix of the corresponding cofactors of

element s of |𝐴𝐴|.

If A= [𝑎𝑎11𝑎𝑎12𝑎𝑎13

𝑎𝑎21𝑎𝑎22𝑎𝑎23

𝑎𝑎31𝑎𝑎32𝑎𝑎33] then Adj A = [𝐴𝐴11𝐴𝐴12𝐴𝐴13

𝐴𝐴21𝐴𝐴22𝐴𝐴23

𝐴𝐴31𝐴𝐴32𝐴𝐴33]Ԣ

= [𝐴𝐴11𝐴𝐴21𝐴𝐴31

𝐴𝐴12𝐴𝐴22𝐴𝐴32

𝐴𝐴13𝐴𝐴23𝐴𝐴33], Where A ij denotes the cofactor of a ij in |𝐴𝐴|.

Eg. 1. If A = [1−24

Ͳ21

−453] find adj A

Sol: |𝐴𝐴|= อ1−24

Ͳ21

−453อ

The cofactors of the elements of the |𝐴𝐴|are given by,

A11 = ቚ21

53ቚ =1; A12 =ቚͲ1

−43ቚ = -4; A13 = ቚͲ2

−45ቚ= 8;

A21 = ቚ−2453ቚ =-26; A22 =ቚ14−43ቚ = 19; A23 =ቚ1−2−45ቚ= 3; munotes.in

## Page 6

6 APPLIED MATHEMATICS

A31 = ቚ−2421ቚ =-10; A32 =ቤቚ14Ͳ1ቚቤ = -1; A33 =ቚ1−2Ͳ2ቚ= 2

adj A = [1−48

−2193

−1Ͳ−12]Ԣ = [1−2−1Ͳ

−419−1

832]

1.3 Operations on Matrices:

1. Addition of Matrices: Let A and B be two comparable matrices, each of

order (m x n). Then their sum (A + B) is a matrix of order (m x n), obtained

by adding the corresponding elements of A and B.

Eg. Let A = [1−

542] and B = [5−3−

1−3Ͳ]

Here, Matrix A and Matrix B both are 2 x 3 matrices .

A and B are comparable matrices. A + B is defined.

A + B = 51ሺ−3ሻሺ−ሻሺ−ሻ

514ሺ−3ሻ2Ͳ൨ = [11−2−13

12]

Properties of Addition of Matrices:

The basic properties of addition for real numbers also hold true for matrices.

Let A, B and C be m x n matrices.

1. Matrix addition is commutative. i.e. A + B = B +A for all comparable

matrices A and B.

2. Matrix addition is commutative. i.e. (A + B) + C = A + (B + C)

3. If O is an m x n null matrix, then A + O = O + A = A

Students can solve proof of these properties as exercise.

2. Scalar Multiplication: If A be a matrix and k be a number then the matrix

obtained by multiplying each element of A by k is called the scalar multiple

of A by k, denoted by kA.

If A is an (m X n) matrix then kA is also an (m X n) matrix.

If A = [5

3−2

−54], Find i) 4A, ii) ଵ

ଶ 𝐴𝐴, iii) -3A

Sol: 4A =[2Ͳ24

12−8−2Ͳ1], ii) ଵ

ଶ 𝐴𝐴 =[ ହ

ଶ3

ଷ

ଶ−1

ିହଶ2] , iii) -3A= [−15−18

−915−12] munotes.in

## Page 7

7Chapter 1: Matrices

3. Multiplication of Matrices: For two given matrices A and B, multiplication

of two matrices AB exists only when number of rows in A is equals the

number of columns in B .

Let A = [a ij]m x n and B = [bjk]n x p be two matrices such that the number of

columns in A equals the number of rows in B.

Then, AB exists and it is an (m x p) matrix, given by

AB = [ C ik] m x p where C ik = (a i1b1k + a i2b2k + …. + aimbmk) = σ𝑎𝑎݆ܾ݆݇݅

ୀଵ

= sum of the products of corresponding elements of ith row of A and

kth column of B.

Properties of Matrix Multiplication:

1. Matrix multiplication is not commutative in general.

Let A and B be two matrices.

a. If AB exists then it is quite pos sible that BA may not exist.

b. Similarly, if BA exists then AB may not exist.

c. If AB and BA both exist, they may not be comparable.

2. Associative Law: For any matrices A, B, C for which(AB)C and A(BC)

both exist, we have (AB)C = A(BC)

3. Distributive laws of multiplication over addition:

i) A. (B + C) = (AB + AC)

ii) (A + B).C = (AC + BC)

4. The product of two non -zero matrices can be a zero matrix.

5. If A is a square matrix and I is an identity matrix of same order as A

then we have A.I = I.A = A.

6. If A is a square matrix and 0 is an identity matrix of same order as A

then we have A.0 = 0.A = 0.

Exercise:

Ex 1. If A = [54

23] and B = [351

84], find AB and BA whichever exists.

Ex 2. If A = [1−12

32Ͳ

−2Ͳ1] , B = [31

Ͳ2

−25] and C = [21−3

3Ͳ−1]

Verify (AB)C = A(BC) munotes.in

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8 APPLIED MATHEMATICS

1.4 Elementary Transformation:

Following are three row operations and three column operations on a matrix, which

are called Elementary operations or transformations.

Equivalent Matrices: Two matrices are said to be Equivalent if one is obtained

from the other by one or more elementary operations and we, A ̱ B.

Three Elementary Row Operations:

i. Interchange of any two rows: The interchange of ith and jth rows is denoted

by R i ՞ Rj.

Eg. Let A = [593

−813

−28] Applying R1՞ R2, we get [−813

593

−28]

ii. Multiplication of the elements of a row by a nonzero number: Suppose

each element of ith row of a given matrix is multiplied by a nonzero number

k. Then, we denote it by R i ืkRi

Eg. Let A = [593

−813

−28] Applying R3ื2R3, we get [593

−813

−4141]

iii. Multiplying each element of a row by a nonzero number and then adding

them to the corresponding elements of another row: Suppose each

element of jth row of a matrix A is multiplied by a nonzero number k and

then added to the corresponding elements of ith r ow.

We denote it by Ri ื Ri + k R j

Eg. Let A = [593

−213

−28] Applying R1ื R1+3R2, we get [−11212

−213

−28]

Three Elementary Column Operations:

i. Interchange of any two columns: The interchange of ith and jth columns is

denoted by C i ՞ Cj.

Eg. Let A = [493

−1−29] Applying C2՞ C3, we get [439

−1−29] munotes.in

## Page 9

9Chapter 1: Matrices

ii. Multipl ying each element of a column by a nonzero number: Suppose

each element of ith column of a given matrix is multiplied by a nonzero

number k. Then, we denote it by Ci ืkCi

Eg. Let A = [22

−33

258] Applying C2ื2C2, we get [2142

−3

21Ͳ8]

iii. Multiplying each element of a column by a nonzero number and then

adding them to the corresponding elements of another column : Suppose

each element of jth column of a matrix A is multiplied by a nonzero number

k and then added to the corresponding elements of ith column .

We denote it by Ci ื Ci + k Cj

Eg. Let A = [523

−213

−248] Applying C1ื C1+2C2, we get [923

Ͳ13

48]

1.5 Inverse of Matrix:

Invertible Matrices: A square matrix A of order n is said to be invertible if there

exists a square matrix B of order n such that AB = BA = I

Also, then B is called the inverse of A and we write, A-1 = B

Eg. Let A = [35

12] and B = [2−5

−13] then

AB = [35

12]. [2−5

−13]=[−5−1515

2−2−5]= [1Ͳ

Ͳ1] =I

BA= [2−5

−13][35

12] = [−51Ͳ−1Ͳ

−33−5]= [1Ͳ

Ͳ1]=I

AB = BA = I Hence A-1 = B.

Singular and Non -singular Matrices: A square A is said to be singular if |𝐴𝐴|= 0

and non -singular if |𝐴𝐴|് 0.

Eg. Let A = [12

48] then |𝐴𝐴| = ቚ12

48ቚ = (8 – 8) = 0 A is singular

Let B = [12

38] then |𝐴𝐴| = ቚ12

48ቚ = (8 – 6) = 2 ് 0 A is non -singular.

Note 1: Uniqueness of Inverse: Every invertible squar e matrix has a unique

inverse.

Note 2: A square matrix A is invertible if and only if A is non-singular ,

i.e. A is invertible ֞ |𝐴𝐴|് 0 munotes.in

## Page 10

10 APPLIED MATHEMATICS

1.5.1 Inverse of matrix by Elementary Row Operations:

Let A be a square matrix of order n.

We can write, A = I.A ………………….. (i)

Now, let a sequence of elementary row operations reduce A on LHS of (i) to

I and I on RHS of (i) to a matrix B.

Then, I =BA ฺ I.A-1= (BA) A-1 = B (A A-1) = BI ฺ A-1 = B

We can write above method as given below.

1. Write A = I A

2. By using elementary row operations on A, transform it into a unit matrix.

3. In the same order we apply elementary operations on I to convert it into a

matrix B.

4. Then, A-1 = B

Ex. 1. By using elementary row operations, find the inverse of the matrix

A= [13−2

−3Ͳ−5

25Ͳ]

Sol: [13−2

−3Ͳ−5

25Ͳ] =[1ͲͲ

Ͳ1Ͳ

ͲͲ1]. A

R2 ื R2 – 3R1, R3 ื R3 – 2R1 ฺ[13−2

Ͳ9−11

Ͳ−14] =[1ͲͲ

31Ͳ

−2Ͳ1]. A

R2 ՞R3 ฺ [13−2

Ͳ−14

Ͳ9−11] = [1ͲͲ

−2Ͳ1

31Ͳ]. A

R1 ื R1 +3R 2, R3 ื R3 + 9R2 ฺ [1Ͳ1Ͳ

Ͳ−14

Ͳ925] = [−5Ͳ3

−2Ͳ1

−1519]. A

R2 ื (-1). R 2 ฺ [1Ͳ1Ͳ

Ͳ1−4

ͲͲ25] = [−5Ͳ3

2Ͳ−1

−1519]. A

R3 ื (ଵ

ଶହ) R3 ฺ [1Ͳ1Ͳ

Ͳ1−4

ͲͲ1] = −5Ͳ3

2Ͳ−1

ିଷ

ହଵ

ଶହଽ

ଶହ. A munotes.in

## Page 11

11Chapter 1: Matrices

R1 ื R1 -10R 3, R2 ื R2 + 4R 3 ฺ[1ͲͲͲ1Ͳ

ͲͲ1] = [ 1ିଶହିଷହ

ିଶ

ହସ

ଶହଵଵ

ଶହ

ିଷ

ହଵ

ଶହଽ

ଶହ] . A

Hence, A-1 = [ 1ିଶ

ହିଷ

ହ

ିଶ

ହସ

ଶହଵଵ

ଶହ

ିଷ

ହଵ

ଶହଽ

ଶହ]

1.5.2 Inverse of matrix by Formula:

Formula for finding A-1:

Let A be a square matrix such that |𝐴𝐴| ്0. Then, A-1 =

||. (adj A)

Ex.1 . Find the inverse of the matrix [3−1Ͳ−1

−282

2−4−2]

Sol: Let A = [3−1Ͳ−1

−282

2−4−2] ฺ |𝐴𝐴|= อ3−1Ͳ−1

−282

2−4−2อ

C1 ื C1 +3C 3 and C 2 ื C2 - 10C 3

|𝐴𝐴|= อͲͲ−1

4−122

−41−2อ = (-1). (64 -48) = - 16 ് 0

As |𝐴𝐴| ് 0 therefore A-1 exists.

The cofactors of the elements of |𝐴𝐴| are given by,

A11 = ቚ82

−4−2ቚ= -8; A 12 = ቚ−22

2−2ቚ= 0; A 13 = ቚ−28

2−4ቚ= -8

A21 = ቚ−1Ͳ−1

−4−2ቚ= -16; A 22 = ቚ3−1

2−2ቚ= -4 A 23 = ቚ3−1Ͳ

2−4ቚ= -8

A31 = ቚ−1Ͳ−1

82ቚ= -12; A 32 = ቚ3−1

−22ቚ= -4 A 33 = ቚ3−1Ͳ

−28ቚ= 4

(Adj A) = [−8Ͳ−8

−1−4−8

−12−44]Ԣ = [−8−1−12

Ͳ−4−4

−8−84]

Hence A-1 = ଵ

|| . adj A munotes.in

## Page 12

12 APPLIED MATHEMATICS

= ଵ

ିଵ [−8−1−12Ͳ−4−4

−8−84] = [ ଵଶ1ଷସ

Ͳଵ

ସଵ

ସ

ଵ

ଶଵ

ଶ−ଵ

ସ]

Ex. 2 If A = [32

5], B= [

89], verify that (AB)-1 = B-1A-1

Sol. We have |𝐴𝐴| = ቚ32

5ቚ=15-14 = 1 ് 0

Cofactors of the elements of |𝐴𝐴| are

A11 = 5, A 12 = -7, A 21 = -2, A 22= 3

adj A= [32

5]Ԣ = [3

25]

Hence, A-1 = ଵ

|| adj A = [3

25] [ |𝐴𝐴|=1]

|ܤ|= ቚ

89ቚ=54-56 = -2 ് 0

Cofactors of the elements of |ܤ| are

B11 = 9, B 12 = -8, B 21 = -7, A 22= 6

adj A= [9−8

−]Ԣ = [9−8

−]

Hence, B-1 = ଵ

|| adj B = − ଵ

ଶ[3

25] [ |ܤ|= - 2]

Now, |𝐴𝐴ܤ| = |𝐴𝐴||ܤ| = 1 x -2 = -2 ് 0

adj AB = adj B. adj A

= [9−8

−][3

25] = [94−39

−8234]

AB)-1 = ଵ

|| adj AB = − ଵ

ଶ[94−39

−8234]

B-1 A-1 = − ଵ

ଶ [3

25][3

25] = − ଵ

ଶ [94−39

−8234]

(AB)-1 = B-1 A-1

Exercise:

1) Find the adjoint of given matrix verify A. ሺ ሻ= ሺ ሻA = |𝐴𝐴|.I

1ሻ[3−5

−12] 2) [1−12

31−21Ͳ3] 3)[453

Ͳ129] munotes.in

## Page 13

13Chapter 1: Matrices

[Ans: 1 .[2513]ǡ 2. [33Ͳ−1118−1−14] , 3. [−42−243Ͳ33Ͳ−21−18−5]]

2) If A =[−4−3−3

1Ͳ1

443], show that adj A = A

3) If A =[−4−3−3

1Ͳ1

443], show that adj A = 3A’

1.6 Rank of Matrix

The maximum number of its linearly independent rows (or columns) of a

matrix A is called the rank of Matrix A. If we have a chance of solving a system of

linear equations, when the rank is equals the number of variables, we may be able to find

a unique solution. Rank of a matrix A is denoted by ʌ (A) or R (A)

Note:

a. The rank of a matrix cannot exceed the number of its rows or columns.

b. The rank of a null matrix is zero.

c. 5DQNRIDPDWUL[$PîQȡ$PîQ0LQPQ

d. ȡ,Q QZKHUH,Q XQLWPDWUL[RIRUGHUQ

e. ,Iȡ $ PDQGȡ % QWKHQȡ $%PLQPQ

1.6.1 Echelon or Normal Matrix : a matrix is said to be echelon form if

a. There exists any zero row, they should be placed below the non - zero row

b. Number of zeros before a non – zero element i n a row should increa se

according with row number.

Eg. A = [145

Ͳ54

ͲͲ1] ȡ$ QXPEHURIQRQ – zero row

B =[154

Ͳ54

ͲͲ43

ͲͲͲͲ] ȡ% QXPEHURIQRQ – zero row

Note: To reduce a matrix into its echelon form only elementary row transformations

are applied. munotes.in

## Page 14

14 APPLIED MATHEMATICS

Computing the Rank of a matrix: A common approach for finding the rank of a

matrix is to reduce it to a simpler form, generally row echelon form by elementary

row operations. Row operations do not change the row rank

Ex 1. Find the rank of the matrix A = [134

2−13

−282]

Sol: We have, A = [134

2−13

−282]

To find the rank of a matrix, we will transform the matrix into its echelon form by

row transformation. Then determine the rank by the number of non -zero rows

R2 = R 2- 2R1, R3 = R 3 +2R 1 A = [134

Ͳ−−5

Ͳ141Ͳ]

R3 = R 3 +2R 2 A = [134

Ͳ−−5

ͲͲͲ]

Number of non-zero rows in matrix A = 2 Rank of matrix A, ȡ$

Exercise:

Ex 1. Find the rank of the following matrices

1. A = [121

231

112]; 2. A = [2−13

1Ͳ1

Ͳ2−1

114]

1.7 Linear Equations

To find the solution to the system of equations is a matrix method. The steps to be

followed are:

• All the variables in the equations should be written in the appropriate order.

• The variables, their coefficients and constants are to be written on the

respective sides.

There are two types of system of equations.

1. Consistent system of Equations: A given system of equations is said to be

consistent if it has one or more solutions.

2. Inconsistent system of Equations: A given system of equations is said to

be inconsistent if it has no solution. munotes.in

## Page 15

15Chapter 1: Matrices

Consider the system of equations.

a1x + b 1y + c 1z = d 1; a2x + b 2y + c 2z = d 2; a3x + b 3y + c 3z = d 3

Let A = [1ܾ1ܿ1

𝑎𝑎2ܾ2ܿ2

ܿ1ܿ2ܿ3], X= ቈݔ

ݕ

ݖ and B = [݀1

݀2

݀3]

Then the given system can be written as

[𝑎𝑎1ܾ1ܿ1

𝑎𝑎2ܾ2ܿ2

𝑎𝑎3ܾ3ܿ3] ቈݔ

ݕ

ݖ = [݀1

݀2

݀3]

AX = B

Case1: when |𝐴𝐴| , In this case, A-1 exists.

AX = B ֜ A-1(AX)= A-1B [multiplying both the sides by A-1]

֜ (A-1A)X= A-1B [By associative law]

֜ I.X = A-1B ֜ X= A-1B

Since A-1 is unique, the given system has a unique solution .

Thus, when |𝐴𝐴| WKHQWKHJLYHQV\VWHPLVFRQVLVWHQWDQGLWKDVDXQLTXH

solution.

Case 2: |𝐴𝐴| = 0 and (adj A) B ് 0

In this case, the given system has no solution and hence it is inconsistent.

Case 3: |𝐴𝐴| = 0 and (adj A) B = 0

In this case, the given system has infinitely many solutions.

Ex.1 Use matrix method to show that the system of equations

2x + 5y = 7 , 6x + 15y = 13 is inconsistent

Sol: The given equations are 2x + 5y = 7 ; 6x + 15y = 13

Let A =[25

15], X= [ݔ

ݕ] and B= [

13]

Then the given system in matrix form is AX = B

Now, |𝐴𝐴|= ቚ25

15ቚ= 30 – 30 = 0

The system will be inconsistent if (adj A) B ് 0

The minors of the elements of |𝐴𝐴| are M11 = 15, M 12 = 6, M 21= 5, M 22 = 2

The cofactors of the elements of |𝐴𝐴| are A11 = 15, A 12 = -6, A 21= -5, A 22 = 2 munotes.in

## Page 16

16 APPLIED MATHEMATICS

Adj A = [15−−52]Ԣ = [15−5−2]

֜ (adj A) B = [15−5

−2][

13] = [1Ͳ5−5

−422]= [4Ͳ

−1]് 0

|𝐴𝐴|=0, (adj A) B ് 0. Hence, the given system of equations is inconsistent.

Ex.2 Show that the following system of equations is consistent and solve it

2x + 5y = 1 , 3x + 2y = 7

Sol: The given equations are

2x + 5y = 1 ; 3x + 2y = 7

Let A =[25

32], X= [ݔ

ݕ] and B= [1

]

Then the given system in matrix form is AX = B

Now, |𝐴𝐴|= [25

32]= 4 - 15 = -11് 0

Hence the given system has a unique solution.

The minors of the elements of |𝐴𝐴| are M11 = 2, M 12 = 3, M 21= 5, M 22 = 2

The cofactors of the elements of |𝐴𝐴| are A11 = 2, A12 = -3, A21= -5, A 22 = 2

Adj A = [2−3

−52]Ԣ = [2−5

−32]

֜ A-1 = ଵ

|| adj A = ିଵ

ଵଵ [2−5

−32] =ିଶ

ଵଵହ

ଵଵ

ଷ

ଵଵିଶ

ଵଵ

X = A-1 B

[ݔ

ݕ] = ିଶ

ଵଵହ

ଵଵ

ଷ

ଵଵିଶ

ଵଵ[1

] = ିଶ

ଵଵ ଷହ

ଵଵ

ଷ

ଵଵ−ଵସ

ଵଵ = [3

−1] ֜ x = 3 and y = -1

Exercise:

1) Use matrix method to solve the following system of equations

3x + 4y +2z = 8 ; 2y – 3z = 3 ; x-2y+6z = -2 [ Ans: x= -2, y = 3 and z = 1]

1.8 Linear dependence and linear independence of vectors

A collection of vectors is either linearly independent or linearly dependent. The

vectors v1, v2 ……vk are linearly independent if the equation involving linear

combination. In the theory of vector spaces, a set of vectors is said to be linearly munotes.in

## Page 17

17Chapter 1: Matrices

dependent if there is a nontrivial linear combination of the vectors that equals the

zero vector. If no such linear combination exists, then the vectors are said to be

linearly independent.

A sequence of vectors v 1, v2, …….. , v k from a vecto r space V is said to be linearly

dependent, if there exist scalars a 1, a2, ….., a k not all zero, such that

a1v1+a2v2+…. + akvk = 0, where 0 denotes the zero vector.

Ex. 1 State whether following set of vectors are linearly dependent or linearly

independent. If dependent find the relation between them.

X1 = (1, 2, 3), X 2 = (3, -2, 1), X 3 = (1, -6, 5)

Sol: Here, there are three vectors. For three vectors are take 3 scalars.

Let ߣ1, ߣ2 and ߣ3 be three scalars.

Consider ߣ1 X1 + ߣ2 X2 + ߣ3 X3 = 0 ……………….. (1)

ߣ1(1, 2, 3) + ߣ2 (3, -2, 1) + ߣ3(1, -6, 5) = 0

From these we make three simultaneous equations.

ߣ1 + 3ߣ2 + ߣ3 = 0; 2ߣ1 - 2ߣ2 - 6ߣ3 = 0; 3ߣ1 + ߣ2 + 5ߣ3 = 0

Put them in matrix form

[131

2−2−

315] [ߣ1

ߣ2

ߣ3] = [Ͳ

Ͳ

Ͳ]

A ߣ =B ……………………………………… (2)

Now augmented matrix,

C = [A: B] ֜ =[131Ͳ

2−2−Ͳ

315Ͳ]

Reduced this matrix in echelon matrix by row transformation

R2 = R 2 – 2R1; R3 = R 3 – 3R1 ฺ C = [131Ͳ

Ͳ−8−8Ͳ

Ͳ−82Ͳ]

R3 = R 3 –R2, C = [131Ͳ

Ͳ−8−8Ͳ

ͲͲ1ͲͲ]

Here we cannot further reduce .

From (2) , [131Ͳ−8−8ͲͲ1Ͳ] [ߣ1ߣ2ߣ3] = [ͲͲͲ] munotes.in

## Page 18

18 APPLIED MATHEMATICS

From matrix multiplication,

ߣ1 + 3ߣ2 + ߣ3 = 0 ………………… (3)

-8ߣ2 - 8ߣ3 = 0 ………………………(4)

ߣ3 = 0

Put ߣ3 = 0 in (4) ߣ2= 0

Put ߣ2, ߣ3 (3) ߣ1= 0

ߣ1= ߣ2= ߣ3= 0 i.e all three scalars are 0.

The given vectors are linearly independent and there exists no relationship.

Ex. 2 Test the linear dependency and find the relationship between if it exists for

X1 = (1, 1, 1, 3), X 2 = (1, 2, 3, 4), X 3 = (2, 3, 4, 7 )

Sol: Here, there are three vectors. For three vectors are take 3 scalars.

Let ߣ1, ߣ2 and ߣ3 be three scalars.

Consider ߣ1 X1 + ߣ2 X2 + ߣ3 X3 = 0 ……………….. (1)

ߣ1(1, 1, 1, 3) + ߣ2 (1, 2, 3, 4) + ߣ3(2,3,4,7) = 0

From these we make simultaneous equations.

ߣ1 + ߣ2 + 2ߣ3 = 0; ߣ1 + 2ߣ2 + 3ߣ3 = 0; ߣ1 + 3ߣ2 + 4ߣ3 = 0; 3ߣ1 + 4ߣ2 + 7ߣ3 = 0

Put them in matrix form

[112

123

134

34] [ߣ1

ߣ2

ߣ3] = [Ͳ

Ͳ

Ͳ

Ͳ]

A ߣ =B ……………………………………… (2)

Now augmented matrix, C = [A: B]

= [112Ͳ

123Ͳ

134Ͳ

34Ͳ]

Reduced this matrix in echelon (upper triangular) matrix by row transformation

R2 = R 2 – R1 , R3 = R 3 – R1, R4 = R 4 – 3R1

C = [112Ͳ

Ͳ11ͲͲ22ͲͲ11Ͳ] munotes.in

## Page 19

19Chapter 1: Matrices

R3 = R 3 –2R2, R4 = R 4 – R2 ฺ C = [112ͲͲ11ͲͲͲͲͲ

ͲͲͲͲ]

from (2) , [112Ͳ

Ͳ11Ͳ

ͲͲͲͲ

ͲͲͲͲ] [ߣ1

ߣ2

ߣ3] = [Ͳ

Ͳ

Ͳ]

From matrix multiplication,

ߣ1 + ߣ2 + 2ߣ3 = 0 ………………… (3)

ߣ2 + ߣ3 = 0 ֜ߣ2 = - ߣ3 ………………………(4)

Consider ߣ3 = k where k is non zero constant . ߣ2 =-k

Put ߣ2, ߣ3 in equation (3) ߣ1-k +2k = 0 ֜ ߣ1+k = 0 ߣ1 = -k

All the scalars are non-zero.

The given vectors are linearly dependent and there exists some relationship.

Now we find relationship between them.

We have, ߣ1 X1 + ߣ2 X2 + ߣ3 X3 = 0 ֜ −݇ X1 -k X 2 + ݇ X3 = 0

Divide equation by -k , X1 + X 2 - X3 = 0

This is the required relationship.

1.9 Linear Transformation

Let U (F) and V(F) be two vector spaces.

A mapping f: U ՜ V is called Linear Transformation of U into V if

i) f (x + y) = f(x) + f(y)

ii) f (ax) = a f(x) where x, y ߳ V, a ߳ F, f (x), f (y) ߳ V.

Sometimes linear transformation is also called vector space homomorphism.

Ex. 1 V3 is a vector. A mapping is given as T.V 3(R) ՜ V2(R) by T(x 1, x2, x3) =

(x1- x2) , (x 1+ x 3). Check whether this is linear transformation.

Sol: Let (x 1, x2, x3) = x ߳ V3 and l et (y 1, y2, y3) = y ߳ V3(R)

T (x + y) = T [(x 1, x2, x3) + (y 1, y2, y3)] = T [(x 1 + y 1, x2 + y 2, x3 + y 3)]

= T [(x 1 + y 1- x2 - y2, x1 + y 1+ x 3 + y 3)]

= T [(x 1 - x2 + y 1- y2, x1 + x 3 + y 1+ y3)] munotes.in

## Page 20

20 APPLIED MATHEMATICS

= T [ (x 1, x2, x3)] + T [(y 1, y2, y3)] = T (x) + T (y)

T (ax) = T [a (x 1, x2, x3)] = T [ (ax 1, ax2, ax3)] = (ax 1 - ax2, ax1 + ax 3)

= a [ (x 1 - x2), (x 1 + x 2)]

T (ax) =a T (x) = T (x)

Both the condition of linear transformation are satisfy.

T is linear transformation.

1.9.1 Matrix representation of Linear Transformation:

Let U (F) and V(F) be two vector spaces over F .

T: U ՜ V be a Linear Transformation

Let B = {u 1, u2, u3, ……….. ….. , u n} and

B’ = {v 1, v2, v3, …………….. , v m}

Are two ordered bases for U and V respectively.

Now, if any ߙ ߳ U ֜ T(ߙ )߳ V

Also T( ߙ )can be represented by B’

T(u 1) = B 1 = a 11v1 + a 12v2 +a13v3+…………..+ A 1mvm

T(u 2) = B 2 = a 21v1 + a 22v2 +a23v3+…………..+ A 2mvm

……………………………………………………….

T(u n) = B n = a n1v1 + a n2v2 +an3v3+…………..+ A nmvm

[ ሺ1ሻ

ሺ2ሻ

.

.

ሺሻ] = [ 𝑎𝑎11𝑎𝑎12………….𝑎𝑎1𝑎𝑎

𝑎𝑎21𝑎𝑎22………….𝑎𝑎2𝑎𝑎

…………………..……

………………………….

𝑎𝑎𝑛𝑛1𝑎𝑎𝑛𝑛2………..𝑎𝑎𝑛𝑛𝑎𝑎] [ ݒ1

2

.

.

]

[ 𝑎𝑎11𝑎𝑎12………….𝑎𝑎1𝑎𝑎

𝑎𝑎21𝑎𝑎22………….𝑎𝑎2𝑎𝑎

…………………..……

………………………….

𝑎𝑎𝑛𝑛1𝑎𝑎𝑛𝑛2………..𝑎𝑎𝑛𝑛𝑎𝑎] = [ T: B:B’]

This is matrix of Linear Transformation.

If we have Linear Transformation T: U(F) ՜ V(F)

then matrix form is [T: B], [T] B

For any n dimensions vector spaces , munotes.in

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21Chapter 1: Matrices

standard basis for v 2(R)= {(1, 0), (1, 0)}, v 3(R)= {(1, 0, 0),

(0, 1, 0), (0, 0,1)}, v 3(R)= {(1, 0, 0, 0), (0, 1, 0, 0), (0, 0,1, 0), (0, 0, 0, 1)}

1.10 Characteristics roots and characteristics vectors

Characteristic vector or Eigen vector of a matrix A is a vector represented by a

matrix X such that when X is multiplied with matrix A, then the direction of the

resultant matrix remains same as vector X.

Let A be a square matrix of order n x n, then a number ߣ is said to be eigen value

of a matrix A if there exists a column matrix X of order n x 1 such that AX = ߣX,

where A is any arbit rary matrix, ߣ are eigen values and X is an eigen vector

corresponding to each eigen value.

֜ AX - ߣX = 0 ֜ (A- ܫߣ)X = 0 ………………………. (1)

Equation (1) is called characteristics equation of the matrix.

The roots of the characteristic equation are the eigen values of the matrix A.

Ex.1 Find the eigen value (characteristics roots) and eigen vector ( characteristics

vector ) for the matrix A = [1−2

−54].

Sol: The characteristic equation for matrix A is,

|𝐴𝐴− ɉ |=Ͳ ฺቚ1− ɉ−2

−54− ɉቚ = 0

ฺሺ1−ɉሻ. ሺ4−ɉሻ − ሺ−5ሻ. ሺ−2ሻ= 0 ฺ 4 - ɉ – 4 ɉɉ2 – 10 = 0

ฺ ɉ2 – 5 ɉ−6 = 0 ฺሺɉ−ሻ. ሺɉ1ሻ = 0 ฺ ɉ = 6, ɉ= −1

Eigen value of A are 6 and -1.

Case I: X 1 = [ܺ

ܻ] be the eigen vector of A corresponding to ɉ = 6

Then ሺ−ɉሻX1 = 0

i.e. [1− ɉ−2

−54− ɉ].[ܺ

ܻ] = 0, [1− −2

−54− ].[ܺ

ܻ] = 0 , ɉ = 6

[−5−2

−5− 2].[ܺ

ܻ] = 0

By row transformation,

R2 = R 2 – R1 [−5−2

ͲͲ].[ܺ

ܻ] = 0

-5X – 2Y = 0 ฺ - 5X = 2Y munotes.in

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22 APPLIED MATHEMATICS

ଶ = ିହ = k ሺሻ

X = 2k, Y = -5k for k=1

Eigen vector X 1 = [2

−5]

Case II: Let X 2 = [ܺ

ܻ] be the eigen vector of A corresponding to ɉ = -1

Then ሺ−ɉሻX2 = 0

i.e. [1− ɉ−2

−54− ɉ].[ܺ

ܻ] = 0

1− ሺ−1ሻ −2

−54−ሺ−1ሻ൨.[ܺ

ܻ] = 0 , ɉ = -1

[2−2

−55].[ܺ

ܻ] = 0

By row transformation,

R2 = R 2 + ହ

ଶ R1 ฺ [2−2

ͲͲ].[ܺ

ܻ] = 0

2X – 2Y = 0 ฺ 2X = 2Y

ଵ =

ଵ = k ሺሻ ฺ X = k, Y = k for k=1

Eigen vector X 2= [1

1]

1.11 Properties of characteristic vectors (eigen vector )

Following are the properties of Eigen vector:

1. Corresponding one eigen vector there exists one eigen value.

Let ߣ1 and ߣ2 are two eigen values of A with one eigen vector X ് 0.

By condition of eigen values,

AX = ߣ1X and AX = ߣ2X

ฺ ߣ1X = ߣ2X ฺ (ߣ1- ߣ2) X = 0

As X ് 0, (ߣ1- ߣ2) = 0 ࣅ1 = ࣅ2

So there exists one eigen value for one eigen vector

2. If ߣ is eigen value of the matrix A of order n x n.

a) ߣ2 is an eigen value of A2 munotes.in

## Page 23

23Chapter 1: Matrices

AX = ߣX

Multiplying by A , A2X = ߣAX = ߣ ߣX = ߣ2X

ߣ2 is an eigen value of A2

b) ߣk is an eigen value of Ak, k is positive integer.

c) f (ߣሻ = a 0 ߣ +a 1 ߣ2+……………. + a n ߣn

is an eigen value of F (Aሻ = a 0I + a 1A+……………. + a nAn

d) ݁ఒ, log ߣ ,ߣ are eigen values of ݁, log A, 𝐴𝐴 respectively.

e) ||

ఒ is an eigen value of adj A.

As AX = ߣX

Let ߣ1 is eigen value of adj A

adj AX = adj ߣ1X

Multiplying by A , A adj AX = ߣ1AX

|𝐴𝐴| IX = ߣ1 ߣX [ adj A = |𝐴𝐴| I]

( |𝐴𝐴| - ߣ1 ߣ )X = 0

As X ് 0, |𝐴𝐴| - ߣ1 ߣ =0 ฺ |𝐴𝐴| = ߣ1 ߣ ฺ ߣ1 = ||

ఒ

1.12 Caley Hamilton Theorem:

Consider A - ߣI = [2−ߣ11

Ͳ1−ߣͲ

ͲͲ2−ߣ] ՚characteristic matrix , Where A is a

square matrix

Characteristic polynomial : If we put characteristic matrix in determinant form

and solved then we get polynomial that is called characteristic polynomial.

| − ߣ | = อ2−ߣ11

Ͳ1−ߣͲ

ͲͲ2−ߣอ

| − ߣ | = 0

อ2−ߣ11

Ͳ1−ߣͲ

ͲͲ2−ߣอ = 0

ሺ2−ߣ )[ ሺ1−ߣሻሺ2−ߣሻ – 0] – 1 [0] + 1[0 - 1−ߣ] = 0

ሺ2−ߣ )[ (2-Ȝ-ȜȜ2)] - 1ߣ =0 munotes.in

## Page 24

24 APPLIED MATHEMATICS

ሺ2−ߣ )[ (2-ȜȜ2)] - 1ߣ =0

4 - 6 ߣ +Ȝ2 - ȜȜ2 - Ȝ3 – Ȝ

- Ȝ3 Ȝ2 - Ȝ

Ȝ3 - Ȝ2 Ȝ- 3 = 0

This is called characteristic equation.

Characteristic Roots ሺܖ܍ܑ܍ ܝܔ܉ܞ܍ܛሻ : Roots of characteristic equation is called

characteristic roots.

State and Prove Caley Hamilton Theorem

Statement: Every square matrix A satisfy its own characteristic equation.

Proof: Let A = [a ij]n x n be any square matrix and P (ߣ ) =|𝐴𝐴−ܫߣ| be a

characteristic equation where ߣ be any constant, I is an identity matrix.

Show that |𝐴𝐴−ܫߣ| = 0

|𝑎𝑎11−ߣ𝑎𝑎12𝑎𝑎13………𝑎𝑎1𝑛𝑛

𝑎𝑎21𝑎𝑎21−ߣ𝑎𝑎23………𝑎𝑎2𝑛𝑛

…………………………

𝑎𝑎𝑛𝑛1𝑎𝑎𝑛𝑛2𝑎𝑎𝑛𝑛3………𝑎𝑎𝑛𝑛𝑛𝑛−ߣ|= 0

a0 +a1 ߣ +a 2 ߣ2+ a3 ߣ3«««««D n ߣn

[ Matrix is n x n order therefore polynomial is of order n]

If we put matrix A in place of ߣ then

a0 +a1A + a 2A2+ a 3A3 ««««D n An

We know that , A adj A = |𝐴𝐴|I

(A - ߣI) adj (A - ߣI) = | − ߣ |.I [A = | − ߣ |]

Now each element in (A - ߣI) is a polynomial of degree almost 1.

Hence adj (A - ߣI) has polynomial of degree n -1.

adj (A - ߣI) = B 0 +B1 ߣ +B 2 ߣ2+ B3 ߣ3«««««% n-1 ߣn-1

(A - ߣI) adj (A - ߣI) = (A - ߣI) [B 0 +B1 ߣ +B 2 ߣ2+ ««««% n-1 ߣn-1]

= AB0 + AB 1 ߣ +AB 2 ߣ2+ ««««$% n-1 ߣn-1 -

B0ߣ - B1 ߣ2 - B2 ߣ3- ………… - Bn-1 ߣn

= AB0 + (AB 1-Bo) ߣ +(AB 2-B1) ߣ2+ ……. - Bn-1 ߣn

Now (A - ߣI) adj (A - ߣI) = | − ߣ | munotes.in

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25Chapter 1: Matrices

AB0 + (AB 1-Bo) ߣ +(AB 2-B1) ߣ2+

……. - Bn-1 ߣn = a0 +a1 ߣ +a 2 ߣ2+ a3 ߣ3………… + a n ߣn

Compare coefficient of equal power of ߣ

AB0 = a 0

AB 1-Bo = a 1

AB 2-B1 = a 2

………

AB n-1-Bn-2 = a n-1

-Bn-1 = a n

Multiplying with I, A, A2, ………, An and then add

AB0 = a 0I

A2B1-Bo = a 1A

A3B2-B1 = a 2 A2

………

AnBn-1-A n-1Bn-2 = a n-1An-1

-AnBn-1 = a n An

Hence a 0 +a1A + a 2A2+ a 3A3 +…………. + a n An =0

Hence proved.

Ex 1. Show that the matrix A = [8−8−2

4−3−2

3−41]

satisfies its characteristic equation and hence determine A-1.

Sol: The characteristic matrix of A is,

อ8−ߣ−8−2

4−3−ߣ−2

3−41−ߣอ = 0

Characteristic equation is given by |𝐴𝐴− ܫߣ| = 0

OR

If there is 2 x 2 matrix then ߣ2 -s1 ߣ +|𝐴𝐴| = 0

If there is 3 x 3 matrix then ߣ3 - s1 ߣ2 + s2 ߣ - |𝐴𝐴| = 0

Where s 1= sum of diagonal element of matrix A munotes.in

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26 APPLIED MATHEMATICS

And s 2= sum of minors of diagonal element of matrix A

Here matrix A is 3 x 3

So Characteristic equation is given by

ߣ3 - s1 ߣ2 + s2 ߣ - |𝐴𝐴| = 0

s1 = sum of diagonal element of matrix A = 8 – 3 + 1 = 6

s2 = sum of minors of diagonal element of matrix A

= ቚ−3−2

−41ቚ + ቚ8−2

31ቚ + ቚ8−8

4−3ቚ

= ( -3 - 8 ) + (8 + 6) + ( -24 + 32) = (-11) + 14 + 8 = 11

Now |𝐴𝐴| = อ8−8−2

4−3−2

3−41อ

= 8 (-3 – 8) - (-8) (4 +6) + (-2) (-16+9) = 8(-11)+80 +14 = 6

ߣ3 -6 ߣ2 + 11 ߣ – 6 = 0

Now, in LHS we replace ߣ by A, we get

𝐴𝐴3 -6 𝐴𝐴2 + 11 𝐴𝐴 – 6I

𝐴𝐴2 = [8−8−2

4−3−2

3−41][8−8−2

4−3−2

3−41] = [2−32−2

14−15−4

11−13]

𝐴𝐴3 = A 𝐴𝐴2 = [8−8−2

4−3−2

3−41][2−32−2

14−15−4

11−13] = [4−1Ͳ41Ͳ

4Ͳ−51−2

33−5213]

𝐴𝐴3 -6 𝐴𝐴2+ 11 𝐴𝐴 – 6I =

[4−1Ͳ41Ͳ

4Ͳ−51−2

33−5213] – 6 [2−32−2

14−15−4

11−13] +11[8−8−2

4−3−2

3−41]- 6 [1ͲͲ

Ͳ1Ͳ

ͲͲ1]

= [4−1Ͳ41Ͳ

4Ͳ−51−2

33−5213] - [15−192−12

84−9Ͳ−24

−918] +

[88−88−22

44−33−22

33−4411]- [ͲͲ

ͲͲ

ͲͲ] = [ͲͲͲ

ͲͲͲ

ͲͲͲ]

𝐴𝐴3 -6 𝐴𝐴2+ 11 𝐴𝐴 – 6I = 0 ………. (1)

Thus, A satisfy its characteristic equation.

To find A-1, multiply equation (1) by A-1 munotes.in

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27Chapter 1: Matrices

𝐴𝐴3 A-1 -6 𝐴𝐴2 A-1+ 11 𝐴𝐴 A-1 – 6 I A-1 = 0

𝐴𝐴2 -6 𝐴𝐴+ 11 ܫ – 6 A-1 = 0 ………………… [A A-1 = I, I = 1]

ฺ 6 A-1 = 𝐴𝐴2 - 6 𝐴𝐴+ 11 ܫ

=[2−32−2

14−15−4

11−13] – 6 [8−8−2

4−3−2

3−41] + 11 [1ͲͲ

Ͳ1Ͳ

ͲͲ1] =

[−1111Ͳ

−1Ͳ148

−88] A-1 = ଵ

[−1111Ͳ

−1Ͳ148

−88]

1.13 Similarity of matrices:

Matrix A and B of order n x n are said to be similar to each other if there exists an

invertible n x n matrix P, such that AP = PB i.e. B = P-1AP

For Similar matrices A, B, we have

i. |𝐴𝐴| = |ܤ|

Since A and B are similar, we have B = P-1AP

|| = |ܲିଵ| [ Taking determinant of both the side]

= |ܲିଵ||| || ฺ = |ܲିଵܲ||| ฺ = |ܫ||| ฺ = || [ As |ܫ| = 1]

|𝐴𝐴| = |ܤ|

ii. Characteristic equation for A and B are same.

If A and B are similar to each other then

|𝐴𝐴−ߣܫ| = |ܤ−ߣܫ|, for all real numbers ߣ.

|ܤ−ߣܫ| = |ܲିଵ−ߣܫ| [As B = P-1AP]

= |ܲିଵ−ߣܲିଵ| = |ܲିଵሺ−ߣሻ| = |ܲିଵ||ሺ−ߣሻ|||

= |ܲିଵܲ||ሺ−ߣሻ| = |||ሺ−ߣሻ| = |ሺ−ߣሻ|

Since |ܤ−ߣܫ| = |ሺ−ߣሻ|, the similar matrices A and B have same characteristic

equation.

1.14 Reduction of matrix to a diagonal matrix which has elements

as characteristics values

If a square matrix A of order n has n linearly independent eigen vectors, then a

matrix P can be found such that P-1AP is a diagonal matrix.

Proof: Let A be a square matrix of order 3. munotes.in

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28 APPLIED MATHEMATICS

Let ߣ1 ǡߣ2 ǡߣ3 be its eigen values and

X1 = [ݔଵ

ݕଵ

ݖଵ], X2 = [ݔଶ

ݕଶ

ݖଶ], X3 = [ݔଷ

ݕଷ

ݖଷ] be the corresponding eigen vectors.

Denoting the square matrix [X 1, X2, X3] by P.

P = [ݔଵݔଶݔଷ

ݕଵݕଵݕଷ

ݖଵݖଶݖଷ]

AP = A [X1, X2, X3] = [AX 1, AX 2, AX 3]

We know that, AX = ߣX

AX 1 = ߣ1X1, AX 2 = ߣ2X2, AX 3 = ߣ3X3

AP = [ ߣ1X1, ߣ2X2, ߣ3X3]

=[ߣଵݔଵߣଶݔଶߣଷݔଷ

ߣଵݕଵߣଶݕଵߣଷݕଷ

ߣଵݖଵߣଶݖଶߣଷݖଷ]

= [ݔଵݔଶݔଷ

ݕଵݕଶݕଷ

ݖଵݖଶݖଷ] x [ߣଵͲͲ

ͲߣଶͲ

ͲͲߣଷ]

= P D, where D is diagonal matrix.

P-1AP = P-1PD ֜ P-1AP = D

P constitute eigen vectors of A and is called Modal matrix of A.

D has eigen values as its diagonal elements and is called special matrix of A.

Ex. 1 Reduce the matrix A = [11−4−

−2−5

1Ͳ−4−] into a diagonal matrix.

Sol: We know that, D = P-1AP

The characteristic equation of A is |𝐴𝐴− ܫߣ |= 0

[11− ߣ−4−

−2−ߣ−5

1Ͳ −4−−ߣ] = 0

Here matrix A is 3 x 3

So Characteristic equation is given by ߣ3 - s1 ߣ2 + s2 ߣ - |𝐴𝐴| = 0

After solving for s 1, s2 and |𝐴𝐴| , we get characteristic equation as,

ߣ3 - 3 ߣ2 + 2 ߣ =0 ֜ ߣ( ߣ−1ሻ ( ߣ−2ሻ = 0 munotes.in

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29Chapter 1: Matrices

֜ ߣ =0, ߣ =1 and ߣ =2 are the eigen values.

Now consider [A - ܫߣ] [ X ]= [ 0 ]

[11− ߣ−4−

−2−ߣ−5

1Ͳ −4−−ߣ]ቈݔ

ݕ

ݖ=[Ͳ

Ͳ

Ͳ]

(11− ߣ) x – 4y - 7z = 0 ; 7x + (-2− ߣ)y - 5z = 0 ; 10x - 4y + (-6− ߣ)z = 0

Case i: ߣ1= 0 in above equations

11x - 4y – 7z = 0 ֜ 7x -2y – 5z = 0 ֜ 10x -4y -6z = 0

Now take any two equations. By rule of cross multiplication,

௫

ଶିଵସ = ି௬

ିହହାସଽ = ௭

ିଶଶାଶ଼ ֜ ௫

= ି௬

ି= ௭

֜ ௫

ଵ = ି௬

ିଵ= ௭

ଵ

X1 = (1, 1, 1)’ eigen vector corresponding to ߣ1 = 0

Case ii: ߣ2= 1 in main equations

10x - 4y – 7z = 0 ֜ 7x -3y – 5z = 0 10x -4y -7z = 0

Now take any two equations. By rule of cross multiplication,

௫

ଶିଶଵ = ି௬

ିହାସଽ = ௭

ିଷାଶ଼ ֜ ௫

ିଵ = ି௬

ିଵ= ௭

ିଶ

Divide by -1 , X2 = (1, -1, 2)’ eigen vector corresponding to ߣ2= 1

Case iii: ߣ3 = 2 in main equations

9x - 4y – 7z = 0 ֜ 7x -4y – 5z = 0 ֜ 10x -4y -8z = 0

Now take any two equations. By rule of cross multiplication,

௫

ି଼ = ି௬

ସ = ௭

ି଼

Divide by -4, ֜ ௫

ଶ = ௬

ଵ= ௭

ଶ

Take -1 common

X3 = (2, 1, 2)’ eigen vector corresponding to ߣ3= 2

P = [112

1−11

122]

Now for P-1

We know that, P-1= ଵ

|| adj P

We know how to calculate P-1 and adj P. munotes.in

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30 APPLIED MATHEMATICS

After calculation we get P-1 = 1 and adj P = [−423−1Ͳ13−1−2]

P-1= ଵ

ଵ [−423

−1Ͳ1

3−1−2]

D = P-1AP

= [−423

−1Ͳ1

3−1−2][11−4−

−2−5

1Ͳ−4−][112

1−11

122]

= [−423

−1Ͳ1

3−1−2][Ͳ14

Ͳ−12

Ͳ24] ֜ D = [ͲͲͲ

Ͳ1Ͳ

ͲͲ2]

Whatever the eigen values is appear in the diagonal matrix.

1.15 Summary

In this chapter, we learned about types of matrices, matrix operations and a system

of simultaneous linear equations in matrix form. We now understand what is

adjoint of a matrix, invertible matrix and rank of a matrix and methods finding

these. Students can s olve a system of linear equations by row -reducing its

augmented form . Students differentiated between Characteristics roots and characteristics vectors also able to reduce a matrix to a diagonal matrix .

1.16 References

1. Applied Mathematics II by P. N. Wartikar and J. N. Wartikar

2. Higher Engineering Mathematics by Dr. B. S. Grewal

3. Fundamentals of Matrix Computation by David S. Watkins

1.17 Exercise

Ex 1. If A = [3−5

−42] Show that A2 – 5A – 14I = 0

Ex 2. A = [4−1−4

3Ͳ−4

3−1−3], show that A2 = I munotes.in

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31Chapter 1: Matrices

Ex 36KRZWKDW$%%$LQHDFKRIWKHIROORZLQJFDVHV

a. A = [5−1

] and B = [21

34] b. A = [123

Ͳ1Ͳ

11Ͳ] and B = [−11Ͳ

Ͳ−11

234]

Ex 4. Find the inverse of given matrices

1ሻ[41

23] 2) [125

1−1−1

23−1] 3)[2−33

223

3−22]

Ex 5. If A = [23

5−2], show that A-1 = ଵ

ଵଽ A

Ex 6. If A = [23

5−2], show that A2 +3A +I = 0 and hence find A-1.

Ex 7. Find the rank of the following matrices

1. A = [2−13

1Ͳ1

Ͳ2−1

114]; 2. A = [1−11−1

−11−11

1−11−1

−11−11]

Ex 8. If A = [12−3

232

33−4] , find A-1 and hence solve the system of linear equations:

x + 2y - 3z = -4; 2x + 3y +2z = 2 ; 3x -3y - 4z = 1 [ Ans: x= 3, y = -2 , z = 1]

Ex 9. Use matrix method to show that the following system of equations is

inconsistent: 3x - y +2z = 3:2x + y +3z = 5; x -2y - z = 1

Ex 10. Show that the matrix A = [−22

−23−1

2−13]

satisfies its characteristic equation and hence determine A-1.

Ex 11. Show that the matrix A = [4−3−3

3−2−3

−112]

satisfies its characteristic equation and hence determine A-1.

Ex 12. Reduce the matrix A = [−13

−24] into a diagonal matrix. [ Ans: D = [1Ͳ

Ͳ2] ]

Ex 13. Reduce matrix A = [−19

−421] into a diagonal matrix.[Ans: D = [2Ͳ

Ͳ−5]

munotes.in

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32 APPLIED MATHEMATICS

32Unit I

2 COMPLEX NUMBERS

Unit structure

2.0 Objectives

2.1 Introduction

2.2 Complex number

2.3 Equality of complex numbers

2.4 Graphical representation of complex number (Argand’s Diagram),

2.5 Polar form of complex numbers

2.5.1 Polar form of x+iy for different signs of x,y,

2.6 Exponential form of complex numbers,

2.7 Mathematical operation with complex numbers and their representation

on Argand’s Diagram

2.8 Circular functions of complex angles

2.9 Definition of hyperbolic function

2.10 Relations between circular and hyperbolic functions

2.11 Inverse hyperbolic functions

2.12 Differentiation and Integration

2.13 Graphs of the hyperbolic functions

2.14 Logarithms of complex quality

2.15 j(=i) as an operator (Electrical circuits)

2.16 Summary

2.17 References

2.18 Exercise

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33Chapter 2: Complex Numbers

2.0 Objective

After going through this chapter, students will able to

• Compute sums, products, quotients, conjugate, modulus and argument of

complex numbers.

• Understand the graphical representation of complex numbers

• Write the complex numbers in polar form, exponential form

• Learn about circular, hyperbolic function, inverse hyperbolic function

• Obtain relations between circular and hyperbolic functions

• Learn about graphs of the hyperbolic functions and logarithms of complex

quality

2.1 Introduction:

This chapter is concerned with the representation and manipulation of complex

numbers. It has some introductory ideas associated with complex numbers, their

algebra and geometry, algebraic properties of complex numbers, Argand plane and

polar representation of complex numbers, exponential form of complex numbers,

mathematical operation with complex numbers and their representation on Argand’s diagram, circular functions of complex angles, hyperbolic functions,

relations between circular and hyperbolic functions, Inverse hyperbolic functions,

graphs of the hyperbolic functions. This includes how comp lex numbers add and

multiply, and how they can be represented graphically. Finally, we look the

logarithms of complex quality and application of complex number in electrical

circuit.

2.2 Complex number:

Imaginary Numbers: If the square of a given number is negative then such a

number is called an imaginary number.

Eg. ξെͳ, ξെʹ are imaginary numbers.

We denote ξെͳ as i.

Thus, ξെͶ = 2i, ξെͻ = 3i and ξെͷ = i ξͷ

Powers of i:

i0 = 1, i1 = i, i2 = -1, i3 = i2 x i= (-1) x i = -i, i4 = i2 x i2 = (-1) x (-1) =1 munotes.in

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34 APPLIED MATHEMATICS

Thus,

Complex Numbers: The numbers of the form ( a + ib), where a and b are real

numbers and i = ξെͳ, are known as complex numbers. The set of all complex

numbers is denoted by C.

C = {(a + ib):a , b אܴሽ

Eg. Each of the numbers (5 + 6i), ( -4 + ξ͵i), and (ଷ

ସ - ହ

i)

is a complex number.

For a complex number, z = (a + ib),

a = real part of z, written as Re (z) and b = imaginary part of z,

written as Im (z).

If z = (5 + 6i) then Re(z) = 5 and Im(z) = 9.

Purely Real and Purely Imaginary Numbers:

A complex numbers z is said to be

i. Purely real, if Im(z) = 0

ii. Purely imaginary, if Re(z) = 0

Thus, each of the numbers 2, -8, ξͶ is purely real and 3i, ( ξͷ i), - ହ

i is purely

imaginary.

Conjugate of a Complex Number :

Conjugate of a complex number z = (a + ib) is defined as, ݖҧ = (a - ib).

Eg, ሺ͵Çሻതതതതതതതതതതതത = (3 - 7i)

Modulus of Complex Number :

Modulus of complex number Z = (a + ib), denoted by ȁݖȁ = ξܽଶܾଶ.

Eg. If z = (2 + 3i) then ȁݖȁ = ξʹଶ͵ଶ = ξͳ͵

If z = ( -5 - 4i) then ȁݖȁ = ඥሺെͷሻଶሺെͶሻଶ = ξͶͳ

2.3 Equality of Complex Number :

If z 1 = a 1 +ib 1 and z 2 = a 2 +ib 2 then z1 = z 2 ֞ a1 = a 2 and b 1 = b 2.

Ex. If 2y + (3x - y) i = (5 - 2i), find the values of x and y. i0 = 1, i1 = i, i2 = -1, i3 = -i, i4=1 munotes.in

## Page 35

35Chapter 2: Complex Numbers

Sol: Equating the real and imaginary parts, we get

2y + (3x - y) i = (5 - 2i) 2y = 5 and 3x – y = -2

y= ହ

ଶ and 3x - ହ

ଶ = -2 y= ହ

ଶ and x = ଵ

Hence x = ଵ

and y= ହ

ଶ

Sum and Difference of Complex Number:

If z 1 = (a 1 +ib 1) and z 2 = (a 2 +ib 2) then

i. z1 + z 2 = (a 1 + a 2) + i (b 1 + b 2) ii. z1 - z2 = (a 1 - a2) + i (b 1 - b2)

Ex. i. z 1 = (3 + 5i) and z 2 = (- 5 + 2i) then

z1 + z 2 = {(3 + ( -5)} + i (5 + 2) = ( -2 + 7i)

z1 - z2 = {(3 - (-5)} + i (5 - 2) = (8 + 3i)

Properties of Addition of Complex Numbers:

i. Closure Property: The sum of two complex numbers is always a complex

number.

ii. Commutative Law: Addition of two complex numbers is commutative.

For any two complex numbers z 1 and z 2, z1 + z 2 = z 2 + z 1, for all z 1, z2 א C

iii. Associative Law: Addition of three complex numbers is associative.

For any complex numbers z 1, z2 and z 3,

(z1 + z 2) + z 3 =z1 + (z2 + z 3) for all z 1, z2, z3 א C

iv. Existence of Additive Identity: For any complex numbers z ,

z + 0 = 0 + z = z ,0 is the additive identity for complex number.

v. Existence of Additive Identity: For any complex numbers z ,

z + (-z) = ( -z) + z = 0

Thus, every complex number z has ( -z) as its additive inverse.

Multiplication of Complex Numbers:

Let z 1 = (a 1 +ib 1) and z 2 = (a 2 +ib 2) then z1 z2 = (a 1 +ib 1) (a 2 +ib 2)

= (a1 a2 - b1b2) + i (a 1b2 +b1a2)

z1 z2 = {Re(z 1). Re(z 2) - Im(z 1). Im(z 2)}+ i{Re(z 1).

Im(z 2) - Im(z 1). Re (z2)}

Ex. 1. Let z 1 = (4 + 2i) and z 2 = (6 + 3i) then munotes.in

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36 APPLIED MATHEMATICS

z1 z2= (4.6 – 2.3) + i (4.3 + 6.2) = (24 - 6) + i (12 + 12) = 18 + 24i

Properties of Multiplication of Complex Numbers:

a. Closure Property: The product of two complex numbers is always a

complex number.

b. Commutative Law: Multiplication of two complex numbers is commutative.

For any two complex numbers z 1 and z 2,

z1.z2 = z 2.z1, for all z 1, z2 א C

c. Associative Law: Multiplication of three complex numbers is associative.

For any complex numbers z 1, z2 and z 3,

(z1.z2). z 3 =z1. (z2.z3) for all z 1, z2, z3 א C

d. Existence of Multiplicative Identity: the complex number

(1 + i0) is multiplicative identity in C.

Let z = (a + ib) then

z x 1 = (a + ib). (1 + i0) = {(a.1 -b.0) + i(a.0 + b.1)} = (a +ib) = z

Similarly, z x 1 = 1x z = z for all z א C

Hence, the complex number 1 = (1 + i0) is the multiplicative identity .

e. Existence of multiplicative Identity:

Let z = (a + ib) then

z-1 = ଵ

௭ = ଵ

ሺୟା୧ୠሻ = ଵ

ሺୟା୧ୠሻx ሺୟି୧ୠሻ

ሺୟି୧ୠሻ = ሺୟି୧ୠሻ

మାమ

Clearly, z x z-1 = z-1 x z = 1

Thus, every z = (a + ib) has its multiplicat ive inverse , given by,

z-1 = ଵ

௭ = ሺୟି୧ୠሻ

మାమ= ௭ҧ

ȁ௭ȁమ zz-1 = ȁݖȁ2

Points to remember:

1. z = (a + ib) ֜ ݖҧ = (a - ib) and ȁݖȁ2 = ሺܽଶܾଶ)

2. z = (a + ib) ֜ z-1 = ௭ҧ

ȁ௭ȁమ = ሺୟି୧ୠሻ

మାమ

f. Distributive Laws: For any complex numbers z 1, z2 and z 3,

z1. (z 2+ z 3) = z 1z2 + z 1z3

(z1 + z 2).z3 = z 1z3 + z 2z3 for all z 1, z2, z3 א C munotes.in

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37Chapter 2: Complex Numbers

Division of Complex Numbers:

Let z 1 and z 2 be complex numbers such that z 2 ് 0 then

ଵ

ଶ = z 1. ଵ

ଶ = z 1. z2-1.

Eg. Find ௭ଵ

௭ଶ when z1 = ሺ6+3i) and z 2 = ሺ3 – i)

Sol: We have ଵ

ଶ = z 1. z2-1.

z2-1= ௭ଶതതതതȁ௭ଶȁమ = ሺଷିపሻതതതതതതതതڿଷమାሺିሻమۀ =ሺଷା୧ሻ

ଵ

ଵ

ଶ = z 1. z2-1

= ሺ6+3i). ሺଷା୧ሻ

ଵ = ሺାଷ୧ሻǤሺଷା୧ሻǤ

ଵ = ሺǤଷିଷǤଵሻା୧ሺǤଵାଷǤଷሻǤ

ଵ = ሺଵହାଵହ୧ሻǤ

ଵ = ଵହሺଵା୧ሻǤ

ଵ = ଷሺଵା୧ሻǤ

ଶ

Some Identities on Complex Numbers:

For any complex numbers z1 and z 2,

i. ሺz1 + z 2)2= z 12 + z 22 + 2 z 1z2

ii. ሺz1 - z2)2= z 12 + z 22 - 2 z1z2

iii. ሺz12 - z22) = ሺz1 + z 2)ሺz1 - z2)

iv. ሺz1 + z 2)3= z 13 + z 23 + 3 z 1z2ሺz1 + z 2)

v. ሺz1 + z 2)3= z 13 - z23 - 3 z1z2ሺz1 - z2)

Students can solve these identities as exercise .

2.4 Graphical representation of Complex Number

(Argand’s Diagram):

Complex Plane or Argand Plane:

Let X’OX and YOY’ be the mutually perpendicular lines, known as the x axis and

the y axis respectively. The complex number (x + iy) corresponds to the ordered

pair (x, y) and it can be represented by the point P(x, y) in the x -y plane. The x -y

plane is known as the complex plane or Argand plane. X axis is called the real axis

and y axis is called the imaginary axis.

munotes.in

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38 APPLIED MATHEMATICS

Note that every number on the x axis is a real number, while each on the y axis is

an imaginary number.

The complex numbers represented geometrically in the above diagram are

(2 + 4i), ( -3 + 2i), ( -4 -3i), (3 - 4i), (5 + 0i), ( -4 + 0i), (0 + 3i), (0 - 3i)

Represented by the points, A (2,4), B ( -3, 2), C ( -4, -3), D (3, -4), E (6,0),

F (-3, 0), G (0, 2) and H (0, -3) respectively.

2.5 Polar form of a Complex Number :

Let the complex number z = x + iy be represented by the point P (x, y) in the

complex plane. Let ס XOP = ߠ and ȁܱܲȁ= r > 0.

Then, P (r,ߠ )are called the polar coordinates of P.

We call the origin O as pole.

Clearly, x = r cos ߠ and y = r sinߠ

We have, z = x + iy = r cos ߠ +i r sin ߠ

= r (cos ߠ +i sinߠ.)

This is called the polar form, or trigonometric form, or modulus -amplitude form,

of z.

munotes.in

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39Chapter 2: Complex Numbers

Here, r =ඥݔଶݕଶ ൌȁݖȁ is called the modulus of z.

And ߠ is called the argument, or amplitude of z, written as arg (z), or amp (z).

The value of ߠ such that -ߨ൏ߠߨ is called the principal argument of z.

2.5.1 Polar form of x + iy for different signs of x, y: -

Method for finding the Principal Argument of a Complex Number

Case I When z = (x +iy) lies on one of the axes:

I. When z is purely real. In this case, z lies on the x axis.

i. If z lies on positive side of the x axis, then ߠ =0.

ii. If z lies on negative side of the x axis, then ߠ =ߨ.

II. When z is purely imaginary. In this case, z lies on the y axis.

i. If z lies on the y axis and above the x axis then ߠ =గ

ଶ.

ii. If z lies on the y axis and above the x axis then ߠ =గ

ଶ.

Case II When z = (x +iy) does not lies on any axes:

Step 1. Find the acute angle ߙ by tan ߙ =ቚூሺሻ

ோሺሻቚ.

Step 2. Find the qua drant in which P (x, y) lies.

Then, ߠ =arg (z) may be obtained as under.

i. When z lies in quad I; Then, ߠ =ߙ ֜ arg (z) = ߙ

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40 APPLIED MATHEMATICS

ii. When z lies in quad I I; Then, ߠ =ሺߨെߙ ሻ֜ arg (z) = ሺߨെߙ ሻ

iii. When z lies in quad I II; Then, ߠ =ሺߙെߨሻݎሺߨߙ ሻ֜ arg (z) =

ሺߙെߨሻݎሺߨߙ ሻ

iv. When z lies in quad I V; Then, ߠ =െߙ ֜ arg (z) = - ߙ

Ex. 1. For following complex numbers find the polar form.

i. z = (1+i ξ͵) ii. z = (-1- iξ͵)

Sol. i. Let z = (1+i ξ͵) i.e. x=1 and y = ξ͵

We know that, Polar form = r (cos ߠ݅ߠሻ

We have to find e and ߠ

We know that, r = ඥݔଶݕଶ = =ξͳ͵ = 2

r = 2

Let tan ߙ =ቚூሺሻ

ோሺሻቚ = ቚ௬

௫ቚ = ቚξଷ

ଵቚ = ξ͵

tan ߙൌగ

ଷ [ tan 60 = ξ͵, tan 60 = గ

ଷǤߨൌͳͺͲሿ

ߙൌగ

ଷ

points (1, ξ͵) lies in I quad , ߠ =ߙ ߠ =గ

ଷ

Polar form of z = 2 (cos గ

ଷ + i sin గ

ଷ )

ii. Let z = (-1- iξ͵) i.e. x= - 1 and y = - ξ͵

Sol: We know that, Polar form = r (cos ߠ݅ߠሻ

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## Page 41

41Chapter 2: Complex Numbers

We have to find e and ߠ

We know that, r = ඥݔଶݕଶ = =ξͳ͵ = 2 r = 2

Let tan ߙ =ቚூሺሻ

ோሺሻቚ = ቚ௬

௫ቚ = ቚିξଷ

ିଵቚ = ξ͵

tan ߙൌగ

ଷ [ tan 60 = ξ͵, tan 60 = గ

ଷǤߨൌͳͺͲሿ

ߙൌగ

ଷ

points ( - 1, - ξ͵) lies in I II quad , ߠ =ߙെߨ

ߠ =గ

ଷ – ߨ =గିଷగ

ଷ = ିଶగ

ଷ

Polar form of z = 2 (cos ିଶగ

ଷ + i sin ିଶగ

ଷ )

Exercise:

Ex 1. If z is a non-zero complex number, such that 2 iz2 = ݖҧ the find ȁݖȁ

[Ans: ȁݖȁ=1/2]

Ex. 2 If | z | = 1, then find the value of ଵା௭

ଵା௭ҧ.

[Ans: z]

2.6 Exponential form of Complex Numbers :

We know that if x is a real number, then

݁௫= 1 + x + ௫మ

ଶǨ + ௫య

ଷǨ + ௫ర

ସǨ + …… …………….. (1)

Assuming this is true for all values of x (real or complex)

Let substitute i ߠ for x in equation (1)

݁୧ఏ= 1 + i ߠ +మఏమ

ଶǨ + యఏయ

ଷǨ + రఏర

ସǨ+ ……………………

Put i2 = -1

݁୧ఏ= 1 + i ߠ - ఏమ

ଶǨ - ఏయ

ଷǨ + ఏర

ସǨ+ ……………………

݁୧ఏ = (1 - ఏమ

ଶǨ + ఏర

ସǨ - ఏల

Ǩ……) + i ( ߠ - ఏయ

ଷǨ + ఏఱ

ହǨ - ఏఱ

Ǩ………)

We know that,

sinߠ =ߠ - ఏయ

ଷǨ + ఏఱ

ହǨ - ఏఱ

Ǩ…… and cos ߠ =1 - ఏమ

ଶǨ + ఏర

ସǨ - ఏల

Ǩ…………

݁୧ఏ = (cosߠ +i sinߠ) munotes.in

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42 APPLIED MATHEMATICS

Complex number z = x +iy (cartesian form)

= r (cos ߠ +i sin ߠ) (Polar form)

= r ݁୧ఏ (Exponential form)

Exponential form of x + iy = r ܑࣂ

݁୧ఏ = (cos ߠ +i sin ߠ) and ݁ି୧ఏ = (cos ߠ - i sinߠ)

2.7 Mathematical operation with complex numbers and their representation on Argand’s Diagram :

1. Addition of Complex Numbers:

Let z 1 and z 2 be two complex numbers.

z1= (x 1 +iy 1) and z2= (x 2 +iy 2)

z1 + z 2 = (x 1 + x 2) + i (y 1 + y 2)

Graphical representation (Argand’s diagram):

Represent the complex numbers z 1 and z 2 by vectors ͳሬሬሬሬሬሬሬሬԦ and ʹሬሬሬሬሬሬሬሬԦ

respectively.

Now complete the parallelogram OP 1P3P2.

By properties of parallelograms, opposite sides of parallelogram are equal

and diagonals of parallelogram bisect each other.

0 (0,0) and P (௫ଵା௫ଶ

ଶ, ௬ଵା௬ଶ

ଶ)

We can calculate coordinates of P 3.

Let consider P 3(X, Y)

coordinates of P 3,

ା

ଶൌ ௫ଵା௫ଶ

ଶ ֜ X = (x 1 + x 2)

ା

ଶൌ ௬ଵା௬ଶ

ଶ ֜ Y = (y 1 + y 2)

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43Chapter 2: Complex Numbers

coordinates of P 3(x1 + x 2, y1 + y 2)

If we represent P3 in complex number as z 3

z3 = (x 1 + x 2) + i (y 1 + y 2)

= x 1 + iy1 + x2 + iy 2

z3 = z 1 + z2 ሾ z1= (x 1 +iy 1) and z2= (x 2 +iy 2)]

2. Subtraction of Complex Numbers:

Let z 1 and z 2 be two complex numbers.

z1= (x 1 +iy 1) and z2= (x 2 +iy 2)

z1 - z2 = (x 1 - x2) + i (y 1 - y2)

Graphical representation (Argand’s diagram):

Represent the complex numbers z 1 and z 2 by vectors ͳሬሬሬሬሬሬሬሬԦ and ʹሬሬሬሬሬሬሬሬԦ

respectively.

Take negative of complex number of z2

Now complete the parallelogram OP 3P4 P1.

By properties of parallelograms, opposite sides of parallelogram are equal

and diagonals of parallelogram bisect each other.

o (0,0) and P (௫ଵି௫ଶ

ଶ, ௬ଵି௬ଶ

ଶ)

We can calculate coordinates of P 4.

Let consider P 4(X, Y)

coordinates of P 4,

ା

ଶൌ ௫ଵି௫ଶ

ଶ ֜ X = (x 1 - x2)

ା

ଶൌ ௬ଵି௬ଶ

ଶ ֜ Y = (y 1 - y2)

coordinates of P 3(x1 - x2, y1 - y2)

If we represent P 4 in complex number as z 4

Z4 = (x 1 - x2) + i (y 1 - y2)

= (x1 +iy1) - (x2 + iy2)

Z4 = z 1 - z2 ሾ z1= (x 1 +iy 1) and z2= (x 2 +iy 2)]

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44 APPLIED MATHEMATICS

3. Multiplication of Complex Numbers:

Let z 1 and z 2 be two complex numbers.

z1= (x 1 +iy 1) and z2= (x 2 +iy 2)

z1.z2 = (x 1 - x2). i (y 1 - y2)

= x 1x2 +ix 1y2+ix 2y1 - y1y2 [as i2 = -1]

= (x 1x2 - y1y2) +i (x 1y2+x2y1)

Let consider the complex numbers in polar form.

Let z 1 = (x 1 +iy 1) = r 1(cosߠ 1+ i sin ߠ 1) = r 1 ݁ఏଵ

z2 = (x 2 +iy 2) = r 2(cosߠ 2+ i sin ߠ 2) = r 2݁ఏଶ

Then z1. z2 = r1 ݁ఏଵ. r2݁ఏଶ

= r1r2݁ሺఏଵାఏଶሻ

= r1r2 [cosሺߠͳߠʹ) + i sin ሺߠͳߠʹሻ]

The product of the complex numbers is a complex number whose modulus

is the product of their moduli and whose amplitude is the sum of their

amplitudes.

Graphical representation (Argand’s diagram):

Let P 1 represent z1 = r1(cosߠ1+ i sin ߠ1),

P2 represent z2 = r2(cosߠ2+ i sin ߠ2) and OM = 1 unit

We get ȟ OP 1M.

Construct the ȟ OP 3P2 similar to ȟ OP 1M.

For modulus, ைଷ

ைଵ = ைଶ

ைெ ֜ைଷ

ଵ = ଶ

ଵ ֜ OP 3 = r1r2

To calculate argument,

סXOP 3 = סXOP 2 + סP2OP 3 = ߠ2 + ߠ1 = ߠ1 + ߠ2

P1(r1, ߠ1) represents the complex number r 1(cosߠ1+ i sin ߠ1) and

P2(r2, ߠ2) represents the complex number r 2(cosߠ2+ i sin ߠ2).

Similarly, P3(r1r2, ߠ1+ߠ2) represents r 1r2 [cosሺߠͳߠʹ) + i sin ሺߠͳߠʹሻ]

whose modulus is the product of their moduli and whose amplitude is the

sum of their amplitudes.

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45Chapter 2: Complex Numbers

Hence z 1. z2 giving simple graphical construction for a product.

4. Quotient of Complex Numbers:

The product of two conjugate complex numbers is a real number i.e (x +iy)

(x - iy)= x2 + y2 leads to the following method of division, where the

denomination is always expressed as a real number.

Let z 1 and z 2 be two complex numbers.

z1= (x 1 +iy 1) and z2= (x 2 +iy 2)

Thus, ୶ଵା୧୷ଵ

୶ଶା୧୷ଶ = ୶ଵା୧୷ଵ

୶ଶା୧୷ଶ . ୶ଶି୧୷ଶ

୶ଶି୧୷ଶ

= ሺ୶ଵǤ୶ଶା୷ଵǤ୷ଶሻା୧ሺ୶ଶǤ୷ଵି୶ଵǤ୷ଶሻ

௫ଶమା௬ଶమ

= ሺ୶ଵǤ୶ଶା୷ଵǤ୷ଶሻ

௫ଶమା௬ଶమ + ୧ሺ୶ଶǤ୷ଵି୶ଵǤ୷ଶሻ

௫ଶమା௬ଶమ

But it is more convenient to divide the complex numbers in their polar

forms or better in exponential form.

୶ଵା୧୷ଵ

୶ଶା୧୷ଶ = ୰ଵഇభ

୰ଶഇమ = ୰ଵ

୰ଶ ݁ሺఏଵିఏଶሻ

୶ଵା୧୷ଵ

୶ଶା୧୷ଶ = ୰ଵഇభ

୰ଶഇమ = ୰ଵ

୰ଶ ሼ
ሺߠͳെߠʹሻെ݅ݏ݅݊ሺߠͳെߠʹሻ}

The modulus of the quotient of two complex number s is the quotient of

their moduli and amplitude of the quotient is the difference of their

amplitudes.

Graphical representation (Argand’s diagram):

Let ߠ1 > ߠ2

Let P 1 represent z1 = r1(cosߠ1+ i sin ߠ1),

P2 represent z2 = r2(cosߠ2+ i sin ߠ2) and OM = 1 unit along X axis.

Construct ȟ OP 1 P2 similar to ȟ OP 3M.

ைଵ

ைଷ = ைଶ

ைெ ֜ଵ

ைଷ = ଶ

ଵ ֜ OP 3 = ଵ

ଶ

סXOP 3 = סXOP 1 - סXOP 2 = ߠ1 - ߠ2

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## Page 46

46 APPLIED MATHEMATICS

P1(r1, ߠ1) represents the complex number r 1(cosߠ1+ i sinߠ1) and

P2(r2, ߠ2) represents the complex number r 2(cosߠ2+ i sinߠ2).

We get , P3 = ( ଵ

ଶ, ሺߠ1-ߠ2) gives complex number

ଵ

ଶ [cosሺߠ1-ߠ2) – i sin ሺߠͳെߠʹሻ] which is equal to ௭ଵ

௭ଶ.

ቚ௭ଵ

௭ଶቚ = ȁ௭ଵȁ

ȁ௭ଶȁ and arg ௭ଵ

௭ଶ = arg z 1 -arg z 2

5. Powers of Complex Numbers DeMoivre’s Theorem:

Statement: If n is any real number, one of the values of (cosߠ +i sin ߠ)n is

cos݊ߠ +i sin ݊ߠ.

Proof: Here we consider three cases.

i. n is positive integer ii. n is negative integer and iii. n is a fraction

i. Let n is positive integer:

(cosߠ +i sin ߠ)n = (cos ߠ +i sin ߠ( )cos ߠ +i sin ߠ..……)n times

= cos [ ߠߠ ……. n times] + i sin [ ߠߠ ……. n times]

= cos݊ߠ +i sin ݊ߠ

ii. Let n is negative integer:

Let n = -m, where m is a positive integer

(cosߠ +i sin ߠ)n = (cos ߠ +i sin ߠ)-m

= ଵ

ሺୡ୭ୱఏା୧ୱ୧୬ఏሻౣ [ a -m = ଵ

]

= ଵ

ୡ୭ୱ୫ఏା୧ୱ୧୬୫ఏ [ from ( i)]

= ଵ

ୡ୭ୱ୫ఏା୧ୱ୧୬୫ఏ . ୡ୭ୱ୫ఏି୧ୱ୧୬୫ఏ

ୡ୭ୱ୫ఏି୧ୱ୧୬୫ఏ

= ୡ୭ୱ୫ఏି୧ୱ୧୬୫ఏ

௦మఏା௦మఏ [ i2 = - 1]

=
ߠെߠ [ ݏܿଶ݉ߠݏ݅݊ଶ݉ߠൌͳሿ

=
ሺെሻߠሺെሻߠ

=
݊ߠ݊ߠ

iii. Let n be a fraction:

n =

,where p and q are + ve or -ve integer.

from (i) and (ii) we have,

(cos ఏ

+ i sin ఏ

)q = cos ߠ +i sin ߠ

ሺ
ߠߠሻభ

= cosఏ

+ i sinఏ

ሺ
ߠߠሻ = ሺ
ߠߠሻ

[ n =

] munotes.in

## Page 47

47Chapter 2: Complex Numbers

= [ ሺ
ߠߠሻభ]p = [cosఏ+ i sinఏ]p

= cos

ߠ +i sin

ߠ

ሺ
ߠߠሻ = cos ݊ߠ +i sin n ߠ

Ex 1. ([SUHVVVLQșDQGFRVșLQWHUPVRISRZHUVRIFRVșDQGVLQș

Sol: 8VLQJGH0RLYUH¶VWKHRUHP

FRVși VLQș FRVșLVLQș3

= (cos3 șíFRVșVLQ2 șLVLQșFRV2 șíVLQ3 ș

We can equate the real and imaginary coefficients separately,

LHFRVș FRV3 șíFRVșVLQ2 ș FRV3 șíFRVș

VLQș VLQșFRV2 șíVLQ3 ș = VLQșíVLQ3 ș

([,I] FRVșLVLQșVKRZWKDW]n + ଵ

௭ = 2 cos n șDQG]n - ଵ

௭ = 2 isin n ș

Sol: Let z = FRVșLVLQș

%\GH0RLYUH¶VWKHRUHP

zn FRVșLVLQșn FRVQșLVLQQș

ଵ

௭ = z-n FRVQș– LVLQQș

zn + ଵ

௭ = FRVQșLVLQQșFRVQș – LVLQQș = FRVQș

Also, zn - ଵ

௭ = FRVQșLVLQQș - FRVQș– LVLQQș = VLQQș

Ex 3 Simplify ( ଵାୡ୭ୱଶା୧ୱ୧୬ଶ

ଵାୡ୭ୱଶି୧ୱ୧୬ଶ )30

Sol: Let Z = FRVșLVLQș

As ȁݖȁ = ȁݖȁ2 = zݖҧ = 1, we get ݖҧ = ଵ

௭ FRVș- LVLQș

ଵାୡ୭ୱଶା୧ୱ୧୬ଶ

ଵାୡ୭ୱଶି୧ୱ୧୬ଶ = ଵା௭

ଵାభ

= ሺଵା௭ሻ௭

௭ାଵ = z

(ଵାୡ୭ୱଶା୧ୱ୧୬ଶ

ଵାୡ୭ୱଶି୧ୱ୧୬ଶ )30 = Z30 = FRVșLVLQș30 = FRVșLVLQș

Ex 4. Simplify (1+ i)18

Sol: Let 1 +i = r FRVșLVLQșWKHQZHJHW

r = ξͳଶʹଶ = ξʹ ; ߙ =tan -1 = ( ଵ

ଵ) = గ

ସ

ș ߙ = గ

ସ [ 1 + i lies in the first quadrant]

(1+ i) = ξʹ (cos గ

ସ + i sin గ

ସ)

Raising to power 18 on both sides

(1+ i)18 = ሾξʹ (cos గ

ସ + i sin గ

ସ)]18 = ξʹ 18(cos గ

ସ + i sin గ

ସ)18

By GH0RLYUH¶VWKHRUHP

(1+ i)18 = 29 (cos ଵ଼గ

ସ + i sin ଵ଼గ

ସ) = 29 (cos ଽగ

ଶ + i sin ଽగ

ଶ)

= 29 (cos [Ͷߨ +గଶ ]+ i sin ሾͶߨగଶሿ) = 29 (cos గଶ + i sin గଶ) = 29i = munotes.in

## Page 48

48 APPLIED MATHEMATICSEx 5. Find cube root of unity. Sol: we have to find 11/3 Let z = 11/3 i.e. z3 = 1 In polar form, z3 = 1 can be written as z3 = cos (0 + 2k ߨ + )i sin (0 + 2k ߨ )= ݁ଶ , k = 0, 1, 2,... z = (cos ଶగ

ଷ + i sin ଶగ

ଷ) = ݁మೖഏ

య, k = 0, 1, 2 Taking k = 0, 1, 2 we get, k = 0, z = (cos 0 + i sin 0) = 1 k = 1, z = (cos ଶగ

ଷ + i sin ଶగ

ଷ) = (cos ( ߨെగ

ଷ ) + i sin (ߨെగ

ଷ ) ) = - cosగ

ଷ + i sin గ

ଷ = - ଵ

ଶ + i ξଷ

ଶ k = 2, z = (cos ସగ

ଷ + i sin ସగ

ଷ) = (cos ( ߨగ

ଷ ) + i sin (ߨగ

ଷ ) ) = - cosగ

ଷ - i sin గ

ଷ = - ଵ

ଶ - i ξଷ

ଶ The cube root of unity are 1, ିଵାξଷ

ଶ, ିଵିξଷ

ଶ Exercise: Ex. 1 Simplify ሺെξ͵͵݅ሻ [Ans: 2 ξ͵ (cos ଶగ

ଷ + i sin ଶగ

ଷ) Ex. 2 Simplify ቀగ

݅
గ

ቁ18 [ Ans: 1] 2.8 Circular functions of complex angles: We know that, ݁୧௫ = (
ݔ +i ݔ) and ݁ି୧௫ = (
ݔ - i ݔ) ݁୧௫ + ݁ି୧௫= (
ݔ +i ݔ) +(
ݔ - i ݔ) =2
ݔ And ݁୧௫ - ݁ି୧௫= (
ݔ +i ݔ )- (
ݔ - i ݔ) =2 i ݔ
ݔ =ܑାషܑ

and ݔ ൌܑିషܑ

These are known as exponential values of the sine and cosine. For any non-real quantity z, where the geometrical definitions of sin z, cos z no longer have a meaning, we may regard them as defined as above so that, ܛܑܖࢠൌܑࢠିషܑࢠ

; ܗ܋ܛࢠ =ܑࢠାషܑࢠ

܉ܜܖࢠ =ࢠ

ࢠ =ܑࢠିషܑࢠ

ሺܑࢠାషܑࢠሻ; ܗ܋ܛ܋܍ࢠ =

ܛܑܖࢠ =

ܑࢠିషܑࢠ munotes.in

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49Chapter 2: Complex Numbers܋܍ܛࢠ =܋ܗܛࢠ =ܑࢠାషܑࢠ ; ܋ܜܗࢠ =ܜܖ܉ࢠ =ሺܑࢠାషܑࢠሻܑࢠିషܑࢠ

2.9 Definition of Hyperbolic Function :

Hyperbolic Functions: The hyperbolic functions are the complex analogues of the

trigonometric functions. The analogy may not be immediately apparent and their

definitions may appear at first to be somewhat arbitrary. However, careful examination of their properties reveals the purpose of the definitions. For example,

their close relationship with the trigonometric functions, both in their identities and

their calculus, means that many of the familiar propert ies of trigonometric functions

can also be applied to the hyperbolic functions.

Definitions: The two fundamental hyperbolic functions are cosh x and sinh x,

which, as their names suggest, are the hyperbolic equivalents of cos x and sin x.

They are defined by the following relations.

Hyperbolic cosine of x ,
ݔ =ೣାషೣ

ଶ

Hyperbolic sine of x , ݔ =ೣିషೣ

ଶ

ݔ is an even function and ݔ is an odd function. By analogy with the

trigonometric functions, the remaining hyperbolic functions are,

ݔ =ୱ୧୬୦௫

ୡ୭ୱ୦௫ = ሺೣିషೣሻ

ሺೣାషೣሻ ;
ݔ =ଵ

ୡ୭ୱ୦௫ = ଶ

ሺೣାషೣሻ

ݔ =ଵ

ୱ୧୬୦௫ = ଶ

ሺೣିషೣሻ ;
ݔ =ଵ

୲ୟ୬୦௫ = ሺೣାషೣሻ

ሺೣିషೣሻ

Identities of Hyperbolic function:

1. ሺെݔሻ= - ݔ 2.
ሺെݔሻ=
ݔ

3. ሺെݔሻ= - ݔ 4. 1 - ଶݔ =
ଶݔ

5.
ଶݔ - ଶݔ =1 6. ሺሻ= ݔ
+
ݔ

7.
ሺሻ=
ݔ
+ ݔ

Now, we prove identity 5, rest of the identities can solve by students as exercise.

Prove that
ଶݔ - ଶݔ =1

Proof: L. H. S. =
ଶݔ – ଶݔ

=[ ೣାషೣ

ଶ]2 - [ ೣିషೣ

ଶ]2

= మೣାଶǤೣషೣାషమೣ

ସ - మೣିଶǤೣషೣାషమೣ

ସ munotes.in

## Page 50

50 APPLIED MATHEMATICS

= ଵସ ሾ݁ଶ௫ʹ݁ିଶ௫െ݁ଶ௫ʹെ݁ିଶ௫ሿ

= ଵ

ସ ሾͶሿ = 1= R.H.S

Prove that 1 - ଶݔ =
ଶݔ

Proof: Just now we proved,
ଶݔ - ଶݔ =1

Divide by
ଶݔ

ୡ୭ୱ୦మ௫

ୡ୭ୱ୦మ௫ - ୱ୧୬୦మ௫

ୡ୭ୱ୦మ௫ = ଵ

ୡ୭ୱ୦మ௫

1 - ଶݔ =
ଶݔ

2.10 Relations between Circular and Hyperbolic Functions:

By definitions of sin z and cos z,

ݖൌିష

ଶ and
ݖ =ାష

ଶ

Put z = ix

ሺሻൌ݁୧ሺ୧୶ሻെ݁ି୧ሺ୧୶ሻ

ʹ݅

=ష౮ି౮

ଶ [i2 = -1]

= ିଵ

ሾ౮ିష౮

ଶሿ = మ

ሾ౮ିష౮

ଶሿ = i ݔ

ሺሻൌሺ౮ሻାషሺ౮ሻ

ଶ = ష౮ା౮

ଶ =
ݔ

Thus, we have,

ሺሻ = i ݔ ;
ሺሻ =
ݔ ; ሺሻ = i ݔ

These definitions enable us to deduce the properties of hyperbolic functions from

those of circular functions.

I.
ଶݖ +ଶݖ =1.

ଶݖ +ଶݖ =ሺାష

ଶ )2 + ሺିష

ଶ )2

= ሺమାଶାషమ

ସ ) - ሺమିଶାషమ

ସ ) = ସ

ସ = 1 [i2 = -1]

II.
ଶݔ - ଶݔ =1.

Put z = ix in I

ଶሺሻ + ଶሺሻ =1 ֜
ଶݔ +ሺݏ݄݅݊ݔ)2 =1

֜
ଶݔ + 2ଶݔ =1 ֜
ଶݔ - ଶݔ =1 [i2 = -1]

III. sin (z 1 ± z2) = sin z 1.cos z 2 ± cos z 1. sin z 2 munotes.in

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51Chapter 2: Complex Numbers

Put z 1 = ix and z 2 = iy

sin i (x ± y) = sin (ix). cos (iy) ± cos (ix). sin (iy)

sinh (x ± y) = sinh x. cosh y ± cosh x. sinh y

Similarly, from the expansion of cos (z 1 ± z2), we get,

cosh (x ± y) = cosh x. cosh y ± sinh x. sinh y

We have following formulae for hyperbolic function which can be deduced from

those of circular functions by similar methods as illustrated above.

a. tanh (x ± y) = ୲ୟ୬୦௫േ୲ୟ୬୦௬

ଵേ୲ୟ୬୦௫୲ୟ୬୦௬

b. sinh x + sinh y = 2 sinh ௫ା௬

ଶ . cosh ௫ା௬

ଶ

sinh x - sinh y = 2 sinh ௫ି௬

ଶ . cosh ௫ି௬

ଶ

cosh x + cosh y = 2 cosh ௫ା௬

ଶ . cosh ௫ା௬

ଶ

cosh x - cosh y = 2 sinh ௫ା௬

ଶ . sinh ௫ା௬

ଶ

c.
ଶݔ =ଵ

ଶ (1 + cosh 2x) and ଶݔ =ଵ

ଶ (cosh 2x -1)

2.11 Inverse Hyperbolic Functions:

Let x and y be two complex numbers.

If ݕ =x then y is called the inverse hyperbolic sin of x and is written as y =

ିଵݔ .

ିଵݔ ,
ିଵݔ ,ିଵݔ etc are called inverse hyperbolic function.

1. Prove that ିଵݔ =log (x + ξݔଶͳ )

Proof: Let sinh y = x then y = ିଵݔ

sinh y = x ……………….. (1)

sinh2 y = x2 ………………. (squaring both the sides)

sinh2 y + 1 = x2 +1 ………... (adding 1 toboth the sides)

cosh2 y = x2 +1 ………….... ( cosh2ߠ – sinnh2ߠ=1)

cosh y = ξݔଶͳ…………(2) (Take square root )

Add (1) and (2)

sinh y + cosh y = x + ξݔଶͳ munotes.in

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52 APPLIED MATHEMATICS

If x is real, we have, cosh y = ౯ାష౯ଶ and sinh y = ౯ିష౯ଶ

౯ିష౯

ଶ + ౯ାష౯

ଶ = x + ξݔଶͳ ฺ ౯ିష౯ା౯ାష౯

ଶ = x + ξݔଶͳ

ଶ౯

ଶ = x + ξݔଶͳ ฺ ݁୷ = (x + ξݔଶͳ ) = y = log (x + ξݔଶͳ )

ିଵݔ =log (x + ξݔଶͳ ) ……… ( y = ିଵݔ )

2.
ିଵݔ =log (x + ξݔଶെͳ )

Proof: Let cosh y = x then y =
ିଵݔ

cosh y = x ……………….. (1)

cosh2 y = x2 ………………. (squaring both the sides)

cosh2 y - 1 = x2 -1 ………... (subtracting 1 from both the sides)

sinh2 y = x2 -1 ………….... ( cosh2ߠ – sinnh2ߠ=1)

sinh y = ξݔଶെͳ…………(2) (Take square root )

Add (1) and (2)

cosh y + sinh y = x + ξݔଶെͳ

If x is real, we have, cosh y = ౯ାష౯

ଶ and sinh y = ౯ିష౯

ଶ

౯ାష౯

ଶ + ౯ିష౯

ଶ = x + ξݔଶെͳ

౯ାష౯ା౯ିష౯

ଶ = x + ξݔଶെͳ

ଶ౯

ଶ = x + ξݔଶെͳ ฺ ݁୷ = (x + ξݔଶെͳ )

y = log (x + ξݔଶെͳ )

ିଵݔ =log (x + ξݔଶെͳ ) ……… ( y =
ିଵݔ )

3. Prove that ିଵݔ =ଵ

ଶ log ଵା௫

ଵି௫

Proof: Let tanh y = x then y = ିଵݔ

tanh y = x

౯ିష౯

౯ାష౯ = x ………………. ( tan x= ௦௫

ୡ୭ୱ௫ )

౯ାష౯

౯ିష౯ = ଵ

௫

౯ାష౯ା౯ିష౯

౯ାష౯ି౯ିష౯ = ଵା௫

ଵି௬…….. (if ௫

௬ =

then ௫ା௬

௫ି௬ = ା

ି) munotes.in

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53Chapter 2: Complex Numbers

౯ା౯ష౯ାష౯ = ଵା௫ଵି௬

ଶ౯

ଶష౯ = ଵା௫

ଵି௬

݁௬Ǥ ݁௬= ଵା௫

ଵି௬

݁ଶ௬= ଵା௫

ଵି௬

2y =log ଵା௫

ଵି௬

y = ଵ

ଶ log ଵା௫

ଵି௬

ିଵݔ =ଵ

ଶ log ଵା௫

ଵି௬ …………. ( y = ିଵݔ)

2.12 Differentiation and Integration :

a. y = ݏ݄ܿݔ ,ௗ௬

ௗ௫ = ݏ݄݊݅ݔ , ݏ݄݊݅ݔ݀ݔൌݏ݄ܿݔ

b. y = ݏ݄݊݅ݔ, ௗ௬

ௗ௫ = ݏ݄ܿݔ, ݏ݄ܿݔ݀ݔൌݏ݄݊݅ݔ

c. y = ܽݐ݄݊ݔ ,ௗ௬

ௗ௫ =
ଶݔ ,
ଶݔ݀ݔൌܽݐ݄݊ݔ

d. y = ିଵ௫

, ௗ௬

ௗ௫ ଵ

ξమା௫మ , ௗ௫

ξమା௫మൌ ିଵ௫

e. y =
ିଵ௫

, ௗ௬

ௗ௫ ଵ

ξ௫మାమ , ௗ௫

ξ௫మାమൌ
ିଵ௫

f. y = ିଵ௫

, ௗ௬

ௗ௫

మି௫మ , ௗ௫

మି௫మൌ ଵ

ିଵ௫

g. y =
ିଵ௫

, ௗ௬

ௗ௫ ି

௫ξమା௫మ , ௗ௫

௫ξమା௫మൌ െଵ

ିଵ௫

h. y =
ିଵ௫

, ௗ௬

ௗ௫ ି

௫ξమା௫మ , ௗ௫

௫ξమା௫మൌ െଵ

ିଵ௫

Series for cosh x and sinh x : -

݁௫ = 1 + x + ௫మ

ଶǨ + ௫య

ଷǨ + ………………..

݁ି௫ = 1 - x + ௫మ

ଶǨ - ௫య

ଷǨ + ………………..

cosh x = ଵ

ଶ ( ݁௫݁ି௫ሻ = 1 + ௫మ

ଶǨ + ௫ర

ସǨ + ………………..

sinh x = ଵ

ଶ ( ݁௫െ݁ି௫ሻ = x + ௫య

ଷǨ + ௫ఱ

ହǨ + ………………..

tanh x = ௦௫௦௫ = x - ଵଷǨ ݔଷ + ଶଵହ ݔହ+ ……………….. munotes.in

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54 APPLIED MATHEMATICS

2.13 Graphs of the hyperbolic functions :

First, we draw the graphs of ݁௫ǡ݁ି௫ and -݁ି௫

We know that,

ݔ =ೣାషೣ

ଶ , ݔ =ೣିషೣ

ଶ and ݔ =ೣିషೣ

ೣାషೣ

For
ݔǡ
ݔ =ೣାషೣ

ଶ = ೣ

ଶ + షೣ

ଶ

Note:
ݔ is an EVEN function . It is symmetric about Y axis and cosh ( -ݔ = )

ݔ

Domain: { x ܴ߳ሽand Range: {y ܴ߳ /y 1}

x՜ - λݐ݄݊݁
ݔ՜ λ and

x՜ λ then
ݔ ՜ λ

For ݔǡ

ݔ =ೣିషೣଶ = ೣଶ - షೣଶ

munotes.in

## Page 55

55Chapter 2: Complex Numbers

Note: ݔ is an ODD function and sinh ( -ݔ = )െݔ

Domain: { x ܴ߳ሽand Range: {y ܴ߳}

x՜ - λݐ݄݊݁ݔ՜ - λ and

x՜ λ then ݔ ՜λ

For ݔǡ

ݔ =ೣିషೣ

ೣାషೣ = ୱ୧୬୦௫

ୡ୭ୱ୦௫

Note: ݔ is an ODD function. It is symmetric about origin and tanh (-ݔ = )

െݔ

Domain: { x ܴ߳ሽand Range: {y ߳R / -1 < y < 1}

x՜ - λݐ݄݊݁ݔ՜ - 1and x ՜ λ then ݔ ՜ 1

The values of sinh x, cosh x and tanh x for x = - λ, 0 and + λfrom definition are

as follows

munotes.in

## Page 56

56 APPLIED MATHEMATICSx sinh x cosh x tanh x െλ െλ λ -1 0 0 1 0 λ λ λ 1 2.14 Logarithms of complex quality :

Let z = x + iy

Expressing the complex number in general polar form,

z = r (cosߠ +sin ߠ)

x + iy = r (cosߠ +sin ߠ)( ..………… A)

Equating real and imaginary parts,

x = r cos ߠ( .……………… 1)

y = r sin ߠ( ..……………… 2)

Eq (1)2+ Eq(2)2

x2 + y2 = r2cos2 ߠ +r2 sin2 ߠ

x2 + y2 = r2 ………………….. [ cos2 ߠ +sin2 ߠ =1]

r = ඥݔଶݕଶ

Eq (2) / Eq (1)

୰ୱ୧୬ఏ

୰ୡ୭ୱఏ = ௬

௫

tan ߠ =௬

௫

ߠ =ିଵ௬

௫

Take a log of Eq ( A)

log (x + iy) = log r (cosߠ +sin ߠ) =log r ݁ఏ = log r + log ݁ఏ

= log r + i ߠ [log e = 1 ]

log (x + iy) = log ඥ + i ܖ܉ܜି

1. Prove that log (x + iy) = log ඥݔଶݕଶ + i ିଵ௬

௫ + 2nߨ i

Proof: Let z = x + iy munotes.in

## Page 57

57Chapter 2: Complex Numbers

Expressing the complex number in general polar form,

z = r (cosߠ +sin ߠ)

x + iy = r (cosߠ +sin ߠ( ..…………) A)

Equating real and imaginary parts,

x = r cos ߠ( .……………… 1)

y = r sin ߠ( ..……………… 2)

Eq (1)2+ Eq( 2)2

x2 + y2 = r2cos2 ߠ +r2 sin2 ߠ

x2 + y2 = r2 ………………….. (cos2 ߠ +sin2 ߠ =1)

r = ඥݔଶݕଶ

Eq (2) / Eq (1)

୰ୱ୧୬ఏ

୰ୡ୭ୱఏ = ௬

௫

tan ߠ =௬

௫

ߠ =ିଵ௬

௫

Take a Log of Eq ( A)

Log ( x + iy) = Log r (cosߠ +sin ߠ)

(Take general value of Log)

= log r {cos (2nߨߠ + )i sin (2nߨߠ)}

= log r ݁ሺଶ୬గାఏሻ

= log r + log ݁ሺଶ୬గାఏሻ [log mn=log m + log n ]

= log ඥݔଶݕଶ + i(2n ߨߠ) log e [log mn=n log m ]

Log ( x + iy) = log ඥݔଶݕଶ + i (2nߨିଵ௬

௫) [log e=1 ]

This shows that for different value of n, the logarithm of a complex quantity x + iy

is multivalued

Ex. 1. Prove that Log (1 + i) = ଵ

ଶlog 2 + i(2n ߨగ

ସ)

Sol: we know that,

Log ( x + iy) = log ඥݔଶݕଶ + i(2nߨିଵ௬௫) munotes.in

## Page 58

58 APPLIED MATHEMATICS

L. H. S. = Log (1 + i)

= log ξ + i(2nߨିଵଵ

ଵ)

= log ξ + i(2nߨగ

ସ) (ିଵͳൌగ

ସ)

= log (ʹభ

మ)+ i(2nߨగ

ସ)

= ଵ

ଶlog (2) + i(2nߨగ

ସ)

= R. H. S.

Ex. 2. Prove that Log (-5) = log 5 + i(2n ߨߨ)

Sol: we know that,

Log ( x + iy) = log ඥݔଶݕଶ + i(2n ߨିଵ௬

௫)

L. H. S. = Log (-5)

= log (-5) + 2nߨ i

= log 5 (-1) + 2nߨ i

= log 5 +log (-1) + 2nߨ i

= log 5 +log (cos ߨ +i sin ߨ + )2nߨ i (
ߨߨൌͳሻ

= log 5 +log ݁గ + 2nߨ i

= log 5 +i ߨlog e + 2nߨ i

= log 5 + i (2nߨ +ߨ )

=R.H.S

Exercise:

Ex1. Prove that log (ା

ି ) = 2i tan-1(௫

௬)

Ex. 2 Show that log (1 + ݁ఏ) = log (2 cos ఏ

ଶ) + ଵ

ଶ iߠ ,if - ߨ൏ߠ൏ߨ

2.15 j(=i) as an operator (Electrical circuits)

j operator is a mathematical operator which when multiplied with any vector, rotate

that vector by 900 in anti -clock wise direction. munotes.in

## Page 59

59Chapter 2: Complex Numbers

j operator has assigned a value of ξെͳ. Thus, it is an imaginary number.

When operator j is operated on vector A, will get new vector jA. This new vector

is displaced from displaced the original vector by 900 in anti -clockwise direction.

the magnitude of vector is remains unchanged when the vector is operated by j.

If the j is applied on the vector jA, the new vector j2A will be the 1800 in anti -

clockwise direction. The new vector j2A is in opposite to the or iginal vector A.

Hence j2A= -A.

Similarly, when j2A is operated with j, the new vector so produced j3A will 2700

ahead of the A. Hence, j3A =-jA. In the same way j4A= A

From above, we can say that,

j2 = -1; j3= j2.j = -j; j2 = j2 j2= 1; (1/j) = -j

We know that from Euler’s Formula, ݁୧௫ = (
ݔ +i ݔ)

munotes.in

## Page 60

60 APPLIED MATHEMATICS

Substitute x = గଶ since cos గଶ = 0, sin గଶ = 1, we get, ݁ഏమ = i

If we take a radius vector of length ‘a’ along a horizontal line then

ai = a ݁ഏ

మ = ia; ai2 = a ݁గ = -a; ai3 = a ݁యഏ

మ = -ia; ai4 = a ݁ଶగ = a

Thus, if we take a radius vector of length ‘a’ along a horizontal line the effect of

raising i to a power n is equivalent to turning this radius vector through an angle

nగ

ଶ.

i. Operation of j (=i) on a sin pt:

A sin pt is the projection of vector ܣܱሬሬሬሬሬԦ (= a) on the horizontal line, where pt

is an angle made by it with vertical, as shown in the fig.

Then j (a sin pt) represents the projection of ܣܱԢሬሬሬሬሬሬሬԦ (= a) on the horizontal line,

when ܣܱሬሬሬሬሬԦ is turned through గ

ଶ.

j (a sin pt) = Projection of ܣܱԢ on XOX’ = a cos pt

j (a sin pt) = a cos pt

ii. Operation of (a + jb) on a sin pt:

(a + jb) sin pt = a sin pt + jb sin pt

= a sin pt + b cos pt [from (i)]

(a + jb) sin pt = ξ (sin pt + ࢻ) , Where tan ߙ =

Operation of (a - jb) on a sin pt:

(a - jb) sin pt = a sin pt - jb sin pt

= a sin pt - b cos pt [from (i)]

munotes.in

## Page 61

61Chapter 2: Complex Numbers

(a - jb) sin pt = ξ (sin pt - ࢻ) , Where tan ߙ =

iii. Operation of ଵ

ୟା୨ୠ on a sin pt:

ଵ

ୟା୨ୠ sin pt = ୟି୨ୠ

మାమ sin pt

= ଵ

మାమ ξܽଶܾଶ (sin pt - ߙ) , Where tan ߙ =

[from (ii)]

ଵ

ୟା୨ୠ sin pt = ଵ

ξమାమ (sin pt - ߙ)

Similarly,

ଵ

ୟି୨ୠ sin pt = ଵ

ξమାమ (sin pt + ߙ)

In electrical engineering, j operator has a great significance and application. You

will encounter this operator often in electrical machine, power system, AC Network

etc.

We know that impedance of a circuit is a complex quantity i.e. it i s having real part

and imaginary part. Real part signifies resistive portion whereas imaginary part

denotes reactance part of the impedance.

In an electric circuit containing resistance R, inductance L and capacity C in series.

We know that, if current I flow through the circuit at any time due to applied

hormonic E. M. F. E 0 sin pt, we have,

ER = RI in phase with I

EL = LpI in quadrature with I (leading)

EC = ூ

େ୮ in quaduture with I (l agging )

Where E R, EL and EC are voltage drops across R, L and C respectively.

munotes.in

## Page 62

62 APPLIED MATHEMATICS

As current through reactance either lags or lead the voltage by 900. Therefore, this

reactance is represented by using j operator. The current through resistance remain

in phase with the voltage, hence resistance is taken as reference and reactance (say

E) is rotated with respect to this reference when operated with j opera tor.

The total impendence which impedes the circuit in AC circuit given by addition of

these vectors.

Hence impendence Z is written as Z = (R േ jE). It may be noted that the capacitive

and inductive reactance are ( -j/ܥ) and jL p.

z = R + jLp + (ି

େ୮)

= R + j (Lp -ଵ

େ୮)

If E 0 sin pt be applied voltage, the current I in the circuit is given by,

ୱ୧୬୮୲

୍ = z

I = ୱ୧୬୮୲

I =

ୖା୨ሺ୮ିభ

ి౦ሻ

=

ටோమାሺ୮ିభ

ి౦ሻమ ሺ – ߙ ,)where ߙ =tan -1 ሺ୮ିభ

ి౦

ோ

[ ଵ

ୟା୨ୠ sin pt = ଵ

ξమାమ (sin pt - ߙ)]

2.16 Summary:

Complex Numbers can be presented in rectangular, polar or exponential form with

the conversion between each complex number algebra form including addition,

subtracting, multiplication and division. We learned about introductory ideas

associated with complex numbers, their algebra and geometry, algebraic properties

of complex numbers, Argand plane and polar representation of complex numbers,

mathematical operation with complex numbers and their representation on Argand’s Diagram, circular functions of complex angles, hyperbolic functions,

relations between circular and hyperbolic functions, inverse hyperbolic functions,

graphs of the hyperbolic functions. Finally, we looked the Logarithms of complex

quality and application of complex number in electrical circuit . munotes.in

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63Chapter 2: Complex Numbers

2.17 References:

1. Applied Mathematics II by P. N. Wartikar and J. N. Wartikar

2. Higher Engineering Mathematics by Dr. B. S. Grewal

3. Complex numbers from A to Z by Titu Andreescu and Dorin Andrica

2.17 Exercise:

Ex. 1 If | z1 | = 1, | z2 | = 2, | z3 | = 3 and | 9z1 z2 + 4z1 z3 + z2 z3 | = 12, then

find the value of | z1 + z2 + z3 |. [

Ans: | z1 + z2 + z3 | = 2]

Ex.2 z1, z2, and z3 are complex numbers such that z1 + z2 + z3 = 0 and

| z1 | =| z2 | =| z3 | = 1 then find z 12 + z 22 + z 33

[Ans: 0]

Ex. 3 Find the fourth roots of unity [Ans: 1, i, -1, -i]

Ex. 4 Find all cube root of ሺξ͵݅ሻ

[Ans: 21/3 (cos గ

ଵ଼ + i sin గ

ଵ଼), 21/3 (cos ଵଷగ

ଵ଼ + i sin గ

଼), 21/3 (cos ଶହగ

ଵ଼ + i sin ଶହగ

ଵ଼)]

Ex. 5 Simplify ( ଵାξଷ

ଵିξଷሻ10 [Ans: (cos ଶగ

ଷ + i sin ଶగ

ଷ)]

Ex 6 Prove that Log i = log i +2n ߨ i

Ex 7 Prove that i log (௫ି

௫ା ) = ߨ - 2 tan-1x

Ex 8 Show that tan ( i log ା

ି ) = ଶ

మାమ

munotes.in

## Page 64

64 APPLIED MATHEMATICS

64Unit 2

3 DIFFERENTIAL EQUATION

EQUATION OF THE FIRST ORDER AND OF THE FIRST DEGREE

Unit Structure

3.1 Objectives

͵Ǥʹ Introduction

3.3 Ordinary Differential Equation

͵ǤͶ Separable Variables - Differential Equation

͵Ǥͷ Equations reducible to homogeneous forms

3.6 Existence of a solution for a differential equation

3.7 Homogeneous polynomial

3.7.1 Homogeneous function

3.7.2 Homogeneous Differential Equation

3.7.3 Non Homogeneous Differential Equation

3.8 Exact Differential Equation

3.9 Integrating Factors

3.10 Integrating Factor of a homogeneous equation

3.11 Linear Equation and equation reducible to homogeneous form

3.12 Partial Differential Equation -An Overview

3.13 Summary

3.14 References

3.15 Questions

3.1 Objectives

- recognize and solve problems in ordinary differential equations

- Understand the application of differential equation in physics and engineering branches such as electronics, electrical, mechatronics etc .

- Evaluate first order differential equations including separable, homogeneous, non-homogeneous exact, and linear and partial munotes.in

## Page 65

65Chapter Differential Equation Equation of The First Order and of the First Degree- Identify research problems where differential equations can be used to model the system

- Analyze mathematical models to solve application problems such as circuits, population modeling, orthogonal trajectories, and slopes

3.2 Introduction

In an equation constituting of dependent and independent variable, when the

derivatives of the former can be represented with respect to one or more

independent variables such equations are called Differential Equation. Some of

the differential equations that can be solved by standard procedures are as

follows:

- Differential equation in which variables are separable

- Homogeneous differential equ ations

- Non homogeneous differential equations which can be reduced homogeneous differential equations

- Linear differential equations

- Bernoulli’s differential equations that are nonlinear and can be reduced to

linear form.

- Exact differential equations

A first order differential equation is an equation that can be represented in the

form

F (t, y, dy/dt ) = 0 or in other words ሺǡǡԢሻ

Equation 1

where Ԣ is the first order derivative of y

This equation can also be represented as

ሺǡሺሻǡԢሺሻሻൌͲ for every value of t

Equation 2

and is function of three variables ሺǡǡԢ).

A differential equation’s order is determined by the highest -order derivative

whereas the degree is the highest power to which a variable is raised within an

equatio n. The higher the order of the differential equation, the more arbitrary

constants need to be added to the general solution. Below are a few examples that

depict different scenarios

Examples ԢԢԢԢԢԢ
ൌͲ Equation 3

ԢെൌͶ Equation 4 munotes.in

## Page 66

66 APPLIED MATHEMATICS3.3. Ordinary Differential Equation Ordinary Differential Equation (ODE) is described as the relation having an independent variable x,a dependent variable y and associated derivatives of y. The order of the ordinary differential equation is the order of the highest derivative in that equation. Few examples of ordinary differential equation are as follows: Equation Order Degree ଷଷȀൌͲ 1 3 Equation 5 ଷଷଶȀଶൌͲ 2 4 Equation 6 Example 1

μȀμൌͶെʹμ/4y-2 = μ

නݕ߲ȀሺͶݕെʹሻൌݔ߲

1/4 ȁͶെʹȁ = x +c

ȁͶെʹȁൌͶͶ

Ͷെʹ=ሺെሻ݁ସ௫ାସ

4y = ( െሻ݁ସ௫ାସ+ 2

y = 1/4( െሻ݁ସ௫ାସ+ 1/2

y = (+ -)1/4exp(4(x + c)) + 1/2

Let C = 1/4exp(4c)

y(x) =Ce4x +1/2

డ௬

డ௫ =4Ce4x

4y-2 =4Ce4x

4y = 4Ce4x +2

Substituting for y in the above

4(Ce4x +1/2) = 4Ce4x +2

The two equations are proved equal.

With y(2) where x = 2 and y(2) = 4 the proof is as follows :

y(2) = 1 then

Ce8 + 1/2 = 1

Ce8 = 1/2 or C = 1/2(e-8)

4(1/2e-8 .e8 +2) = 4(1/2e-8.. e8) + 2 = 4 Ans munotes.in

## Page 67

67Chapter Differential Equation Equation of The First Order and of the First Degree

Example 2

ൌμȀμൌ2x3 and y(2) = 3

ൌμȀ2 = x3dx

ൌ1/7߲ݕ/y2 = ݔଷ݀ݔ

ൌെଵ

ିଶାଵ = ௫ర

ସ +c

ൌሺെͳሻିଵൌସ+c

ൌെͳȀሺȀͶሺସ)+c) = ି

ૠ

ା

Putting x =2

͵ൌሺെͳሻȀሺȀͶሺͳ) + c)

3 = െͳȀʹͺ

c= െͺͷȀ͵

y = ሺିሻ

ሺૠ

ሻܠିૡ

Ans

ப୷

ப୶ = డ

డ௫ሺି

ૠ

ା = )௫య

ሺళ

ర௫రାሻమ = ଷൈଶ where y = ିଵ

ళ

ర௫రା

3.4 Separable Variables - Differential Equation

Variables are said to be separable when all the similar terms are on the same side

i.e. x and dx on one side and y and dy on the other side. The general

representation of the equation is as follows :

f(x)dx = g(y)dy (or) f(x)dx + g(y)dy = 0 Equation 7

Consider an example as follows:

(ݕଶ+1)μሺଶ͵ሻμൌͲ

ሺݕଶͳሻμሺଶ͵ሻμൌͲ

+y +

+3x = c

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68 APPLIED MATHEMATICS

Examples

a)

b) μȀμൌ୶ି୷ +ݔଶ݁ି௬ = ݁ି௬(݁௫ݔଶሻ

= μ/݁ି௬ = (݁௫ݔଶሻμ

= ݁௬ݕ߲ൌ݁௫߲ݔݔଶ߲ݔ

ൌ݁௬ൌ ݁௫+௫య

ଷ +c Ans

c) y - x dy/dx = a(y2 +dy/dx)

= y - xdy/dx = ay2 + ady/dx

= y - ay2 = dy/dx(x+a) = dx/(x+a) = dy/y -ay2

ܠ܌Ȁሺܠ܉ሻ =Ȁെ

ଵ

୷ሺଵିୟ୷ሻൌ

୷

ଵିୟ୷

1 = A (1-ay) +By

1 = A - a(Ay) + By

1-A = -y(aA -B)

1-A = -y or y = A -1

B-aA= 1 -A if A= 1 then B - a = 0 or B=a

Ȁሺሻ =݀ݕȀݕെܽݕଶ

1/y(1 -ay) = A/y +B/1 -ay

Upon integrating it is

log y +a( -1/a)log(1 -ay) = log y - log(1 -ay) = log(y/1 -ay) +c

log (x+a) = log(y/1 -ay) +c

log(x+a) -log(y) +log(1 -ay) =log c

log(x+a)(1 -ay)/log y = log c

(x+a)(1 -ay) = cy Ans

munotes.in

## Page 69

69Chapter Differential Equation Equation of The First Order and of the First Degree

Example

ሺͳሻሺͳሻൌͲ

(1+x)dx = -x(1+y)dy

(1+x)dx/x = -(1+y)dy

ௗ௫

௫ + ݔ݀ൌെௗ௬

௬െݕ݀

=ൌെെ

= log |x |+log |y| +x+y = c

=ȁȁൌ

3.5 Equations reducible to homogeneous forms

A function f(x,y) is called Homogeneous of degree n if ),( ),( yxft yxfn

Equation 8

and where t is a nonzero real number. Thus ¸¸¹·¨¨©§

yxandy xy xxy sin... ,2 210 10

Equation 9 are homogeneous function of degree 1, 8 and 0 respectively

A first order differential equation of the form ),(yxfdxdy is said to be

homogeneous if the function f depends only on ratio of (y/x).Thus first order

homogeneous equation are of the form ¸¹·¨©§ xygdxdy Equation 10

and is transformed into an equation that is separable by substituting y = vx and ¸¹·¨©§ dxdvx vdxdy Equation 11

and

g(v) = v +x(dv/dx) and dv/(v -g(v) = െȀ

Equation 12

munotes.in

## Page 70

70 APPLIED MATHEMATICSȌ (x-y)డ௬డ௫ = x+3y ௫ାଷ௬

௫ି௬ = డ௬

డ௫ let y = hx; డ௬

డ௫ = h +xడ

డ௫ = ௫ାଷ௩௫

௫ି௩௫ = ଵାଷ

ଵି = xడ

డ௫ = ଵାଷ

ଵି - h = ଵାଷ୦ି୦ା୦మ

ଵି = ଵାଶାమ

ଵି = డ௫

௫ = ሺଵିሻ

ሺଵାሻమ = ȁȁ
ൌଵି୦

ሺଵା୦ሻమμ Putting m = 1+h = ଶି

మ݉݀ൌ ʹడ

మ -డ

= - ȁȁ - ଶ

= ିଶ

ሺଵାሻ݈݃ȁͳ݄ȁ = ିଶ

ሺଵାሻ݈݃ȁͳ݄ȁ = ȁȁ
= ିଶ

ଵା

ೣ - log|1+௬

௫| = ȁȁ
+ ଶ௫

௫ା௬ + log|௫ା௬

௫ | = 0 = log|x+y| +ଶ௫

௫ା௬ = c Ans Ȍ Solve : ሺͻെሻൌሺʹ͵െሻ

ൌሺͻെሻ

ሺʹ͵െሻ =ௗ௬

௫ାଽ௬ି = ௗ௫

ଶ௫ାଷ௬ି ൌǡൌ here h and k can be solved for their values Equations to be considered are as follows: ሺ݄ͻ݇െሻ and ሺʹ݄͵݇െ) that are solved to get h = ଵଵ

ହ and k = ଼

ଵହ డ௬

డ௫ = ଡ଼ାଽଢ଼ାሺ୦ାଽ୩ିሻ

ଶଡ଼ାଷଢ଼ାሺଶ୦ାଷ୩ିሻ డ௬

డ௫ = ൌப୷

ப୶ൌப୦

ப୶ i.e.equal to ାଽ

ଶାଷ = ப୴

ப୶ = ାଽ

ଶାଷ = ଵାଽ

ଶାଷ i.e. ப୴

ப୶ = ଵାଽ

ଶାଷ - h = ଵା௩ିଷ௩మ

ଶାଷ௩ = ப୴

ப୶

ʹ͵ݒ

ͳݒെ͵ݒଶ߲ݒൌ߲ܺ

ܺ

නʹ͵ݒ

ͳݒെ͵ݒଶ߲ݒൌ݈݃ȁܺȁܿ ʹడ௩

ଵା௩ିଷ௩మ +͵௩డ௩

ଵା௩ିଷ௩మ = ݈݃ȁܺȁܿ ʹሺͳെ͵ଶ) +3vlog( ͳെ͵ଶ) +ଷ

ሺଵା௩ିଷ௩మሻ௩మ

ሺଶሻ = ݈݃ȁܺȁܿ munotes.in

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71Chapter Differential Equation Equation of The First Order and of the First Degreeܗܔሺૠ܇܆ - 3(܇܆)2 + 3()log(1+7(ሻ - 3(܇܆)2 + ሺሻሺାૠሺ

ሻିሺ

ሻሻሺሻ = ȁȁ Ans

Example

ܿ݁ݏଶx tan y μ
ଶy tan x μൌͲ

ܿ݁ݏଶ μ = - ୱୣୡమ୷୲ୟ୬୶ப୷

௧௬ = -௧௫డ௬

௦௬௦௬ = ௦మ௫డ௫

௧௫ =- డ௫

௦௫௦௫ = డ௬

௦௬௦௬

Upon integrating

డ௫

௦௫௦௫ = െడ௬

௦௬௦௬ + log|c| = ሺ௦మ௫ା௦మ௫ሻడ௫

௦௫௦௫

= െሺ௦మ௬ା௦మ௬ሻడ௬

௦௬௦௬ + c = 0

௦௫

௦௫ ݔ߲+௦௫

௦௫ ݔ߲+௦௬

௦௬ݕ߲ +௦௬

௦௬ ݕ߲=c

= log (secx) + log (sinx) +log (secy)+log(sin y) = log(c)

=log (tanx)+log(tany) = log(c)

=log (tanxtany) = log(c)

= tan x tan y = c Ans

Example

The cost of producing x socks is 6 + 10x í 6x2 . The total cost of producing a pair

is INR 100. Find the function representing total and average cost.

Cost = 6 + 10x í 6x2

డ

డ௫ = 6 + 10x í 6x2 = ߲ܿൌሺ6 + 10x í 6x2 )μ + k

C = 6x +10௫మ

ଶ - 6௫య

ଷ + k = C = 6x +5x2 - 2x3 +k

When x = 2 and C = 100 then K = 84

Hence Average Cost if there are x units of socks is ܠାܠିܠାૡ

Ans

Example

A curve passes through points (1, 2) and lines to the curve pass through the point

(1,0). Formulate the equation of the curve using differential equation

Slope of a line given by y = mx + c μȀൌ = ୷ଶି୷ଵ௫ଶି௫ଵ = ௬ି௫ିଵ = ப୷௬ = ப୶௫ିଵ munotes.in

## Page 72

72 APPLIED MATHEMATICS

Slope of a normal at any given point P(x, y) is represented by െப୶ப୷ = ௬ି௫ିଵ

= ݕ߲ݕ =െሺݔെͳሻݔ߲ܿ =- ௫మ

ଶ + x +c

= ௬మ

ଶ = - ୶మ

ଶ + x +c

Passing through points (1, 2) we have c = ହ

ଶ - 1 = ଷ

ଶ

Putting c = ଷ

ଶ = ௬మ

ଶ + ୶మ

ଶ -x - ଷ

ଶ = y2 + x2 - 2x - 3 = y2 = 2x - x2 + 3 Ans

Example

A sum of INR 4,000 is compounded at a 10% per a nnum rate of interest . In how

many years will the amount be double the original principal? (log e 2 = 0.69)

Principal = P, Rate of Interest = 10 percent per annum, Sum = P+ P*(ଵ

ଵ)

డ

డ௧ൌଵ

ଵP = డ

= ଵ

ଵ߲ݐܿൌ݈݃ȁܲȁൌሺͲǤͳݐሻܿൌܲൌ݁Ǥଵ௧݁=

c’ ݁Ǥଵ௧ = 4000 = c’ when t = 0 and e = 1 , = 8000 = 4000e.1t, 2 = e.1t, t/10 = log 2

.69 = t/ 10, t = 6.9 years. Ans

3.6 Existence of a solution for a differential equation

The general solution of the equation dy/dx = h(x, y) and has the form f(x, y, C) =

0, C being a constant. Below is the theorem that presents the scenario :

A general solution of dy/dx = h(x, y) exists ove r a region S of points (x, y) based

on certain conditions

a) h(x, y) is continuous and single -valued over S

b) J\H[LVWVDQGLVFRQWLQXRXVDWDOOSRLQWVRI S

The general solution f( x, y, C) = 0 of a differential equation dy/dx = h(x, y) over

some region S consists of set of curves, where each curve represent s a particular

solution, such that through each point in S there passes one and only one curve

for different values of C.

The di fferential equation associates with each point (x 0, y0) in the region S a

direction that is given by

m = డ௬

డ௫|x,y = h(x,y)

The direction at each point of S is the tangent to that curve of the family f(x, y, C)

= 0 that passes through the point. munotes.in

## Page 73

73Chapter Differential Equation Equation of The First Order and of the First Degree

A region S in which a direction is associated with each point is called a direction

field . For an equation such as y = x2 +c the direction would be 2x. The curves or

parabolas can be represented as shown in the diagram

3.7 Homogeneous polynomial

A polynomial whose terms sum to the same degree with respect to all the

variables taken together. Thus

m2 + 2mn - 2n2 degree 2 homogeneity

2m3n + 3 m2n2 + 5n4 degree 4 homogeneity

2m + 5n degree 1 homogeneity

3.7.1 Homogeneous function

A function is said to be homogeneous when we can take a function: f(x, y ) and

multiply each of the variable so that the function is of the form n: f (nx,ny) and

represent it in the form zn f(x, y) .Thus

2m2 ln(

)+ 4n2 is homogeneous of degree 2

m2 n + n3 sin (

) is homogeneous of degree 3

3.7.2 Homogeneous Differential Equation

A homogeneous equation is a differential equation of the form

M(p, q) dp + N(p, q) dq = 0 Equation 13

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## Page 74

74 APPLIED MATHEMATICS

where M(p, q) and N( p, q) are homogeneous functions of the same degree. Here

variables can be separated by substitution by introducing a new variable p = sq

(or q = sp), where s is a new variable.

Note. Differentiating p = sq gives d p = s dq + q ds, a quantity that must be

substituted for d p when sq is substituted for p.

Example

Solve the equation

(x2 -y2)dx + 2xy dy = 0

Solution Separation of variables though not possible the can be represented as

homogeneous function as follows . Substituting

y = vx and dy = v dx + x dv

we get

(x2 - v2x2)dx +2x(vx)(vdx+xdv) = 0

x2dx - v2x2dx +2v2x2dx + 2x3vdv = 0

x2dx+v2x2dx + 2x3vdv = 0

(1+v2)x2dx = -2x3vdv

dx/x = -2v dv/(1+v2)

Upon integrating

ௗ௫

௫ = -2௩ௗ௩

ሺଵା௩మሻ

= -log (x)+ log C = log(1+v2)

=x(1+v2) = C

Since y = vx

=x(1+(௬

௫)2) = C

=x(ሺ௫మା௬మ

௫మ) = C

= x2 + y2 = C Ans

3.7.3 Non Homogeneous Differential Equation

These can be represented in the form as follows:

డ௬

డ௫ = ௫ା௬ା

ᇲ௫ାᇲ௬ାᇱ Equation 14 munotes.in

## Page 75

75Chapter Differential Equation Equation of The First Order and of the First DegreeWe can now replace x = X + h and y = Y + k ሺାሻାሺାሻା

ᇱሺାሻାᇱሺାሻାᇱ = ୮ሺଡ଼ሻା୮ሺ୦ሻା୯ሺଢ଼ሻା୯ሺ୩ሻା୰

୮ᇱሺଡ଼ሻା୮ᇱሺ୦ሻା୯ᇱሺଢ଼ሻା୯ᇱሺ୩ሻା୰ᇱ = ሺሻାሺሻାାା

ᇱሺሻାᇱሺሻାᇱାᇱାᇱ

=0; ԢԢԢ = 0; =-r; ԢԢ = - Ԣ ;

h = ᇲିᇱ

ᇱିᇱ and ܓൌܘᇱܚିܘܚᇱ

ܙܘᇱିܘᇱܙ and subject to the condition that the term pq’ -

S¶TWKHHTXDWLRQWDNHVWKHIRUPDV follows:

డ

డ = ା

ᇱାᇱ that is transformed into a homogeneous equation.

If ԢെԢൌͲ

Ǥ If ԢെԢൌͲthen

ᇲൌ

ᇱ = ଵ

LHS¶ QSDQGT¶ QTDQGWKHGLIIHUHQWLDOHTXDWLRQEHFRPHV

డ௬

డ௫ = ௫ା௬ା

ᇲ௫ାᇲ௬ାᇱ = డ௬

డ௫ = ௫ା௬ା

௫ା௬ାᇱ = ௫ା௬ା

ሺ௫ା௬ሻାᇱ

ൌൌப୴

ப୶ = p +q డ௬

డ௫ = డ௬

డ௫ = ሺങೡ

ങೣି୮ሻ

= ௩ା

௩ାᇱ = F(v) = μൌ

ப୴

୮ା୯ሺሺ୴ሻሻ that which can be integrated.

Example

ሺെͶͳሻப୷

ப୶ = ሺ͵െʹͳሻ = డ௬

డ௫ = ଷ୶ିଶ୷ାଵ

ଶሺଷ୶ିଶ୷ሻାଵ

Following the above transformation let v = ͵െʹ, డ௩

డ௫ = ͵െʹப୷

ப୶ = డ௬

డ௫

= ଷିങೡ

ങೣ

ଶ = ௩ାଵ

ଶ௩ାଵ = ͵െப୴

ப୶ = ଶ௩ାଶ

ଶ௩ାଵ = 3 - ଶ௩ାଶ

ଶ௩ାଵ ൌ୴ାଷିଶ୴ାଵ

ଶ୴ାଵൌப୴

ப୶ൌସ୴ାସ

ଶ୴ାଵ

ൌμൌሺଶ୴ାଵሻ

ସ୴ାସμ

= ଵ

ଶ{v + ଵ

ସ ORJY` [F¶

= {v +ଵ

ଶ log (4v +1)} = 2x + 2c

= {v +ଵ

ଶ log Y` [F¶ZKHUHF¶ FYଵ

ଶlog Y [F¶

= 3x - 2y = v and substituting for v in terms of x and y we get value RIF¶

= 3x-2y + ଵ

ଶ log {4(3x - \` [F¶

= x -2y + log {4(3x - 2y) +1} = c’ Ans munotes.in

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76 APPLIED MATHEMATICS

3.8 Exact Differential Equation

The total differential of a function u(x, y) is, by definition

Equation 15

The exact differential is given as follows:

= 0 Equation 16

or

M (x, y) dx + N (x, y) dy = 0

For example to see if this equation is exact or not

(3x2y - y)dx + (x3 - x + 2y)dy = 0

(M) (N)

ࣔ

ࣔ =3x2 - 1 ࣔ

ࣔ =3x2 - 1 Ans

In the above the left hand side is an exact differential of the right side of the

equation hence the differential is said to be an exact differential or in other words

a relevant factor u(x, y) known as integrating factor has been appended to the

given dif ferential equation.

ሺଷെʹሻൌଷെଶ upon differentiating with respect to x

gives ͵ଶെ
. Here the integrating factor is y.

Similarly considering another example

2ydx + xdy = 0 This cannot be considered as an exact differential equation but if

it multiplied by x then it gets transformed into an exact equation

= (2xy) dx + (x2) dy = 0 = M (x, y) dx + N (x, y) dy = 0

= డெ

డ௬ = 2x and డே

డ௫ = 2x and also

ݔଶμൌݔଶ

ǤǤሺʹሻ,here “y” is again an integrating factor.

munotes.in

## Page 77

77Chapter Differential Equation Equation of The First Order and of the First Degree
ൌሺʹെ͵ଶሻμሺଶെʹሻμ With respect to y and

x= ப

ப୷ൌʹப

ப୶ൌʹ
Ǥ

ሺǡሻ

Ǥ

μand that is [ -y2]. ݔ݀ܯ =ሺʹെ͵ଶሻݔ݀

The final result is as follows : ܡ-x3 - y2 = c the general solution

Example

= ሺଶሻμሺଶሻൌͲ

= M = ሺଶሻ, N = ଶǡ

డெ

డ௬ = 2xy,డே

డ௬ = 2xy

Integrating μμ we get

=ܯ߲ݔൌݕଶݔଶ/2 + ௫

ଶଶ+ c(y) ,

Differentiating with respect to y this previous equation

f(x,y) = x2y +dc/dy and dc/dy = 0

Hence the generalized equation becomes x2 y2 +x2 = c is the general

solution.

Example

ሺ୷ୡ୭ୱ୶ାୱ୧୬୷ା୷ሻ

ሺ௦௫ା௫௦௬ା௫ሻ = డ௬

డ௫

M =ycosx + siny +y , డெ

డ௬ = cosx +cosy +1

N = sinx + xcosy + x,డே

డ௫ = cosx+cos y +1 hence the equations are exact

ൌሺ
ሻμሺ
ሻμൌ

= ysinx +(siny +y)x = c Ans

Example

= (1+2xycosx2 - 2xy) dx +(sinx2 - x2)dy = 0

=డெ

డ௬=డே

డ௫= 2xcosx2 - 2x I.e. equation is exact

=ሺܠܡܛܗ܋ܠെܡሻሻܠሺܕܚ܍ܜܛܗۼܖܜܗܜܖܗ܋ܑ܉ܖܑܖܠሻൌ܋ Ans munotes.in

## Page 78

78 APPLIED MATHEMATICS

Example

2xydx + ( ଶ͵ଶሻൌͲ

ൌʹǡൌଶ͵ଶǡப

ப୷ = 2x and డே

డ௫ = 2x

Hence these equations are exact

Now to find function u(x, y) we have ݔ߲ܯൌݔଶݕܿሺݕሻ

Substituting the expression for u(x, y) in the 2nd equation

We have u(x, y) = డ

డ௬(ݔଶݕܿሺݕሻ) i.e. x2 +c’(y) = x2 + 3ଶ .

Hence c’(y) = 3y2

Hence the integral of last equation in the above is given as ͵ݕଶݕ߲ =y3

The final form is +y3 = C Ans

Example

(6x2 - y +3) dx +(3y2 - x -2)μx

=డெ

డ௬ = -1 and డே

డ௫ = -1 Hence the equations are exact

Now to find function u(x, y) we have ۻܠൌ2x3 -xy + 3x2 +k(y) .

Now f/y = -k(y) + (-x) = 3y2 - x -2

So the final equation becomes : 2x3 -xy +3x+3y2 - 2 = c

Example

(3x2 +4xy)dx +(2x2 + 2y)dy = Mdx + Ndy = 0

dM/dy = 4x, dN/dx = 4x and hence the equations are exact

Integrating M(x,y)dx = ሺ͵ݔଶ+4xy) μx = x3 +2x2y +k(y) = f(x,y)

Differentiating with respect to y

df/dy = 2x2 +μ/μy(k(y) = 2x2 + μk/μy = 2x2 + 2y

So μk/μy = 2y Upon integrating k(y) = y2 +c’

f(x,y) =x3 + 2x2y +y2 +c’ = c” = x3 + 2x2y +y2 = c is the general solution. munotes.in

## Page 79

79Chapter Differential Equation Equation of The First Order and of the First Degree

3.9 Integrating Factors

The equation μμ cannot be represented as an exact differential equation, then there exists a multiplying factor Ɋ that is a function of x and y that

makes the equation exact. This factor is otherwise known as the Integrating

Factor.

A given differential equation assumes the form as follows;

ࣔ

ࣔ +M(x)y = N(x) Equation 17

then the integrating factor Ɋ is defined as follows:

ૄൌ܍ۻሺܠሻܠ Equation 18

:KHUH0[WKHIXQFWLRQRI[LVDPXOWLSOHRI\DQGȝGHQRWHVLQWHJUDWLQJIDFWRU

OR

ࣔ

ࣔ +P(y) = Q

ࣔ

ࣔ ݁డ௫ +y(݁డ௫P) = Q݁డ௫

Upon integration

y܍۾ܠൌࣔ dx +c

Equation 19

For example consider the function

ሺܠെܡሻܠܠܡ = 0 or ܠܡൌെሺܠെܡሻܠ or ࣔ

ࣔ =ି

where ି

is

considered as M.

The steps for the integrating factor are as follows : The differential equation is

represented in the above form and the value of M(x) is found out. The integration

IDFWRUQHHGVWREHFDOFXODWHGLHȝ7KHHTXDWLR n at the next step needs to be

represented as follows:

ȝࣔ

ࣔ ȝ0[\ ȝ1[

On the left -hand side of the equation, a particular differential is obtained as

follows:

ࣔࣔሺࣆǡሻൌࣆሺሻ munotes.in

## Page 80

80 APPLIED MATHEMATICS

In the end, integration of the expression needs to follow and the required solution

to the given equation is represented as: ȝ\ ȝ1[ + C.

Now considering this equation ሺെሻμμ = 0 here M = x -y and N = x

hence ࣔ

ࣔ =-1 and ࣔ

ࣔ =1 and the equations are not exact.

In order to make the equations exact the ࣆ should be such that ࣔ

ࣔ should be equal

to ࣔ

ࣔ .

Hence if we multiply M and N with

then డெ

డ௬ = ି

and డே

డ௫ =

then

differentiating M and N gives us the results as follows: డெ

డ௬ = - ଵ

௫మ and డே

డ௫ = - ଵ

௫మ

hence ȝ -

ܠ and this becomes the integrating factor.

Example

Solve the equation: ሺଶെʹଷሻሺ͵െʹଶሻൌͲǤ

The given equation is not exact , because

డெ

డ௬ = 2xy - ݕଶ,డே

డ௫ = 2y2 and the equations are not exact We try to find the general solution of the equation using an integrating factor. Calculate the difference i.e. ࣔ

ࣔ - ࣔ

ࣔ =2xy -4y2

( ࣔ

ࣔ - ࣔ

ࣔ = )ܠܡିܡ

ܠܡିܡ =

and the integrating factor is ȝ that is dependent on y.

ȝ =

,ࣔૄ

ࣔ =-2,

ૄ( ࣔૄ

ࣔ = )-

Upon integrating

-2 ࣔ

=ln|y| = ȝ =( +-)

Now the exact equation is got i.e.

ሺଶെʹଷሻȀଶμሺ͵െʹଶሻȀଶμൌሺെʹሻμሺଷ

୷మെ2x) = 0

ࣔ

ࣔ =-2, ࣔ

ࣔ =-2. Now to find u from the above

ࣔ

ࣔ =x-2y and ࣔ

ࣔ =

ܡെ2x ,u(x,y) = ሺെʹ)dx = ݔଶ - 2yx + (this is from

the first equation and from the second equation

డ

డ௬( ݔଶ - 2yx + = )ଷ

୷మെ2x = -2x + ('y) = ଷ

୷మെ2x and ('y) = ଷ

୷మ , (y)

= ଷ

୷మμ = - ଷ

௬

Hence the final equation becomes - 2yx - =c with y = 0 Ans munotes.in

## Page 81

81Chapter Differential Equation Equation of The First Order and of the First Degree

Example

y(log y ) μx + (x -log y) μy = 0

μx/μy +x/(y log y) = 1/y which is a leibnitz’s equation in x

Integrating Factor = ݁భങ

= ݁ሺ௬ሻ = log y

Thus the solution is as follows: x (I.F.) = ଵ

௬(Integrating Factor) μy + c

= x log y = ଵ

௬(Integrating Factor) μy + c = (1/2) (log y)2 + c

= x = (1/2) log y + c/log y Ans

Example

Solve (x + 1)ࣔ

ࣔ - ye3x(x+1)2

డ௬

డ௫ - y/(x+1) = e3x(x+1) here P = - 1/(x+1) and ݔ߲ܲ

=െப୶

୶ାଵ= -log (x+1)

Here integrating factor is as follows: ݁ങೣ = ݁ሺ௫ାଵሻଵషభ = 1/x+1

As per the above equation y. (Integrating Factor) =

݁ଷ௫(x+1)( Integrating factor ).μ

y(1/(x+1) = ܠ +c or y = (x+1)(c+1/3 )

Integrating Factor found by Inspection

Example

y (2xy +ex)dy +2xy2 dx = 0

Dividing by 1/y2 that is the Integrating Factor then equation becomes

μሺ
ሻൌ

=

+x2 = c Ans

3.10 Integrating Factor of a homogeneous equation

If Mdx +Ndy =0 be a homogeneous equation then its integrating factor is

0[1\DQG0[1\

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82 APPLIED MATHEMATICS

Example

(x2y - 2xy2)dx - (x3 -3x2y)dy = 0 .

It is in the homogeneous form Integrating Factor = 1/x2y2

Multiplying the equation with IF , the equation becomes exact in the form

(1/y-2/x)dx - (x/y2- 3/y)dy =0 and is exact

Applying μሺ
ሻൌ

= x/y -2logx +3logy = c Ans

3.11 Linear Equation and equation reducible to homogeneous

form

A differential equation is said to be linear if its differential coefficient occur in the

first degree and is not multiplied together and is represented as follows:

ࣔ

ࣔ +P(y) = Q where P and Q are functions of x.

Here when Q = 0 then డ௬

௬ + P ߲x = 0

Upon integration ࣔ

+P ࣔ =log y +P ࣔ =log c or y/c = - Pࣔ =y/c =

ିࣔ and the rest is the same as the liebnitz equation.

Bernoulli Equation can be represented as follows:

ࣔ

ࣔ +P(y) = Q yn where P and Q are functions of x and upon solving gives

ࣔࢠ

ࣔ (+1-n)P z = (1 -n)Q

Another equation that can be linear in the form is f’(y) μȀμ +Pf(y) = Q

Then dz/dx + P(z )= Q where f(y) = z

Example

Solve

0 22 2 dyx xy y

222

xxy y

dxdy

dxdvx vdxdy for y =vx

v vxxvx xv

dxdvxv 22 2

22 2

munotes.in

## Page 83

83Chapter Differential Equation Equation of The First Order and of the First Degree

v vdxdvx 32 ³ ³ xdx

vvdv

3

³ ³ ¸¹·¨©§

xdxdvv v 31 1

31

c x v v log log 3 log31log31

xc

vv 1 log33log

3

1 log3log »¼º«¬ª »¼º«¬ªxc

vv

3 313

3 xc

xc

vv »¼º«¬ª

33 xc

xyxy

»»»¼º«««¬ª

)3 (3 ycyx Ans

Example

Solve (2x -5y )dx +(4x -y)dy = 0

dy/dx = ଶ௫ିହ௬

ସ௫ି௬

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## Page 84

84 APPLIED MATHEMATICS

Example

Solve xడ௬

డ௫ + y = x3y6

= Dividing by xy6 we have డ௬

డ௫y-6 + y-5 = x2

Let y-5 = z = =5y-6 = dz/dx or -dz/dx -(5/x)z = -5x2 which is linear in z

Applying Integration Factor

i.e. ష

ࣔ =e-5logx = x-5

= z * (Integrating Factor) = c +Q(Integrating factor ) dx

= z* x-5 = ሺെͷݔଶx-5dx +c

= y-5 x-5 = -5x-2/(-2) +c Ans

Example

tan y ሺy/ܠሻ +tan x = cos y cos2x

Dividing by cos y it gives

secy tan y ሺμy/μሻ+secy tanx = cos2x

Let sec y = z then μȀμ = secy tan y μy/μ

μz/μ + z tan x = cos2x

It is in the linear form hence the integrating factor I.F

݁௧௫డ௫ = ݁௦௫ = sec x

So z * (Integrating Factor) = c +Q (Integrating factor ) dx

= z*(sec x) = c+ ( cos2xsecx) dx

=z*(sec x) = c+ Integration of (cosx) dx

secy secx = c+ sin x

secy = (c+sinx)cosx Ans

The DE is not homogeneous.

0 ) ( ) (2 2 2 1 1 1 dycybxa dxcybxa

It can be reduced to homogeneous form

Type -1 If 21

21bb

aaz

then the transformation is as follows: x = X + h, y = Y + k

0 ) () (

2 2 2 2 21 1 1 1 1

dYckbhaYbXadXckbhaYbXa munotes.in

## Page 85

85Chapter Differential Equation Equation of The First Order and of the First Degree

Type 2

If 21

21bb

aa

then put ybxaz1 1 and the given equation will reduce to a

separable equation .

Example

Solve dy/dx = (2x+y+1)/(x -2y+3)

munotes.in

## Page 86

86 APPLIED MATHEMATICS

Ans

Example

Solve dy/dx = 3x -4y-2/3x-4y-3

munotes.in

## Page 87

87Chapter Differential Equation Equation of The First Order and of the First Degree

3.12 Partial Differential Equation - An overview

A differential equation that constitutes of partial derivatives is known as a partial

differential equation. The differential equation presented below is a partial

differential equation since a derivative can result with respect to both x and y.

Example Consider an equation of the form F(x,y).A partial differential equation

that can be represented is as follows :

Ȁሺ ሺǡሻ with respect to x otherwise written as ܨ௫(x,y) or μȀμ where x is

allowed to vary. Upon finding the derivative of the same function with respect to y the representation is as follows:

ܨ௫௬(x,y) i.e. μȀμሺμȀμሻ which is equivalent to ߲ଶf/μμ.Few examples of the

partial differential equation are as follows for ready reference and a basic understanding: μȀμሺμȀμሻμȀμሺμȀμሻൌͲ Equation 20 μȀμሺμȀμሻμȀμሺμȀμሻሺμȀμሺμȀμሻൌଶଶ Equation 21

3.13 Summary

This chapter discuss es on the concepts of differential equation and their solving methodologies, as differential equation formulation and representation with respect to heat conduction, oscillation in mechanical and electrical systems and

circuitry take a centre stage in all modern scientific and engineering studies. In

applied mathematics generally, the study of differential equation constitutes of

modelling the equation, solving the equation using different criterion and conduction as rules of separation, reduction, multiplication by a certain integration factor to make it exact. Here even mechanisms to find certain integrating factors by inspection or of a homogeneous equation or represented in

a complex format to find a general solution to the real world problems. This chapter introduces the students to the fundamental problem solving in the segment of first order and first degree equations that are moderately complex to

model and solve.

munotes.in

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88 APPLIED MATHEMATICS

3.14 References

1. Higher Engineering Mathematics B.S. Grewal,43rd Edition,Khanna Publishers

2. Differential Equation, Shepley L Ross Wiley Publications, 3rd Edition

3. https://byjus.com/maths/differential -equation/

4. https://abdullahsurati.github.io/bscit

5. ISC Mathematics, O.P. Malhotra, S. Chand Publications

3.15 Questions

1. Given the differential equation dp/dq=p4íq4(p2+q2)pq the degree of differential equation.

2. Solve (m2 + n2 + m) dm + mndn = 0.

3. Solve the following equations by the method of inspection

a) y(3yx + ex) dx - ex dy = 0

b) ydx -xdy+lnxdx=O fr all x,y>O.

c) (xy - 2y2) dx - (x2 - 3xy)dy = 0.

4. Solve the homogeneous equation : Solve (x2y - 2xy2) dx - (x3 - 3x2y) dy = 0

using Integrating Factor.

5. Solve (p4+y4)dp - py3dy = 0. (Hint When bp - ay # 0 and the different

equation a(p,y) dy + b(p,y) dp = 0 can be written in the form qf,(p y)dp +

pf,(p,y)dy = 0 with I as an integrating factor).

6. Check for exactness of the equation :

Solve y(x2y2 + 2) dx + x(2 - 2x2 y2) dy = 0

7. Solve for exactness and find the integrating factor

(3x2y4+2xy) dx + (2xY3 -x2) dy = 0.

munotes.in

## Page 89

89Unit 2

4 DIFFERENTIAL EQUATION

OF THE FIRST ORDER OF

A DEGREE HIGHER THAN THE FIRST

Unit Structure

4.0 Objectives

4.1 Introduction

4.1.1 Equations solvable for x

4.1.2 Equations solvable for y

4.2 Equations not containing dependent/independent variable

4.3 Clairaut’s Form of the Equation

4.3.1 Equations reducible to Clairaut’s form

4.4 Summary

4.5 References

4.6 Questions

Please note two conventions of differentials have been used (ܡ

ܠ,dy/dx)

4.0 Objectives

Here nonlinear equations are considered where the derivatives are of first order

and of higher degree. The equations are not solvable by any structured methodology. Here, some typical types of equations are considered to describe

the techniques of solution of such equations. One will able to solve differential

equation of first order and higher degree solvable for solvable for x, solvable for y

and the Clairaut’s form of the equation. Also obtain the solution of the differential

equati ons in which x or y is absent

4.1 Introduction

Isaac Newton (1642 -1727), the English mathematician and scientist, classified

differential equations of the first order then known as fluxional equation which munotes.in

## Page 90

90 APPLIED MATHEMATICS

was published in 1736. Then Count Jacopo Riccat i (1676 -1754), an Italian

mathematician, contributed towards advancing the theory of differential equations

with reduction of an equation of the second order in y to an equation of first

order in p. In 1723, he exhibited the solution of an equation to whi ch the name of

Riccati is attached. Later the French mathematician Alexis Claude Clairaut (1713 -

1765) pioneered the idea of differentiating a given differential equations in a

specific form to solve them.

These equations are described as equations constituting of dependent and independent variable, that are solvable using the following : equations that are

solvable for p,y,x and the Clairaut’s form, the techniques and methodologies of

which are described in the succeeding section.

Equations that are solvable for p

( ܘൌܡ

ܠ) and for y : y = f (x, p) and for x = f (y,p )

For Clauriat’s form of equation it is a follows : y = p(x) + f (p )

The equations that are solvable for p of the first order and the nth degree is

represented as follows:

= pn + f1(x,y)pn-1 + f2(x,y)pn-2 +f3(x,y)pn-3 +……+ f n-1(x,y)p +f n(x,y) = 0 Now the left hand side of the above equation is split up into n linear representative equations as follows :

[p - ࣂ(x,y)], [p - ࣂ(x,y)]………[ p - ࣂ(x,y)] and these are of first order and first

degree. Each individual solution to the above can be represented in the form as

follows : f1(x1,y1,c) = f 2(x2,y2,c)….f n(xn,yn,c) = 0 and these together form the

solution for the above equation as follows.

Example

Solve m2 +m (݁௫ ଵ

ೣ ) + 1 = 0

= m (m+݁௫) + ଵ

ೣ(m + ݁௫) = 0

= (m+݁௫) (m+ଵ

ೣ) = 0

= y + ݁௫+ k’ = 0, y+ ݁ି௫ +k” = 0 Ans

Here k’ and k” can be replaced by k and the final equation constitutes of

first degree and first order representation.

=(y + +k)(y+ ି +k) = 0 Ans

munotes.in

## Page 91

91Chapter 4 Differential Equation of the First Order of A Degree Higher Than The First

Example

Solve ௗ

ௗ௫ - ௗ௫

ௗ =

௫ - ௫

= q - ଵ

=

௫ - ௫

= q2 - 1 = q(

௫ - ௫

)

= q2 - 1 - q(

௫ - ௫

) = 0

= q(q -

௫ ) + ௫

(q -

௫) = 0

= (q - ௫

)(q -

௫ ) =0

= q = -

,

=q = ௗ

ௗ௫ = -x/p , Upon integrating ݀ൌെݔ݀ݔ

=

+

= c i.e. p2 + x2 = c is the first solution

When q = ୮

୶

Then ௗ

ௗ௫ - ୮

୶ = 0

= ௗ

- ୢ୶

୶ = 0

= ln(p) - ln(x) = 0

=ln(p/x) = ln (c), p = xc is the required solution

Example

Solve p2 + 2 py cot x = y2

The square root of p will be equal to

=(-b(+-)square root of ((b2 -4ac))/2a

=(1/2)( -2ycotx (+ -)ඥͶݕଶݐܿଶݔെͶݕଶ

í\FRW[\FRVHF[

= డ௬

డ௫ = -ycotx +ycosecx

= డ௬

డ௫ = y(cosecx - cotx)

= డ௬௬ = (cosecx -cotx)߲ݔ munotes.in

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92 APPLIED MATHEMATICS

=డ௬௬=ሺ
െ
ሻݔ߲

=log y = log tanx(௫

ଶ) - log(sinx) +log(c)

= y (1+cosx) = c

Similarly for the equation

డ௬

డ௫ = -ycotx - ycosecx

= y íFRV[ F

= y (1(+-) cosx) = c Ans

Example

Solve xyp3 + +(x 2 í\2 Sí[\S

= p [xyp2 + (x2 - 2y2)p -2xy] = 0

= p (xp- 2y)(yp+x) = 0

= ( p = 0 , y -c = 0),(xp - 2y = 0), (yp +x = 0)

= (xp - 2y) = 0, let p = ௗ௬

ௗ௫ = x ௗ௬

ௗ௫ = 2y, or ௗ௬

௬ = 2 ௗ௫

௫ , y = cx2

= yp + x = 0, with p = ௗ௬

ௗ௫ , ydy +xdx = 0, x2 +y2 - 2c = 0

So the final equation becomes (y-c)(y-cx2)(x2 + y2 - 2c) = 0 Ans

4.1.1 Equations solvable for x

Let there be equation of the form x= f(y,p)

Differentiating with respect to y it can be represented as follows:

S G[G\ ș\SGSG[

The solution that can be deduced is as follows: F(y,p,c) = 0 that can be shown

through the following example.

Example

Solve x = 4(p+p2)

dx/dy = 1/p = pdx = dy Differentiating with respect to y

= 1/p =4(1 +2p)dp/dy

= dy = 4p(1 + 2p)dp

Integrating we have

ݕ݀ൌͶሺͳʹሻ݀

=y = 2p2 + (8/3)p3 +c Ans munotes.in

## Page 93

93Chapter 4 Differential Equation of the First Order of A Degree Higher Than The First

Example

Solve y = 2 px + y2 p3

= y - y2p3 = 2px

= ୷ି୷మ୮య

ଶ = x

= y/2p - y2p2/2 = x

Differentiating the above with respect to y

The first component is ଵ

ൌଶିଶ௬ങ

ങ

ସమ

and the second component is ଶ௬మା௬మଶ

ଶ

= 0 = ଶିଶ௬ങ

ങ

ସమ - ଶ௬మା௬మଶ

ଶ - ଵ

=0 = ( ப୮

ப୷ሻ(yp +ଵ

ଶమ) = 0

= (ப୮

ப୷ሻ = c = log(py) = logc

= py = c

Eliminating p from the main equation

We have y2 = 2cx +c3 as the solution

Example

Solve: y2p 2 í[S\

The differential equation is of the form x = f (y,p),

where f (y,p) = (1/3)(y/p+y 2p).

Differentiating with respect to y we get

ୢ୶

ୢ୷(3) = 3(1/p)

= (1/p) -(y/p2)(dp/dy) +2yp +y2(dp/dy)

Simplifying we get

2p +y(dp/dy) = 0 so (dp/p) +2(dy/y) = 0

= p =

௬మ

Hence y3 - 2cx + c2 = 0 x then becomes p +1/p

(dx/dy) = (dp/dy) - (1/p2)(dp/dy)

Integrating

ሺെͳȀሻdp = ሺെͳȀሻ +c

= y = (p2/2) - log p +c and x = p +1/p Ans

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## Page 94

94 APPLIED MATHEMATICS

Example

Solve y2p2 - 3xp + y = 0

The equation can be represented in the form

x = f(y,p) = p = tan-1(p +

ା )

= ࣔ

ࣔ =ଵ

= (ଵ

ଵାଶሻడ

డ௬ + ଵାమିଶమ

ሺଵାమሻଵమ

= డ௬

డ = ଶ

ሺଵାమሻଵమ

Upon integration

y = c (1 +p2) - 1 from where y cannot be removed. Ans

4.1.2 Equations solvable for y

A differential equation of first order and higher degree takes the form y = f (x, p).

Differentiating the equation w.r.t x , we have p = ௗ௬

ௗ௫ ș\SGSG[7KHVROXWLRQ

for the same will be in the form of : F(x,p,c) = 0. Now taking into consideration y

= f(x, p) and solution being F(x, p, c) the (x, y) variables in the equation can be

represented as x = F 1(p, c) and y = F 2(p, c) respectively as the solution.

Example

Solve y = p x +a p(1-p)

We differentiate the above with respect to x.

dy/dx = p + x dp /dx + a(dp/dx) - a(2p)dp/dx

dy/dx = p + dp/dx[x +a -2ap]

p = p + dp/dx[x +a -2ap]

0 = dp/dx[x +a -2ap]

Here p is a constant hence the equation becomes p = 1/2a(x +a)

y = (1/2a)(x + a)[x + a(1/2a(x + a))(1 -(1/2a)(x + a))

Example

Solve x+ 2(xp - y) + p2

The equation to be represented as y = f(x, p) and hence is solvable for y

Representing in the form with y on the left hand side the equation takes the

form = y= ௫

ଶ + xp + మ

ଶ munotes.in

## Page 95

95Chapter 4 Differential Equation of the First Order of A Degree Higher Than The First

= ௗ௬ௗ௫ = p = ଵଶ + p + xௗௗ௫ + p ௗௗ௫

= (x + p)ௗ

ௗ௫ + ଵ

ଶ = 0

Let (x + p) = u , 1 + ௗ

ௗ௫ = ௗ௨

ௗ௫, ଶ௨

ଶ௨ିଵ μ= μ

ൌሺͳଵ

ଶ௨ିଵ ) μ ൌݔ߲ +c = 0

= u + ଵ

ଶ log (2u -1 ) = x+ c

Replacing p with x + u we have

= 2p + ଵ

ଶlog (2(x+p) - 1) = x + c

= 2x + 2p - 1

= e2p - c = x = (

)ି +1 - p and y =

+ xp +

Ans

Example

Solve p2 - py +x

= y = (x + p2)/p

= x/p + p

= xp-1 + p

=dp/dx +1/p - xp-2(dp/dx) = p = dy/dx

Solving this equation

= ௗ௫

ௗ + ௫

ሺାଵሻሺିଵሻ =

మିଵ

The integrating factor is

݁

మషభ = ݁ቂభ

మሺషభሻାభ

మሺశభሻିభ

ቃௗ

=݁ሾሺశభሻሺషభሻభ

మ

= ሺమିଵሻభ

మ

Hence the final solution is

x(ሺమିଵሻభ

మ

) =

మିଵሺమିଵሻభ

మ

dp = ௗ

ඥమିଵ = c +cosh-1 p

= x = p(c + cosh-1p) (p2 -1)1/2 Ans

Example

Solve y = 2px +pn

Differentiating with respect to x

p = 2p +2x(dp/dx) +npn-1(dp/dx) munotes.in

## Page 96

96 APPLIED MATHEMATICS

0 = p + 2x(dp/dx) +npn-1(dp/dx)

0 = p +2x(dp/dx) +(np/p)(dp/dx)

-p =(dp/dx)(2x ) +npn-1(dp/dx)

= - ௗ௫

ௗ[p] = 2x +npn-1

0= డ௫

డ[p] + [2x] +[

మ]

Integrating factor ݁ଶ = p2

Solution is

xp2 = - ߲݊ +c

= xp2 = - n శభ

ାଵ +c

= x = - n pn+1-2 + cp-2

= x = -npn-1 +cp2

Then substitute for y in the given equation

y = 2px +pn and the solution is as follows :

y =

+ା

ି Ans

4.2 Equations not containing dependent/independent variable

Sometimes the equations do not contain dependent/independent variable and

either it contains y or x and not both such equations can be represented in the

form as follows: f(x,p) = 0 or f(y,p) = 0. For example y = ଵ

ଵାమ is one of form of

equation where the x is missing as an independent variable from the equation.

Type I

In the former scenario equations do not contain independent variable

This equation can be represented as follows after differentiating with respect to x

as follows :

S G\G[ ș\

In order to seek clarity lets consider the below example

y = 3p +6p2 This equation is already in the form y = f(p)

p = 3 (ௗௗ௫ ) + 12pௗௗ௫ munotes.in

## Page 97

97Chapter 4 Differential Equation of the First Order of A Degree Higher Than The First

p = (ௗௗ௫) (3 + 12 p)

dx = (3 + 12p)/p (dp)

=x = 3 ln(p) +12p +c and y = 3p +6p2 Ans

Example

Solve y2 = a2 (1 +p2)

The above equation is in y and p only.It can be written as follows :

= p2 =

- 1

= p =( + -)ට௬మ

మ - 1

= డ௬

డ௫ = ඥ௬మିమ

= a ln ȁඥଶ-a2| = x +c

= a ln ȁඥଶ-a2| (-+)x-c Ans

Type II

Equations not containing a “y” as the dependent variable

Let the equation be as follows:

x= ଵ

ଵାమ

= 1/p = (1 +p2)-1

= 1/p = (1+p)-2(-2p)(dp/dy)

= dy =ିଶ

ሺଵାሻమ

= ݕ߲ =ିଶమ

ሺଵାమሻమ݀

=ݕ߲ =ʹሾିଵ

ሺଵାమሻ + ଵ

ሺଵାమሻమ] dp

= y = tan-1 p + 2

We will use the substitution [ WDQșLPSO\LQJWKDW dx=sec2șGș

, VHF2ș GșWDQ2ș2

Note that 1+tan2ș VHF2ș

munotes.in

## Page 98

98 APPLIED MATHEMATICS

Integrating Factor VHF2șGșVHF4ș GșVHF2ș FRV2șGș

Recall that FRVș FRV2șíVR cos2ș FRV2ș

Integrating Factor FRV2șGșGș

Integrating Factor =1/4sin2șș&

From [ WDQș we see that ș DUFWDQ[

We see that 1/4sin2ș VLQșFRVș

Also, since WDQș [IRUDULJKWDQJOHWULDQJOHZLWKWKHVLGHRSSRVLWH ș being x, the

adjacent side being 1, and the hypotenuse ¥[2.

Thus, VLQș [¥[2 and FRVș ¥[2:

Integrating Factor VLQșFRVșDUFWDQ[&

Integrating Factor [¥[2¥[2)+arc tanx/2+c

Integrating Factor =x/2(1+x2)+arctanx/2+c Ans

4.3 Clairaut’s Form of the Equation

When an equation is of fir st degree in x and y, it is solvable for both independent

and dependent x and y variables both and hence it can be put in the following

forms:

y = xf 1(p) +f 2(p) or

x = yg 1(p) + g 2(p) and these can be solved normally.

But if f 1S SWKHQLWWDNHVWKH&ODLUDXW¶VIRUPDVIROORZV : y = xp + f(p) and

these equations can be non linear in nature. Here f(p) is a known function that

does not contain an x or y.

Instances , like y = px + p2 and y = x + eq DUHH[DPSOHVRI&ODLUDXW¶VHTXDWLRQ

whereas equations y = xy2 + p or y - x2p2 + yp2 are not of the Clairaut's form.

Let there be an equation of the form y = px + f(p) where y is the dependent

variable and (p,x) are the independent var iable.

= ௗ௬

ௗ௫ = p +xௗ

ௗ௫ I¶Sௗ

ௗ௫

= p = p +xௗ

ௗ௫ I¶Sௗ

ௗ௫

= 0 = xௗ

ௗ௫ I¶Sௗ

ௗ௫

Now with ୢ୮

ୢ୶ = 0 and p = c we have y = cx + f (c) which is the general solution

RI&ODLUDXW¶VHTXDWLRQ munotes.in

## Page 99

99Chapter 4 Differential Equation of the First Order of A Degree Higher Than The First

Example

Solve y = mx +

Since the said equation is exactly in the form of a

Clairaut’s representation hence there is no need to solve it further.

Exam ple

Solve q = log(qx - y )

eq = qx -y

y = qx - eq Replacing q with c the equation becomes

y = cx - ec and this equation is in the required Clairaut’s form.

Example

Solve y = xy’ +(y’ )2

let y’ = p

y = x (p) + (p)2

p (dx) = x (dp) + p(dx) + 2pdp

0= x(dp) +2pdp

= dp(x + 2p) = 0

= dp =0; p = c ; x = -2p; p = c

= x = -2p; y = xp + p2

= p = -

= -

(Eliminating p )Ans

Example

Solve y = xy’ + ඥሺܡᇱሻ

Let y’ = q

y = xq + ඥሺଶሻͳ

dy = xdq + qdx + ௗ

ඥమାଵ

0= xdq + ୯ୢ୯

ඥ୯మାଵ

Now dq = 0 and p =c

So y = cx + ඥሺ܋ሻ

munotes.in

## Page 100

100 APPLIED MATHEMATICS

The other equation is as follows :

x = ି

ඥା and y = (ି

ඥା)x + ඥሺܙሻ

y =

ඥା

Elimination of p happens by putting the equation in the form of x2+ y2 = 1 Ans

Example

Solve: e4xSíH2ypp2 = 0.

The differential equation is not in WKH&ODLUDXW¶VIRUPEXWE\WDNLQJ

e2x = u and e2y = v and can be convert ed LWLQWR&ODLUDXW¶VIRUP

v = u d v/ du + (dv/du)2 and now this is in the &ODLUDXW¶VIRUP

dv/ du = c = ֜ v = uc +c 2 =֜ e 2y = ce2x +c 2 is the general solution.

4.2.1 Equations reducible to Clairaut’s form

Many differential equations of the first order but of the higher degree can be

UHGXFHGWR&ODLUDXW¶VIRUPZLWKVXEVWLWXWLRQV

Example

Transform and solve the following equation i.e. x2(y-px) = p2y is transformed into

&ODLUDXW¶VIRUP

Here x2 and y2 can be considered as u and v respectively i.e.2xdx = du,2ydy = dv

ௗ௩

ௗ௨ = (ௗ௩

ௗ௬/ௗ௨

ௗ௫)/(ௗ௬

ௗ௫) .

Let p = ௗ௬

ௗ௫ = ௫

௬ௗ௩

ௗ௨ = (௨

௩)1/2ௗ௩

ௗ௨

= (u)(v1/2 - (௨

௩ሻଵȀଶ(u)1/2ௗ௨

ௗ௩) = ௨

௩( ௗ௩

ௗ௨)2(v)1/2

= v = u

(+

)2 Ans

Example

(ௗ௬

ௗ௫) + 4x ( ௗ௬

ௗ௫ ) - 4y = 0

Let ௗ௬

ௗ௫ = p then p +4xp -4y = 0

2U\ S[SDQGWKLVLVLQWKH&ODLUDXW¶V(TXDWLRQ

Differentiating with respect to x

S SS¶[SS¶

S¶[SS¶

Assuming p =c , y = cx +(c2/4)

Eliminating p we have y(x) = -x2 as it satisfies y = (p +4xp)/4 munotes.in

## Page 101

101Chapter 4 Differential Equation of the First Order of A Degree Higher Than The First

4.4 Summary

There are equations where the left-hand side of the equation can be resolved into

rational factors of the first degree and also there are equations where the left -hand

side of the equations cannot be factorized. Equations that cannot be factorized in

addition to exact and homogeneous are summarized below. Differen tial equations

of the first order but of a higher degree can be solved by one or more of the

following four methods :

- Equations solvable for p , i.e. p = ௗ௬

ௗ௫ where the general solution can be

represented as p - fi (x,y) = 0 and F i(x,y,c) = 0

- Equations solvable for y i.e. y = f ( x,p ), solution for the same can be

represented as f(x,p,c) and the elimination of p if not possible then x =

f1(p,c) and y = f 2(p,c) are combined to form the solutions.

- Equations solvable for x i .e. x = f ( y,p ), solution for the same can be

represented as f(y,p,c) and the elimination of p if not possible then x =

f’1(p,c) and y = f’2(p,c) are combined to form the solutions.

Clairaut’s equation takes the form y = px + f(p) .The general solution for

the same is obtained by replacing p by c . Some complex differential

equations can be reduced to Clairaut’s form with the help of appropriate

substitutions.

4.5 References

1. Differe ntial equations with Application and Programs. S. Balachandra Rao

and H. R. Anuradha, University Press (India) Limited 1996.

2. Lecture notes on Differential Equation by Dr. B. Patel,Department of

Mathematics,Gujarat University

4.6 Problems Find for the below problems whether they are solvable for x,y and p

1. py2 - 2pyx(tan2ș\2sec2ș- x2tan2ș

2. Given p3 í[\S\2 = where p = dy/dx

3. Given y = p tan p + log(cos p)

4. *LYHQ\ S[íS1/2 munotes.in

## Page 102

102 APPLIED MATHEMATICS

5. xp2 - 2yp + x+ 2y = 0

6. y = x +ctan-1q

7. x = tan-1q + q/(1+q2)

8. yq2 + (x-y)q -x = Hint [((x -y+c)(x2 + y2 +constant) = 0) the r.h.s of the

equation is the answer Ans]

9. Represent in the Clairaut’s form and solve the following :

a) y = 2px + 6y2 p2 ( y = v3)

b) sin qx cosy = cos qx siny +q

c) e4x(p-1) +e2yp2 = 0

munotes.in

## Page 103

103Unit 2

5 LINEAR DIFFERENTIAL EQUATIONS WITH

CONSTANT COEFFICIENTS

Unit Structure

5.0 Objectives

5.1 Introduction

5.2 The Differential Operator

5.3 Linear Differential Equation f (D) y = 0

5.3.1 Solution of f (D) y = 0:

5.4 Different cases depending on the nature of the root of the equation f(D) = 0

5.5 Linear differential equation f (D) y = X

5.6 The complimentary Function

5.7 The inverse operator 1/f(D) and the symbolic expiration for the particular

integral 1/f(D) X; the general method s

5.8 Particular integral: Short methods

5.9 Particular integral: Other methods

5.10 Differential equations reducible to the linear differential equations with

constant coefficients

5.11 Summary

5.12 References

5.13 Questions

5.0 Objectives

After going through this chapter, students will able to learn

• The Differential Operator

• Properties of operators

• Linear Differential Equation f(D) y = 0 and solution Of f(D) y = 0

• Different cases depending on the nature of the root of the equation f(D) = 0

• Linear differential equation f(D) y = X

• The complimentary Function

• The inverse operator 1/f(D)

• Particular Integral munotes.in

## Page 104

104 APPLIED MATHEMATICS

5.1 Introduction

A linear equation or polynomial, with more than one term, constituting of the

derivatives of the dependent variable with regard to one or more than one

independent variable is known as a linear differential equation.

A differential equation which comprises of the differential coefficients and the

dependent variable in the first degree, that does not include the product of a

derivative with another derivative or with dependent variable, and in which the

coefficients are as constants is called a linear differential equation with constant

coefficients.

The general form of such a differential equation of order "n" is

b0ௗ௬

ௗ௫ + b 1ௗషభ௬

ௗ௫షభ + b 2ௗషమ௬

ௗ௫షమ + ………. + b n-1ௗ௬

ௗ௫ + b 2y = X …………………

(Equation)

Here b 0, b1, b2 … are constants. Above equation is a nth order linear differential

equation with constant coefficients.

E.g. when n = 3 is put in the equation we get

b0ௗయ௬

ௗ௫య + b 1ௗమ௬

ௗ௫మ + b 2ௗ௬

ௗ௫ + b 3y = X

which is a 3rd order linear differential equat ion with constant coefficients.

Using the differential operator D as ௗ

ௗ௫ i.e. Dy = ௗ௬

ௗ௫; D2 y =ௗమ௬

ௗ௫మ, …… Dn y = ௗ௬

ௗ௫ ,

the above equation will take the form

b0 Dn y + b 1 Dn–1 y + b 2 Dn–2 y + … + b n–1 Dy + b n y = X

OR

(b0 Dn + b 1 Dn–1 + b 2 Dn–2 + … + b n–1 D + b n)y = X

…………………………..(Equation)

in which each term in the parenthesis is multiplied to y and the results are added

to form the equation.

/HWI'ŁE 0 Dn + b 1 Dn–1 + b 2 Dn–2 + … + b n–1 D + b n

f (D) is called as nth order polynomial in D.

Then the above equation can be written as f(D) y = f(x) … (Equation)

If in equation (1), if b 0, b1, b2 … … bn are functions of x then it is called nth order

linear differential equation. munotes.in

## Page 105

105Chapter 5 Linear Differential Equations with Constant Coefficients

5.2 The Differential Operator D

It is appropriate to present the symbol D to denote the operation of differentiation

with respect to x.

D2 designate differentiation twice.

D3 designate differentiation three times.

In general, let Dk designate differentiation k times.

LH'Łௗ

ௗ௫ , so that

ௗ௬

ௗ௫ = Dy; ௗమ௬

ௗ௫మ = D2y; ௗయ௬

ௗ௫య = D3y; ……; ௗ௬

ௗ௫ = Dn y

and ௗ௬

ௗ௫ + ay = (D + a) y

The differential operator D or (Dn) correlates to the algebraic laws.

Properties of the operator D

Suppose y 1 and y 2 are differenti able functions of x and "b" is a constant and p, q

are positive integer then the following holds true

Ǥ Dp (Dq ) y = Dq (Dp ) y = Dp+q y

Ǥ (D – p1) (D – p2) y = (D – p2) (D – p1) y

Ǥ (D – p1) (D – p2) y = [D2 – (p1 + p 2) D + p 1 p2] y

Ǥ D (bu ) = b · D(u); Dn (bu) = b · Dn (u)

Ǥ D (y 1 + y 2) = D (y 1) + D (y 2); Dn (y 1 + y 2) = Dn (y1) + Dn (y2).

5.3 Linear Differential Equation f(D) y = 0

Consider f (D) y = 0 …………………… (Equation)

where, f(D) = b 0 Dn + b 1 Dn–1 + b 2 Dn–2 + … + b n–1 D +……… b n

is nth order polynomial in D and D obeys the laws of algebra, f(D) can be

factorized into n linear factors as follows :

f(D) = (D – p1 ) (D – p2 ) (D – p3 ) … (D – pn) where p 1 , p2 , p3 , … p n are the

roots of the algebraic equation f(D) = 0

Therefor the equation can be written as follows :

f(D) y = (D – p1 ) (D – p2 ) (D – p3 ) … (D – pn) y = 0 … (Equation)

The equation f (D) = 0 is called as an auxiliary equation for the above equations. munotes.in

## Page 106

106 APPLIED MATHEMATICS

e.g. ௗమ௬ௗ௫మ +ܦ +12y = 0

By using operator D for ௗ

ௗ௫ ,

we have (D2 + D = 12) y = 0

f (D) = D2 +(4D -3D) -12 = 0 is an auxiliary equation.

(D2 + D - 12) y = (D + 4) (D -3) y =0

5.3.1 Solution of f(D) y = 0

Being nth order Differential Equation, the above equations will ha ve exactly n

constants in its general solution.

The equation (5) will be satisfied by the solution of the equation

(D – pn) y = 0

i.e. ௗ௬

ௗ௫ – pn y = 0

On solving this first order and first degree differential equation by separating

variables, we get y = c n epnx, where, c n is an arbitrary constant.

Similarly, since the factors in equation can be taken in any order, the equation

will be satisfied by indep endently solving each of these equations (D – p1) y = 0,

(D – p2 ) y = 0 … etc., that is by y = c 1 e p1x , y = c 2 e p2x ………… etc.

It can, therefore, easily be proved that the sum of these individual solutions is the

sum of n arbitrary constants , i.e. y = c1 e p1x + c 2 e p2x + … + c n e pnx … where the

original equation is of terms containing till the nth order and so also are the

constants for the above said equations.

The general solution of the equation f (D) y = 0 is,

y = c 1 e p1x + c 2 e p2x + … + c n e pnx

where p 1 , p2 , … p n are the roots of the auxiliary equation f(D) = 0.

Example :

Solve ௗయ௬

ௗ௫య - 6ௗమ௬

ௗ௫మ + 11ௗ௬

ௗ௫ - 6y = 0

Solution: Let D stand for ௗ

ௗ௫ and the given equation can be written as

(D3 – 6D2 + 11D – 6) y = 0.

Here auxiliary equation is D3 – 6D2 + 11D – 6 = 0

i.e. (D – 1) (D – 2) (D – 3) = 0

֜ p1 = 1, p 2 = 2, p 3 = 3, are roots of auxiliary equation. The general solution is y = c 1 e x + c 2 e 2x + c 3 e3x munotes.in

## Page 107

107Chapter 5 Linear Differential Equations with Constant Coefficients

5.4 Different cases depending on the nature of the root of the

equation f(D) = 0

Ǥ The Case of Real and Different Roots

If roots of f (D) = 0 be p 1, p2, p3 ….… p n, all are real and different, then the

solution of f (D) y = 0 will be

y = c 1 e p1x + c 2 e p2x + … + c n epnx

Ǥ The Case of Real and Repeated Roots

Let p 1 , p2, p3 , p4 … p n be the roots of f(D) = 0, then the part of solution

corresponding to p 1 and p 2 will look like c 1 e p1x + c 2 e p1x (p1 = p 2 )

= (c 1 + c 2) ep1x = c'ep1x

But this means that number of arbitrary constants now in the solution will

be n – 1 if 2 p’s are the same. Hence it is no longer the general solution. The

rectification of the anomaly is done as follows:

Pertaining to p 1 = p 2, the part of the equation will be (D – p1 ) (D – p1 ) y =

0

Put (D – p1) y = t, then we have

(D – p1) t = 0

t = c 1 ep1x

Hence putting value of t in (D – p1) y = t,

we have (D – p1 ) y = c 1 e p1x

or (ௗ௬

ௗ௫ – p1) y = c 1 e p1x which is a linear differential equation.

Its I.F. = e – S1dx = e – p1x and hence solution is

y (e – p1x F1 e p1x · e –p1x dx + c 2 = c 1 x + c2

y = (c1 x + c2) ep1x

If p1 = p2 are real, and the remaining roots p 3, p4, p5, ….., p n are real and

different then solution of f(D) y = 0 is

y = (c 1 x + c 2) ep1x + c 3 e p3x + c 4 e p4x «F n e pnx

Similarly, when three roots are repeated.

i.e. if p 1 = p 2 = p 3 are real, and the remaining roots p 4 , p5 , … p n are real and

different then solution of f(D) y = 0 is

y = (c 1 x2 + c 2 x + c 3) ep1x + c 4 e p4x «F n e pnx munotes.in

## Page 108

108 APPLIED MATHEMATICS

If p 1 = p 2 = p 3 = … = p n

i.e. n roots are real and equal then solution of f(D) y = 0 is

y = (c 1 xn–1 + c 2 xn–2 + … + c n–1 x + c n) ep1x

Example

For (D2 – 6D + 9) y = 0

Auxiliary Equation = (D – 3)2 y= 0

and solution takes the form (c 1x +c 2)e3x

and the final representation is as follows:

y = (c 1 x + c 2) e3x Ans

Example

For (D – 1)3 (D + 1) y = 0

solution is y = (c 1 x2 + c 2 x + c 3 ) ex + c 4 e –x where p 1 = p 2 =p3

Example

For (D – 1)2 (D + 1)2 y = 0 where p 1 = p 2 = p 3 = p 4

solution is y = (c 1 x + c 2 ) ex + (c 3 x + c 4 ) e–x.

c. The Case of Imaginary or the Complex Roots

The coefficients of the auxiliary equation that are real will have the imaginary roots that will occur in conjugate pairs.

/HWĮLȕEHRQHVXFKSDLU

p1 ĮLȕS 2 Į– Lȕ

Then the solution of the equation f(D) y = 0 takes the form as follows :

y = Pe ĮLȕ[ + Q e Į– Lȕ[

= eĮ[ [P e Lȕ[ + Qe –Lȕ[]

= eĮ[ >3FRVȕ[LVLQȕ[4FRVȕ[ – LVLQȕ[@

= eĮ[ >34FRVȕ[L3 – 4VLQȕ[@

y = eĮ[ [c1 FRVȕ[F 2 VLQȕ[@

where, c 1 = P + Q and c 2 = i (P – Q) are arbitrary constants.

y = C eĮ[ FRVȕ[ș ZKHUH&șDUHDUELWUDU\FRQVWDQWV

using c 1 &FRVșF 2 = – VLQș munotes.in

## Page 109

109Chapter 5 Linear Differential Equations with Constant Coefficients

Example:

Solve (D2 + 2D + 5) y = 0.

Solution: The auxiliary equation is D2 + 2D + 5 = 0

whose roots are D = – 1 ± 2i which are both imaginary.

+HUHĮ – ȕ

Hence the solution is y = e–x [P cos 2x + Q sin 2x]

Example:

Solve ௗర௬

ௗ௫ర - 5ௗమ௬

ௗ௫మ + 12 ௗ௬

ௗ௫ + 28y = 0

Solution: The auxiliary equation is D4 – 5D2 + 12D + 28 = 0

Roots are D = – 2, –2, 2 ± ξ͵ i.

+HUHĮ ȕ +HQFHWKHVROXWLRQLV

y = (c 1 x + c 2) e–2x + e2x [P cos ξ͵ x + Q sin ξ͵ x]

Example:

Solve For (D2 \ ' L+HUHĮ ȕ

֜ y = P cos 2x + Q sin 2x.

d. The Case of Repeated Imaginary Roots

If the imaginary roots p 1 ĮLȕDQGS 2 Į– LȕRFFXUWZLFHWKHQWKHSDUW

of solution of f (D) y = 0 will be

y = (P x + Q) ep1x + (Rx + S) ep2x «E\XVLQJFDVH

= (P x + Q) eĮLȕ x + (Rx + S) eĮ– Lȕ x

= eĮ[ [(P x + Q) eLȕ[ + (Rx + S) e–Lȕ[ ]

= eĮ[ >3[4^FRVȕ[LVLQȕ[`5[6^FRVȕ[ – LVLQȕ[`@

= eĮ[ >3[45[6FRVȕ[L (Px + Q – Rx – 6VLQȕ[@

\ HĮ[>F[FFRVȕ[F[FVLQȕ[@ZLWK

constants as c 1 , c2 , c3 and c 4 .

Example:

Solveௗల௬

ௗ௫ల + 6ௗర௬

ௗ௫ర + 9 ௗమ௬

ௗ௫మ = 0

Solution : The auxiliary equation D6 + 6D4 + 9D2 = 0 has roots munotes.in

## Page 110

110 APPLIED MATHEMATICS

D = 0, 0, ± i ξ͵ , ± i ξ͵ where the imaginary roots ± i ξ͵ are seen to

occur in a recurrent manner.

Hence the solution is

y = c 1 x + c 2 + (c 3 x + c 4) cos ξ͵ x + (c 5 x + c 6) sin ξ͵ x

Example:

Solve (D4 + 2D2 + 1) y = 0.

Solution: The auxiliary equation D4 + 2D2 + 1 = 0 has roots

D = ± i, ± i, recurring imaginary roots. Hence the solution is

y = (c 1x + c 2) cos x + (c 3x + c 4) sin x.

Summary of four cases

Case 1: Real & Distinct Roots:

Auxiliary Equation ֜ (D – p1) (D – p2) (D – p3) … (D – pn) = 0

Solution is y = c 1 e p1x + c 2 e p2x + c 3 e p3x + … + c n e pnx

Case 2: Repeated Real Roots

For p1 = p2 ֜ Auxiliary Equation ֜

(D – p1) (D – p2) (D – p3) … (D – pn) = 0

Solution is y = (c1 x + c 2) e p1x + c3 e p3x + … + c n e pnx

For p 1 = p 2 = p 3 ֜ A.E. ֜ (D – p1) (D – p1) (D – p1) (D – p4) … (D – pn) =

0

Solution is y = (c 1 x2 + c 2 x + c 3) ep1x + c4 e p4x + … + c n e pnx

Case 3: Imaginary Roots

)RU' ĮLȕ

Solution is y = eĮ[ [c1 FRVȕ[F 2 VLQȕ[@

Case 4: Repeated Imaginary Roots

)RU' ĮLȕEHUHSHDWHGWZLFH

Solution is y = eĮ[ [(c1 x + c 2FRVȕ[F 3 x + c 4VLQȕ[@

5.5 Linear differential equation f (D) y = X

The general solution of the equation f (D)y = X can be represented as

y = Y c +Y p munotes.in

## Page 111

111Chapter 5 Linear Differential Equations with Constant Coefficients

i.e. General solution = Complementary function + Particular integral

Yc is the solution of the given equation with X = 0 that is of equation f (D)y = 0

and is called the complementary function. It involves n arbitrary constant s and is

denoted by Complementary f unction ( C.F).

By definition of Y c, f(D) Y c = 0.

Yp is any function of s, which satisfies the equation f(D)y = X, so that f(D) Y p =

X. Y p is called the particular integral and is denoted by particular integral(P.I). It

does not contain any arbitrary constants.

Thus, on substituting y = Y c +Y p in f (D) y, we have

f(D) [Y c +Y p ] = f(D) Y c + f(D) Y p

= 0 + X …………… [by definition of Y c and Y p ]

= X

y = Y c +Y p satisfies the equation f (D)y = X and it contains n arbitrary

constants, is the general (or complete) solution of the equation.

5.6 The Complimentary Function

The solution where the order of the differential equation matches the number of

arbitrary constants is called the complementary function (C.F.) of a Differential

equation.

Method of Finding Complementary Function (C.F)

Step I: Find auxiliary equation (Auxiliary .Equation.)

Step II: Find the roots of the equation. i.e. values of p. Let the roots are p 1,

p2,…… , p n .

Step III: Required C.F. is obtained as per the roots stated below.

Rules of finding C.F

If all roots p 1, p2 ,…… , p n are real and distinct of auxiliary equation then

complementary function will be c 1 e p1x + c 2 e p2x + … + c n epnx .

If p 1 = p 2, but other roots are real and distinct then complementary function will

be (c 1 x + c2) c 1ep1x + c 3 e p3x + c 4 e p4x + … + c n e pnx.

,IURRWVDUHLPDJLQDU\ĮLȕWKHQFRPSOHPHQWDU\IXQFWLRQZLOOEHHĮ[ [c1 cos

ȕ[F 2 VLQȕ[@ munotes.in

## Page 112

112 APPLIED MATHEMATICSIf roots are imaginary and repeated twice then complementary function will be eĮ[ [(c1 x + c 2FRVȕ[F 3 x + c 4VLQȕ[@

Example

Solve (D2- 3D - 4) y = 0.

Solution: Here Auxiliary equation is (D2- 3D - 4) = 0.

D2- 3D - 4 = 0

(D - 4). (D +1) = 0

D = 4, -1

Hence roots are 4 and -1, real and different

Complementary Function is y = c 1݁ 4ݔ +c 2݁ – ݔ

Example

Solve (D3 - 8) y = 0.

Solution: Here Auxiliary equation is (D3 - 8) = 0.

D3 - 8 = 0.

(D - 2). (D2 + 2D + 4) = 0

D = 2, D = -L¥

Hence roots are 2, and -L¥

one is real and the rest is a pair of imaginary roots.

Complementary Function is,

y = c 1 e2x + e-x (c2 FRV¥[F 3 VLQ¥[

5.7 The inverse operator 1/f(D) and the symbolic expiration for

the particular integral 1/f(D) X

To find the Particular Integral, it is essential to specify the inverse operator

ଵ

ሺୈሻ.So If X is any function of x, then ଵ

ሺୈሻ X is that function of x that is free from

arbitrary constant which when operated by f (D) gives the function X.

The order of operator f (D) and ଵ

ሺୈሻ can be interchanged.

f (D) { ଵ

ሺୈሻ X} = ଵ

ሺୈሻ f (D) X = X

General Method of finding the Particular Integral munotes.in

## Page 113

113Chapter 5 Linear Differential Equations with Constant Coefficients

Factor Method

To evaluate ଵ

ሺୈሻ X, where X is a function of x, resolve f (D) into factors of the

type (D - a), then operate on X successively by the reciprocal of these factor in

any order using the formula

ଵ

ሺୈିୟሻ X = ݁௫ ܺ݁ି௫dx

If X = ݁௫

ଵ

ሺୈሻ ݁௫ = ೌೣ

ሺୟሻ ; ሺ

= ௫ೌೣ

ᇱሺୟሻ ; f ’ሺD ௫మೌೣ

ᇱᇱሺୟሻ ; f ’’ሺDDQGVRRQ

Same formula is applicable for sin ሺax +b) and cos ሺax +b)

Similarly when there are functions like ଵ

ሺୈሻ ݁௫ ξݔ =݁௫ ଵ

ሺୈାୟሻξݔ we can use

the above methodology

Method of partial fractions

Resolve ଵ

ሺୈሻ into partial fractions and then operate on X by each of these

fractions.

To find the value of ଵ

ሺୈሻ ݔ, p is any positive integer, then

ଵ

ሺୈሻ ݔ= [f ሺሻሿ-1ݔ

Since D is an operator, which can be manipulated as expanding [f ሺሻሿ-1 by the

Binomial theorem in ascending power of D as far as the result of expanding Dp+1

on ݔ is 0.Then operating upon ݔ with each term of the expansion.

Examples

Example :

Solve ௗమ௬

ௗ௫మ - y = 3 + 6x

Solution: Auxiliary equation is D2 – 1 =0

Roots are 1 and -1 and

C.F. is,

Yc= C 1 ݁௫ + C 2 ݁ି௫

The P. I of the equation is given by,

ଵሺమିଵሻ [3 +6x] = ଵଶ { ଵିଵ - ଵାଵ } [3 +6x] munotes.in

## Page 114

114 APPLIED MATHEMATICS

= ଵଶ { ଵିଵ [3 +6x] - ଵାଵ [3+6x]}

= ଵ

ଶ {݁௫ ݁ି௫ [3 +6x] dx - ݁ି௫ ݁௫ [3 +6x] dx}

= ଵ

ଶ [- 6 – 12x]

= -3 – 6x =-3[1-2x].

P. I. is, Yp = - 3[1– 2x]

The complete solution of the equation is

Y = Y c + Y p

Y = C 1 ݁௫ + C 2 ݁ି௫ - 3 – 6

5.8 Particular integral

Short method for finding Particular integral (P.I.):

,I;LQHTXDWLRQ a0ௗ௬

ௗ௫ + a 1ௗషభ௬

ௗ௫షభ + a 2ௗషమ௬

ௗ௫షమ «««D n-1ௗ௬

ௗ௫ + a 2y = X

then P.I. = ଵ

ሺୈሻ X

Following are the methods for finding particular integral

Rules for finding Particular Integral : Types of function What to do Corresponding P.I. X = ݁௫ Put D = a in f(D) ଵሺୈሻ ݁௫SURYLGHGID

If f(a) = 0 then (D -a) is one of the factor of f(D).This facto r is solved by using the

formula ଵ

ሺୈିୟሻ X = ݁௫

݁ି௫ݔ݀ܺ. And rest is solved by the above method given here. X = ݔ Put [f(D)]- 1ݔ Expand [f(D)]-1 using binomial expansions and if

(D-a) remains in the denominator then take rationalization of denominator and place D in

the numerator as derivative munotes.in

## Page 115

115Chapter 5 Linear Differential Equations with Constant Coefficientsof the corresponding function. X = = ݁௫v First operate on ݁௫ on ଵሺୈሻ then operate ݁௫ଵሺୈାୟሻv, then solve for v by above method X = sin ax (or cos ax) Put D2= -a2 in f(D) ଵሺିమሻ sin ax (or cos ax) , provided ଵ

ሺିమሻ ് 0 or otherwise use following formula: ଵ

మାమ sin ax = -

ଵ

ଶୟ cos ax or ଵ

మାమ cos ax =

௫ଶୟ sin ax

Example:

Solve (D2+ 4D+3)y = ݁ିଶ௫

Solution: Here auxiliary equation is (D2+ 4D+3) = 0

D2+ 4D+3 = 0

(D+3) (D+1) = 0

D = -3,-1

Hence roots are -3 and -1, real and different.

Therefore C.F. is

C.F = C 1݁-3x C2݁ –x

Now to find P.I:

P.I = ଵ

ሺୈሻ X

= ଵ

ሺୈሻ X

= ଵ

ୈమାସୈାଷ ݁ିଶ௫

Here X = ݁௫ therefore put D = a = -2

= ଵ

ሺିଶሻమାସሺିଶሻାଷ ݁ିଶ௫

P. I. = - ݁ିଶ௫

Hence the general solution is y = C.F.+ P.I.

Y = C 1݁-3x C2݁ –x - ݁ିଶ௫ munotes.in

## Page 116

116 APPLIED MATHEMATICS

5.9 Particular Integral: Other methods

Method of Variation by Parameters

The method of Variation of Parameters is a generalized method that can be used

in many more cases. However, there are two disadvantages to the method. First,

the complementary solution is required to solve the problem. Secondly, in order

to complete the method a couple of integrals need to be solved.

In some cases we may not be able to actually find the solutions if the integrals

are too difficult or if we are unable to find the complementary solution.

Example:

Solve by Method of Variation by Parameters [D2 + 4] = tan 2x

Solution: The Auxiliary Equation is p2 + 4 = 0

p2 = - 4

p = ± 2i

Complimentary Function is represented as follows :

Yc= C 1
ʹݔ +C 2 ʹݔ

Particular Integral = - y1 ௬మ

ௐ dx + y 2 ௬భ

ௐ dx

y1 = cos2x, y 2 = sin2x; X = tan 2x

and for W, by Wronskian determinant,

W = ฬݕͳݕʹ

ݕͳԢݕʹԢฬ

= ቚݏܿʹݔݏ݊݅ʹݔ

െʹݏ݊݅ʹݔʹݏܿʹݔቚ

= 2cos22x + 2 sin2 2x

= 2 [cos22x + sin2 2x]

= 2

Particular Integral = - cos2x ௦ଶ௫Ǥ௧ଶ௫

ଶ dx + sin2x ௦ଶ௫Ǥ௧ଶ௫

ଶ dx

= - cos2x ௦మଶ௫

ଶ௦ଶ௫ dx + sin2x ௦ଶ௫

ଶ dx [ tanx = ௦௫Ǥ

ୡ୭ୱ௫ଶ ]

= - cos2x ଵି௦మଶ௫

ଶ௦ଶ௫ dx + ௦ଶ௫

ଶ [ି௦ଶ௫

ଶ ]

= - ௦ଶ௫ଶ {ଵ௦ଶ௫ dx - ௦మଶ௫௦ଶ௫ dx} + [ ௦ଶ௫ଶ ି௦ଶ௫ଶ ] munotes.in

## Page 117

117Chapter 5 Linear Differential Equations with Constant Coefficients

= - ௦ଶ௫ଶ {ܿ݁ݏʹݔ dx - ܿݏʹݔ݀ݔ } - ௦ଶ௫௦ଶ௫ସ

= - ௦ଶ௫

ଶ {୪୭ሾ௦ଶ௫ା୲ୟ୬ଶ௫ሿ

ଶ - ௦ଶ௫

ଶ} - ௦ଶ௫௦ଶ௫

ସ

= - ௦ଶ௫

ସ ሾܿ݁ݏʹݔʹݔሿ + ௦ଶ௫௦ଶ௫

ସ - ௦ଶ௫௦ଶ௫

ସ

Particular I ntegral= - ௦ଶ௫

ସ ሾܿ݁ݏʹݔʹݔሿ

The complete solution of the equation is

Y = Y c + Y p

Y = C 1
ʹݔ +C 2 ʹݔ - ௦ଶ௫

ସ ሾܿ݁ݏʹݔʹݔሿ

By the method of variation of parameters,

solve the following differential equation:

డమ௬

డ௫మ + 4y = 4tan2x

= y “+4y = 0

= p2 + 4 = 0

= p ሺെሻʹൌͲ

=p1 = - 2 ; p 2 = 2

=yc = c 1cos2x + c 2sin2x

Now let y 1 = cos2x and y 2 = sin2x

y1’ = - 2sin2x and y 2’ = 2cos2x

W = y 1 * y 2’ - y2 *y1’ = 2[cos2x +sin2x] = 2

A’ =ି௬ଶכସ௧ଶ௫

ௐ

B’ =௬ଵכସ௧ଶ௫

ௐ

A = ିଶ௦మଶ௫

௦ଶ௫μ

B = ʹʹμ

= A = -log (sec2x +tan 2x) +sin2x +c 1,B = -cos2x +c 2

= y = Acos2x +Bsin2x , we put A and B in this equation

and get the final result.

munotes.in

## Page 118

118 APPLIED MATHEMATICS

5.10 Differential equations reducible to the linear differential

equations with constant coefficients

Linear differential Eq uation:

X Linear:

ୢ୶

ୢ୷+ P(y).x = Q(y)

Integrating Factor (I. F.) = ୮ሺ୷ሻୢ୷

x. IF = (y)dy + c

Y Linear:

ୢ୷

ୢ୶+ P(x).y = Q(x)

Integrating Factor (I. F.) = ୮ሺ୶ሻୢ୶

y. IF = (x)dx + c

Example:

Ex 1: (1 + y2) + (x - ିଵ) ୢ୷

ୢ୶ = 0

Sol: Multiply by ୢ୶

ୢ୷

(1 + y2)ୢ୶

ୢ୷ + x - ିଵ = 0

(1 + y2)ୢ୶

ୢ୷ + x = ିଵ

ୢ୶

ୢ୷ +୶

ሺଵା୷మሻ = ୲ୟ୬షభ୷

ሺଵା୷మሻ ,

P = ଵ

ሺଵା୷మሻ and Q = ୲ୟ୬షభ୷

ሺଵା୷మሻ

IF = ୮ሺ୷ሻୢ୷

= భ

ሺభశ౯మሻୢ୷

= ୲ୟ୬షభ୷

x. IF = (y)dy + c

x.୲ୟ୬షభ୷ = ୲ୟ୬షభ୷୲ୟ୬షభ୷

ሺଵା୷మሻ dy + c

Let ିଵ = t

ଵ

ሺଵା୷మሻ = dt

x.୲ = ୲ + c munotes.in

## Page 119

119Chapter 5 Linear Differential Equations with Constant Coefficients

= ୲ + c

= t ୲- ୲ + c

Put t = ିଵ

x.୲ୟ୬షభ୷ = ିଵ ୲ୟ୬షభ୷- ୲ୟ୬షభ୷ + c

Reducible to Linear differential Equation

ୢ୷

ୢ୶+ P(x). y = Q(x) y n

Dividing by y n

ଵ

୷ୢ୷

ୢ୶+ P(x) ଵ

୷షభ = Q(x) ……….. (1)

Let ଵ

୷షభ = t

Differentiating. with respect to y

(-n+1) ଵ

୷ୢ୷

ୢ୶ = ୢ୲

ୢ୶

ଵ

୷ୢ୷

ୢ୶ = ଵ

ଵି୬ ୢ୲

ୢ୶

Eq (1) becomes

ଵ

ଵି୬ ୢ୲

ୢ୶ + P(x) = Q(x)

This is a linear equation in t.

Example:

Ex1. ୢ୷

ୢ୶ - y tan x = - y 2 sec x

Sol: ୢ୷

ୢ୶ - y tan x = - y 2 sec x

Dividing by y2

ଵ

୷మୢ୷

ୢ୶ - ଵ

୷tan x = - sec x

Let ଵ

୷= t

െଵ

୷మୢ୷

ୢ୶ = ୢ୲

ୢ୶

െୢ୲

ୢ୶ - t tan x = - sec x

ୢ୲

ୢ୶ + t tan x = sec x

This is a linear equation in t munotes.in

## Page 120

120 APPLIED MATHEMATICS

P = tan x and Q = sec x

IF = ୲ୟ୬୶ୢ୶

= ୪୭ୱୣୡ୶

= sec x

t. IF = ሺሻ + c

t. sec x =
ଶ + c

Ans:

ଵ

୷ sec x = tan x + c

5.11 Summary

This chapter provides the students with an understanding of linear differential

equation of higher order and degree with constant coefficients and goes on to

explain the concepts of complimentary functions and integral values and their

usage in solving the problems that constitute the above. Students are made to use

the concept of inverse operator and the case of real, repeated and imaginary roots

to solve complex differential equations of the higher order and higher degree. The

techniques of using substitution methods to solve the differential equations by

using the con cept of reduction is also dealt with in this chapter.

5.12 References

“Higher Engineering Mathematics” by B.V.Raamna,Tata McGraw -Hill

Publication, New Delhi.

a. “Schaum’s Outline of Differential Equations” by Richard Bronson and

Gabriel Costa.

b. Applied Mathema tics II by P. N. Wartikar and J. N. Wartikar.

c. https://www.library.gscgandhinagar.in/assets/admin/images/MAT -

102(UNIT1,2).pdf

d. http://www.math.utah.edu/~zwick/Classes/Fall2013_2280/Lectures/Lecture

6_with_Examples.pdf

e. http://www.rahu landmaths.com/bsc -students/differential -equations

5.13 Questions

Solve ୢమ୷

ୢ୶మ + 4y = 0

Solve ୢర୷ୢ୶ర - 16y = 0 munotes.in

## Page 121

121Chapter 5 Linear Differential Equations with Constant Coefficients

Solve (D2 - 3D - 4) y = 0.

Solve (D3 - 8) y = 0

Solve (D2 -3D+2)y=e5x

Solve(D3 -3D2+4)y=e3x

3. ୢయ୷

ୢ୶యെൌଶ୶

4. (D2-2D+1)y = e3x

5. (D2-2D+1)y = ex

6. Assuming that the rate of growth of any organism is directly proportional to

N(t) present at time t, so to find the value of N(t) given that N(0) = 100 and

after (t+1 with t = 0), the size of the organism has grown to 200.

Solution :

In this case t = 0, N(0) = 100. The solution of the problem is given by

N(t) = 100 exp (kt), t >=0

Determine m from the additional condition

N(l) = 200 (N(1) = size of I at time t = 1).

Hence 200 = 100 exp (k) ,k = 1n2

Hence the solution is

N(t) = 100 exp (t ln2) = 100 exp (ln2t) or N(t) = (100) 2t . So the equation

can be represented as shown here.

Ǥ Applications - Electrical circuits

E= Lୢ୧

ୢ୲ + R i

Ǥ $UHVLVWDQFHRIȍDQGDQ inductance of 0.1H are connected in series with

EDWWHU\RI9)LQGFXUUHQWLQFLUFXLWDWDQ\WLPHµW¶

Ǥ In a network circuit of R -/VHULHV5 ȍDQG/ +DFRQVWDQWYROWDJH

150V is applied at t=0 by closing the switch. Find the current in the circu it

at t=0.10sec.

munotes.in

## Page 122

122 APPLIED MATHEMATICS

Unit 3

6 THE LAPLACE TRANSFORM

Unit Structure

6.0 Objectives

6.1 Introduction

6.2 Definition

6.3 Table of Elementary Laplace Transform

6.4 Theorems on Important Properties of Laplace Transformation

ǤͶǤͳ Flow Chart of Gamma Function

ǤͶǤʹ Beta Function

ǤͶǤ͵ Properties of Beta Function:

ǤͶǤͶ Problem based on Beta Function

ǤͶǤͷ Duplication Formula of Gamma Functions

6.5 Additional Problems

6.6 Exercise

6.7 Summary

6.8 References

6.0 Objectives

After going through this unit, you will be able to:

• Understand the concept of Laplace Transformation, Theorems on Important

Properties of Laplace Transformation

• Solve the problem based on Elementary Laplace Transforms with its type.

• Understand the concept of First shifting and Second shifting theorem

• Understand Convolution Theorem Laplace Transform of an Integral and

Derivatives munotes.in

## Page 123

123Chapter 6: The Laplace Transform

6.1 Introduction

In mathematics, the Laplace transform, named after its inventor Pierre -Simon

Laplace, is an integral transform that converts a function of a real variable t (often time) to a function of a complex variable s. It is an essential part of

mathematical background required of engineers and scientists. This method has

advantage of directly giving the solution of differential equations with given

boundary values without the necessity of finding the general solution and then

evaluating from it the arbitrary constants. It also provide ready tables of Laplace

transforms which reduce the problem of solv ing differential equations to plain

algebraic manipulations.

Whenever a mathematical operator works on a function, the function is changed

or transformed into another function . For example when the differential operator

ܦቀௗ

ௗ௫ቁworks on f(x) = tan,x ,

it produces a new function ߶ሺݔሻؠܦ݂ሺݔሻൌܿ݁ݏଶݔǤ

6. 2 Definition

Laplace transform is yet another operational tool for solving constant coeffi cients

linear diff erential equations. The process of solution consists of three main steps:

i) The given “ hard" problem is transformed into a “ simple" equation.

ii) This simple equation is solved by purely algebraic manipulations.

iii) The solution of the simple equation is tr ansformed back to obtain the

solution of the given problem.

In this way the Laplace transformation reduces the problem of solving a differential equation to an algebraic problem. The third step is made easier by

tables, whose role is similar to that of integral tables in integration.

݂ሺݐሻ
ݐǡනିୱ୲݂ሺݐሻ݀ݐǡஶ

ǡ
ݏǡ݂ሺݏሻǤ

݂ሺݐሻ݂ഥሺݏሻǡ݂ሺݐሻǡ

munotes.in

## Page 124

124 APPLIED MATHEMATICS
ݐ
݂ሺݏሻǡ
ݏǤ ݂ሺݐሻ
ǡ
ݐ

Ͳǡ݂ҧሺݏሻ

ǡݏ ǡ

Ǥ

݂ሺݐሻ݂ሺݏሻǡሺሻൌන܍ିܜܛሾሺሻሿஶ

െെെെെሺሻ

ࣦሼ݂ሺݐሻሽൌ݂ሺݏሻǡ݂݀݊ܽҧሺݏሻ is called the Laplace

transform of f(t).

ࣦሼܣܨଵሺሻܤܨଶሺሻሽൌܣࣦሼܨଵሺݐሻሽ

ܤࣦሼܨଶሺݐሻሽ

6. 3 Table of Elementary Laplace Transform f(t) f(s) 1 െǡܽ

Sin at

Cos at െ

sinh at െ

cosh at െ ሺሻ ሺെሻ ሺሻ ሺሻǨା

6.4 Theorems on Important Properties of Laplace

Transformation

۷Ǥܖܑۺ܍ܑܚ܉ܜܡ۾ܚܘܗܡܜܚ܍

ǡǡ
ǡǡ
ǡ

ሼሺሻሺሻെ
ሺሻሽൌሼሺሻሽሼሺሻሽെ
ሼሺሻሽǡ munotes.in

## Page 125

125Chapter 6: The Laplace Transform

۷۷Ǥܜܛܚܑ۴ܐ܁ܑܑܜܖ܂ܐܗ܍ܚܕ܍

ࣦሼ݂ሺݐሻሽൌ݂ҧሺݏሻǡࣦሼ݁ି௧݂ሺݐሻሽൌ݂ҧሺݏܽሻ

ࣦሼ݁ି௧݂ሺݐሻሽൌන݁ି௦௧ሼ݁ି௧݂ሺݐሻሽ݀ݐஶ

ൌන݁ିሺ௦ାሻ௧݂ሺݐሻ݀ݐஶ

ൌන݁ି௧݂ሺݐሻ݀ݐሺݓ݄݁ݎ݁ൌݏܽሻൌ݂ሺሻൌ݂ҧஶ

ሺݏܽሻ

ǣ݊݅ܨ݀ݐ݄݈݈݁ܽܽܿ݁ݐ݂ݏ݊ܽݎݎ݂݉݅ሻ݁ି௧ݐܽݏܿ

܁ܗܝܔܗܑܜܖࣦሺ
ܽݐሻൌݏ

ݏଶܽଶ

ࣦሼ݁ି௧ݐܽݏܿሽൌݏܾ

ሺݏܾሻଶܽଶ

ǣ݊݅ܨ݀ݐ݄݈݈݁ܽܽܿ݁ݐ݂ݏ݊ܽݎݎ݂݉݅ሻݐଶ݁ଷ௧

܁ܗܝܔܗܑܜܖࣦሺݐଶሻൌʹǨ

ݏଷ

ࣦሼݐଶ݁ଷ௧ሽൌʹǨ

ሺݏെ͵ሻଶ

ǣ݊݅ܨ݀ݐ݄݈݈݁ܽܽܿ݁ݐ݂ݏ݊ܽݎݎ݂݉݅ሻʹݐ͵ݐ݅݅ሻݏܿଶʹݐ݅݅݅ሻݏ݊݅ଷʹݐ ܁ܗܝܔܗܑܜܖ

ሻ
ݏ݊݅ʹݐݏ݊݅͵ݐൌଵ

ଶሾ
െ
ͷሿ

ࣦሼʹݐ͵ݐሽൌͳ

ʹሾࣦሺ
ሻെࣦሺ
ͷሻሿ

ൌͳ

ʹቂ

ଶͳଶെ

ଶͷଶቃ

ൌͳʹ

ሺଶͳሻሺଶʹͷሻ

ሻ
ݏܿଶʹݐൌͳ

ʹሺͳݏܿͶݐሻ munotes.in

## Page 126

126 APPLIED MATHEMATICSࣦሼݏܿଶʹݐሽൌͳʹሾࣦሺͳሻࣦሺ
Ͷݐሻሿൌͳʹ൬ͳݏݏݏଶͳ൰

ሻ
ൌ͵ʹݐെͶݏ݊݅ଷʹݐݎݏ݊݅ଷʹݐൌ͵

Ͷʹݐെͳ

Ͷݐ

ࣦሼݏ݊݅ଷʹݐሽൌ͵

Ͷሾࣦሺʹݐሻെͳ

Ͷࣦሺݐሻሿ

ൌ͵

ͶǤʹ

ݏଶʹଶെͳ

ͶǤʹ

ݏଶଶ

ൌͶͺ

ሺݏଶͶሻሺݏଶ͵ሻ

ǣܨ݊݅݀ݐ݄݈݈݁ܽܽ݁ܿݐݏ݊ܽݎ݂ݎ݂݉

݅ሻିଷ୲ሺʹ
ͷെ͵ͷሻ݅݅ሻ݁ଶ௧ݏܿଶݐ

ܖܗܑܜܝܔܗ܁ ሻࣦሼିଷ୲ሺʹ
ͷെ͵ͷሻሽ

ൌʹࣦሺିଷ୲
ͷሻെ͵ࣦሺିଷ୲ͷሻ

ൌʹǤ͵

ሺ͵ሻଶͷଶെ͵Ǥͷ

ሺ͵ሻଶͷଶ

ൌʹെͻ

ଶ͵Ͷ

ሻ
ࣦሼݏܿଶݐሽൌͳ

ʹࣦሺͳ
ʹݐሻൌͳ

ʹ൜ͳ

ݏݏ

ݏଶͶൠ

ǡࣦሼଶ୲ݏܿଶݐሽൌͳ

ʹ൜ͳ

ݏെʹݏെʹ

ሺݏെʹሻଶͶൠ

ǣܫ݂ࣦሼ݂ሺݐሻሽൌ݂ҧሺݏሻݏ݄ݓݐ݄ݐܽ

݅ሻࣦሾሺݐܽሻ݂ሺݐሻሿൌͳ

ʹሾ݂ҧሺݏെܽሻെ݂ҧሺݏܽሻሿ

݅݅ሻࣦሾሺ
ݐܽሻ݂ሺݐሻሿൌͳ

ʹሾ݂ҧሺݏെܽሻ݂ҧሺݏܽሻሿ

ܪ݁ܿ݊݁ݒ݈݁ݑݐܽ݁ሺ݅ሻʹݐ͵ݐሺ݅݅ሻ
͵ݐ
ʹݐ

ܖܗܑܜܝܔܗ܁ ǣࣦሼሺݐܽሻ݂ሺݐሻሽൌࣦ൜ͳ

ʹሺ݁௧െ݁ି௧ሻ݂ሺݐሻൠ

ൌͳ

ʹሾࣦሼ݁௧݂ሺݐሻሽെࣦሼ݁ି௧݂ሺݐሻሽሿ

ൌͳ

ʹൣ݂ҧሺݏെܽሻെ݂ҧሺݏܽሻ൧ǡݕܾݏ݄݊݅ݐ݂݅݃ݕݐݎ݁ݎ munotes.in

## Page 127

127Chapter 6: The Laplace Transformݕ݈ݎ݈ࣦܽ݅݉݅ܵሼሺ
ܽݐሻ݂ሺݐሻሽൌͳʹሾࣦሼ݁௧݂ሺݐሻሽࣦሼ݁ି௧݂ሺݐሻሽሿ

ൌͳ

ʹൣ݂ҧሺݏെܽሻ݂ҧሺݏܽሻ൧ǡܾݕݏ݄݅ݐ݂݅݃݊ݎݕݐݎ݁

ሺ݅ሻݓ݄݁ܽݒࣦ݁ሺ͵ݐሻൌ͵

ݏଶ͵ଶ

ࣦሺʹݐ͵ݐሻൌͳ

ʹ൜͵

ሺݏെʹሻଶ͵ଶെ͵

ሺݏʹሻଶ͵ଶൠ

ൌͳʹݏ

ݏସͳͲݏଶͳͻ

ሺ݅݅ሻݓ݄݁ܽݒࣦ݁ሺ
ʹݐሻൌݏ

ݏଶʹଶ

ࣦሺ
͵ݐ
ʹݐሻൌͳ

ʹ൜ݏെ͵

ሺݏെ͵ሻଶʹଶݏ͵

ሺݏ͵ሻଶʹଶൠ

ൌʹݏሺݏଶെͷሻ

ݏସെͳͲݏଶͳͻ

ǤǤ܌ܖܗ܋܍܁ܖܑܜܑܐ܁ ܂ܗ܍ܐܚ܍ܕ

۷۷۷Ǥ܌ܖܗ܋܍܁ܖܑܜܑܐ܁ ܂ܗ܍ܐܚ܍ܕ

ࣦሼ݂ሺݐሻሽൌ݂ҧሺݏሻ ሺሻൌሼ௧ழሺ௧ିሻ௧வࣦሼܨሺݐሻሽൌ݁ି௦݂ҧሺݏሻ

ࣦሼܨሺݐሻሽൌන݁ି௦௧ܨሺݐሻ݀ݐஶ

ൌන݁ି௦௧ܨሺݐሻ݀ݐන݁ି௦௧ܨሺݐሻ݀ݐஶ

ൌන݁ି௦௧ሺͲሻ݀ݐ

න݁ି௦௧݂ሺݐെܽሻ݀ݐஶ

ൌන݁ି௦௧݂ሺݐെܽሻ݀ݐஶ

ൌන݁ି௦ሺ௨ାሻ݂ሺݑሻ݀ݑǡሾݑൌݐെܽሿஶ

ൌ݁ି௦න݁ି௦௨݂ሺݑሻ݀ݑൌஶ

݁ି௦݂ҧሺݏሻ

ǡखሼሺሻሽൌିതሺሻǤ۴ሺܜሻൌሼழሺିሻவ

ǣܨࣦ݀݊݅ሼܨሺݐሻሽ݂ݎ ሺሻൌሼழ௧ழଵሺିሻவଵ

ܖܗܑܜܝܔܗ܁ ݂ሺݐሻൌݐଷǡ
݂ҧሺݏሻൌ͵Ǩ

ݏସ munotes.in

## Page 128

128 APPLIED MATHEMATICSݕܤ݁ݒܾܽݐ݄݁ݎ݁݉ݏܽܽൌͳǡࣦሼܨሺݐሻሽൌ͵Ǩ݁ି௦ݏସ

܄۷Ǥ۷खሼሺሻሽൌതሺܛሻǡܖ܍ܐܜखሼሺሻሽൌሺെሻ

തሺሻǡ

ൌǡǡǥ

ܚܚܗ۱Ǥ܂܍ܐܛ܍ܚܔܝ۷ܜखሼሺሻሽൌെ

തሺሻൌെᇱሺሻ

െ

ૠǣ݅ܨࣦ݊݀ሼܨሺݐሻሽ݂ݎሺ݅ሻݐ

ʹܽ݊݅ݏ݄ܽݐሺ݅݅ሻݐଶ
ܽݐ

܁ܜܝܔܗܑܖܗሺ݅ሻ݂ሺݐሻൌܽݐ

ʹܽǡ݂ሺݏሻൌͳ

ʹͳ

ݏଶെܽଶ

ࣦ൜ݐͳ

ʹܽܽݐൠൌሺെͳሻ݀

݀ݏ൜ͳ

ʹǤͳ

ݏଶെܽଶൠ

ൌሺെͳሻͳ

ʹെʹݏ

ሺݏଶെܽଶሻଶ

ൌݏ

ሺݏଶെܽଶሻଶ

ሺ݅݅ሻ݂ሺݐሻൌ
ܽݐǡതሺܛሻൌݏ

ݏଶܽଶ

ࣦሼݐଶ
ܽݐሽൌሺെͳሻଶ݀ଶ

݀ݏଶቄݏ

ݏଶܽଶቅ

ൌʹݏሺݏଶെ͵ܽଶሻ

ሺݏଶܽଶሻଷ

ૡǣ݅ܨ݀݊ݐ݄݁ܮ݈ܽܽܿ݁ݐݏ݉ݎ݂ݏ݊ܽݎ ݂

ሺ݅ሻݐ
ܽݐሺ݅݅ሻݐଶܽݐሺ݅݅݅ሻݐଷ݁ିଷ௧ሺݒ݅ሻݐ݁ି௧͵ݐ

ሺ݅ሻܵ݁ܿ݊݅ǡࣦሺ
ܽݐሻൌݏ

ݏଶܽଶ

ࣦሺ
ܽݐሻൌെ݀

݀ݏቀݏ

ݏଶܽଶቁൌെሺݏଶܽଶെݏǤʹݏሻ

ሺݏଶܽଶሻ

munotes.in

## Page 129

129Chapter 6: The Laplace Transformൌെሺܽଶെݏଶሻሺݏଶܽଶሻൌሺݏଶെܽଶሻሺݏଶܽଶሻ

ሺ݅݅ሻܵ݅݁ܿ݊ǡࣦሺݐܽሻൌܽ

ݏଶܽଶ

ࣦሺଶݐܽሻൌሺെͳሻଶ݀ଶ

݀ݏଶቀܽ

ݏଶܽଶቁ

ൌ݀

݀ݏ൜െʹܽݏ

ሺݏଶܽଶሻଶൠൌʹܽሺ͵ݏଶെܽଶሻ

ሺݏଶܽଶሻଷ

ሺ݅݅݅ሻܵ݅݁ܿ݊ǡࣦሺ݁ିଷ௧ሻൌͳ

ݏ͵

ࣦሺଷ݁ିଷ௧ሻൌሺെͳሻଷ݀ଷ

݀ݏଷ൬ͳ

ݏ͵൰ൌሺെͳሻଷǤ͵Ǩ

ሺݏ͵ሻଷାଵൌ

ሺݏ͵ሻସ

ሺݒ݅ሻܵ݅݁ܿ݊ǡࣦሺ͵ݐሻൌ͵

ݏଶ͵ଶ

ࣦሺݐ͵ݐሻൌെ݀

݀ݏቀݏ

ݏଶ͵ଶቁൌݏ

ሺݏଶͻሻଶ

ܷ݊݅ݏ݃ݏ݄݂݅ݐ݅݃݊ݎ݁ݎݐݕǡݓ݁ݐ݁݃

ࣦሺ݁ି௧ݐ͵ݐሻൌሺݏͳሻ

ሾሺݏͳሻଶͻሿଶൌሺݏͳሻ

ሺݏଶʹݏͳͲሻଶ

܄Ǥ۷खሼሺሻሽൌതሺܛሻǡܖ܍ܐܜखቊሺሻ

ቋ

ൌනതሺሻஶ

ǡܕܑܔ

՜ାሺሻ

ૢǣܨ݀݊݅ݐ݄݁ܮ݈ܽ݁ܿܽݎ݂ݏ݊ܽݎݐ ݉ݏ݂

ሺ݅ሻሺͳെ݁௧ሻ

ݐሺ݅݅ሻ
ݐܽെ
ܾݐ

ݐݐݐܽ

ሺ݅ሻܵ݅݁ܿ݊ǡࣦሺͳെ݁௧ሻൌࣦሺͳሻെࣦሺ݁௧ሻൌͳ

ݏെͳ

ݏെͳ

ࣦቆͳെ݁௧

ݐቇൌන൬ͳ

ݏെͳ

ݏെͳ൰݀ݏൌȁݏെሺݏെͳሻȁஶஶ

௦ munotes.in

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130 APPLIED MATHEMATICSൌቚቀݏݏെͳቁቚஶൌെͳͳെሺͳݏሻΤ൨ൌ൬ݏെͳݏ൰

ሺ݅݅ሻܵ݅݁ܿ݊ǡࣦሺ
ݐܽെ
ܾݐሻൌݏ

ݏଶܽଶെݏ

ݏଶܾଶ

ࣦܽ݊݀ሺݐܽሻൌݏ

ݏଶܽଶ

ࣦ൬
ݐܽെ
ܾݐ

ݐ൰ࣦሺݐܽሻ

ൌනቀݏ

ݏଶܽଶെݏ

ݏଶܾଶቁݏ݀ஶ

ୱെ

ቀݏ

ݏଶܽଶቁ

ൌฬͳ

ʹሺݏଶܽଶሻെͳ

ʹሺݏଶܾଶሻฬ

ୱஶ

െ൬െʹݏ

ሺݏଶܽଶሻଶ൰

ൌͳ

ʹ

ୱ՜ஶݏଶܽଶ

ݏଶܾଶെͳ

ʹݏଶܽଶ

ݏଶܾଶʹܽݏ

ሺݏଶܽଶሻଶ

ൌͳ

ʹ൬ͳͲ

ͳͲ൰െͳ

ʹቆݏଶܽଶ

ݏଶܾଶቇʹܽݏ

ሺݏଶܽଶሻଶ

ൌቆݏଶܽଶ

ݏଶܾଶቇଵଶൗ

ʹܽݏ

ሺݏଶܽଶሻଶെെെെെെ

െሺͳൌͲሻ

۷܄Ǥ۷खሼሺሻሽൌതሺܛሻǡܖ܍ܐܜखሼሺሻሽ

ൌ

തቀ

ቁሺሻ

ǣ݂ܫࣦሼ݂ሺݐሻሽൌͺͳʹݏെʹݏଶ

ሺݏଶͶሻଶǡ݂ࣦ݀݊݅ሼ݂ሺʹݐሻሽ

ܝܔܗ܁ܜܗܑܖǣ ǡ

ࣦሼ݂ሺʹݐሻሽൌͳ

ʹە۔ۓͺͳʹቀݏ

ʹቁെʹቀݏ

ʹቁଶ

൬ቀݏ

ʹቁଶ

Ͷ൰ଶۙۘۗ

ൌͶሺͳͳʹݏെݏଶሻ

ሺݏଶͳሻଶ

۷۷܄Ǥ܉ܚ܂ܗܛܖܕܚܗ۳ܚܗܚܚ۴ܖܝ܋ܖܗܑܜ

ܹ݁݇݊ݓ݁ݎ݂൫ξݔ൯ൌʹ

ξߨන݁ି௧మݐ݀ξ௫

munotes.in

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131Chapter 6: The Laplace Transformࣦ൛݂ݎ݁൫ξݐ൯ൟൌͳݏξݏെͳ

ǤǤ܍ܐ܂ܖܗܑܜܝܔܗܞܖܗ܋ ܍ܐ܂ܗܚ܍ܕ

۷۷܄Ǥ܍ܐ܂ܖܗܑܜܝܔܗܞܖܗ܋ ܍ܐ܂ܚܗ܍ܕ

ሺሻ

തሺሻ
ǡ തሺሻ
ǤǤǤ തሺሻൌ݂ଵሺݏሻ݂ଶሺݏሻ

ݓ݄݁ݎ݂݁ଵഥሺݏሻ݂݀݊ܽଶഥሺݏሻܽ݁ݎݐ݂ݏ݊ܽݎݎ݉ݏ݂݇݊ݓ݂݊ݑ݊ܿݐ݅ݏ݂݊ଵሺݐሻ݂݀݊ܽଶሺݐሻ

݄ܶ݁ݐ݄݁݁ݎ݉ݏ݁ݐܽݐݏݐ݄ܽݐ

खቐනሺെሻሺሻ

ቑൌሺሻሺሻൌതതതሺሻതതതሺሻ

ൌखනሺሻሺെሻ

݄ܶݏ݅ݐ݄݁݁ݎ݉ݏ݅ݑ݂݁ݏݑ݈ݐ݂݅݀݊݊݅ݒ݁ݏݎ݁ݐ݂ݏ݊ܽݎݎ݉ܽݐ݅݊Ǥ

ǣܸ݂݅ݎ݁ݕݐ݄݁݊݅ݐݑ݈ݒ݊ܿ ݐ݄݁݁ݎ݉ݎ݂݄݁ݎ݂݂݅ܽݑ݊ܿݐ݅ݏ݊

݂ଵሺݐሻൌݐǡ݂ଶሺݐሻൌ݁௧

ǣ݂ଵഥሺݏሻൌͳ

ݏଶǡ݂ଶഥሺݏሻൌͳ

ݏെܽ

݂ଵഥሺݏሻ݂ଶഥሺݏሻൌͳ

ݏଶሺݏെܽሻ

ܰݓǡන݂ଵሺݑሻ݂ଶሺݐെݑሻݑ݀௧

ൌනݑǤ݁ሺ௧ି௨ሻݑ݀௧

ൌെݑ

ܽ݁ሺ௧ି௨ሻെͳ

ܽଶ݁ሺ௧ି௨ሻ൨

௧

ൌͳ

ܽଶሾ݁௧െܽݐെͳሿ खቐන݂ଵሺݑሻ݂ଶሺݐെݑሻݑ݀௧ቑൌख൜ͳܽଶሾ݁௧െܽݐെͳሿൠ munotes.in

## Page 132

132 APPLIED MATHEMATICSൌͳܽଶͳݏെܽെܽݏଶെͳݏ൨

ൌͳ

ݏଶሺݏെܽሻൌ݂ଵഥሺݏሻ݂ଶഥሺݏሻ

ǤǤ

ݕܤ݂݁݀݊݅ݐ݅݊݅ǡ

खቐන݂ሺݑሻ݀ݑ௧

ቑൌන݁ି௦௧න݂ሺݑሻ݀ݑ௧

ஶ

ݐ݀െെെെെെെെെሺ݅ሻ

ݏ݊݅ܿ݁݀

ݐ݀න݂ሺݑሻ݀ݑ௧

ൌ݂ሺݐሻǡ

ݓ݁݃ݐܾ݁ݕ݃݁ݐ݊݅ݎ݃݊݅ݐܾܽݕݏݐݎܽǡݐ݄݁ݑݏ݁ݎ݈ݐሺ݅ሻܽݏ

ࣦቐන݂ሺݑሻ݀ݑ௧

ቑൌെͳ

ݏ݁ି௦௧න݂ሺݑሻ݀ݑ௧

ஶ

ͳ

ݏන݁ି௦௧݂ሺݐሻݐ݀ஶ

ൌͳ

ݏ݂ҧሺݏሻ

݄ܶݑݏǡखቐනሺሻ

ቑൌ

തሺሻ

݅Ǥ݁Ǥ݂݊݅ݐܿ݊ݑݏ݅݃݁ݐ݊݅ݎ݀݁ݐܽ݊ݒݎ݁ሺͲǡݐሻݐ݄݁ݎ݂ݏ݊ܽݎݐ ݂݉݃݊݅ݐ݄݁݃݁ݐ݊݅ݎ݈ܽݏ݅ ݀݁݊݅ܽݐܾ ܾݕ݅݀ݒ݊݅݀݅݃ݐ݄݁ݎ݂ݏ݊ܽݎݐ ݂݉ݐ݄݂݁݊݅ݐܿ݊ݑܾݕݏǤ

ܫ݂ࣦሺݐሻൌ݂ҧሺݏሻǡݐ݄ࣦ݊݁ቐන݂ሺݑሻ݀ݑ௧

ቑൌͳ

ݏ݂ҧሺݏሻǡ

ܮݐ݁ሺݐሻൌන݂ሺݑሻ݀ݑ௧

ǡݐ݄݊݁ᇱሺݐሻൌ݂ሺݐሻܽ݀݊ሺͲሻൌͲ

ࣦሼᇱሺݐሻሽൌݏഥሺݏሻെሺͲሻ

ܱݎഥሺݏሻൌͳ

ݏࣦሼᇱሺݐሻሽ݅Ǥ݁Ǥࣦቐන݂ሺݑሻ݀ݑ௧

ቑൌͳ

ݏ݂ҧሺݏሻ munotes.in

## Page 133

133Chapter 6: The Laplace Transformǣݕ݂݅ݎܸࣦ݁ቐනݑଶ݁ି௨݀ݑ௧ቑൌͳݏखሼݐଶ݁ି௧ሽ

ǣනݑଶ݁ି௨݀ݑ௧

ൌሾെሺݑଶʹݑʹሻ݁ି௨ሿ

ൌʹെሺݐଶʹݐʹሻ݁ି௧

ࣦቐනݑଶ݁ି௨݀ݑ௧

ቑ

ൌखሼʹെሺݐଶʹݐʹሻ݁ି௧ሽെെെെെെെെെെሺሻ

ࣦሺݐଶ݁ି௧ሻൌሺെͳሻଶ݀ଶ

݀ݏଶ൬ͳ

ݏͳ൰ൌʹ

ሺݏͳሻଷ

ࣦሺʹݐ݁ି௧ሻൌʹǤሺെͳሻ݀

݀ݏ൬ͳ

ݏͳ൰ൌʹ

ሺݏͳሻଶ

ሺሻǡ

ࣦቐනݑଶ݁ି௨݀ݑ௧

ቑൌʹ

ݏെʹ

ሺݏͳሻଷʹ

ሺݏͳሻଶʹ

ሺݏͳሻଵ൨

ൌʹ

ݏሺݏͳሻଷൌͳ

ݏࣦሼݐଶ݁ି௧ሽ

ǣ݈ܽݒܧݑܽ݁ݐݐ݄݂݈݈݁ݓ݅݃݊ǣ

ሺ݅ሻනݐ݁ିଷ௧ݐ݀ݐሺ݅݅ሻஶ

න݉ݐ

ݐ݀ݐஶ

ሺ݅݅݅ሻන݁௧൬
ݐܽെ
ܾݐ

ݐ൰݀ݐஶ

ሺݒ݅ሻࣦቐනି୲ݐ

ݐ݀ݐ௧

ቑ

ሺ݅ሻනݐ݁ିଷ௧ݐ݀ݐൌනݐ݁ି௦௧ሺݐݐሻ݀ݐݓ݄݁ݎ݁ݏൌ͵ஶ

ஶ

ൌࣦሺݐݏݐ݊݅ሻǡܾݕ݀݊݅ݐ݂݅݊݅݁

ൌሺെͳሻ݀݀ݏ൬ͳݏଶͳ൰ munotes.in

## Page 134

134 APPLIED MATHEMATICSൌʹݏሺݏଶͳሻଶൌʹൈ͵ሺ͵ଶͳሻଶൌ͵ͷͲ

ሺ݅݅ሻනݐ݉

ݐ݀ݐஶ

ࣦሺݐ݉ሻൌ݉

ሺݏଶ݉ଶሻൌ݂ሺݏሻǡሺݕܽݏሻ

ࣦ൬ݐ݉

ݐ൰ൌන݂ሺݏሻ݀ݏൌන݉݀ݏ

ݏଶ݉ଶஶ

௦ஶ

௦ൌቚܽݐ݊ିଵݏ

݉ቚ

௦ஶ

ܾݕ݁݀݅ݐ݂݅݊݅݊න݁ି௦௧ݐ݉

ݐ݀ݐஶ

ൌ

െܽݐ݊ିଵݏ

݉

ܰݓǡ

௦՜ܽݐ݊ିଵሺݏ݉ሻൌͲ݂݅݉Ͳݎߨ݂݅݉൏Ͳ Τ

݄ܶݑݏܽݐ݈݇݅݃݊ݐ݅݉݅ݏܽݏ՜Ͳǡ݁ݓݐ݁݃

නݐ݉

ݐ݀ݐൌߨ

ʹஶ

݂݅݉Ͳݎെߨʹ݂݅݉൏Ͳ Τ

ሺ݅݅݅ሻන݁௧൬
ݐܽെ
ܾݐ

ݐ൰݀ݐஶ

ܹ݁݇ݓ݊ݐ݄ݐࣦܽሺ
ݐܽሻൌݏ

ሺݏଶܽଶሻǡࣦሺ
ܾݐሻൌݏ

ሺݏଶܾଶሻ

ࣦ൬
ݐܽെ
ܾݐ

ݐ൰ൌන൬ݏ

ሺݏଶܽଶሻെݏ

ሺݏଶܾଶሻ൰ஶ

௦݀ݏ

ൌͳ

ʹቊ݈݃ቆݏଶܽଶ

ݏଶܾଶቇቋ

௦ஶ

ൌͳ

ʹ݈݃ቆݏଶܾଶ

ݏଶܽଶቇ

݄ܶ݅ݏ݈݅݉݅ݏ݁න݁ି௦௧ஶ

൬
ݐܽെ
ܾݐ

ݐ൰݀ݐൌͳ

ʹ݈݃ቆݏଶܾଶ

ݏଶܽଶቇ

ܶ݃݊݅݇ܽݏൌͳǡ݁ݓݐ݁݃න݁ି௧ஶ

൬
ݐܽെ
ܾݐ

ݐ൰݀ݐ

ൌͳ

ʹ݈݃ቆͳܾଶ

ͳܽଶቇ

ሺݒ݅ሻࣦቐනି୲ݐ

ݐ݀ݐ௧

ቑ ࣦܵ݊݅ܿ݁൬ݐݐ൰ൌන݀ݏݏଶͳஶൌܽݐ݊ିଵݏൌߨʹെܽݐ݊ିଵݏൌݐܿିଵݏ munotes.in

## Page 135

135Chapter 6: The Laplace Transformࣦ൜݁௧൬ݐݐ൰ൠൌܿݐିଵሺݏെͳሻെെെെെݕܾݏ݄݂݅ݐ݃݊݅ݎ݁ݎݕݐ

ࣦන൜݁௧൬ݐ

ݐ൰ൠ݀ݐ௧

ൌͳ

ݏܿݐିଵሺݏെͳሻ

ǤǤ

ሺሻ

ൌͲ

ሺǤǤ
՜ͲሻǤ

ࣦሾ݂ሺݐሻሿൌ݂ҧሺሻሺሻ

ቂ݅Ǥ݁Ǥ

՜ஶ݁ି௦݂ሺ݉ሻൌͲǡݎ݂ݏቃǡݐ݄݁݊खሼᇱሺሻሽൌതሺሻെሺሻ

ܹ݄݁ݎ݂݁ሺͲሻ݅ݏݐ݄݁ݑ݈ܽݒ݂݂݁ሺݐሻݐܽݐൌͲǤ

खሼᇱሺሻሽൌതሺሻെሺሻെെെെെെെെെെെሺሻ

െ

ܫ݂ࣦሼ݂ሺݐሻሽൌ݂ҧሺݏሻݐ݄ࣦ݁݊ሼ݂ᇱԢሺݐሻሽൌݏଶ݂ҧሺݏሻെݏ݂ሺͲሻെ݂ᇱሺͲሻ

ܮݐ݁ܨሺݐሻൌ݂ᇱሺݐሻݐ݄݁݊

ࣦሼ݂ᇱԢሺݐሻሽൌࣦሼܨᇱሺݐሻሽ

ൌݏࣦሼܨሺݐሻሽെܨሺͲሻെെെെെെݕܾሺሻ

ൌݏࣦሼ݂ᇱሺݐሻሽെ݂ᇱሺͲሻ

ൌݏൣ݂ҧሺݏሻെ݂ሺͲሻ൧െ݂ᇱሺͲሻെെെെെെݕܾሺሻ

ൌݏଶ݂ҧሺݏሻെݏ݂ሺͲሻȂ݂ᇱሺͲሻെെെെെെݕܾሺሻ

ܤݕ݃݊݅ݏݑ݉ݐ݄݈ܽ݁ܽܿ݅ݐܽ݉݊݅ݐܿݑ݀݅݊ǡݓ݁ܿ݊ܽݏ݄ݓݐ݄ݐܽ

6.5 Additional Problems

ǣܨ݀݊݅ݐ݄݈݈݁݁ܿܽܽݐ݉ݎ݂ݏ݊ܽݎ ݂݄݂݁ܿܽݐ݄݁݃݊݅ݓ݈݈݂ ݏ݊݅ݐܿ݊ݑ݂ ሺ݅ሻ
ݐ
ʹݐሺ݅݅ሻݐଶെ͵ݐͷሺ݅݅݅ሻݐଶݐܽሺ݅ݒሻ݁ସ௧ܿݏ݄ͷݐ

ሺ݅ሻࣦሼ
ݐ
ʹݐሽൌࣦ൜ͳ

ʹሺ
͵ݐ
ݐሻൠ ൌͳʹሼࣦሺ
͵ݐሻࣦሺ
ݐሻሽ

munotes.in

## Page 136

136 APPLIED MATHEMATICSൌͳʹ൜ݏݏଶሺ͵ሻଶݏݏଶሺͳሻଶൠ

ൌቊݏሺݏଶͷሻ

ሺݏଶͳሻሺݏଶͻሻቋ

ሺ݅݅ሻࣦሼݐଶെ͵ݐͷሽ

ܹ݁ݓ݊݇ǡࣦሼݐିଵሽൌሺ݊െͳሻǨ

ݏ

ࣦሼݐଶሽൌʹǨ

ݏଷൌʹ

ݏଷǡ݊ൌ͵

ࣦሼݐሽൌͳǨ

ݏଶൌͳ

ݏଶǡ݊ൌʹ

ࣦሼݐଶሽൌͳ

ݏǡ݊ൌͳ

ࣦሼݐଶെ͵ݐͷሽൌࣦሺݐଶሻെ͵ࣦሺݐሻͷࣦሺͳሻ

ൌʹ

ݏଷെ͵

ݏଶͷ

ݏൌͷݏଶെ͵ݏʹ

ݏଷ

ሺ݅݅݅ሻݐଶܽݐ

ࣦሼݐଶܽݐሽൌሺെͳଶሻ݀ଶ

݀ݏଶǤቄݏ

ݏଶܽଶቅൌʹܽሺ͵ݏଶെܽଶሻ

ሺݏଶܽଶሻଷ

ሺݒ݅ሻ݁ସ௧ܿݏ݄ͷݐ

ࣦሼ݁ସ௧ܿݏ݄ͷݐሽൌݏെͶ

ሺݏെͶሻଶെͷଶൌݏെͶ

ݏଶെͺݏെͻ

ǣܨࣦ݀݊݅ሼ݂ሺݐሻሽǡ݂݂݅ሺݐሻൌሼǡழఈୡ୭ୱሺ௧ିఈሻǡ௧வఈ

ݕܤ݀݊݅ݐ݂݅݊݅݁ ǡ

ࣦሼ݂ሺݐሻሽൌන݁ି௦௧݂ሺݐሻ݀ݐஶ

ൌන݁ି௦௧ሺͲሻ݀ݐන݁ି௦௧
ሺݐെߙሻ݀ݐஶ

ఈఈ

ൌන݁ି௦ሺ௨ାఈሻ
ݑ݀ݑሾݓ݄݁ݎ݁ሺݑൌݐെߙሻሿஶ

ൌ݁ିఈ௦න݁ି௦௨
ݑ݀ݑஶ

ൌ݁ିఈࣦሼ
ݑሽൌ݁ି௦ఈݏݏଶͳ munotes.in

## Page 137

137Chapter 6: The Laplace Transformǣ

ሺ݅ሻݐହଶΤሺ݅݅ሻ݁ିଷ௧ݐିଵଶΤሺ݅݅݅ሻξݐ

ሺ݅ሻݐହଶΤ

ݓ݄݁ݒࣦܽ݁ሺݐሻൌሺ݊ͳሻǨ

ݏାଵ

ࣦ൫ݐହଶΤ൯ൌ൫ʹൗ൯Ǩ

ݏଶൗൌͳͷ

ͺ൫ͳʹൗ൯Ǩ

ݏଶൗൌͳͷ

ͺටߨ

ݏ

ሺ݅݅ሻ݁ିଷ௧ݐିଵଶΤ

ࣦ൫ݐିଵଶΤ൯ൌ൫ͳʹൗ൯Ǩ

ݏଵଶൗൌටߨ

ݏ

ࣦ൫݁ିଷ௧ݐିଵଶΤ൯ൌටߨ

ݏ͵

ሺ݅݅݅ሻξݐ

ݕܤ݂݀݁݅ݐ݂݅݊݅݊ݎݎ݁ݎ݂ݑ݊ܿ݊݅ݐ

ݎ݂݁൫ξݐ൯ൌʹ

ξߨන݁ି௫మݔ݀ξ௧

ൌͳ

ξߨනݑିଵଶൗ݁ି௨݀ݑሾݔଶൌݑሿ௧

ࣦ൛ݎ݂݁൫ξݐ൯ൟൌͳ

ξߨࣦቐනݑିଵଶൗ݁ି௨݀ݑ௧

ቑ

ൌͳ

ξߨͳ

ݏࣦቄݑିଵଶൗ݁ି௨݀ݑቅെܮܽǤܶܽݎǤ݊ܫ݁ݐ݊݃ܽݎ݈ݏ

ൌͳ

ξߨͳ

ݏ൫ͳʹൗ൯Ǩ

ሺݏെͳሻଵଶൗൌͳ

ݏξݏെͳ

ͳǣܩ݊݁ݒࣦ݅ቐʹඨݐ

ߨቑൌͳ

ݏଷଶൗݏ݄ݓݐ݄ܽݐࣦ൜ͳ

ξߨݐൠൌͳ

ξݏ munotes.in

## Page 138

138 APPLIED MATHEMATICSݐ݁ܮ݂ሺݐሻൌʹඨݐߨ

ܨᇱሺݐሻൌʹ

ξߨͳ

ʹݐିଵଶൗൌͳ

ξߨݐ

ࣦሺ݂ᇱሺݐሻሻൌࣦͳ

ξߨ

ൌݏ݂ҧሺݏሻെ݂ሺͲሻെെെെܽܮǤݎܶܽǤ݊ܦݎ݁ݒ݅ݐܽݒ݅ݏ݁

ൌݏࣦቐʹඨݐ

ߨቑൌݏǤͳ

ݏଷଶൗୀͳ

ξݏ

ͳͺǣܧܽݒݑ݈݁ݐܽනݐ݁ିଷ௧ݐ݀ݐஶ

නݐ݁ିଷ௧ݐ݀ݐൌන݁ି௦௧ሺݐݐሻ݀ݐሺஶ

ஶ

ݓ݄݁ݎ݁ݏൌ͵ሻ

ൌࣦሼݐݐሽെݕܾ݀݊݅ݐ݂݅݊݅݁

ൌሺെͳሻ݀

݀ݏ൜ͳ

ݏଶͳൠ

ൌʹݏ

ሺݏଶͳሻଶ

ൌʹכ͵

ሺሺ͵ሻଶͳሻଶൌ͵

ͷͲሾ݈ܽ݁ݎܿ݊݅݃ݏൌ͵ሿ

6.6 Exercise

ͳǤ
ǣ

ሺ݅ሻሺݐଶͳሻଶቆܛܖۯସͶଶʹͶ

ହቇ

ሺሻሺͳሻଶ݁௧ቆܛܖۯଶͳ

ሺെͳሻଷቇ

ሺ݅݅݅ሻݐଶ
݇ݐቆܛܖۯʹሺଶെ͵ଶሻ

ሺଶଶሻଷቇ ሺሻݏ݊݅ଷݐ൬ܛܖۯሺଶͳሻሺଶͻሻ൰ munotes.in

## Page 139

139Chapter 6: The Laplace Transformሺݒሻ
ݐܽݐܽቆܛܖۯሺଶെʹଶሻସͶସቇ

ሺሻሺʹݐെ
ʹݐሻଶ൬ܛܖۯ

ܛെͶ

ሺଶͳሻ൰

ʹǤ ǡ
݂ሺݐሻǡ

ሺ݅ሻ݂ሺݐሻൌሼହǡ୲வସ୲ǡழ୲ழସቆܛܖۯሺͳെିୠୱሻ

ቇ

ሺ݅݅ሻ݂ሺݐሻൌሼǡ୲வୠୟǡழ୲ழୠ൬ܛܖۯͳ

ଶ൬ͳ

െͳ

ଶ൰ିସୱ൰

ሺ݅݅݅ሻ݂ሺݐሻൌሼǡழ୲ழଵሺ୲ିଵሻమǡ୲வଵ൬ܛܖۯʹିୱ

ଷ൰

͵Ǥ
݂ሺݐሻǣ

ሺ݅ሻͳ

ݐሺͳെ
ݐܽሻቆܛܖۯͳ

ʹቈଶଶ

ଶቇ

ሺ݅݅ሻͳ

ݐሺ
ݐܽെ
ܾݐሻቆܛܖۯͳ

ʹቈଶଶ

ଶଶቇ

ሺ݅݅݅ሻݏ݄݅݊ݐ

ݐ൬ܛܖۯͳ

ʹͳ

െͳ൨൰

ሺ݅ݒሻන݁௧௧

ݐ

ݐ݀ݐ൬ܛܖۯͳ

ିଵሺെͳሻ൰

ͶǤࣦ݂ሺݐሻൌݏଶെݏͳ

ሺʹݏͳሻଶሺݏെͳሻǡ݂ࣦ݀݊݅ሺʹݐሻቆܛܖۯݏଶെʹݏͶ

Ͷሺݏͳሻଶሺݏെʹሻቇ

ͷǤǣ

ሺሻනݐଷ݁ି௧ஶ

ݐ݀ݐሺܛܖۯͲሻ

ሺሻන݁ିଶ௧ஶ

ଷݐ݀ݐ൬ܛܖۯ

ͷ൰

ሺሻනݐ݁ିଷ௧ஶݐ݀ݐ൬ܛܖۯ͵ͷͲ൰ munotes.in

## Page 140

140 APPLIED MATHEMATICS

6.7 Summary

In this unit we learn Laplace Transform definition, Elementary Laplace

Transforms, Theorems on Important Properties of Laplace Transformation

ሺሻൌන܍ିܜܛሾሺሻሿஶ

ǡࣦሼ݂ሺݐሻሽ

ൌ݂ሺݏሻǡ݂݀݊ܽҧሺݏሻ
݂ሺݐሻ

ࣦሼܣܨଵሺሻܤܨଶሺሻሽ

ൌܣࣦሼܨଵሺݐሻሽܤࣦሼܨଶሺݐሻሽ

Table of Elementary Laplace Transform

f(t) f(s) 1 ͳݏ ݁௧ ͳݏെܽǡݏܽ

Sin at ܽݏଶܽଶ

Cos at ݏݏଶെܽଶ

sinh at ܽݏଶെܽଶ

cosh at ݏݏଶെܽଶ ݐʹܽ݅ݏ݊ܽݐ ݏሺݏଶܽଶሻଶ ͳʹܽଷሺ݅ݏ݊ܽݐെܽݐܿݏܽݐሻ ͳሺݏଶܽଶሻଶ ݐ ሺ݊ͳሻǨݏାଵ

ܚܑ۴ܜܛܐ܁ܑܑܜܖ܂܍ܐܗ܍ܚܕࣦሼ݂ሺݐሻሽൌ݂ҧሺݏሻǡࣦሼ݁ି௧݂ሺݐሻሽ

ൌ݂ҧሺݏܽሻ

܁܍܋ܗ܌ܖܐ܁ܑܑܜܖ܂܍ܐܗ܍ܚܕࣦሼ݂ሺݐሻሽൌ݂ҧሺݏሻ ሺሻ

ൌሼ௧ழሺ௧ିሻ௧வࣦሼܨሺݐሻሽൌ݁ି௦݂ҧሺݏሻ

ࣦሼ݂ሺݐሻሽൌ݂ҧሺሻǡࣦሼݐ݂ሺݐሻሽൌሺെͳሻ݀ݏ݂݀ҧሺݏሻǡݓ݄݁ݎ݁݊ൌͳǡʹǡ͵ǥ munotes.in

## Page 141

141Chapter 6: The Laplace Transform

ࣦሼ݂ሺݐሻሽൌ݂ҧሺሻǡࣦቊ݂ሺݐሻ

ݐቋൌන݂ҧሺݏሻ݀ݏஶ

௦ǡݎ݅ݒ݀݁݀

௧՜ା݂ሺݐሻ

ݐ݁ݔݏݐݏ݅

ࣦሼ݂ሺݐሻሽൌ݂ҧሺሻǡࣦሼ݂ሺܽݐሻሽ

ൌͳ

݂ܽҧቀݏ

ܽቁሺሻ

ܕܚܗܛܖ܉ܚ܂ ܗ۳ܚܗܚܚܝ۴ܖܑܜ܋ܖܗࣦ൛݂ݎ݁൫ξݐ൯ൟൌͳ

ݏξݏെͳ

܍ܐ܂ܖܗ܋ܞܗܔܖܗܑܜܝܕ܍ܚܗ܍ܐ܂ :

ࣦቐන݂ଵሺݐെݑሻ݂ଶሺݑሻ݀ݑ௧

ቑൌ݂ଵሺݏሻ݂ଶሺݏሻൌ݂ଵഥሺݏሻ݂ଶഥሺݏሻ

ൌࣦන݂ଵሺݑሻ݂ଶሺݐെݑሻ݀ݑ௧

ǣࣦቐන݂ሺݑሻ݀ݑ௧

ቑൌͳ

ݏ݂ҧሺݏሻ

ࣦሼ݂ᇱሺݐሻሽൌݏ݂ҧሺݏሻെ݂ሺͲሻ

6.8 References

1. A Text Book of Applied Mathematics Vol I - P. N. Wartikar and J. N.

Wartikar

2. Applied Mathematics II - P. N. Wartikar and J. N. Wartikar

3. Higher Engineering Mathematics - Dr. B. S. Grewal

munotes.in

## Page 142

142 APPLIED MATHEMATICS

142Unit 3

7

INVERSE LAPLACE TRANSFORM

Unit Structure

7.0 OBJECTIVES

7.1 Introduction: Inverse Laplace Transform

7.1.1 Shifting Theorem

7.1.2 Partial fraction Methods

7.1.3 Use of Convolution Theorem

7.2 Exercise

7.3 Summary

7.4 References

7.0 Objectives

After going through this unit, you will be able to:

• Understand the concept of Inverse Laplace Transformation, shifting

theorem and use of Convolution Theorem

• Solve the problem based on Ordinary Linear Differential Equations with

Constant Coefficients

• Understand the concept Solution of Simultaneous Ordina ry Differential

Equations, • Understand Laplace Transformation of Special Function, Periodic Functions, Heaviside Unit Step Function, Dirac -delta Function

7.1 Introduction: Inverse Laplace Transform

Having find the Laplace Transforms of few functions , let us now determine the

inverse transforms of given functions. We are now in a position to find the Laplace

transform ݂ҧሺݏሻ
݂ሺݐሻǤ

We shall now consider the inverse problem, i.e. given ݂ҧሺݏሻǡ
݂ሺݐሻ
݂ሺݏሻ
Ǥ munotes.in

## Page 143

143Chapter 7: Inverse Laplace Transform

Definition: ࣦሼ݂ሺݐሻሽൌ݂ሺݏሻǡ݂ሺݐሻ is called the inverse Laplace Transform

of ݂ҧሺݏሻ.

खି൛തሺሻൟൌሺሻ ࣦିଵͳݏ൨ൌͳ ࣦିଵͳݏെܽ൨ൌ݁௧ ࣦିଵͳݏ൨ൌݐିଵሺ݊െͳሻǨǡ݊ൌͳǡʹǡ͵ǤǤ ࣦିଵͳሺݏെܽሻ൨ൌ݁௧ݐିଵሺ݊െͳሻǨ ࣦିଵͳݏଶܽଶ൨ൌͳܽ݅ܵ݊ܽݐ ࣦିଵቂݏݏଶܽଶቃൌܿݏܽݐ ࣦିଵͳݏଶെܽଶ൨ൌ݄݅ܵ݊ܽݐ ࣦିଵቂݏݏଶെܽଶቃൌܿݏ݄ܽݐ ࣦିଵͳሺݏെܽሻଶܾଶ൨ൌͳܾ݁௧ܾ݅ܵ݊ݐ ࣦିଵݏെܽሺݏെܽሻଶܾଶ൨ൌ݁௧ܥݏܾݐ ࣦିଵݏሺݏଶܽଶሻଶ൨ൌͳʹܽݐ݅ܵ݊ܽݐ ࣦିଵͳሺݏଶܽଶሻଶ൨ൌͳʹܽଷሺ݅ܵ݊ܽݐെܽݐܿݏܽݐሻ

ࣦିଵͳ

ሺݏଶܽଶሻଶ൨ൌͳ

ʹܽݐ݅ܵ݊ܽݐ

ࣦሺݐ݅ܵ݊ܽݐሻൌʹܽݏ

ሺݏଶܽଶሻଶࣦܽ݊݀ሺݐܿݏܽݐሻൌݏଶെܽଶ

ሺݏଶܽଶሻଶ

ݐ݅ܵ݊ܽݐൌʹࣦܽିଵݏ

ሺݏଶܽଶሻଶ൨ܪࣦ݁ܿ݊݁ିଵͳ

ሺݏଶܽଶሻଶ൨ൌͳ

ʹܽݐ݅ܵ݊ܽݐ

ݐܿݏܽݐൌࣦିଵቈݏଶെܽଶ

ሺݏଶܽଶሻଶൌࣦିଵቈሺݏଶܽଶሻെʹܽଶ

ሺݏଶܽଶሻଶ

ݐܿݏܽݐൌࣦିଵͳ

ሺݏଶܽଶሻ൨െʹܽଶࣦିଵͳ

ሺݏଶܽଶሻଶ൨

ݐܿݏܽݐൌͳ

ܽ݅ܵ݊ܽݐെʹܽଶࣦିଵͳ

ሺݏଶܽଶሻଶ൨

ࣦିଵͳ

ሺݏଶܽଶሻଶ൨ൌͳ

ʹܽଷሺ݅ܵ݊ܽݐെܽݐܿݏܽݐሻ

munotes.in

## Page 144

144 APPLIED MATHEMATICSǣ ሺ݅ሻͳݏͶሺ݅݅ሻʹݏݏଶͶ

ܖܗܑܜܝܔܗ܁ ሺ݅ሻͳ

ݏͶǡࣦିଵͳ

ݏͶ൨ൌ݁ିସ௧ǡ݂ሺݐሻൌ݁ିସ௧

ሺ݅݅ሻʹݏ

ݏଶͶൌʹݏ

ݏଶͶ

ݏଶͶൌʹݏ

ݏଶͶ͵ʹ

ݏଶͶ

ࣦሺ
ʹݐሻൌݏ

ݏଶͶǡࣦሺʹݐሻൌʹ

ݏଶͶ

ࣦିଵʹݏ

ݏଶͶ൨ൌʹࣦିଵቄݏ

ݏଶͶቅ͵ࣦିଵ൜ʹ

ݏଶͶൠ

ൌʹ
ʹݐ͵ʹݐ

ǣ ሺ݅ሻݏଶെ͵ݏͶ

ݏଷሺ݅݅ሻݏʹ

ݏଶെͶݏͳ͵

ܖܗܑܜܝܔܗ܁ ሺ݅ሻݏଶെ͵ݏͶ

ݏଷǡ

ࣦିଵቈݏଶെ͵ݏͶ

ݏଷൌࣦିଵͳ

ݏ൨െ͵ࣦିଵͳ

ݏଶ൨Ͷࣦିଵͳ

ݏଷ൨

ൌͳെ͵ݐͶǤݐଶ

ʹǨൌͳെ͵ݐʹݐଶ

ሺ݅݅ሻݏʹ

ݏଶെͶݏͳ͵ǡ

ࣦିଵݏʹ

ݏଶെͶݏͳ͵൨ൌࣦିଵݏʹ

ሺݏെʹሻଶͻ൨

ൌࣦିଵݏെʹͶ

ሺݏെʹሻଶ͵ଶ൨

ൌࣦିଵݏെʹ

ሺݏെʹሻଶ͵ଶ൨Ͷࣦିଵͳ

ሺݏെʹሻଶ͵ଶ൨

ൌ݁ଶ௧
͵ݐͶ

͵݁ଶ௧͵ݐ

Ǥ

munotes.in

## Page 145

145Chapter 7: Inverse Laplace TransformૠǤǤܖܑܜܑܐ܁܂ܐܕ܍ܚܗ܍

ሺሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻǡݐ݄ࣦ݊݁ିଵ൛݂ҧሺݏെܽሻൟൌ݁௧݂ሺݐሻൌ݁௧ࣦିଵ൛݂ҧሺݏሻൟ

ሺሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻ݂ܽ݊݀ሺͲሻൌͲǡݐ݄ࣦ݊݁ିଵ൛ݏ݂ҧሺݏሻൟൌ݀

ݐ݀ሼ݂ሺݐሻሽൌ݂ᇱሺݐሻ

݅Ǥ݁Ǥ ݂ҧሺݏሻ ݏǡ

݂ሺݐሻ

ǡࣦିଵ൛ݏ݂ҧሺݏሻൟൌൌ݀

ݐ݀ሼ݂ሺݐሻሽǡ ݂ሺͲሻൌ݂ᇱሺͲሻൌڮ

ൌ݂ିଵሺͲሻൌͲ

ࣦሼݐ݂ሺݐሻሽൌሺെͳሻ݀

݀ݏ݂ҧሺݏሻൌሺെͳሻ݂ҧሺሻሺݏሻ

ࣦିଵ൛݂ሺሻሺݏሻൟൌሺെͳሻݐ݂ሺݐሻ

ሺሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻݐ݄ࣦ݊݁ିଵቊ݂ҧሺݏሻ

ݏቋൌන݂ሺݐሻݐ݀௧

݈ܣݏࣦିଵቊ݂ҧሺݏሻ

ݏଶቋൌන௧

ቐන݂ሺݐሻݐ݀௧

ቑݐ݀

ࣦିଵቊ݂ҧሺݏሻ

ݏଶቋ

ൌන௧

ቐනቌන݂ሺݐሻ௧

ݐ݀ቍ௧

ݐ݀ቑݐ݀Ǥ

ሺሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻݐ݂ሺݐሻൌࣦିଵ൜Ȃ݀

ݏ݀ൣ݂ҧሺݏሻ൧ൠǡ

ࣦሺݐ݂ሺݐሻሻൌȂ݀

ݏ݀ൣ݂ҧሺݏሻ൧

ሺሻࣦቆ݂ሺݐሻ

ݐቇൌන݂ҧሺݏሻݏ݀ǡ݂ሺݐሻ݂ሺݏሻǡஶ

௦

න݂ҧሺݏሻ
ஶ

௦ munotes.in

## Page 146

146 APPLIED MATHEMATICSǣ݀݊݅ܨ݂ሺݐሻǡ݂݂݅ҧሺݏሻൌݏݏଶʹݏͷ

ܖܗܑܜܝܔܗ܁

ǡ

ݏଶʹݏͷൌሺݏͳሻଶሺʹሻଶ

ǡ݂ҧሺݏሻൌݏ

ݏଶʹݏͷൌሺݏͳሻ

ሺݏͳሻଶሺʹሻଶ͵ʹ

ሺݏͳሻଶሺʹሻଶ

ࣦሺ
ʹݐሻൌݏ

ݏଶሺʹሻଶǡࣦሺʹݐሻൌʹ

ݏଶሺʹሻଶ

ǡ

ࣦିଵቊሺݏͳሻ

ሺݏͳሻଶሺʹሻଶቋൌ݁ି௧
ʹݐ

ࣦିଵ൜ʹ

ሺݏͳሻଶሺʹሻଶൠൌ݁ି௧ʹݐ

݂ሺݐሻൌࣦିଵ൜ݏ

ݏଶʹݏͷൠ

ൌࣦିଵ൜ሺݏͳሻ

ሺݏͳሻଶሺʹሻଶൠ͵ࣦିଵ൜ʹ

ሺݏͳሻଶሺʹሻଶൠ

ൌ݁ି௧
ʹݐ͵݁ି௧ʹݐ

ǣ
ǣ

ሺሻݏଶ

ሺݏെʹሻଷሺ݅݅ሻሺݏʹሻଶ

ሺݏଶͶݏͺሻଶ

ܖܗܑܜܝܔܗ܁ ሺ݅ሻݏଶൌሺݏെʹሻଶͶሺݏെʹሻͶ

ݏଶ

ሺݏെʹሻଷൌͳ

ሺݏെʹሻͶ

ሺݏെʹሻଶͶ

ሺݏെʹሻଷ

ࣦିଵቊݏଶ

ሺݏെʹሻଷቋ

ൌࣦିଵ൜ͳ

ሺݏെʹሻൠͶࣦିଵ൜ͳ

ሺݏെʹሻଶൠͶࣦିଵ൜ͳ

ሺݏെʹሻଷൠ

ൌ݁ଶ௧Ͷ݁ଶ௧ݐʹ݁ଶ௧ݐଶሺሻ munotes.in

## Page 147

147Chapter 7: Inverse Laplace Transformሺ݅݅ሻࣦିଵቊሺݏʹሻଶሺݏଶͶݏͺሻଶቋ

ൌࣦିଵቊሺݏʹሻଶ

ሺݏଶͶݏͶͶሻଶቋ

ൌࣦିଵቊሺݏʹሻଶ

ሺሺݏʹሻଶͶሻଶቋ

ൌ݁ିଶ௧ࣦିଵቊݏଶ

ሺݏଶͶሻଶቋൌ݁ିଶ௧ࣦିଵቊݏଶͶെͶ

ሺݏଶͶሻଶቋ

ൌ݁ିଶ௧ࣦିଵ൜ͳ

ሺݏଶͶሻെͶ

ሺݏଶͶሻଶൠ

ൌ݁ିଶ௧ʹݐ

ݐെͶ݁ିଶ௧൜ͳ

Ͷ൬ʹݐ

Ͷെݐ
ʹݐ

ʹ൰ൠ

ൌ݁ିଶ௧൜ʹݐ

ʹെʹݐ

Ͷݐ
ʹݐ

ʹൠ

ൌ݁ିଶ௧൜ʹݐ

Ͷݐ
ʹݐ

ʹൠ

ૠǤǤܔ܉ܑܜܚ܉۾܉ܚ܋ܑܜܖܗ܍ۻܜܐܗܛ܌

݂ҧሺݏሻ
ܨഥሺݏሻ

ܩҧሺݏሻ

ܨഥሺݏሻܩҧሺݏሻ

ܣ

ሺܽݏܾሻǡܣ

ሺܽݏଶܾݏܾሻሺݎൌͳǡʹǡǥሻ

ǡࣦିଵ൛݂ҧሺݏሻൟ

ǣ
ǣ

ሺሻʹݏଶെݏͷ

ݏଷെݏଶͳͳݏെሺ݅݅ሻͶݏͷ

ሺݏെͳሻଶሺݏʹሻ

ሺሻݏଷെʹͳݏଶʹͲݏെ

ሺݏͳሻሺݏെʹሻଷሺݒ݅ሻݏଶʹݏെͶ

ሺݏଶʹݏͷሻሺݏଶʹݏʹሻ munotes.in

## Page 148

148 APPLIED MATHEMATICSܖܗܑܜܝܔܗ܁ሺ݅ሻʹݏଶെݏͷݏଷെݏଶͳͳݏെǡݏଷെݏଶͳͳݏെ

ݏଷെݏଶͳͳݏെ

ൌሺݏെͳሻሺݏെʹሻሺݏെ͵ሻ

ʹݏଶെݏͷ

ݏଷെݏଶͳͳݏെൌʹݏଶെݏͷ

ሺݏെͳሻሺݏെʹሻሺݏെ͵ሻ

ൌܣ

ሺݏെͳሻെܤ

ሺݏെʹሻܥ

ሺݏെ͵ሻ

ܣൌሾʹכͳଶെכͳͷሿ

ሺͳെʹሻሺͳെ͵ሻൌͳ

ʹ

ܤൌሾʹכʹଶെכʹͷሿ

ሺʹെͳሻሺʹെ͵ሻൌെͳ

ܥൌሾʹכ͵ଶെכ͵ͷሿ

ሺ͵െͳሻሺ͵െʹሻൌͷ

ʹ

ʹݏଶെݏͷ

ݏଷെݏଶͳͳݏെൌͳʹൗ

ሺݏെͳሻെͳ

ሺݏെʹሻͷʹൗ

ሺݏെ͵ሻ

ࣦିଵ൜ͳ

ሺݏെͳሻൠൌ݁௧ǡࣦିଵ൜ͳ

ሺݏെʹሻൠ

ൌ݁ଶ௧ǡࣦିଵ൜ͳ

ሺݏെ͵ሻൠൌ݁ଷ௧

݂ሺݐሻൌࣦିଵቊʹݏଶെݏͷ

ሺݏെͳሻሺݏെʹሻሺݏെ͵ሻቋ

ൌͳ

ʹࣦିଵ൜ͳ

ሺݏെͳሻൠെࣦିଵ൜ͳ

ሺݏെʹሻൠͷ

ʹࣦିଵ൜ͳ

ሺݏെ͵ሻൠ

ൌͳ

ʹ݁௧െ݁ଶ௧ͷ

ʹ݁ଷ௧

ሺ݅݅ሻͶݏͷ

ሺݏെͳሻଶሺݏʹሻǡ ሺݏെͳሻଶሺݏʹሻ

ܮݐ݁Ͷݏͷ

ሺݏെͳሻଶሺݏʹሻൌܣ

ሺݏെͳሻܤ

ሺݏെͳሻଶͶሺെʹሻͷ

ሺെʹെͳሻଶሺݏʹሻ

ሺݏെͳሻଶሺݏʹሻǡ munotes.in

## Page 149

149Chapter 7: Inverse Laplace TransformሺݏെͳሻଶሺݏʹሻሺͶݏͷሻൌܣሺݏെͳሻሺݏʹሻܤሺݏʹሻെͳ͵ሺݏെͳሻଶ

ݏൌͳ ǡǡͻൌ͵ܤǡൌ͵

ݏଶǡͲൌܣെͳ

͵ǡ

ൌͳ

͵

ࣦିଵ൜Ͷݏͷ

ሺݏെͳሻଶሺݏʹሻൠ

ൌͳ

͵ࣦିଵ൬ͳ

ሺݏെͳሻ൰͵ࣦିଵͳ

ሺݏെͳሻଶ൨െͳ

͵ࣦିଵͳ

ሺݏʹሻ൨

ൌͳ

͵݁௧͵ݐ݁௧െͳ

͵݁ିଶ௧

ሺ݅݅݅ሻݏଷെʹͳݏଶʹͲݏെ

ሺݏͳሻሺݏെʹሻଷൌʹ

ሺݏͳሻͶ

ሺݏെʹሻ͵

ሺݏെʹሻଶെͳ

ሺݏെʹሻଷ

ࣦିଵ൜ͳ

ሺݏെܽሻൠ

ൌݐିଵ

ሺ݊െͳሻǨ݁ଷ௧ǡ

ǡ݂ሺݐሻൌࣦିଵቊݏଷെʹͳݏଶʹͲݏെ

ሺݏͳሻሺݏെʹሻଷቋ

ൌʹࣦିଵ൜ͳ

ሺݏͳሻൠͶࣦିଵ൜ͳ

ሺݏെʹሻൠ͵ࣦିଵ൜ͳ

ሺݏെʹሻଶൠ

െࣦିଵ൜ͳ

ሺݏെʹሻଷൠ

ൌʹ݁ି௧Ͷ݁ଶ௧͵ݐ݁ଶ௧െݐଶ

ʹǨ݁ଶ௧

ൌʹ݁ି௧ሺͶ͵ݐെͳ

ʹݐଶሻ݁ଶ௧

ሺݒ݅ሻݏଶʹݏെͶ

ሺݏଶʹݏͷሻሺݏଶʹݏʹሻǡ

ݑݍܽݎ݀݅ݐ݂ܽܿݐܿܽݏݎݐܾ݊݊ܽܿ݁ݏ݁ݎ݈݀݁ݒ݅ݐ݊ܽ݁ݎ݈݂ݐܿܽݏݎݓݐ݄݅ ܽ݁ݎ݈݊ݑ݉ݏݎܾ݁ǡ݄݁ܿ݊݁ munotes.in

## Page 150

150 APPLIED MATHEMATICSݏଶʹݏെͶሺݏଶʹݏͷሻሺݏଶʹݏʹሻൌ͵ሺݏଶʹݏͷሻെͶሺݏଶʹݏʹሻ

ൌቀ͵

ʹቁǤʹ

ሺݏͳሻଶሺʹሻଶെʹ

ሺݏͳሻଶሺͳሻଶ

ǡ݂ሺݐሻൌࣦିଵቊݏଶʹݏെͶ

ሺݏଶʹݏͷሻሺݏଶʹݏʹሻቋ

ൌ͵

ʹࣦିଵ൜ʹ

ሺݏͳሻଶሺʹሻଶൠെʹࣦିଵ൜ͳ

ሺݏͳሻଶሺͳሻଶൠ

ൌ͵

ʹ݁ି௧ʹݐെʹ݁ି௧ݐ

ሺሻܫ݂ࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻ݂݀݊ܽሺͲሻൌͲǡݐ݄݁݊खି൛തሺሻൟൌ

ሼሺሻሽൌᇱሺሻ

݅Ǥ݁Ǥ ݂ҧሺݏሻݏǡ

݂ሺݐሻ

ܫ݊݃݁݊݁ݎ݈ܽǡࣦିଵ൛ݏ݂ҧሺݏሻൟൌൌ݀

ݐ݀ሼ݂ሺݐሻሽǡ

ݎݒ݂݀݁݀݅ሺͲሻൌ݂ᇱሺͲሻൌڮൌ݂ିଵሺͲሻൌͲ

ܵ݉݅ݐ݁݉ݏ݈݁ܽ݊݃ݓݐ݄݅ݐ݄ܾ݁ܽ݁ݒݎ݁ݑݏݐ݈ݓ݁݁ݎݍ݁ݎ݅ݑݐ݁ݏݑ݂݃݊݅ݓ݈݈

ࣦሼݐ݂ሺݐሻሽൌሺെͳሻ݀

݀ݏ݂ҧሺݏሻൌሺെͳሻ݂ҧሺሻሺݏሻݓ݄݄ܾܿ݅݊ܽܿ݁ݔ݁ݎ݁ݏݏ݁݀ܽݏ

खି൛ሺሻሺሻൟൌሺെሻሺሻ

ǣ
ǣ

ሺሻݏଶ

ሺݏଶܽଶሻଶሺ݅݅ሻݏଶ

ሺݏܽሻଷ

ሺሻ݈݃ቆͳܽଶ

ݏଶቇሺ݅ݒሻݐ݊ܽିଵ൬ʹ

ݏଶ൰

ܝܔܗ܁ܜܑܗܖሺ݅ሻ݂ҧሺሻൌͳ

ݏଶܽଶࣦିଵ൬ͳ

ݏଶܽଶ൰ൌͳ

ܽܽݐൌ݂ሺݐሻ

ܰݓǡ݂ҧᇱሺሻൌെʹݏ

ሺݏଶܽଶሻଶǡܷ݃݊݅ݏखି൛ሺሻሺሻൟ

ൌሺെሻሺሻݓ݁݃ݐ݁ munotes.in

## Page 151

151Chapter 7: Inverse Laplace Transformࣦିଵ൜െʹݏሺݏଶܽଶሻଶൠൌሺെͳሻሺଵሻݐଵͳܽܽݐ

ࣦିଵ൜ݏ

ሺݏଶܽଶሻൠൌͳ

ʹܽܽݐ

ǡखି൛തሺሻൟൌ

ሼሺሻሽǡ

ࣦିଵቊݏଶ

ሺݏଶܽଶሻଶቋൌࣦିଵቊݏǤݏଶ

ሺݏଶܽଶሻଶቋ

ൌ݀

݀ݐ൜ͳ

ʹܽܽݐൠൌͳ

ʹܽሺܽ
ܽݐሻ

ሺ݅݅ሻݏଶ

ሺݏܽሻଷ

ݓܰǡ݂ҧሺሻൌͳ

ሺݏܽሻǡ݂ሺݐሻൌ݁ି௧

ݓܰǡ݂ҧᇱሺሻൌͳ

ሺݏܽሻଶǡ݂ҧᇱᇱሺሻൌʹ

ሺݏܽሻଷ

ܷ݅ݏ݃݊खି൛ሺሻሺሻൟൌሺെሻሺሻ݁ݓݐ݁݃

ࣦିଵ൜ʹ

ሺݏܽሻଶൠൌሺെͳሻଶݐଶ݁ି௧

ࣦିଵ൜ͳ

ሺݏܽሻଷൠൌͳ

ʹݐଶ݁ି௧

ǡखି൛തሺሻൟൌ

ሼሺሻሽǡ

ࣦିଵቊݏଶ

ሺݏܽሻଷቋൌ݀ଶ

݀ݐଶ൜ͳ

ʹݐଶ݁ି௧ൠൌͳ

ʹሾܽଶݐଶെͶܽݐʹሿ݁ି௧

ሺሻ݂ሺݏሻൌ݈݃ቆͳܽଶ

ݏଶቇൌሺݏଶܽଶሻെʹݏ

݂ഥᇱሺݏሻൌʹݏ

ݏଶܽଶെʹ

ݏൌܨሺݏሻ

ܷ݅ݏ݃݊खି൛ሺሻሺሻൟൌሺെሻሺሻ݁ݓݐ݁݃

ࣦିଵ݂ഥᇱሺݏሻൌʹͻ
ܽݐെͳሻൌെݐ݂ሺݐሻ

݂ሺݐሻൌʹ

ݐሺͳെ
ܽݐሻ munotes.in

## Page 152

152 APPLIED MATHEMATICSሺݒ݅ሻ݊ܽݐିଵ൬ʹݏଶ൰

݂ሺݏሻൌ݊ܽݐିଵ൬ʹ

ݏଶ൰ǡ

݂ഥᇱሺݏሻൌͳ

ͳͶ

ݏସ൬െͶ

ݏଷ൰ൌെͶݏ

ݏସͶ

ൌെͶݏ

ሺݏଶെʹݏʹሻሺݏଶʹݏʹሻ

ൌെͳ

ሺݏଶെʹݏʹሻȂͳ

ሺݏଶʹݏʹሻ൨

ൌെͳ

ሺݏെͳሻଶͳȂͳ

ሺݏͳሻଶͳ൨

ࣦିଵ݂ഥᇱሺݏሻൌെሾ݁௧ݐെ݁ି௧ݐሿ

ൌെቈ݁௧െ݁ି௧

ʹʹݐൌെʹݐݐ

ࣦିଵ݂ᇱഥሺݏሻൌെݐ݂ሺݐሻൌെʹݐݐ

ࣦିଵ݊ܽݐିଵ൬ʹ

ݏଶ൰ൌ݂ሺݐሻൌʹ

ݐݐݐ

ૠǤǤ܍ܛ܃ܗ۱ܗܞܖܗܑܜܝܔܗܖܐ܂ܗ܍ܚ܍ܕ

݂ഥሺݏሻǡ ǡ

ܨഥሺݏሻ

כܩҧሺݏሻǡ ܨതሺݏሻܩഥሺݏሻǡ

۷खି۴തሺሻൌሺሻǡखି۵ഥሺሻൌሺሻഥሺሻൌഥሺሻכഥሺሻ

൛खିതሺሻൟൌखିሼഥሺሻכഥሺሻሽൌනሺെሻሺሻ

ǣ
खି൬

൰ൌखିതሺሻൌሺሻ

ܮݐ݁ܨഥሺݏሻൌͳ

ܩҧሺݏሻൌ݂ഥሺݏሻǡ݄݁݊݁ܿ݉ݎ݂ܾܽ݁ݒݏ݁ݎ݈ݑݐݓ݁݃ݐ݁ǡ खିቊሺሻቋൌනǤሺሻǣܨሺݐሻ݊ܽ݀ܩሺݐሻ݁ݎܽ݅ܿݎ݁ݐ݄ܾ݈݊ܽ݁݃݊ܽ݁munotes.in

## Page 153

153Chapter 7: Inverse Laplace Transform

ǣ݊݅ܽݐܾܱݐ݄݁݊݅ݒ݁ݏݎ݁ܮ݈ܽܽܿ݁ݏ݊ܽݎݐ݂ݎ݂݉ݐ݄݈݈݂݁ݓ݊݅݃ǣ

ሺሻ

ሺሻሺሻ

ሺሻ

ǣሺ݅ሻͳ

ݏଶሺݏͳሻଶൌͳ

ݏଶͳ

ሺݏͳሻଶ

ࣦିଵ൜ͳ

ݏଶൠൌݐൌܨሺݐሻࣦ݊ܽ݀ିଵ൜ͳ

ሺݏͳሻଶൠൌݐ݁ି௧ൌܩሺݐሻ

ܷ݃݊݅ݏǡ൛खିതሺሻൟൌखିሼഥሺሻכഥሺሻሽൌනሺെሻሺሻ

ࣦିଵ൜ͳ

ݏଶሺݏͳሻଶൠൌනሺݐെݑሻݑ݁ି௨ݑ݀௧

ൌሾെሺݑݐെݑଶሻ݁ି௨െሺݐെʹݑሻ݁ି௨െሺെʹሻ݁ି௨ሿ௧

ൌݐ݁ି௧ʹ݁ି௧ݐെʹ

खିቊሺሻ

ቋൌනǤሺሻ

ǡݑ݃݊݅ݏ݈݀݁ݐܽ݁݁ݎ ݕǡࣦିଵ൜ͳ

ሺݏͳሻଶൠ

ൌݐ݁ି௧ൌ݂ሺݐሻ

ࣦିଵ൜ͳ

ݏଶͳ

ሺݏͳሻଶൠ

ൌනݑ݁ି௨ݑ݀ൌሾെݑ݁ି௨െ݁ି௨ሿ௧ൌͳെሺݐͳሻ݁ି௧௧

ࣦିଵ൜ͳ

ݏଶሺݏͳሻଶൠൌࣦିଵ൜ͳ

ݏଶͳ

ሺݏͳሻଶൠ

ൌනሾͳെሺݑͳሻ݁ି௨ሿݑ݀

ൌሾݑሺݑͳሻ݁ି௨݁ି௨ሿ௧ൌݐ݁ି௧ʹ݁ି௧ݐെʹ

ሺሻݏ

ሺݏଶܽଶሻଶൌͳ

ݏଶܽଶǤݏ

ݏଶܽଶ

ࣦିଵቄݏ

ݏଶܽଶቅൌݏܿݐܽൌܨሺݐሻ munotes.in

## Page 154

154 APPLIED MATHEMATICSࣦିଵ൜ͳݏଶܽଶൠൌݐܽܽൌܩሺݐሻ

۷खି۴തሺሻൌሺሻǡखି۵ഥሺሻൌሺሻഥሺሻ

ൌഥሺሻכഥሺሻ

൛खିതሺሻൟൌखିሼഥሺሻכഥሺሻሽ

ൌනሺെሻሺሻ

ࣦିଵ൜ݏ

ሺݏଶܽଶሻଶൠൌࣦିଵ൜ͳ

ሺݏଶܽଶሻǤݏ

ሺݏଶܽଶሻൠ

ൌන
ܽሺݐെݑሻܽݑ

ܽ݀ݑ௧

ൌͳ

ʹܽනሾݐܽሺʹܽݑെݐܽሻሿ݀ݑ௧

ൌͳ

ʹܽݑݐܽെͳ

ʹܽ
ሺʹܽݑെݐܽሻ൨

௧

ൌͳ

ʹܽݐݐܽ

ૠǣ ǣ

ሺ݅ሻͳ

ሺݏെʹሻସሺݏ͵ሻ

ǣሺ݅ሻࣦିଵ൜ͳ

ሺݏെʹሻସሺݏ͵ሻൠൌࣦିଵ൜ͳ

ሺݏെʹሻସሺݏെʹͷሻൠ

ൌ݁ଶ௧ࣦିଵ൜ͳ

ݏସሺݏͷሻൠ

ࣦିଵ൜ͳ

ݏସൠൌݐଷ

͵Ǩǡࣦିଵ൜ͳ

ݏͷൠൌ݁ିହ௧

ࣦିଵ൜ͳ

ݏସሺݏͷሻൠൌනݑଷ

݁ିହሺ௧ି௨ሻ݀ݑ௧

ൌ݁ିହ௧

නݑଷ݁ହ௨݀ݑ௧

ൌ݁ିହ௧

൬ͳ

ͷݑଷെ͵

ʹͷݑଶ

ͳʹͷݑെ

ͳʹͷ൰݁ହ௨݀ݑ൨

௧

ൌͳ൬ͳͷݐଷെ͵ʹͷݐଶͳʹͷݐെͳʹͷ൰െ݁ିହ௧ʹͷ munotes.in

## Page 155

155Chapter 7: Inverse Laplace Transform

7.2 Solution of Ordinary Linear Differential Equations with Constant Coefficients

The Laplace transform method of solving differential equations yields particular

solution without the necessity of first finding the general solution and then

evaluating the arbitrary c onstant.

This is specially useful for solving linear differential equations with constant

coefficients.

Procedure to solve a Linear Differential Equations with Constant Coefficients by

transform method.

1. Take the Laplace transform of both sides of the diffe rential equation using

Laplace Transform of derivative (From Previous chapter) and the given

initial conditions

2. Transpose the terms with minus signs to the right.

3. Divide by the coefficient of y ࡄ ,getting y ࡄ as a known function of s

4. Resolve this function of s into partial fractions and take the inverse transforms of both sides.

This gives y as a function of t which is the desired solution satisfying the given

conditions.

܍ܔܘܕ܉ܠ۳ ૡǣ ǡ

ݕᇱᇱᇱʹݕᇱᇱെݕᇱെʹݕൌͲ݊݁ݒ݅݃ݕሺͲሻൌݕᇱሺͲሻൌͲܽ݊݀ݕᇱᇱሺͲሻൌ

ܗ܁ܔܖܗܑܜܝǣ

ݕᇱᇱᇱʹݕᇱᇱെݕᇱെʹݕൌͲ

ሾݏଷݕഥെݏଶݕሺͲሻെݏݕᇱሺͲሻെݕᇱᇱሺͲሻሿʹሾݏଶݕതെݏݕሺͲሻെݕᇱሺͲሻሿെሾݏݕഥെݕሺͲሻሿെʹݕഥൌͲ

ǡ

ሺݏଷʹݏଶെݏെʹሻݕതൌ

ݕതൌ

ሺݏଷʹݏଶെݏെʹሻ

ൌ

ሺݏെͳሻሺݏͳሻሺݏʹሻൌ

ሺݏെͳሻ

ሺݏͳሻ

ሺݏʹሻ munotes.in

## Page 156

156 APPLIED MATHEMATICSൌሺݏെͳሻሺሻሺെʹሻሺݏͳሻ͵ሺݏʹሻ

ǡݕ

ൌࣦିଵ൬ͳ

ሺݏെͳሻ൰െ͵ࣦିଵ൬ͳ

ሺݏͳሻ൰ʹࣦିଵ൬ͳ

ሺݏʹሻ൰

ݎܱݕൌ݁௧െ͵݁ି௧ʹ݁ିଶ௧
Ǥ

܍ܔܘܕ܉ܠ۳ ૢǣ

݀ଶݔ

݀ݐଶെʹ݀ݔ

݀ݐݔൌ݁௧݅ݓݐ݄ݔൌʹǡ݀ݔ

݀ݐൌെͳܽݐݐൌͲ

ܗ܁ܔܖܗܑܜܝǣ

݀ଶݔ

݀ݐଶെʹ݀ݔ

݀ݐݔൌ݁௧

ሾݏଶݔҧെݏݔሺͲሻെݔᇱሺͲሻሿെʹሺݏݔҧെݔሺͲሻሻݔҧൌͳ

ݏെͳ

ǡ

ሺݏଶെʹݏͳሻݔҧൌͳ

ݏെͳʹݏെͷൌʹݏଶെݏ

ݏെͳ

ݔҧൌʹݏଶെݏ

ሺݏെͳሻሺݏଶെʹݏͳሻൌʹݏଶെݏ

ሺݏെͳሻሺݏെͳሻଶൌʹݏଶെݏ

ሺݏെͳሻଷ

ݔҧൌʹ

ݏെͳെ͵

ሺݏെͳሻଶͳ

ሺݏെͳሻଷ

ǡݔ

ൌʹࣦିଵ൬ͳ

ݏെͳ൰െ͵ࣦିଵ൬ͳ

ሺݏെͳሻଶ൰ࣦିଵ൬ͳ

ሺݏെͳሻଷ൰

ൌʹ݁௧െ͵݁௧Ǥݐ

ͳǨ݁௧Ǥݐଶ

ʹǨൌʹ݁௧െ͵ݐ݁௧ͳ

ʹݐଶ݁௧

munotes.in

## Page 157

157Chapter 7: Inverse Laplace Transform܍ܔܘܕ܉ܠ۳ሺܦଶ݊ଶሻݔൌܽ݅ݏ݊ሺݐ݊ߙሻǡݔൌܦݔൌͲݐൌͲ

ܗ܁ܔܖܗܑܜܝǣ

ሺܦଶ݊ଶሻݔൌܽ݅ݏ݊ሺݐ݊ߙሻ

ሾݏଶݔҧെݏݔሺͲሻെݔᇱሺͲሻሿ݊ଶݔҧൌࣦܽሼݐ݊Ǥ
ߙ
ݐ݊Ǥߙሽ

ሺݏଶ݊ଶሻݔҧൌܽܿݏߙǤ݊

ݏଶ݊ଶܽߙǤݏ

ݏଶ݊ଶ

ݔҧൌܽ݊ܿݏߙǤͳ

ሺݏଶ݊ଶሻଶܽߙǤݏ

ሺݏଶ݊ଶሻଶ

ǡ

ݔൌܽ݊ܿݏߙǤͳ

ʹ݊ଷሺݐ݊െݐ݊
ݐ݊ሻܽߙǤݐ

ʹ݊ݐ݊

ൌܽሼݐ݊
ߙെݐ݊
ሺݐ݊ߙሻሽ

ʹ݊ଶ

܍ܔܘܕ܉ܠ۳ ሺܦଷെ͵ܦଶ͵ܦെͳሻݕൌݐଶ݁௧

ݕሺͲሻൌͳǡݕᇱሺͲሻൌͲǡݕᇱᇱሺͲሻൌെʹ

ܗ܁ܔܖܗܑܜܝǣ
ǡݐ

ሾݏଷݕതെݏଶݕሺͲሻെݏݕᇱሺͲሻെݕᇱᇱሺͲሻሿെ͵ሾݏଶݕതെݏݕሺͲሻെݕᇱሺͲሻሿ͵ሾݏݕതെݕሺͲሻሿെݕത

ൌʹ

ሺݏെͳሻଷ

ǡ

ݕതൌݏଶെ͵ݏͳ

ሺݏെͳሻଷʹ

ሺݏെͳሻൌሺݏെͳሻଶെሺݏെͳሻെͳ

ሺݏെͳሻଷʹ

ሺݏെͳሻ

ݕതൌͳ

ሺݏെͳሻͳ

ሺݏെͳሻଶെͳ

ሺݏെͳሻଷʹ

ሺݏെͳሻ

ǡ

ࣦିଵሼݕതሽൌࣦିଵ൜ͳ

ሺݏെͳሻൠࣦିଵ൜ͳ

ሺݏെͳሻଶൠെࣦିଵ൜ͳ

ሺݏെͳሻଷൠʹࣦିଵ൜ͳ

ሺݏെͳሻൠ

ݕൌ݁௧൬ͳെݐെͳ

ʹݐଶͳ

Ͳݐହ൰

munotes.in

## Page 158

158 APPLIED MATHEMATICS܍ܔܘܕ܉ܠ۳݀ଶݔ݀ݐଶͻݔൌ
ʹݐǡ݂݅ݔሺͲሻൌͳǡݔቀߨʹቁൌെͳ

ܗ܁ܔܖܗܑܜܝǣ
ݔᇱሺͲሻǡݔᇱሺͲሻൌܽ

݀ଶݔ

݀ݐଶͻݔ

ൌ
ʹǡ

ࣦሺݔᇱᇱሻͻࣦሺݔሻൌࣦሺ
ʹݐሻǡ݅Ǥ݁ǤሾݏଶݔҧെݏݔሺͲሻെݔᇱሺͲሻሿͻݔҧൌݏ

ݏଶͶ

ሺݏଶͻሻݔҧൌݏܽݏ

ݏଶͶݔҧൌݏܽ

ሺݏଶͻሻݏ

ሺݏଶͶሻሺݏଶͻሻ

ݔҧൌݏܽ

ሺݏଶͻሻݏ

ሺݏଶͶሻሺݏଶͻሻ

ݔҧൌܽ

ሺݏଶͻሻͳ

ͷǤݏ

ሺݏଶͶሻͶ

ͷǤݏ

ሺݏଶͻሻ

ǡ

ࣦିଵሼݔҧሽൌࣦିଵ൜ܽ

ሺݏଶͻሻൠͳ

ͷǤࣦିଵ൜ݏ

ሺݏଶͶሻൠͶ

ͷǤࣦିଵ൜ݏ

ሺݏଶͻሻൠ

ݔൌܽ

͵ݏ݅݊͵ݐͳ

ͷܿݏʹݐͶ

ͷܿݏ͵ݐ

ܹ݄݁݊ݐൌቀߨ

ʹቁǡെͳൌെܽ

͵െͳ

ͷܴܱܽ

͵ൌͶ

ͷݏ݅݊݁ܿݔቀߨ

ʹቁൌെͳ

ݔൌͳ

ͷሺ
ʹݐͶ͵ݐͶ
͵ݐሻ

ૠǤܗ܁ܔܖܗܑܜܝܗܑ܁ܕܝܔܜ܉ܖ܍ܛܝܗ܌ܚ۽܉ܖܑܡܚ۲ܔ܉ܑܜܖ܍ܚ܍ܑ ܙ۳ܝ܉ܛܖܗܑܜ The Laplace transform method is applicable to solve two or more simultaneous ordinary differential equations.

܍ܔܘܕ܉ܠ۳

݀ݔ

݀ݐͷݔെʹݕൌݐǡ݀ݕ

݀ݐʹݔݕൌͲܾ݊݅݁݃݅݃ݒ݊݁ݔൌݕൌͲݓ݄݁݊ݐൌͲǤ

munotes.in

## Page 159

159Chapter 7: Inverse Laplace Transformܖܗܑܜܝܔܗ܁ǣ
ǡ

ݔ݀

ݐ݀ͷݔെʹݕൌݐǡ݀ݕ

ݐ݀ʹݔݕൌͲǡ݁ݓ݃ݐ݁

ሾݏݔҧݔሺͲሻሿͷݔҧെʹݕതൌͳ

ݏଶሾݔሺͲሻൌͲሿ

݅Ǥ݁Ǥሺݏͷሻݔҧെʹݕതൌͳ

ݏଶെെെെെሺ݅ሻ

ሾݏݕതെݕሺͲሻሿʹݔҧݕതൌͲሾݕሺͲሻൌͲሿ

݅Ǥ݁ǤʹݔҧሺݏͳሻݕതൌͲെെെെെሺ݅݅ሻ

ሺ݅ሻ݊ܽ݀ሺ݅݅ሻݎ݂ǡ݁ݓ݃ݐ݁

ݔҧൌฬͳݏଶΤെʹ

Ͳݏͳฬቚݏͷെʹ

ʹݏͳቚ

ݔҧൌݏͳ

ݏଶሺݏ͵ሻଶ

ݔҧൌͳ

ʹݏͳ

ͻݏଶെͳ

ʹሺݏ͵ሻെʹ

ͻሺݏ͵ሻଶ

ݔҧሺ݅݅ሻǡ

ݕതൌെʹ

ݏଶሺݏ͵ሻଶൌͶ

ʹݏെʹ

ͻݏଶെͶ

ʹሺݏ͵ሻെʹ

ͻሺݏ͵ሻଶ

ǡ

ݔൌͳ

ʹݐ

ͻെͳ

ʹ݁ିଷ௧െʹ

ͻݐ݁ିଷ௧ǡ݊ܽ݀ݕൌͶ

ʹെʹݐ

ͻെͶ

ʹ݁ିଷ௧െʹ

ͻݐ݁ିଷ௧

ܘܕ܉ܠ۳ܔ܍
ሺݔǡݕሻ

݀ݕ

ݐ݀ʹݔൌݏ݊݅ʹݐǡݔ݀

ݐ݀െʹݕൌܿݏʹݐǡሺݐͲሻǤ

ݐൌͲǡݔൌͳݕൌͲǡ ǡ

ͶݔଶͶݕݔͷݕଶൌͶ

ܖܗܑܜܝܔܗ܁ ǣܶܽ݇݃݊݅ݐ݄݁ܮ݈ܽ݁ܿܽݎ݂ݏ݊ܽݎݐ ݉ݏ݂ݐ݄݁݃݅ݒ݊݁݁ݑݍݏ݊݅ݐܽǡ

݊ܽ݀݃݊݅ݐ݊ݐ݄ܽݐݕሺͲሻൌͲǡݔሺͲሻൌͳǡ݁ݓ݃ݐ݁ munotes.in

## Page 160

160 APPLIED MATHEMATICSሾݏݕഥȂݕሺͲሻሿʹݔҧൌʹݏଶʹଶݎʹݔҧݏݕതൌʹݏଶͶ݀݊ܽെሺ݅ሻ ሾݏݔഥȂݔሺͲሻሿെʹݕതൌଶ

௦మାଶమݎݏݔҧെʹݕതൌଶ

௦మାସ

ͳെെെሺ݅݅ሻ

ሺ݅ሻܾݕݏሺ݅݅ሻܾݕʹ
ǡ

ሺݏଶͶሻݕതൌെʹݎݕതൌെʹ

ሺݏଶͶሻ

ǡݕൌെʹࣦିଵͳ

ሺݏଶͶሻ൨ൌെʹݐ

ǡ

ʹݔൌݏ݊݅ʹݐെݕ݀

ݐ݀ൌݏ݊݅ʹݐെ݀

ݐ݀ሺെݏ݊݅ʹݐሻ

ݎʹݔൌݏ݊݅ʹݐʹܿݏʹݐݎͶݔଶ

ൌሺݏ݊݅ʹݐʹܿݏʹݐሻଶെെെെെെെെെെሺ݅݅݅ሻ

݈ܣݏͶݔݕൌሺݏ݊݅ʹݐʹܿݏʹݐሻሺെʹݏ݊݅ʹݐሻ

ൌെʹሺݏ݊݅ଶʹݐʹݏ݊݅ʹݐܿݏʹݐሻെെെെെെሺݒ݅ሻ

݀݊ܽͷݕଶൌͷݏ݊݅ଶʹݐെെെെെെሺݒ݅ሻ

݀݀ܽ݅݃݊ሺ݅݅݅ሻǡሺݒ݅ሻǡ݀݊ܽሺݒሻǡ݁ݓ݊݅ܽݐܾ

ͶݔଶͶݔݕͷݕଶ

ൌݏ݊݅ଶʹݐͶݏ݊݅ʹݐܿݏʹݐͶܿݏଶʹݐെʹݏ݊݅ଶʹݐ

െͶݏ݊݅ʹݐܿݏʹݐͷݏ݊݅ଶʹݐ

ൌͶݏ݊݅ଶʹݐͶܿݏଶʹݐൌͶ

܍ܔܘܕ܉ܠ۳

ܦଶݔ͵ݔെʹݕൌͲǡܦଶݔܦଶݕ͵ݔͷݕൌͲ

where D=d/dt .

ݔൌͲǡݕൌͲǡݔൌ͵ǡݕൌʹݐൌͲǡݔݕݐൌͳ

ʹǤ

ܗ܁ܔܖܗܑܜܝǣ
ǡ

ሾݏଶݔҧെݏݔሺͲሻെݔᇱሺͲሻሿ͵ݔഥെʹݕതൌͲ

݅Ǥ݁Ǥǡሺݏଶ͵ሻݔഥെʹݕഥൌ͵െെെെെെെെെെሺ݅ሻ

݀݊ܽሾݏଶݔҧെݏݔሺͲሻെݔᇱሺͲሻሿሾݏଶݕതെݏݕሺͲሻെݕᇱሺͲሻሿെ͵ݔҧͷݕതൌͲ munotes.in

## Page 161

161Chapter 7: Inverse Laplace Transform݅Ǥ݁Ǥǡሺݏଶെ͵ሻݔഥെሺݏଶͷሻݕഥൌͷെെെെെെെሺ݅݅ሻ

ݒ݈ܵ݅݃݊ሺ݅ሻܽ݊݀ሺ݅݅ሻ݂ݎݔҧܽ݊݀ݕതǡݓ݁݁݃ݐ

ݔҧൌቚ͵െʹ

ͷݏଶͷቚฬݏଶ͵െʹ

ݏଶെ͵ݏଶͷฬൌ͵ݏଶʹͷ

ሺݏଶͳሻሺݏଶͻሻ

ݔҧൌͳͳ

ͶǤͳ

ݏଶͳͳ

ͶǤͳ

ሺݏଶͻሻ

ݕഥൌቚݏଶ͵͵

ݏଶെ͵ͷቚฬݏଶ͵െʹ

ݏଶെ͵ݏଶͷฬൌʹݏଶʹͶ

ሺݏଶͳሻሺݏଶͻሻ

ݕതൌͳͳ

ͶǤͳ

ݏଶͳ͵

ͶǤͳ

ሺݏଶͻሻ

ܱ݊݅ݏݎ݁ݒ݊݅݊ǡݓ݁݁݃ݐ

ݔൌͳͳ

Ͷݐͳ

ͳʹ͵ݐǡݕൌͳͳ

Ͷݐെͳ

Ͷ͵ݐ

ૠǤ܍܋܉ܔܘ܉ۺܚ܂܉ܕܚܗܛܖ܉ܖܗܑܜܗ܁ܔ܉ܑ܋܍ܘܝ۴ܖܗܑܜ܋ܖ

In some physical and engineering problems, it is required to find the solution of a

differential equation of the system which it is ac ted on by

(i) a periodic force or periodic voltage

(ii) a impulsive force or voltage acting instantaneously at a certain time, or a

concentrated load acting at a point,

(iii) a force acting on a part of the system or voltage acting for finite interval

of time •

• ۾ܗܑܚ܍܌܋ܑܝ۴ܛܖܗܑܜ܋ܖ

• ۶܉܍ܞ܍܌ܑܛܑ܃ܜܑܖ܁ܘ܍ܜܝ۴ܖܗܑܜ܋ܖ

• ۲܋܉ܚܑെ܉ܜܔ܍܌ܝ۴ܖܗܑܜ܋ܖሺ܃ܜܑܖ۷܍ܛܔܝܘܕܝ۴ܖܗܑܜ܋ܖሻ

ૠǤǤ۾ܗܑܚ܍܌܋ܑܝ۴ܛܖܗܑܜ܋ܖ

݂ሺݐሻܶ

݂ሺݐܶሻൌ݂ሺݐሻǡܶͲെെെെെെെെሺܫሻ

ǤǤሺሻ݂ሺݐሻൌݐ
ܶൌʹɎǡ

݂ሺݐܶሻൌሺݐʹɎሻൌൌሺሻ ݂ሺݐܶሻൌݏ݊݅ሺݐʹߨሻൌݏ݊݅ݐൌ݂ሺݐሻ munotes.in

## Page 162

162 APPLIED MATHEMATICS ǤǤሺሻSquare Wave Function

݂ሺݐሻൌͳǡͲͳ൏ܽ

ൌെͳǡܽ൏ݐ൏ʹܽ݅ݓݐ݄ݎ݁݅݀ʹܽ

݂ሺݐሻ݂݅݁݀݊݀݁ݕܾሺܫሻ݅ݏ݃݅ݒ݊݁ݕܾ

݂ҧሺݏሻൌࣦሼ݂ሺݐሻሽ

ൌන݁ି௦௧݂ሺݐሻݐ݀ஶ

ൌͳ

ͳെ݁ି௦்න݁ି௦௨݂ሺݑሻݑ݀െെെെെെሺܫܫሻ݂ݎݎ்݁݅݀ܶ

܍ܔܘܕ܉ܠ۳ ̶Square Wave Function ̶ʹܽ

݂ሺݐሻൌͳǡͲݐ൏ܽ

ൌെͳǡܽ൏ݐ൏ʹܽ

Laplace transform of IW

ܗ܁ܔܖܗܑܜܝǣ݂ҧሺݏሻൌࣦሼ݂ሺݐሻሽൌන݁ି௦௧݂ሺݐሻݐ݀ஶ

ǡ

ൌͳ

ͳെ݁ି௦்න݁ି௦௨݂ሺݑሻݑ݀െെെെെെ݂ݎݎ்݁݅݀ܶ

ࣦሼ݂ሺݐሻሽൌͳ

ͳെ݁ିଶන݁ି௦௨݂ሺݑሻݑ݀െെെെെെ݂ݎݎ݁݅݀ܶൌʹܽଶ

ൌͳ

ͳെ݁ିଶන݁ି௦௨ͳǤݑ݀න݁ି௦௨ሺെͳሻǤݑ݀ଶ

ൌͳ

ݏሺͳെ݁ି௦ሻଶ

ሺͳെ݁ିଶ௦ሻൌሺͳെ݁ି௦ሻ

ሺͳ݁ି௦ሻൌͳ

ݐ݄݊ܽܽݏ

ʹ

munotes.in

## Page 163

163Chapter 7: Inverse Laplace TransformૠǤǤ܍܌ܑܛܑܞ܉܍۶ܖ܃ܑܜ܁܍ܜܘܝ۴ܖ܋ܑܜܗܖ

There are some fractions of which the inverse transform can not be determined

from the formulae so far derived.To over come the such cases,the Unit Step

Function ሺƍ
ሻ
Ǥ

Ǥܖ܃ܑܜ܁܍ܜܘܝ۴ܖ܋ܑܜܗܖሺ܍܌ܑܛܑܞ܉܍۶ᇱܛܖ܃ܑܜܝ۴ܖ܋ܑܜܗܖሻ
ݑሺݐെܽሻǣ

ݑሺݐെܽሻൌቄͲ൏ܽͳ ǡܽǤܪሺݐെܽሻǤ

Fig (Unit Step Function )

ሻܚ܂ܛܖ܉ܗܕܚܗܜܑܖܝܖܗܑܜ܋ܖܝ Ǥ

ࣦሼݑሺݐെܽሻሽൌන݁ି௦௧ݑሺݐെܽሻ݀ݐൌஶ

ൌන݁ି௦௧ǤͲ݀ݐන݁ି௦௧Ǥͳ݀ݐൌஶ

Ͳቤ݁ି௦௧

െݏቤ

ஶ

ࣦሼݑሺݐെܽሻሽൌ݁ି௦

െݏ

݂ሺݐሻݑሺݐെܽሻൌ൜Ͳݎ݂ݐ൏ܽ

݂ሺݐሻݎ݂ݐܽ

݂ሺݐെܽሻǤݑሺݐെܽሻ ݂ሺݐሻ

munotes.in

## Page 164

164 APPLIED MATHEMATICS܌ܖܗ܋܍܁ܛܑܜܑܐܖܚܘܗܘ܍ܚܜܡ

ࣦሼ݂ሺݐሻሽൌ݂ҧሺݏሻǡݐ݄ࣦ݊݁ሼ݂ሺݐെܽሻǤݑሺݐെܽሻሽൌ݁ି௦݂ҧሺݏሻ

ࣦሼ݂ሺݐെܽሻǤݑሺݐെܽሻሽൌන݁ି௦௧݂ሺݐെܽሻݑሺݐെܽሻ݀ݐஶ

ൌන݁ି௦௧݂ሺݐെܽሻሺͲሻ݀ݐ

න݁ି௦௧݂ሺݐെܽሻ݀ݐஶ

ሾݐെܽൌݑሿ

ൌන݁ି௦ሺ௨ାሻ݂ሺݑሻ݀ݑஶ

ൌ݁ି௦න݁ି௦௨݂ሺݑሻ݀ݑஶ

ൌ݁ି௦݂ҧሺݏሻ

ܕ܉ܠ۳ܘܔ܍
ሺ ሻ

ܗ܁ܔܝܑܜܖܗǣ

݂ሺݐሻൌ൝ͲǡͲ൏ݐ൏ͳ

ݐെͳǡͳ൏ݐ൏ʹ

Ͳǡݐʹ

܀۽݂ሺݐሻൌሺݐെͳሻሾݑሺݐെͳሻെݑሺݐെʹሻሿݑሺݐെʹሻ

ൌሺݐെͳሻݑሺݐെͳሻെሺݐെʹሻݑሺݐെʹሻ

ǡࣦሼ݂ሺݐെܽሻǤݑሺݐെܽሻሽൌ݁ି௦ࣦሼ݂ሺݐሻሽ

ۯܗܛܔࣦሼ݂ሺݐሻሽൌࣦሺݐሻൌͳ

ݏଶ

ࣦሼሺݐെͳሻݑሺݐെͳሻሽൌ݁ି௦ͳ

ݏଶࣦܽ݀݊ሼሺݐെʹሻݑሺݐെʹሻሽൌ݁ିଶ௦ͳ

ݏଶ

۶܍܋ܖ܍ࣦሼ݂ሺݐሻሽൌࣦሼሺݐെͳሻݑሺݐെͳሻെሺݐെʹሻݑሺݐെʹሻሽൌ݁ି௦െ݁ିଶ௦

ݏଶ

munotes.in

## Page 165

165Chapter 7: Inverse Laplace Transform܍ܔܘܕ܉ܠ۳ૡ
ǡ

݂ሺݐሻൌ൝ݐǡͲݐ൏ߨ

ʹݐǡߨݐ൏ʹߨ

͵ݐݐʹߨǡ

ܗ܁ܔܖܗܑܜܝ

݂ሺݐሻൌݐሾݑሺݐെͲሻെݑሺݐെߨሻሿ

ʹݐሾݑሺݐെߨሻെݑሺݐെʹߨሻሿ͵ݐǤݑሺݐെʹߨሻ

ൌݐሺʹݐെݐሻݑሺݐെߨሻሺ͵ݐെʹݐሻݑሺݐെʹߨሻ

ࣦሾ݂ሺݐെܽሻݑሺݐെܽሻሿൌ݁ି௦݂ҧሺݏሻࣦሾܽݐሿൌܽ

ݏଶܽଶ

ࣦሾ݂ሺݐሻሿൌࣦሾݐሿࣦሾሺʹݐെݐሻǤݑሺݐെߨሻሿ

ࣦሾሺ͵ݐെʹݐሻǤݑሺݐെʹߨሻሿ

ൌͳ

ݏଶͳ݁ିగ௦൬ʹ

ݏଶͶെͳ

ݏଶͳ൰݁ିଶగ௦൬͵

ݏଶͻെʹ

ݏଶͶ൰

܍ܔܘܕ܉ܠ۳ ૢሺܑሻ
ሺ ሻ

Ǥ

ሺܑܑሻ
݁ି௧ሾͳെݑሺݐെʹሻሿǤ

ܗ܁ܔܖܗܑܜܝሺܑሻ

݂ሺݐሻൌቄݐെͳǡͳ൏ݐ൏ʹ

͵െݐǡʹ൏ݐ൏͵

܀۽݂ሺݐሻൌሺݐെͳሻሼݑሺݐെͳሻെݑሺݐെʹሻሽሺ͵െݐሻሼݑሺݐെʹሻെݑሺݐെ͵ሻሽ

ൌሺݐെͳሻݑሺݐെͳሻെʹሺݐെʹሻݑሺݐെʹሻሺݐെ͵ሻݑሺݐെ͵ሻ
ࣦሼ݂ሺݐെܽሻǤݑሺݐെܽሻሽൌ݁ି௦݂ҧሺݏሻ

munotes.in

## Page 166

166 APPLIED MATHEMATICSࣦሼ݂ሺݐሻሽൌ݁ି௦Ǥͳݏଶെʹ݁ିଶ௦Ǥͳݏଶ݁ିଷ௦Ǥͳݏଶሾ݂ሺݐሻൌݐሿ

ൌ݁ି௦ሺͳെ݁ି௦ሻଶ

ݏଶ

ሺܑܑሻࣦሼ݁ି௧ሾͳെݑሺݐെʹሻሿሽൌࣦሼ݁ି௧ሽെࣦሼ݁ି௧ݑሺݐെʹሻሽ

ൌͳ

ݏͳെ݁ିଶࣦ൛݁ିሺ௧ିଶሻݑሺݐെʹሻൟ

݂ሺݐሻൌ݁ି௧ǡ݂ҧሺݏሻൌͳ

ݏͳ

ࣦሼ݂ሺݐെܽሻǤݑሺݐെܽሻሽൌ݁ି௦݂ҧሺݏሻ

ࣦ൛݁ିሺ௧ିଶሻݑሺݐെʹሻൟൌ݁ିଶ௦Ǥͳ

ݏͳ

ǡࣦ݁ି௧ሼͳെݑሺݐെʹሻሽൌ൛ͳെ݁ିଶሺ௦ାଵሻൟ

ݏͳ

܍ܔܘܕ܉ܠ۳
ǡ

න݁௧ሺͳʹݐെݐଶݐଷሻܪሺݐെͳሻ݀ݐஶ

ܗ܁ܔܜܝܑܖܗࣦሼሺͳʹݐെݐଶݐଷሻܪሺݐെͳሻሽ

ൌ݁ି௦ࣦሼͳʹሺݐͳሻെሺݐͳሻଶሺݐͳሻଷሽ

ൌ݁ି௦ࣦሼ͵͵ݐʹݐଶݐଷሽ

ൌ݁ି௦൬͵Ǥͳ

ݏ͵Ǥͳ

ݏଶʹǤʹǨ

ݏଷ͵Ǩ

ݏସ൰

ൌ݁ି௦൬͵

ݏ͵

ݏଶͶ

ݏଷ

ݏସ൰

ǡ

න݁ି௦௧ሺͳʹݐെݐଶݐଷሻܪሺݐെͳሻ݀ݐஶ

ൌ݁ି௦൬͵

ݏ͵

ݏଶͶ

ݏଷ

ݏସ൰

ൌͳǡ munotes.in

## Page 167

167Chapter 7: Inverse Laplace Transformන݁௧ሺͳʹݐെݐଶݐଷሻܪሺݐെͳሻ݀ݐஶൌ݁ିଵሺ͵͵Ͷሻൌͳ݁

܍ܔܘܕ܉ܠ۳ ሺሻࣦିଵቊ݁ି௦െ͵݁ଷି௦

ݏଶቋሺሻࣦିଵ൜ݏ݁ି௦

ݏଶെ߱ଶൠǡܽͲ

ܗ܁ܔܖܗܑܜܝࣦିଵ൜݁ି௦Ǥͳ

ݏଶൠൌቄݐെͳǡݐͳ

Ͳǡݐ൏ͳቅൌሺݐെͳሻݑሺݐെͳሻ

ࣦିଵ൜݁ିଷ௦Ǥͳ

ݏଶൠൌቄݐെ͵ǡݐ͵

Ͳǡݐ൏͵ቅൌሺݐെ͵ሻݑሺݐെ͵ሻ

ሺሻࣦିଵቊ݁ି௦െ͵݁ଷି௦

ݏଶቋൌࣦିଵ൜݁ି௦

ݏଶൠെ͵ࣦିଵቊ݁ିଷ௦

ݏଶቋ

ൌሺݐെͳሻݑሺݐെͳሻെ͵ሺݐെ͵ሻݑሺݐെ͵ሻ

ሺሻࣦିଵ൜ݏ݁ି௦

ݏଶെ߱ଶൠǡࣦିଵቄݏ

ݏଶെ߱ଶቅൌܿݏ݄߱ݐ

ࣦିଵ൜ݏ݁ି௦

ݏଶെ߱ଶൠൌ൜ܿݏ݄߱ሺݐെܽሻǡݐܽ

Ͳǡݐ൏ܽ

ൌܿݏ݄ɘሺݐെܽሻݑሺݐെܽሻǡ

܍ܔܘܕ܉ܠ۳

ሺሻݏ݁ି௦ଶΤߨ݁ି௦

ݏଶߨଶሺሻ݁ି௦

ݏଶሺݏܽሻǡሺܿͲሻ

ܗ܁ܔܖܗܑܜܝሺܑሻࣦିଵቄݏ

ݏଶߨଶቅൌܿݏߨݐǡࣦିଵቄߨ

ݏଶߨଶቅൌ݊݅ݏߨݐ

ࣦିଵ൛݁ି௦݂ҧሺݏሻൟൌ݂ሺݐെܽሻǤݑሺݐെܽሻ

ࣦିଵቊݏ݁ି௦ଶΤߨ݁ି௦

ݏଶߨଶቋ

ൌࣦିଵቄ݁ି௦ଶΤǤݏ

ݏଶߨଶቅࣦିଵቄ݁ି௦Ǥߨ

ݏଶߨଶቅ

ൌ
ߨ൬ݐെͳ

ʹ൰Ǥݑ൬ݐെͳ

ʹ൰ߨሺݐെͳሻǤݑሺݐെͳሻ

ൌߨݐǤݑ൬ݐെͳ

ʹ൰െߨݐǤݑሺݐെͳሻ ൌ൜ݑ൬ݐെͳʹ൰െݑሺݐെͳሻൠߨݐ munotes.in

## Page 168

168 APPLIED MATHEMATICSሺܑܑሻࣦିଵ൜݁ି௦ݏଶሺݏܽሻൠൌࣦିଵ൜݁ି௦൬െͳܽଶǤͳݏͳܽǤͳݏଶͳܽଶǤͳݏܽ൰ൠ

ࣦିଵ൛݁ି௦݂ҧሺݏሻൟൌ݂ሺݐെܽሻǤݑሺݐെܽሻ

ࣦିଵ൜݁ି௦

ݏଶሺݏܽሻൠ

ൌെͳ

ܽଶሼͳǤݑሺݐെܿሻሽͳ

ܽሼሺݐെܿሻǤݑሺݐെܿሻሽͳ

ܽଶ൛݁ିሺ௧ିሻǤݑሺݐെܿሻൟ

ൌͳ

ܽଶ൛ܽሺݐെܿሻെͳ݁ିሺ௧ିሻൟݑሺݐെܿሻ

ૠǤǤ۲ܑ܋܉ܚെܜܔ܍܌܉ܖܝ۴ܑܜ܋ܖܗሺ܃ܜܑܖܘܕ۷ܔܝܛ܍ܖܝ۴ܑܜ܋ܖܗሻ

The idea of a very large force acting for a very short time is of frequent

occurrence in mechanics. This Unit impulse ( Dirac Delta) function is useful in

this case.

Ǥ܃ܜܑܖܑ܍ܛܔܝܘܕሺ۲ܑ܋܉ܚ۲ܜܔ܍܉ሻ

ߜᖡሺݐെܽሻൌͳᖡΤǡᖡ

ൌͲǡݐ݄݅ݓݎ݁ݏ݁

ᖡ՜ͲǤ
ᖡ՜Ͳǡ

Ǥ

ߜሺݐെܽሻǣ ߜሺݐെܽሻൌλݐൌܽǢൌͲݐ്ܽǡ

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## Page 169

169Chapter 7: Inverse Laplace Transform

නߜሺݐെܽሻஶ

݀ݐൌͳሺܽͲሻ

ǡݓ
ݔൌܽ

ݓᖡ Τ ݔൌܽݔൌܽᖡǤ

ݓሺݔሻൌݓᖡΤǡ൏൏ᖡ

ൌͲǡݐ݄ݎ݁ݓ݁ݏ݅

݅Ǥ݁Ǥݓሺݔሻൌݓߜሺݔܽሻ

Ǥܕܚܗܛܖ܉ܚ܂ ܗܝܖܜܑܕܑܘܝܔܛ܍ሺ۲܉ܚܑ܋۲܉ܜܔ܍ሻܖܗܑܜ܋ܖܝ Ǥ

݂ሺݐሻ
ݐݐൌܽǡ

න݂ሺݐሻஶ

ߜᖡሺݐെܽሻǤ݀ݐൌන݂ሺݐሻୟାᖡ

ᖡǤ݀ݐ

ൌሺܽᖡെܽሻ݂ሺߟሻͳ

ᖡൌ݂ሺߟሻǡݓ݄݁ݎ݁ܽ൏ߟ൏ܽᖡǡ

ᖡ՜ǡන݂ሺݐሻஶ

ߜሺݐെܽሻ݀ݐൌ݂ሺܽሻǤ

ǡ݂ሺݐሻൌ݁ି௦௧ǡࣦሼߜሺݐെܽሻሽൌ݁ି௦ ǡ݂ሺܾሻെ݂ሺܽሻܾെܽ

ൌ݂ᇱሺܿሻሺܿǡܽ൏ܿ൏ܾሻ

݂ ܽ൏ݔ൏ܾǡ
ܽݔܾ

munotes.in

## Page 170

170 APPLIED MATHEMATICS܍ܔܘܕ܉ܠ۳ǣሺሻන݊݅ݏʹݐஶߜሺݐെߨͶሻΤ݀ݐሺሻࣦ൜ͳݐߜሺݐെܽሻൠ

ܗ܁ܔܖܗܑܜܝǣሺሻන݂ሺݐሻߜሺݐെܽሻ݀ݐൌ݂ሺܽሻஶ

න݊݅ݏʹݐஶ

ߜሺݐെߨͶሻΤ݀ݐൌሺʹǤߨͶሻΤൌሺߨʹሻΤൌͳ

ሺሻࣦሼߜሺݐെܽሻሽൌ݁ି௦

ࣦ൜ͳ

ݐߜሺݐെܽሻൠൌනࣦሼߜሺݐെܽሻሽஶ

௦݀ݏൌන݁ି௦݀ݏஶ

௦

ൌฬ݁ି௦

െܽฬ

௦ஶ

ൌͳ

ܽ݁ି௦

۳܍ܔܘܕ܉ܠǣ ܧߜሺݐሻ

ܮǡܴǡܥ
Ǥ݅

ݐ՜Ͳ

ܗ܁ܔܖܗܑܜܝǣ
݅

ܮ݀݅

݀ݐܴ݅ͳ

ܥන݅݀ݐൌܧߜሺݐሻ݅ൌͲǡݐൌͲ௧

ǡ

ܮሾݏଓҧെ݅ሺͲሻሿܴଓҧͳ

ܥͳ

ݏଓҧൌܧሺ ሻ

ܚ۽൬ݏଶܴ

ܮݏͳ

ܥܮ൰ଓҧൌܧ

ܮݏܚ۽ሺݏଶʹܽݏܽଶܾଶሻଓҧൌܧ

ܮݏ

ܴ

ܮൌʹܽͳ

ܥܮൌܽଶܾଶ

ܚ۽ଓҧൌܧ

ܮሺݏܽሻെܽ

ሺݏܽሻଶܾଶൌܧ

ܮቊሺݏܽሻ

ሺݏܽሻଶܾଶെܽͳ

ሺݏܽሻଶܾଶቋ

ǡ

ଓҧൌܧ

ܮቄ݁ି௧ܿݏܾݐെܽ

ܾ݁ି௧݊݅ݏܾݐቅ

ݐ՜Ͳǡ݅՜ܧܮΤ

munotes.in

## Page 171

171Chapter 7: Inverse Laplace Transform
݅ൌͲǡ

ݐൌͲǤ
ܧȀܮ

7.5 Exercise

ͳǤ
ǣ

ሺ݅ሻͳ

ሺݏെͳሻହሺͳ

ʹͶݐସ݁ି௧ሻ

ሺ݅݅ሻͶݏͳͷ

ͳݏଶെʹͷ൬ͳ

Ͷܿݏ݄ͷ

Ͷݐ͵

Ͷݏ݄݅݊ͷ

Ͷݐ൰

ሺ݅݅݅ሻ͵ሺݏଶെͳሻଶ

ʹݏହ൬͵

ʹെ͵

ʹݐଶͳ

ͳݐସ൰

ሺݒ݅ሻͳ

ݏଷଶΤቌʹඨݐ

ߨቍ

ሺݒሻͳ

ξʹݏ͵൬ͳ

ξʹߨݐ݁ିଷ௧ଶൗ൰

ʹǤ
ǣ

ሺ݅ሻͶݏͳʹ

ݏଶͺݏͳሺͶ݁ିସ௧ሺͳെݐሻሻ

ሺ݅݅ሻ͵ݏ

ݏଶെʹݏെ͵ሺͶ݁ଷ௧െ݁ି௧ሻ

ሺ݅݅݅ሻݏଶͳ

ݏଷ͵ݏଶʹݏ൬ͳ

ʹെʹ݁ି௧ͷ

ʹ݁ିଶ௧൰

ሺݒ݅ሻݏʹͻ

ሺݏͶሻሺݏଶͻሻ൬݁ିସ௧െܿݏ͵ݐͷ

͵ݏ݅݊͵ݐ൰

ሺݒሻݏʹ

ݏଷሺݏെͳሻଶሺሺ͵ݐെͺሻ݁௧ݐଶͷݐͺሻ

ሺ݅ݒሻͳ

ݏଷሺݏଶͳሻ൬ͳ

ʹݐଶܿݏݐെͳ൰

ሺ݅݅ݒሻݏଶെܽଶ

ሺݏଶܽଶሻଶሺݐܿݏܽݐሻ

ሺ݅݅݅ݒሻݏ

ሺݏଶͳሻሺݏଶͶሻ൬ͳ

͵ሺܿݏݐെܿݏʹݐሻ൰

ሺݔ݅ሻͳ

ሺݏଶܽଶሻଶ൬ͳ

ʹܽଷሺݏ݅݊ܽݐെܽݐܿݏܽݐሻ൰ ሺݔሻݏݏସͶܽସ൬ͳʹܽଶሺݏ݅݊ܽݐݏ݄݅݊ܽݐ൰ munotes.in

## Page 172

172 APPLIED MATHEMATICS͵Ǥ
݉ݎ݂ݏ݊ܽݎݐ݂݄݂݁ܽܿݐ݄݂݈݈݁݅ݓ݊݃

ሺ݅ሻܽݏሺݏെܽሻሺܛܖۯୟ୲െͳሻ

ሺ݅݅ሻͳ

ݏሺݏଶܽଶሻ൬ͳ

ܽଶሺͳെݏܿݐܽሻ൰

ሺ݅݅݅ሻͳ

ݏξݏͶ൬ͳ

ʹ൫ʹξݐ൯൰

ሺ݅ݒሻͳ

ݏ݈݃ሺݏ͵ሻ

ሺݏʹሻቌන݁ିଶ௫െ݁ିଷ௫

ݔ௧

݀ݔቍ

ͶǤ ሺͲሻݐ݅ݓ݄݊݁ݒ݈݅݃ܽ݅ݐ݈݅݊݅ܽݒݑݏ݁

ሺ݅ሻሺܦͳሻଶݕൌݐǡݐ݅ݓ݄ݕൌ݀ݕ

݀ݐൌͳݐܽݐൌͲ

൬ܛܖۯൌͷ

ʹି୲͵

ʹି୲െͳ

ʹ
൰

ሺ݅݅ሻሺܦͳሻଶݕൌݐǡݐ݅ݓ݄ݕൌ݀ݕ

݀ݐൌͳݐܽݐൌͲ

൬ܛܖۯൌͷ

ʹି୲͵

ʹି୲െͳ

ʹ
൰

ሺ݅݅ሻሺܦଶͶܦͺሻݕൌͳǡݐ݅ݓ݄ݕൌͲǡܦ௬ൌͳݐܽݐ

ൌͲ

൬ܛܖۯൌͳ

ͺሺͳെିଶ୲
ʹെ͵ିଶ୲ʹሻ൰

ሺ݅݅݅ሻሺܦͳሻݕൌݐଶ݁ି௧ǡ݊݁ݒ݅݃ݕൌ͵ݓ݄݁݊ݐൌͲǡ

൭ܛܖۯሺሻൌି୲ቆଷ

͵͵ቇ൱

ሺ݅ݒሻ݀ଶݕ

݀ݐଶെ݀ݕ

݀ݐെݕൌʹݐ݅ݓ݄ݕሺͲሻൌͳݕᇱሺͲሻൌͲǡ

൬ܛܖۯൌെͳ

͵ͺ

ͳͷଷ୲Ͷ

ͷିଶ୲൰

ሺݒሻ݀ଶݕ

݀ݐଶݕൌݐݐ݅ݓ݄ݕሺͲሻൌͳݕᇱሺͲሻൌെͳ

ʹǡ

൬ܛܖۯൌ൬ͳെ

ʹ൰
൰

ͷǤܵ݁ݒ݈ݐ݄݁ݏݑ݁݊ܽݐ݈ݑ݉݅ݏ ݍ݁ݏ݊݅ݐܽݑ

ሺ݅ሻ݀ݔ

݀ݐൌʹݔെ͵ݕ

݀ݕ݀ݐൌݕെʹݔൢݑݏ݆ܾݐܿ݁ݐݏ݊݅ݐ݅݀݊ܿ ݔሺͲሻൌͺǡݕሺͲሻൌ͵ǡ ሺܛܖۯݔൌͷ݁ି௧͵݁ସ௧ǡݕൌͷ݁ି௧െʹ݁ସ௧ሻ munotes.in

## Page 173

173Chapter 7: Inverse Laplace Transformሺ݅݅ሻሽݔ݀ݐ݀െݕ݀ݐ݀ʹݕൌ
ʹݐݔ݀

ݐ݀ݕ݀

ݐ݀െʹݔൌʹݐൢ݆ܾܿ݁ݑݏݐݐܿ݊݅ݐ݅݀݊ݏݔሺͲሻൌͲǡݕሺͲሻൌെͳǡ

൬ܛܖۯݔൌͳ

ʹ݁௧ሺ
ݐݐሻെͳ

ʹ
ʹݐǡݕ

ൌെ݁௧ሺ
ݐെݐሻെʹݐ൰

ሺ݅݅ሻ݀ଶݔ

݀ݐଶݔݕൌͲ

Ͷ݀ଶݔ

݀ݐଶെݔൌͲۙۘۗݔሺͲሻൌെʹܽǡݕሺͲሻൌܽǡ

ݔᇱሺͲሻൌʹܾǡݕᇱሺͲሻൌെܾ

൬ܛܖۯݔൌʹܽ
ݐξʹʹξʹܾݐξʹ൰

7.6 Summary

In this unit we learn Inverse Laplace Transform definition, Shifting Theorem,

Partial fraction Methods, Use of Convolution Theorem .

खି൛തሺሻൟൌሺሻ ࣦିଵͳݏ൨ൌͳ ࣦିଵͳݏെܽ൨ൌ݁௧ ࣦିଵͳݏ൨ൌݐିଵሺ݊െͳሻǨǡ݊ൌͳǡʹǡ͵ǤǤ ࣦିଵͳሺݏെܽሻ൨ൌ݁௧ݐିଵሺ݊െͳሻǨ ࣦିଵͳݏଶܽଶ൨ൌͳܽܵ݊݅ܽݐ ࣦିଵቂݏݏଶܽଶቃൌݏܿܽݐ ࣦିଵͳݏଶെܽଶ൨ൌ݄ܵ݊݅ܽݐ ࣦିଵቂݏݏଶെܽଶቃൌݏ݄ܿܽݐ ࣦିଵͳሺݏെܽሻଶܾଶ൨ൌͳܾ݁௧ܵ݊݅ݐܾ ࣦିଵݏെܽሺݏെܽሻଶܾଶ൨ൌ݁௧ܥݏݐܾ ࣦିଵݏሺݏଶܽଶሻଶ൨ൌͳʹܽݐܵ݊݅ܽݐ ࣦିଵͳሺݏଶܽଶሻଶ൨ൌͳʹܽଷሺܵ݊݅ܽݐെܽݐݏܿܽݐሻ

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174 APPLIED MATHEMATICSܖܑܜܑܐ܁܂ܐܕ܍ܚܗ܍ǣ

ሺሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻǡݐ݄ࣦ݊݁ିଵ൛݂ҧሺݏെܽሻൟൌ݁௧݂ሺݐሻൌ݁௧ࣦିଵ൛݂ҧሺݏሻൟ

ሺሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻ݂ܽ݊݀ሺͲሻൌͲǡݐ݄ࣦ݊݁ିଵ൛ݏ݂ҧሺݏሻൟൌ݀

ݐ݀ሼ݂ሺݐሻሽ

ൌ݂ᇱሺݐሻ

ሺሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻݐ݄ࣦ݊݁ିଵቊ݂ҧሺݏሻ

ݏቋൌන݂ሺݐሻݐ݀௧

ሺሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻݐ݄݊݁ݐ݂ሺݐሻൌࣦିଵ൜Ȃ݀

ݏ݀ൣ݂ҧሺݏሻ൧ൠ

ሺሻࣦቆ݂ሺݐሻ

ݐቇൌන݂ҧሺݏሻݏ݀ஶ

௦

܉۾ܑܜܚܔ܉܉ܚ܋ܖܗܑܜۻܗܐܜ܍ܛ܌

݂ҧሺݏሻ
ܨഥሺݏሻ

ܩҧሺݏሻ

ܨഥሺݏሻܩҧሺݏሻ

ܣ

ሺܽݏܾሻǡܣ

ሺܽݏଶܾݏܾሻሺݎൌͳǡʹǡǥሻ

ǡࣦିଵ൛݂ҧሺݏሻൟ

ࣦିଵ൛ݏ݂ҧሺݏሻൟൌൌ݀

ݐ݀ሼ݂ሺݐሻሽǡݎ݅ݒ݂݀݁݀ሺͲሻൌ݂ᇱሺͲሻൌڮൌ݂ିଵሺͲሻൌͲ

ࣦሼݐ݂ሺݐሻሽൌሺെͳሻ݀

݀ݏ݂ҧሺݏሻൌሺെͳሻ݂ҧሺሻሺݏሻݓ݄݄ܾ݅ܿ݊ܽܿ݁ݔ݁ݎ݁ݏݏ݁݀ܽݏ

खି൛ሺሻሺሻൟൌሺെሻሺሻ

܃܍ܛܗ۱ܖܗܞܔܗܝܖܗܑܜ܂ܐܕ܍ܚܗ܍

۷खି۴തሺሻൌሺሻǡखି۵ഥሺሻൌሺሻഥሺሻൌഥሺሻכഥሺሻ

൛खିതሺሻൟൌखିሼഥሺሻכഥሺሻሽൌනሺെሻሺሻ

ǣ
खି൬

൰ൌखିതሺሻൌሺሻ

खିቊሺሻ

ቋൌනǤሺሻ

ۺ܉ܘ܍܋܉ܔܚ܂ܛܖ܉ܕܚܗܑܜ܉ܗܖܗ܁܋܍ܘܑܔ܉۴ܖܗܑܜ܋ܖܝ munotes.in

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175Chapter 7: Inverse Laplace Transform• ܋ܑ܌ܗܑܚ܍۾۴ܖܝ܋ܜܗܑܛܖ • ۶܍ܞ܉܍܌ܑܛܑ܃ܜܑܖ܁܍ܜܘ۴ܖܗܑܜ܋ܖܝ

• ۲܋܉ܚܑെ

܍܌ܔ܉ܜ۴ܖܗܑܜ܋ܖܝሺ܃ܜܑܖܕ۷܍ܛܔܝܘ۴ܖܗܑܜ܋ܖܝሻ

܋ܑ܌ܗܑܚ܍۾ ۴ܛܖܗܑܜ܋ܖܝ

݂ሺݐሻܶ

݂ሺݐܶሻൌ݂ሺݐሻǡܶͲ

݂ҧሺݏሻൌࣦሼ݂ሺݐሻሽ

ൌන݁ି௦௧݂ሺݐሻ݀ݐஶ

ൌͳ

ͳെ݁ି௦்න݁ି௦௨݂ሺݑሻ݀ݑെെെെെݎ݂ݎ்݁݅݀ܶ

۶܍ܞ܉܍܌ܑܛܑ܃ܜܑܖ܁܍ܜܘ۴ܖܗܑܜ܋ܖܝ

ݑሺݐെܽሻǣ

ݑሺݐെܽሻൌቄͲ൏ܽ

ͳ

ǡܽǤܪሺݐെܽሻǤ

۲܋܉ܚܑെ܍܌ܔ܉ܜ۴ܖܗܑܜ܋ܖܝሺ܃ܜܑܖܕ۷܍ܛܔܝܘ۴ܖܗܑܜ܋ܖܝሻ

܃ܜܑܖ܍ܛܔܝܘܕܑ ሺ۲܋܉ܚܑ۲܉ܜܔ܍ሻ

ߜᖡሺݐെܽሻൌͳᖡΤǡᖡ

ൌͲǡݐ݄݁ݎݓݏ݅݁ ᖡ՜ͲǤ
ᖡ՜Ͳǡ
Ǥ

7.7 References

1. A Text Book of Applied Ma thematics Vol I - P. N. Wartikar and

J. N. Wartikar

2. Applied Mathematics II - P. N. Wartikar and J. N. Wartikar

3. Higher Engineering Mathematics - Dr. B. S. Grewal

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176 APPLIED MATHEMATICS

176Unit 4

8 MULTIPLE INTEGRALS

Unit St ructure

8.0 Objectives

8.1 Double Integral: Introduction and Notation

8.2 Change of the order of the integration

8.3 Double integral in polar co -ordinates

8.4 Triple integrals

8.5 Summary

8.6 Exercises

8.7 References

8.0 Objectives

After reading this chapter, you should be able to:

1. Understand double integrals & notations.

2. Solve problems based on double integrals .

3. Understand double integral in polar co -ordinates ,

4. Know the concept of triple integrals ,

8.1 Double Integral: Introduction and Notation

It is presumed that the students are familiar with “ the limit of a sum as an integ er.”

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177Chapter 8: Multiple Integralsఋ௫՜ݔߜݕ௫ୀ௫ୀනݕ݀ݔ

Thus

ఋ௫՜ݔߜݕ௫ୀ

௫ୀൌනݕ݀ݔ

Let us now consider the integration of a function of two variables over a given

area.

To make the idea clear, we shall consider a plane lamina in the xOy plane, the

surface density ߪ of which is a function of the position of the p oint P( ݔǡݕ .)Thus

surface density ߪൌ݂ሺݔǡݕ.)

To find the mass of the lamina, we shall take a small area ߜ about the point

P(ݔǡݕ.)

The mass of this elementary area is ݂ሺݔǡݕሻߜ. To find the total mass of the lamina,

we shall find out expression s such as ሺݔǡݕሻߜ ,all over the lamina, form the sum

݂ሺݔǡݕሻߜ ,and to be more accurate, ߜmust be taken a small as possible.

That is

The mass of the lamina =

ఋ՜݂ሺݔǡݕሻߜ ( … …8.1)

where summation extends all over the lamina.

Let us take ߜ in a more convenient way so that the summation in ( 8.1) can be

carried out.

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178 APPLIED MATHEMATICS

Divide the lamina by a system of straight lines parallel to the x and y axis into a

mesh of elementary rectangles. Take the rectangle with one corner at P( x,y).

Then the area of rectangle PQRS ߜA= ݔߜǤݕߜ

And the mass of the elementary rectangle = ݂ሺݔǡݕሻݕߜݔߜ .

By (8.1) the mass of the lamina M is

ൌ

ఋ୶՜

ఋ௬՜σσ݂ሺݔǡݕሻݕߜݔߜǥǥሺͺǤʹሻ

We shall evaluate the expression on the R.H.S. of the ( 8.2) in a systematic way.

Taking the sum of ݂ሺݔǡݕሻݕߜݔߜ over the strip ABCD , we have for the mass of

the elementary strip ABCD

ൌ

ఋ௬՜݂ሺݔǡݕሻݕߜݔߜୈ

ǥǥሺͺǤ͵ሻ

Where in this summation we note that x and ݔߜ are constants. We can therefore

write ( 8.3) as

ൌݔߜ

ఋ௬՜݂ሺݔǡݕሻݕߜ௬ీ

௬ఽǥǥሺͺǤͶሻ

And by introductory remarks on the limit of the sum as an integral we write ( 8.4)

as

ൌݔߜන݂ሺݔǡݕሻ݀ݕ௬మሺ௫ሻ௬భሺ௫ሻǥǥሺͺǤͷሻ

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179Chapter 8: Multiple IntegralsWhere y1(x) and y2(x) are the values of y at A and D and both depend on the position of the ordinate, that is on x.

It is to be remembered in the integral of ( 8.5) that x is to be regarded as a

constant in the integration w.r.t. y and since the limits of the integral are the

functions of x,

න݂ሺݔǡݕሻ݀ݕ௬మሺ௫ሻ

௬భሺ௫ሻ
ݔǡሺݔሻǤ

න݂ሺݔǡݕሻ݀ݕ௬మሺ௫ሻ

௬భሺ௫ሻൌሺݔሻǥǥሺͺǤሻ

So that from ( 8.5) , we can write the mass o f the elementar y strip ABCD as

[ሺݔሻ. ߜݔ ]

Next taking the mass of each strip such as ABCD parallel to the y - axis ,over the

area of the lamina, we have

ൌ

ఋ௫՜ሺݔሻߜݔǤ௫ୀ

௫ୀ

ൌනሺݔሻ݀ݔǤ௫ୀ

௫ୀǥሺͺǤሻ

Substituting for ሺݔሻ from ( 8.6) in ( 8.7), we get

ൌනቐන݂ሺݔǡݕሻ݀ݕ௬మሺ௫ሻ

௬భሺ௫ሻቑ݀ݔ௫ୀ

௫ୀǥሺͺǤͺሻ

The expression on the R.H.S . of the equatio n (8.8) is called a double integral for

obvious reason and is written in various ways as follows

නݏ

න݂ሺݔǡݕሻ݀ݕ݀ݔǤ௬మሺ௫ሻ

௬భሺ௫ሻǥሺͺǤͻܽሻ

or

නݏ

න݂ሺݔǡݕሻ݀ݔ݀ݕǤ௬మሺ௫ሻ

௬భሺ௫ሻǥሺͺǤͻܾሻ

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180 APPLIED MATHEMATICSනݔ݀න݂ሺݔǡݕሻ݀ݕǤ௬మሺ௫ሻ௬భሺ௫ሻǥሺͺǤͻܿሻ

This last way of writing the integral is more convenient, as it expresses clearly the

order in which the integration is performed i.e. we first integrate w.r.t. y considering

x as a constant and then we integrate w.r.t. x. It may also be noted that when w e

take the elementary strips parallel to the y-axis, we first integrate w.r.t. y.

If instead of taking the elementary strip parallel to the y-axis we take it parallel

to the x-axis such as EFGH shown in the adjacent figure, we have by a similar

reasoni ng to the above

ൌන݀ݕ

න݂ሺݔǡݕሻݔ݀௫మሺ௬ሻ

௫భሺ௬ሻǥሺͺǤͳͲሻ

In which we have to first integrate w.r.t. x and then w.r.t. y, thus changing the order

of the integration. Both the integrals ( 8.9) and ( 8.10) represent the mass of the

lamina and so are equal. The total area of the lamina is known as the region of

integration.

The function f(x,y) was considered as the surface density of the lamina, just for the

sake of understanding clearly the idea of doubl e integral. However f(x,y) may be

any function of the position of a point in the loop -area, and the double integral of

this function over the area of the loop is given by ( 8.9) or ( 8.10) that is

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181Chapter 8: Multiple Integrals

… (8.11)

8.2 Change of the order of integration ; Evaluation of Double

integra ls

The method of evaluating the double integrals ( 8.11) is actually clear from the

theory developed in the previous section. We note that in the evaluation of the

double integrals, we integrate first w.r.t. one variable ( y or x depending upon the

limits, and the elementary strip) and considering the other variable as constant and

then integrate with respect to the remaining variable.

If the limit s of integration are the constants such as in the region of integration

being a rectangle, then the change in the order of integration does not require

change of the limits of integration.

Thus from the adjacent figure, we see that

නݔ݀

න݂ሺݔǡݕሻ݀ݕௗ

ൌන݀ݕௗ

න݂ሺݔǡݕሻݔ݀

ǥሺͺǤͳʹሻ

But if the limits be the variable as i n the general case taken in section 8.1 then in

changing the order of integration a corresponding change is to be made in the limits

of integr ation as seen from (8.11). Sometimes in changing the order of integration

we are required to split up the region of integration and the new integral is

expressed as a sum of a number of double integrals. The examples solved below

make this ideas clear. The change of the order of integration is sometimes

convenient in the evaluation of the double integrals. This is also illustrated in

problems solved below. In changing the order of integration, it is convenient to

නݔ݀න݂ሺݔǡݕሻ݀ݕ௬మሺ௫ሻ௬భሺ௫ሻන݀ݕන݂ሺݔǡݕሻݔ݀௫మሺ௬ሻ௫భሺ௬ሻ munotes.in

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182 APPLIED MATHEMATICS

draw rough sketch of the region o f integration, which will help to fix up the new

limits of integration.

Example 1. Evaluate ሺݔଶെݕଶሻܣ݀ over the area of the triangle whose vertices

are the points (0,1),(1,1) and (1,2).

The equations of the sides of the triangle whose vertices are at A(0,1), B(1,1),

C(1,2) are x = 1, y = 1 and x = y -1 … (i) as shown in the figure 6.

If we take an elementary strip parallel to the x-axis, we will be integrating the given

function with respect to x. The ends of this strip are bounded by the lines x = y - 1

and x = 1, so that these are the limits of integration with respect to x. Next we

integrate w.r.t. y from y = 1 to y = 2, which then covers the whole area of the

triangle ABC.

Thus if ൌሺݔଶെݕଶሻܣ݀ taken over the area of the triangle ABC

Then,

ൌන݀ݕଶ

ଵනሺݔଶെݕଶሻ݀ݔଵ

௬ିଵǤǤǤǤሺሻ

To evaluate th e first integral, we regard y as a constant,

ൌන݀ݕଶ

ଵቈݔଷ

͵െݕଶݔ

௬ିଵଵ

ൌනቊͳ

͵െݕଶെሺݕെͳሻଷ

͵ݕଶሺݕെͳሻቋ݀ݕଶ

ଵ

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183Chapter 8: Multiple Integralsൌනቊͳ͵െʹݕଶെሺݕെͳሻଷ͵ݕଷቋ݀ݕଶଵ

ൌቈݕ

͵െʹݕଷ

͵െሺݕെͳሻସ

ͳʹݕସ

Ͷ

ଵଶ

ൌʹ

͵െͳ

͵െͳ

ͳʹͶെͳ

͵ʹ

͵െͳ

Ͷ൨

ൌെʹ

͵

It will be interesting to try the above example by taking strips parallel to the y -axis,

which is left to the students as an exercise leading to the same result as above.

Example 2 . Evaluate

න݀ݕ

නݔݕሺݔܽሻ

ሺݔെܽሻଶ݀ݔିඥమି௬మ

In the integral as it stands, the integration is first w.r.t. x and this integration, as is

clear is complicated. As integration w.r.t. y is simple, we therefore change the

order of integration, for which sake we find out the region of integration for the

given problem.

In the given Interval where the integration is first w.r.t. x, the elementary strips are

parallel to the x-axis and these strips extend from x = 0 (i.e. the y – axis) to

x = ܽെඥܽଶെݕଶ i.e. to the boundary of the circle ሺݔെܽሻଶݕଶൌܽଶ.

Moreover as x = a minus ඥܽଶെݕଶ, it extends upto the side (i) of the circle and

not upto (ii) for which x = a plus ඥܽଶെݕଶ.

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184 APPLIED MATHEMATICS

An elementary strip such as this is shown in the figure 7 by AB. Next we int egrate

w.r.t. y from y = 0 to y = a and so the strips such as AB, bounded on one side by

the y-axis and on the other by the circumference of the circle are taken from y = 0

to y = a . Thus the region of integration is the shaded part in the figure.

If we ch ange the order of integration, integrating first w.r.t. y then the elementary

strip is parallel to the y -axis, such as BC in the figure which extends from

circumference of the circle ሺݔെܽሻଶݕଶൌܽଶ i.e. ݕൌξʹܽݔെݔଶ to the

line y = a . These are therefore the limits of integration w.r.t. y. To have same region

of integration as in the given integral. We must take such strips from x = 0 to

x = a , which are the limits of integration w.r.t. x. Thus changing the order of

integration, the g iven integral say I, can be written as

ൌන݀ݔ

නݔݕሺݔܽሻ

ሺݔെܽሻଶ݀ݕ

ξଶ௫ି௫మ

Integrating w.r.t. y considering x as constant, we have

ൌන݀ݔ

ݔሺݔܽሻ

ሺݔെܽሻଶቈݕଶ

ʹ

ξଶ௫ି௫మ

ൌͳ

ʹන݀ݔ

ݔሺݔܽሻ

ሺݔെܽሻଶሾܽଶെʹܽݔݔଶሿ

ൌͳ

ʹනݔሺݔܽሻ݀ݔ

This can be integrated by parts, with log ( x + a) as a part to be differentiated

which gives

ൌܽଶ

ߠሾʹܽͳሿ

Example 3. Change the order of Integration in

නݏଶ

න݂ሺݔǡݕሻ݀ݔ݀ݕξଶ௫

ξଶ௫ି௫మ

The order of integration in the given integral is first w.r.t. y and then w.r.t. x

The elementary strips here are parallel to the y-axis (such as A B) and extend from

y = ξʹܽݔെݔଶ, [ i.e. the circle ݔଶݕଶെʹܽݔൌ 0. with centre at ( a,0) and radius

a] to y = ξʹܽݔ [i.e. the parabola ݕଶൌʹܽݔ] and such strips are taken from x = 0 to munotes.in

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185Chapter 8: Multiple Integrals

x = 2a. The shaded area between the parabola and the circle is therefore the region

of integration.

In changing the order of integration, we integrate first w.r.t. x, with elementary

strips parallel to the x-axis, such as CD. In covering the same region as above, the

ends of these strips extend to different curves. We therefore divide the region by

the line y = a into three parts (I),(II),(III) as shown in the figure.

For the region (I), the strip extend from the parabola ݕଶൌʹݔܽ i.e. ݔൌ௬మ

ଶ to the

straight line

x = 2a, so these are the limits of integration w.r.t. x. Such strips are to be taken

from y = a to y = 2a, to cover the region (I) completely. So the part of the integral

in this region I 1 is

ଵൌන݀ݕଶ

න݂ሺݔǡݕሻ݀ݔଶ

௬మ

ଶǤǤǤǤሺሻ

From the region (II), the strips extend from the parabola ݕଶൌʹݔܽ i.e. ݔൌ௬మ

ଶ to

the circle

ݔଶݕଶെʹݔܽ =0 i.e. x = a േඥܽଶെݕଶ in which we take the negative sign

with the radical as i s obvious from the figure, so the limits of integration w.r.t. x

are ݔൌ௬మ

ଶ to x = a െ ඥܽଶെݕଶ and such strips are taken from y = 0 to y = a, to

cover this region completely. The contribution to the integral from this region I 2

is therefore

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186 APPLIED MATHEMATICS

ଶൌනݕ݀න݂ሺݔǡݕሻ݀ݔୟିඥమି௬మ௬మ

ଶǤǤǤǤሺሻ

For the region (III), the strips extend from the circle ݔଶݕଶെʹܽݔ =0

[i.e. x = a േඥܽଶെݕଶ; in this we have to take the positive sign with the radical as

is clear from the figure] to the line x=2a, so that the limits of integration w.r.t x

are x = a ඥܽଶെݕଶ to x=2a; and such strips are to be taken from y = 0 to y = a,

which covers in the integration the region (III) Denoting this part of integra l by I 3,

we have

ଷൌනݕ݀

න݂ሺݔǡݕሻ݀ݔଶୟ

ୟାඥమି௬మǤǤǤǤሺሻ

Thus if we change the order of integration, we have to divide the region of

integration, and the given integral is equal to I 1 + I2 + I3 or from (i), (ii), (iii)

න݀ݔଶ

න݂ሺݔǡݕሻݕ݀ξଶ௫

ξଶ௫ି௫మൌනݕ݀ଶ

න݂ሺݔǡݕሻ݀ݔଶ

௬మ

ଶනݕ݀

න݂ሺݔǡݕሻ݀ݔୟିඥమି௬మ

௬మ

ଶ

නݕ݀

න݂ሺݔǡݕሻ݀ݔଶ

ୟାඥమି௬మ

This example illustrates that in changing the order of integration sometimes not

only limits ar e to be changed, but it is necessary to split up the region of

integration.

Example 4. Change the order of integration for the integral

නݏ

නݔݕ݀ݔݕ݀ଶି௫

௫మ

and evaluate the same with reversed order of integration.

The given integral is

න݀ݔ

නݕݕ݀ଶି௫

௫మ

ǤǤǤǤሺ݅ሻ

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187Chapter 8: Multiple Integrals

In this the integration is first w.r.t. y with strips such as AB, parallel to the y -axis

with extremities lying on the parabola ݕൌ௫మ

and the straight line y = 2 a – x. These

strips are taken from x = 0 to x = a, that gives the region of integration, the

curvilinear triangle OPQ, shaded i n the figure 9.

In changing the order of integration, the integration is to be taken first w.r.t. x with

elementary strip parallel to x axis, such as CD, and that needs dividing the region

of integration by the line y = a, i.e. the line PR, into two parts the triangle PQR and

the curvilinear triangle OPR denoted in the figure by (I) and (II) respectively.

For the region (I), the limits of integration w.r.t. x are x = 0 to x = 2a – y and the

limits of the next integration w.r.t. y are y = a to y = 2a, so the contribution to the

given integral from region (I) is

ଵൌන݀ݕଶ

නݔݕ݀ݔଶୟି௬

ǤǤǤǤሺሻ

For the region (II), the limits of integrat ion w.r.t . x are x = 0 to x = ඥܽݕ and

those w.r.t. y are y = 0 to y = a , so the contribution to the given integral from the

region (II) is

ଶൌන݀ݕ

නݔݕ݀ݔξ௬

ǤǤǤǤሺሻ

Hence, reversing the order of integration, from (i), (ii) and (iii),

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188 APPLIED MATHEMATICSනݔ݀නݔݕ݀ݕൌଶି௫௫మ

න݀ݕଶනݔݕݔ݀ଶୟି௬

න݀ݕ

නݔݕݔ݀ξ௬

ǤǤǤǤሺሻ

Now, with usual method of evaluating the double integral

න݀ݕଶ

නݔݕݔ݀ଶୟି௬

ൌන݀ݕǤݕଶ

ቈݔଶ

ʹ

ଶି௬

ൌͳ

ʹනݕሺʹܽെݕሻଶ݀ݕଶ

ൌͷ

ʹͶܽସǤǤǤǤሺሻ

and

න݀ݕ

නݔݕݔ݀ξ௬

ൌන݀ݕǤݕ

ቈݔଶ

ʹ

ξ௬

ൌͳ

ʹනܽݕଶ݀ݕ

ൌͳ

ܽସǤǤǤǤሺሻ

From (iv), (v) and (vi),

නݔ݀

නݔݕݔ݀ଶୟି௫

௫మ

ൌͷ

ʹͶܽସͳ

ܽସൌ͵

ͺܽସ

8.3 Double integral in polar co -ordinates

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189Chapter 8: Multiple Integrals

In case we use polar co -ordinates, divide the region of integration by curves

r = const. (which are circles) and Ș FRQVWZKLFKDUHVWUDLJKW - lines)

This gives a mesh of the form shown, where the elementary area is ݎߜǤݎ ߠߜ

Thus if f (r,ߠ )be a function of position, we have over the wedge PQ, the sum as

ఋ՜ߠߜ݂ሺݎǡߠሻǤݎݎߜொ

ൌߠߜන݂ሺݎǡߠሻݎ݀ݎమሺఏሻ

భሺఏሻǥǥሺͺǤͳ͵ሻ

Where ݎଵሺߠሻ and ݎଶሺߠሻ are equations of the two parts of curves where Ʌ is kept

constant, while integrating w.r.t . r. Finally summing for all the wedges between

Ʌൌן and ɅൌȾ, we get

ఋ՜ߜɅ݂ሺݎǡߠሻݎ݀ݎమሺఏሻ

భሺఏሻఉ

ఈ

ൌනߠ݀ఉ

ఈන݂ሺݎǡߠሻݎ݀ݎమሺఏሻ

భሺఏሻǥǥሺͺǤͳͶሻ

The order of integration may be changed with appropriate changes in the limits.

Example 1. Evaluate

ඵݎ݀ݎ݀Ʌ

ξܽଶݎଶ

over one loop of the lemniscate ݎଶൌܽଶcos 2 Ʌ

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190 APPLIED MATHEMATICSൌන݀Ʌగସିగ

ସනݎ݀ݎξܽଶݎଶξ௦ଶ

ൌන݀Ʌగ

ସ

ିగ

ସቂඥܽଶݎଶቃ

ξ௦ଶ

ൌන݀Ʌగ

ସ

ିగ

ସ൛ܽξͳݏܿʹɅหെܽൟ ൌܽනൣξʹȁ
Ʌെͳ൧݀Ʌగ

ସ

ିగ

ସ ൌܽൣξʹȁɅെɅ൧ିగ

ସగ

ସ

ൌܽቂʹെߨ

ʹቃൌʹܽቂͳെߨ

Ͷቃ Example 2 . Evaluate

න݀

නݕ݀

ඥܽଶെݔଶെݕଶξమି௫మ

ξ௫ି௫మǡ by changing to polar coordinates.

Here the elementary strips, such as AB are parallel to the y axis and extend from

y = ξܽݔെݔଶ

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191Chapter 8: Multiple Integrals

[which is the circle ݔଶݕଶെܽݔ =0,

with centre at ቀ

ଶǡͲቁ and radius

ଶ ] to

y = ξܽଶെݔଶ [ i.e. the circle ݔଶݕଶൌܽଶ , with center at the origin and ra dius

a.] such strips are taken from ݔ =0 to ݔ =a, and so the area between the two circles,

is the region of integration.

To change the given integral to polar coordinates, we substitute ݔ =ݎ cosɅ, ݕ =ݎ

sinɅ, and ݀ݔ݀ݕ by its equivalent elementary area in polar coordinates ݎ݀ݎ݀Ʌ. The

equations of the circle in polar coordinates are ݎ =ܽ cosɅ and ݎ =ܽ and the ends

of the elementary wedge, such as CD along the radius vector lies on these circles

and so give the limits of integration w.r.t. ݎ and to cover the same region of

integration as in given integral. 0 varies from 0 to గ

ଶ.

Thus the transformed integral I is

ܫൌන݀Ʌగ

ଶ

නݎ

ξܽଶെݎଶୟ

ୡ୭ୱ݀ݎ

ൌන݀Ʌగ

ଶ

ቂെඥܽଶെݎଶቃ

ୡ୭ୱ

ൌනܽ݊݅ݏɅ݀Ʌൌܽగ

ଶ

8.4 Triple integrals :-

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192 APPLIED MATHEMATICS

Let ݂ሺݔǡݕǡݖሻ be any function of the position of a point ሺݔǡݕǡݖሻ in space [say the

density of the body]. Divide the body by a system of planes into small rectangular

blocks. The element of volume at ሺݔǡݕǡݖሻ is then d ݀ݕ݀ݖ.

The mass of the elementary cuboid at ൌ݂ሺݔǡݕǡݖሻ. ݀݀ݕ݀ݖ

Then

ௗ௭՜݂ሺݔǡݕǡݖሻ݀݀ݕ݀ݖ௭మ

௭భ

ൌ݀݀ݕන݂ሺݔǡݕǡݖሻ݀ݖ௭మሺ௫Ǥ௬ሻ

௭భሺ௫ǡ௬ሻǥǥሺͺǤͳͷሻ

where ݖଵሺݔǡݕሻ and ݖଶሺݔǤݕሻ are the equations of the lower and upper surfaces of

the bounding volume. The result (8.15 ) gives the mass of the el ementary column

on݀݀ݕ in the xOy plane as the base. In the integral (8.15 ), x, y are constants .

We now have to sum for all the columns standing on the area in the xOy plane

vertically below the surface. Taking first all the columns in a slice parallel to the

y-z plane which means integration w.r.t. y while keeping x constant, we get

නݏ௬మሺ௫ሻ

௬భሺ௫ሻቐන݂ሺݔǡݕǡݖሻ݀ݖ௭మሺ௫Ǥ௬ሻ

௭భሺ௫ǡ௬ሻቑ݀ݕ݀ݔǥǥሺͺǤͳሻ

and finally summing for all the slices from x = a to x = b , we have

ǥሺͺǤͳሻ

The evaluation of a space or volume integral involves three successive integration

and so is called a triple integral. The order of integration may be changed with

appropriate changes in the limits.

In polar co -ordinates the volume of an elementary cuboid

dv = r2 sinɅݎ݀ߠ݀݀

and the integral (8.17 ) takes the form ම݂ሺݎǡߠǡሻݎଶߠݎ݀ߠ݀݀ න݂ሺݔǡݕǡݖሻݒ݀ൌන݀ݔන݀ݕ௬మሺ௫ሻ௬భሺ௫ሻන݂ሺݔǡݕǡݖሻ݀ݖ௭మሺ௫Ǥ௬ሻ௭భሺ௫ǡ௬ሻ௩ munotes.in

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193Chapter 8: Multiple Integrals

And in cylindrical co -ordinates , the elementary volume is

dv =ߩ݀ߩ݀ݖ݀

and the integral (8.17 ) takes t he from

ම݂ሺߩǡǡݖሻߩ݀ߩ݀ݖ݀

with appropriate limits.

Example 1 . Show that the volume bounded by the cylinder y2 = z , y = x2

And the planes z = 0 , x + y +z = 2 is equal to

නݏଵ

නݏξ௫

௫మනݔ݀ݕ݀ଶି௫ି௬

ݖ݀

and evaluate it.

The cylinder stands on the area common to the parabolas with generators parallel

to the z -axis, and the volume required is the portion of this cylinder cut -off by the

planes z = 0 and x+y+z =2 i.e. z = 2 – x - y

Integrating first w.r.t. z we obtain th e volume of the elementary column,

on dx dy as the base, where limits for z are z = 0 to z = 2- x- y.

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194 APPLIED MATHEMATICS

Thus the volume of elementary column on the dxdy as the base is

ݕ݀ݔ݀නݖ݀ଶି௫ି௬

ǤǤǤǤሺሻ

Taking a slice parallel to the yOx plane , of all such columns, leads on the

integration w.r.t. y from y = x2 to y = ξݔ (ref. fig.14 ), we thus have the volume of

an elementary slice parallel to the yOz plane as

නݕ݀ξ௫

௫మනݖ݀ଶି௫ି௬

ǥǥሺሻ

Summing the volumes of such slices, bounded by the curves y = x2, y = ξݔ ,from

x = 0 to x = 1, gives the total volume of the cylinder in question and is

නݔ݀ଵ

නݕ݀ξ௫

௫మනݖ݀ଶି௫ି௬

ǥǥሺሻ

which is the same as the given integral. To evaluate it we use the same principles

as used in the evaluation of a double integral. Thus

නݔ݀ଵ

නݕ݀ξ௫

௫మනݖ݀ଶି௫ି௬

ൌනݔ݀ଵ

නݕ݀ξ௫

௫మሾݖሿଶି௫ି௬

ൌනݔ݀ଵ

නሺʹെݔെݕሻݕ݀ξ௫

௫మ

ൌනݔ݀ଵ

ቈሺʹെݔሻݕെݕଶ

ʹ

௫మξ௫

ൌනቊሺʹെݔሻξݔെݔ

ʹെሺʹെݔሻݔଶݔସ

ʹቋݔ݀ଵ

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## Page 195

195Chapter 8: Multiple Integrals

ൌ

െ

െ

െ

ൌ

8.5 Summary

The eight chapter of this book introduces the students with concepts of double

integral , evaluation of d ouble integrals : change of the order of the integration and

double integral i n polar co -ordinates with notations, which is important in

understanding, implementation in application areas of integrals. Triple integrals is

also explained with solved problems and illustrations.

8.6 Exercises

Evaluate the following Integrals

ͳǤඵݕ݀ݔ݀ݕ

ሻ ݕൌݔଶݔݕൌʹ

ii) the area bounded by x = 0, y = x2 and x + y = 2 in the first quadrant.

ʹǤඵݕݔሺݔݕሻݕ݀ݔ݀ ݔଶൌݕ݊ܽ݀ݕଶ

ൌȂݔǤ

͵Ǥሻඵሺݔଶݕଶሻݕ݀ݔ݀

ሻඵݔଶݕݕ݀ݔ݀

ݔଶ

ܽଶݕଶ

ܾଶൌͳǤ

ͶǤඵݕ݀ݔ݀

ݔସݕଶݔͳݕݔଶ

Change the order of integrals and evaluate

ͷǤනݏଶ

නݕ݀ݔ݀ݕݔ௫మ

ସ

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## Page 196

196 APPLIED MATHEMATICSǤනݏଵනݕ݀ݔ݀ݕݔξ௬௬

Ǥනݏ

නሺݔଶݕଶሻݕ݀ݔ݀ට௫

௫

ͺǤනݔ݀ଵ

නݔ

ඥݔଶݕଶξଶି௫మ

௫െݕ݀

Answers ͳǤሻ͵ͷሻͳͷ ʹǤͳͳͶͶʹͲ ͵ǤሻܾܽߨͳሺܽଶܾଶሻሻܽସܾଶʹͶ ͶǤߨͶ ͷǤͳ͵ ǤͳʹͶ Ǥସቂమଵହቃ ͺǤቂͳെଵξଶቃ

Show the region of integration and change the order of integration

ͻǤනݏ

ିන݂ሺݔǡݕሻݔ݀ݕ݀௬మ

ͳͲǤනݏଵ

ିଶන݂ሺݔǡݕሻݕ݀ݔ݀ଶି௫

௫మ

ͳͳǤනݕ݀

න݂ሺݔǡݕሻݔ݀௬ା

ඥమି௬మͳʹǤනݔ݀

න݂ሺݔǡݕሻݕ݀ݔ݀మ

௫

Evaluate

ͳ͵Ǥඵݎ݀ݎ݀Ʌݕ݀ݔ݀
ݎൌͳ
Ʌ

ͳͶǤඵݎଷ݀ݎ݀Ʌݕ݀ݔ݀
ݎ

ൌʹɅݎൌͶɅ

ͳͷǤඵݎସܿݏଷɅ݀ݎ݀Ʌ
ݎൌʹ
Ʌ

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197Chapter 8: Multiple Integrals

Express the following integrals in polar coordinates, showing the region of

integration and evaluate.

ͳǤනݏ

නݕଶඥݔଶݕଶ݀ݔ݀ݕඥమି௬మ

ͳǤනݏସ

නݔଶെݕଶ

ݔଶݕଶ݀ݔ݀ݕ௬

௬మ

ସ

ͳͺǤනݏଶ

න݀ݔ݀ݕ

ሺݔଶݕଶሻଶଵାξଶ௫ି௫మ

ଵିξଶ௫ି௫మ

ͳͻǤනݏ

නݔ݀ݔ݀ݕ

ݔଶݕଶ

௬

Change to polar coordinates and evaluate

ʹͲǤඵݔଶെݕଶ

ሺݔଶݕଶሻଷ

ଶ݀ݔ݀ݕ
ݔଶݕଶൌʹܽݔ

in the first quadrant.

ʹͳǤඵݕଶ݀ݔ݀ݕ
ݔଶݕଶെܽݔൌͲ

ݔଶݕଶെʹܽݔൌͲǤ

ʹʹǤ ඵ݀ݔ݀ݕ

ሺͳݔଶݕଶሻଷ

ሺݔଶݕଶሻଶൌݔଶെݕଶ

Answers

ͳ͵Ǥଷగ

ଶ ͳͶǤସହగ

ଶ ͳͷǤగ

ସܽହ ͳǤగఱ

ଶ ͳǤͺܽଶቀగ

ଶെହ

ଷቁ

ͳͺǤߨ ͳͻǤగ

ସ ʹͲǤଶ

ଷ ʹͳǤଵହగ

ସܽସ ʹʹǤగିଶ

ସ

ʹ͵Ǥම݀ݔ݀ݕݖ݀

ሺݔݕݖͳሻଶൌͳ

ʹ൬ʹെͷ

ͺ൰ǡ ܾ݃݊݅݁ݐܽ݇݊݁

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198 APPLIED MATHEMATICS

8.7 References

1. A Text B ook of Applied Mathematics Vol I - P. N. Wartikar and J. N. Wartikar

2. Applied Mathematics II - P. N. Wartikar and J. N. Wartikar

3. Higher Engineering Mathematics - Dr. B. S. Grewal

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## Page 199

199Chapter 9: Applications of Integration

199Unit 4

9 APPLICATIONS OF INTEGRATION

Unit St ructure

9.0 Objectives

9.1 Introduction

9.2 Areas

9.3 Volumes of solids

9.4 Summary

9.5 Exercises

9.6 References

9.0 Objectives

After reading this chapter, you should be able to:

1. Know the concept Areas & volume of solids .

2. Formulae of these in terms of integrals .

3. Single & multiple integrals & their use in examples

4. Solve problems based on area & volume integrals

9.1 Introduction

In this chapter we shall study the applications of integral calculus to the problems

involving areas, volumes and surface of solids , centre of gravity, hydrostatic centre

of pressure, moment of inertia, mean and root mean square values etc. Formulae

for these in terms of integrals, single a nd multiple are developed and their use in

the example on these topics is illustrated.

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200 APPLIED MATHEMATICS

9.2 Areas

The area A, included by the curve y = f(x) the x-axis and the ordinates x = a

and x = b is given by

Similarly the area A', included by the curve y = f(x) , the y-axis, y = c and y = d is

Ԣൌݔݕ݀ௗ

ǥǥሺͻǤʹሻ

In case of a loop as shown in figure 2, the area of an elementary rectangle at

P (x , y) is dxdy and so the area of the loop is given by

ൌනݏන݀ݔݕ݀௬మሺ௫ሻ௬భሺ௫ሻ ….. (9.1)

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201Chapter 9: Applications of Integration

If the equation of curve is given in polar coordinates by r = f (Ʌ), then as ߜɅ՜Ͳ,

the area of the elementary triangle OPQ is ଵ

ଶݎଶߜɅ [ for dropping PR

perpendicular to OQ, PQ = ݎߜɅ can be taken as the base of the ο of which

the height is r, and its area is ଵ

ଶݎଶߜɅ; as ߜɅ՜Ͳ,

ο՜ο] and so

… ( 9.4)

Example 1. Trace the curve ݕଶܽସൌݔହሺʹܽെݔሻand show that its area is equal

to ହగ

ସܽଶ.

The tracing done by the methods of curve -tracing gives the curve as a

symmetrical loop on the x – axis between x = 0 and x = 2a.

݀ݕݔଶ

gives the area of the upper half of the loop and so the area A of the loop

is

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202 APPLIED MATHEMATICS

ൌʹݔ݀ݕଶǥሺ݅ሻ

From the equation of the curve ݕൌ௫ఱ

మሺଶି௫ሻభ

మ

మ, subst ituting this in (i),

ܣൌʹනݔହ

ଶሺʹܽെݔሻଵ

ଶ

ܽଶݔ݀ଶ

ǥǥሺ݅݅ሻ

For integration, we put ݔൌʹܽଶɅ, so that ݔ݀ൌͶܽݏ݅݊Ʌ
Ʌ݀Ʌ and

when x = 0, ɅൌͲ and when ݔൌʹܽ ,Ʌൌగ

ଶ.

ܣൌͶܽଶǤනɅ
ଶɅ݀Ʌగ

ଶ

ǥǥሺ݅݅݅ሻ

By the reduction formulae, we can write the value of this integral, so

ܣൌͶܽଶǤሺͷǤ͵ǤͳሻǤሺͳሻ

ͺǤǤͶǤʹߨ

ʹൌͷߨ

Ͷܽଶ

Example 2. Trace the curve ܽଶݔଶൌݕଷሺʹܽെݕሻ and show that its area is equal

to ߨܽଶ.

Here the loop is on the y-axis, and so we use the formula (9.2) for the area.

Thus the area of the loop is

ൌʹනݕ݀ݔଶ

ൌʹනݕଷ

ଶሺʹܽെݕሻଵ

ଶ

ܽଶ

ݕ݀

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## Page 203

203Chapter 9: Applications of IntegrationݕൌʹܽଶɅǡൌ͵ʹܽଶනସɅ
ଶɅ݀Ʌగ

ଶ

ൌ͵ʹܽଶǤሺ͵ǤͳሻǤሺͳሻ

ǤͶǤʹߨ

ʹൌߨܽଶǤ

Example 3 . Prove that the area of the loop of the curve

ݔହݕହൌͷܽݔଶݕଶହ

ଶܽଶǤ

From the equation of the curve, it is clear that the loop does not lie on the x or y

axis and so is inclined to them. In case of inclined loop, we change the equation

to polar co -ordinates with ݔൌݎ
Ʌ , ݕൌݎɅ.

The equation of the curve in polar coo rdinates is

ݎൌହୱ୧୬మୡ୭ୱమ

ୱ୧୬ఱାୡ୭ୱఱǥሺሻ

r is zero when ɅൌͲగ

ଶ , so the loop of the curve lies between these two

limits. Using formula ( 9.4), the area A of the loop is

ൌͳ

ʹනݎଶ݀Ʌగ

ଶ

ǥǥሺ݅݅ሻ

Substituting for r from (i) in (ii),

ൌʹͷܽଶ

ʹනସɅ
ସɅ

ሺହɅ
ହɅሻଶ݀Ʌగ

ଶ

Dividing the numerator and denominator by
ଵɅ ,

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204 APPLIED MATHEMATICS

ൌʹͷܽଶ

ʹන
ଶɅǤସɅሺͳହɅሻଶǤ݀Ʌగଶ

Put z = 1+ ହɅ, dz = 5
ଶɅ ସɅ ݀Ʌ. When ɅൌͲ, z = 1 and

when Ʌൌగ

ଶ, z = λ ,

ൌͷܽଶ

ʹනݖ݀

ݖଶൌͷܽଶ

ʹെͳ

ݖ൨

ଵஶ

ൌͷܽଶ

ʹஶ

ଵǤ

Example 4. In the cycloid ݔൌܽሺɅɅሻǡൌܽሺͳെ
Ʌሻ find the area

between its base and portion of the curve from cusp to cusp.

The sketch of the curve is shown in the figure with cusps at P and Q and the

base PQ.

The area required is that of the curvilinear figure POQ.

ൌʹනݔ݀ݕ

ൌʹනݔ݀ݕ

݀Ʌ݀Ʌగ

ǥǥሺ݅ሻ

From the equation of the cycloid ݔൌܽሺɅɅሻǡௗ௬

ௗൌܽɅsubstituting

in (i)

ൌʹනܽଶሺɅɅሻɅ݀Ʌగ

ൌʹܽଶනሾɅɅଶɅሿగ

݀Ʌǥǥሺ݅݅ሻ

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205Chapter 9: Applications of IntegrationනɅɅగ݀ɅൌሾെɅ
ɅɅሿగൌߨǥǥሺ݅݅݅ሻ

නଶɅగ

݀ɅൌʹනଶɅ݀Ʌగ

ଶ

ൌʹǤߨ

Ͷ

ൌߨ

ʹǥǥሺݒ݅ሻ

Substituting these values of the integrals in (ii)

ൌʹܽଶቂߨߨ

ʹቃൌ͵ߨܽଶǤ

Example 5 . Find the area between ݕଶൌ௫య

ି௫ and its asymptote. The nature of the

curve is shown in the figure with asymptote x = a [Asymptote is the line to which

the curve approaches]

The required area A is :

ൌʹනݕ݀ݔ

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206 APPLIED MATHEMATICSൌʹනݔଷଶሺܽെݔሻଵ

ଶ݀ݔ

with ݔൌܽଶɅ,

ൌͶܽଶනସɅ݀గ

ଶ

Ʌ

ൌͶܽଶǤ͵

ͶǤͳ

ʹߨ

ʹൌ͵

ͶߨܽଶǤ

Example 6 . Find the area of the loop of the curve

ݎൌܽ
͵Ʌܾ͵ɅǤ

Let ןൌିଵ

, so that ൌξܽଶܾଶ sinߙ,

ൌξܽଶܾଶ cosߙ so that the equation of the curve can be written as

ൌξܽଶܾଶሺsinߙ cos ͵Ʌ + cos ߙ sin ͵Ʌ).

or ൌξܽଶܾଶሺ͵Ʌߙሻǥǥሺሻ

To find the position of the loop, we have when r = 0, ͵Ʌߙ =݊ߨ

(where n is an integer).

Taking consecutive values of n as 0 and 1,

one of the loop lie between Ʌൌെఈ

ଷ and Ʌൌగିఈ

ଷǤ

ൌൌͳ

ʹනݎଶ݀Ʌగିఈ

ଷ

ିఈ

ଷǥǥሺሻ

Substituting for r from (i)

ൌሺܽଶܾଶሻ

ʹනଶሺ͵Ʌߙሻగିఈ

ଷ

ିఈ

ଷ݀Ʌ

In this put ൌ͵Ʌߙ ;so that

ൌሺܽଶܾଶሻ

නଶగ

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207Chapter 9: Applications of Integration

ൌሺܽଶܾଶሻ͵නଶగଶ݀

ൌሺܽଶܾଶሻ

͵Ǥͳ

ʹǤߨ

ʹൌߨ

ͳʹሺܽଶܾଶሻǤ

Example 7. Find by double integration the area included between the curves

ݕൌ͵ݔଶെݔെ͵

and ݕൌെʹݔଶͶݔ.

The abscissa of the points of intersection of the two parabolas, a rough sketch of

which is given in the adjacent diagram are given by

͵ݔଶെݔെ͵ൌെʹݔଶͶݔ

ǤǤݔଶെݔെʹൌͲ

ݔൌെͳǡʹǤ

Taking the elementary strip parallel to the y-axis, such as AB, bounded by the

two parabolas we integrate first w.r.t. y, and then integrating w.r.t. x from x = -1

to x = 2, gives for the area A required .

ൌනݏଶ

ିଵන݀ݔ݀ݕିଶ௫మାସ௫ା

ଷ௫మି௫ିଷ

ൌනݏଶ

ିଵ݀ݔሾݕሿଷ௫మି௫ିଷିଶ௫మାସ௫ାൌͷනሺെݔଶݔʹሻ݀ݔଶ

ିଵ

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208 APPLIED MATHEMATICSൌͷቈെݔଷ͵ݔଶʹʹݔିଵଶ ൌͷ൜െͺ͵ͶʹͶെ൬ͳ͵ͳʹെʹ൰ൠ

ൌͷ

ʹǤ Example 8. Find by double integration the area included between the curve ݎൌܽሺ
Ʌߠݏܿሻ and its asympotote r = a secɅ.

By transforming the equations to cartesian, coordinates, the curves are easily traced, as shown in figure. Taking a wedge such as AB, its extremities lie on the curve r = a secߠ and ݎൌܽሺ
Ʌ
Ʌሻand to get the area between the asymptote and the curve, Ʌvaries from -గ

ଶݐగ

ଶ ; or by symmetry the area A required is :

ൌʹනݏగ

ଶ

න ݎ݀ݎ݀ߠሺୱୣୡఏା௦ఏሻ

௦ఏ

ൌනݏሾݎଶሿୱୣୡሺୱୣୡఏା௦ఏሻగ

ଶ

݀ߠ ൌܽଶනݏగ

ଶ

ሼሺ
Ʌ
Ʌሻଶെ
ଶɅሽ݀ߠൌܽଶනሾʹ
ଶɅሿగ

ଶ

݀ߠ

ൌܽଶቂߨߨ

Ͷቃൌͷߨ

Ͷܽଶ

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209Chapter 9: Applications of Integration

9.3 VOLUMES OF SOLIDS

Let z = f(x,y) be the equation of the surface, of which the orthogonal projection in

the xOy plane is the contour PQR, whose equation is f(x,y) = 0. The volume of an

elementary parallellopiped on dxdy bounded by the surface, z= f(x,y) and sides

parallel to the z axis is

zdxdy = f(x,y) dx dy.

The summation of all such terms over the area of closed curve PQR gives the

volume of the solid cylinder bounded by the given surface and the plane xOy with

generators parallel to the z- axis as

ൌඵ݂ሺݔǡݕሻ݀ݔ݀ݕǥǥሺͻǤሻ

to be taken on the area of the contour PQR.

To express the volume of a solid as a triple integral, we note that the volume

of an elementary cuboid is dx dy dz ; and so the volume of the solid is given by

ൌම݀ݔ݀ݕݖ݀ǥǥሺͻǤሻ

Where the limits of integra tion w.r.t. z (if we integrate first w.r.t. z) are z1 and z2

obtained from its equations to the top and bottom of the given surface and then

the double integration is w.r.t. x and y is performed over the area of projection of

the given solid on the xOy plane.

If ߩൌ݂ሺݔǡݕǡݖሻ is the density of the solid at the point P( x,y,z), then the mass of

the solid is

ම݂ሺݔǡݕǡݖሻ݀ݔ݀ݕݖ݀ǥǥሺͻǤͺሻ

with appropriate limits of integrations.

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210 APPLIED MATHEMATICS

Example 1 . Find by double integration the volume of the sphere

ݔଶݕଶݖଶൌܽଶcut off by the plane ݖ =0 and the cylinder ݔଶݕଶൌܽݔ.

Taking the polar co -ordinate in the xOy plane, elementary area at P ሺݎǡߠሻ is

݀ݎ݀ݎɅ. If the line at P drawn parallel to the z – axis has length z, the volume of

the elementary parallelopi ped at P zrdrd Ʌ, and the volume of the cylinder on the

circle ݔଶݕଶൌܽݔ ,z = 0 bounded at the top by the surface of the sphere

ݔଶݕଶݖଶൌܽଶwith proper limits of integration.

ඵݖݎݎ݀݀Ʌǥǥሺሻ

As ݔଶݕଶൌݎଶ so the equation of the sphere is ݖଶݎଶൌܽଶ or ݖൌξܽଶെݎଶ.

The region of integration is the circle ݔଶݕଶെܽݔൌͲ which has its center at

ቀ

ଶǡͲǡͲቁ and radius is

ଶ. Its polar equation is ݎൌܽ
Ʌ. So the limits of

integration w.r.t. r are 0 and ܽ
Ʌ and w.r.t. Ʌ are െగ

ଶ݊ܽ݀గ

ଶ. With these

considerations and using (i), the volume V required is (by symmetry)

ൌʹන݀Ʌగ

ଶ

නඥܽଶെݎଶୡ୭ୱ

ȁݎ݀ݎǥǥሺሻ

To evaluate the first integral put ݐଶൌܽଶെݎଶ, so we have

නඥܽଶെݎଶȁݎ݀ݎୡ୭ୱ

ൌെනݐଶݐ݀ୱ୧୬

ൌെቈݐଷ

͵

ୱ୧୬

ൌͳ͵ܽଷሾͳെଶɅሿǥǥሺሻ

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211Chapter 9: Applications of Integration

Using this in (ii), the volume required is

ൌʹܽଷ

͵නሾͳെଶɅሿ݀Ʌగ

ଶ

ൌܽଷ

ͳሺ͵ߨെͶሻ

Example 2. Find the volume bounded by the cylinder ݔଶݕଶൌͶand the planes

ݕݖൌͶ and ݖ =0.

From Fig. 14 it is self -evident that ݖൌͶെݕ is to be integrated over the circle

ݔଶݕଶൌͶin the xy - plane.

To cover the shaded half of this circle, x varies from 0 to ඥሺͶെݕଶሻ and y varies

from -2 to 2.

ൌʹනݏଶ

ିଶන ݖ݀ݔ݀ݕඥሺସି௬మሻ

ൌʹනݏଶ

ିଶන ሺͶെݕሻ݀ݔ݀ݕඥሺସି௬మሻ

ൌʹනሺͶെݕሻሾݔሿඥሺସି௬మሻ݀ݕଶ

ିଶൌʹනሺͶെݕሻඥሺͶെݕଶሻ݀ݕଶ

ିଶ

ൌʹනͶඥሺͶെݕଶሻ݀ݕଶ

ିଶെʹනݕඥሺͶെݕଶሻ݀ݕଶ

ିଶ

ൌͺනඥሺͶെݕଶሻ݀ݕଶ

ିଶ

ൌͺ௬ඥሺସି௬మሻ

ଶସ

ଶିଵ௬

ଶ൨

ିଶଶ

= ʌ.

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## Page 212

212 APPLIED MATHEMATICS܍ܔܘܕ܉ܠ۳Ǥ ݔଶܽଶݔଶܾଶݔଶܿଶൌͳǤ

Let OABC be the positive octant of the given ellipsoid which is bounded by the

planes OAB ( z = 0), OBC ( x = 0), OCA ( y = 0), and the surface ABC, i.e.

ݔଶ

ܽଶݕଶ

ܾݖଶ

ܿଶൌͳǤ

Divide this region R into rectangular parallelopipeds of volume ߜݔߜݕߜݖ .Consider

such an element at ሺݔǡݕǡݖሻ (Fig. 15)

ൌͺම݀ݔ݀ݕ݀ݖ௦

ோ

In this region R,

(i) z varies from 0 to MN , where

ൌܿඨͳെݔଶ

ܽଶെݕଶ

ܾଶ

(ii) y varies from 0 to EF, where

ൌܾඨͳെݔଶ

ܽଶ

ǡǤǤ௫మ

మ௬మ

మൌͳǤ

(iii) x varies from 0 to OA = a.

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213Chapter 9: Applications of Integration

Hence the volume of the whole ellipsoid

ൌͺනݏ

නݏටଵି௫మ

మ

නݔ݀ݕ݀ݖ݀ටଵି௫మ

మି௬మ

మ

ൌͺනݔ݀

නݕ݀ටଵି௫మ

మ

ሾݖሿටଵି௫మ

మି௬మ

మ

ൌͺܿනݔ݀

නඨቆͳെݔଶ

ܽଶെݕଶ

ܾଶቇݕ݀ටଵି௫మ

మ

ൌͺܿ

ܾනݔ݀

නඥሺߩଶെݕଶሻݕ݀ఘ

ߩൌܾඨͳെݔଶ

ܽଶ

ൌͺܿ

ܾනݔ݀

ݕඥሺߩଶെݕଶሻ

ʹߩଶ

ʹିଵݕ

ߩ൩

ఘ

ൌͺܿ

ܾනܾଶ

ʹሺͳെݔଶ

ܽଶሻߨ

ʹݔ݀

ൌʹߨܾܿනቆͳെݔଶ

ܽଶቇݔ݀

ൌʹߨܾܿቈݔെݔଷ

͵ܽଶ

ൌͶߨܾܿܽ

͵Ǥ

9.4 Summary

The ninth chapter of this book discusses the applications of integral calculus to the

problems involving areas, volumes and surface of solids . Formulae of these

concepts in single and multiple integrals are developed and their use in the

example s are illustrated with diagrams . At the end, unsolved problems as exercise

are left to students for practice.

9.5 Exercises

ͳǤ

ݔଶൌͶܽݕݔଶͶܽଶൌ଼య

௬

ʹǤ
ݔଶݕଶ

ൌܽଶሺݕଶെݔଶሻ Ǥ

͵Ǥ

ݕଶൌݔଶെݔ͵ݕൌʹݔെͻ munotes.in

## Page 214

214 APPLIED MATHEMATICSͶǤ
ݕଶൌͶܽଶሺʹܽെݔሻݔ

ͷǤ
ݕଶെͶݔ݀݊ܽʹݔെ͵ݕͳ

ʹൌͲǤ

Ǥ

ͻݔݕൌͶ݀݊ܽʹݔݕൌʹǤ

Ǥ
ݔଶ

ܽଶݕଶ

ܾଶൌͳ݀݊ܽݔଶ

ܾଶݕଶ

ܽଶൌͳǤ

8. Find the double integration area included between the curves

ݕଶൌͶܽሺݔܽሻݕଶൌͶܾሺܾെݔሻǤ

ͻǤ
ݎൌ

ܽ
ߠమ

ସ
ǡǤ

ൌܽൌܽ
ͷߠǤ

ͳͲǤ
ൌʹܽ
͵ߠ

ൌܽǤ

ͳͳǤ

ൌܽ
ߠǡܾܽǤ

ͳʹǤ
ݎଶൌܽଶ
ʹߠǤ

ͳ͵Ǥ

ݕଶሺܽݔሻൌݔଶሺ͵ܽെݔሻthe area between the curve and its asymptot

ͳͶǤ

ݕଶൌͶܽݔݔଶൌͶܽݕݏ݅ଵ

ଷܽଶ

ͳͷǤ
ݔݕଶൌܽଶሺܽെሻ

ሺܽെሻݕଶൌܽଶݔݏ݅ሺɎെʹሻܽଶǤ

ͳǤ munotes.in

## Page 215

215Chapter 9: Applications of Integrationݔଶܽଶݔଶܾଶൌͳǡሺܾሻ

ܾଶݔଶൌሺܽଶെܾଶሻݕܽଵ

ଷܾଶ݁݊݅ݏିଵ݁

Ǥ

ͳǤ

ሺሻ
ߠൌ
ʹߠܽଶሺͶെɎሻ

ʹǤ

ሺሻൌߠ
ߠɎܽଶ

ͻሺɎଶെሻǤ

ͳͺǤ

ଶሺʹ
ଶ
ߠെʹܽ
ߠܿߠݏܽଶଶߠሻൌܽଶ
ଶɎܽ

Answers

ͳǤଶమ

ଷሺ͵ߨെʹሻʹǤͶܽଶ ͵ǤͳͲଶ

ଷ ͶǤͶߨܽଶ

ͷǤଵ

ଶ Ǥଵ

ଷെସ

ଽ݈݃ʹ ǤͶܾܽݐ݊ܽିଵ

ͺǤ଼

ଷሺܾܽሻξܾܽ

ͻǤെǤమ

ସǢെమ

ଶǡଷ

ସɎܽଶ ͳͲǤܽଶሺ

ଷξଷ

ଶሻ

ͳͳǤɎቀܽଶଵ

ଶܾଶቁǤ ͳʹǤଶ

ͳͻǤIf the density at a point varies as the square of the distance of the point from

the xy- plane, find the mass of the volume common to the sphere

ݔଶݕଶݖଶൌܽଶ and cylinder ݔଶݕଶൌܽݔ.

ʹͲǤ
ൌ
ሺͳെݔ

ܽሻሺͳെݕ

ܾሻ

ݔଶ

ܽଶݔଶ

ܾଶൌͳǡݖൌͲ

ʹͳǤ

ൌͶെݔଶെͳ

ʹݕଶൌ͵ݔଶͳ

ʹݕଶǤ

ʹʹǤ
ݔଶݕଶ

ൌܽଶǡݔଶݖଶൌܽଶǤ munotes.in

## Page 216

216 APPLIED MATHEMATICSʹ͵Ǥ
ܽʹܽ

ݖൌͲǡൌܽǤ
ݔଶݕଶൌܽݔǤ

ݔଶݕଶൌݖଶǤ

Answers

ͳͻǤଶఱ

ଵହሺɎെଵ

ଵହሻ ʹͲǤୟୠୡ

ସሺɎെଵଷ

ሻ ʹͳǤͶξʹɎʹʹǤଵ

ଷܽଷ

ʹ͵Ǥܽଷ

͵ሺͻɎെͳሻ

9.6 References

1. A Text B ook of Applied Mathematics Vol I - P. N. Wartikar and J. N. Wartikar

2. Applied Mathematics II - P. N. Wartikar and J. N. Wartikar

3. Higher Engineering Mathematics - Dr. B. S. Grewal

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## Page 217

217Chapter 10: Beta and Gamma Functions

Unit 5

10 BETA AND GAMMA FUNCTIONS

Unit Structure

10.0 OBJECTIVES

10.1 Introduction

10.2 Gamma Functions

10.3 Applications of Gamma Functions:

10.4 Properties of Gamma Functions:

10.5 Flow Chart of Gamma Function

10.6 Beta Function

10.7 Properties of Beta Function :

10.8 Problem based on Beta Function

10.9 Duplication Formula of Gamma Functions

10.10 Exercise

10.11 Summary

10.12 References

10.0 Objectives

After going through this unit, you will be able to:

• Understand the concept of Gamma function , properties of Gamma function

• Solve the problem based on Gamma function with its type.

• Understand the concept of Beta function , properties of Beta function

• Understand the relation between Gamma and Beta Function

• Know the concept of Duplication formula

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## Page 218

218 APPLIED MATHEMATICS

10.1 Introduction

At this stage students are well versed with elementary methods of integration and

evaluation of real definite integrals. In this chapter we introduce some advanced

techniques. Beta and Gamma integrals or typically called Beta and Gamma

functions are the special kind of integrals which find their applications in theory of

probability, integral transforms, fluid mechani cs and so on. Certain kind of real

definite integrals can be evaluated by using Beta and Gamma Functions. Their use

is prominent in evaluation of multiple integrals. In this chapter we shall discuss

some properties of Beta and Gamma Functions and Duplication formula. Leonhard Euler

Historically, the idea of extending the factorial to non -integers was considered by Daniel Bernoulli and Christian Goldbach in the 1720s.It was solved by Leonhard Euler at the end of the same decade.

Euler discovered many interesting

properties, such as its reflection formula:.

ȞሺݔሻȞሺͳെݔሻൌߨ

ሺߨെݔሻ James Stirling, contemporary of Euler, also tried to extend thefactorial and came

up with the Stirling formula, which gives

a good approximation of n! but it is not

exact. Lat er on, Carl Gauss, the prince of mathematics, introduced the Gamma function for complex numbers using the

Pochhammer factorial. In the early 1810s,

it was Adrien Legendre who first used the

ī symbol and named the Gamma function.

10. 2 Gamma Functions

Consider the definite integral ݁ି௫ݔିଵ݀ݔஶ

LWLVGHQRWHGE\WKHV\PEROVī (n)

[ we read is as Gamma ‘n’ ] and is called as Gamma Function of n. Thus īQ ିିஶ ( n > 0) ------------------------------ (1)

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219Chapter 10: Beta and Gamma Functions

Gamma Function is also called as Euler’s Integral of the second kind . It defines a

function of n for positive values of n.

10.3 Applications of Gamma Functions:

In a Gamma distribution, the gamma function is used to dete rmine time based

occurrences such as

1. The time between occurrences of earthquakes .

2. Life length of electronic component.

3. Waiting time between any two consecutive events.

4. Gamma function arises in various probability distribution function.

10.4 Properties of Gamma Functions:

1. īQ ିିஶ

Proof: īQ ݁ି௫ݔିଵ݀ݔஶ

Put x = t2 , dx = 2t dt

= ݁ି௧మݐଶିଶʹݐ݀ݐஶ

= ʹ݁ି௧మݐଶିଵ݀ݐஶ

īQ ିିஶ ---------------------------------- (2)

[ It may be borne in mind that variable of integration is immaterial in a definite

integral ]

Relations (1) and (2) are both considered as definitions of Gamma functions.

2. ī

Proof: By definition īQ ݁ି௫ݔିଵ݀ݔஶ

put n =1

ī1) = ݁ି௫ݔ݀ݔஶ

= ݁ି௫݀ݔൌஶ

ሾെ݁௫ሿஶ=(-e + e0 ) = 0 + 1 = 1 ī

x 0 t 0 munotes.in

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220 APPLIED MATHEMATICS

3. Reduction Formulae for Gamma Function :

īQ QīQ

Proof: By definition īQ ݁ି௫ݔିଵ݀ݔஶ

Replace n by n+1

īQ+1 ) = ݁ି௫ݔ݀ݔஶ

Now, integrating by parts

īQ+1 ) = ሾݔሺെ݁௫ሻሿஶെ݊ݔିଵሺ݁ି௫ሻ݀ݔஶ

Now,

௫՜ஶ௫

ೣൌͲǡͲǡ௫

ೣൌͲݔൌͲቂ௫

ೣቃ

ஶ

= 0

īQ+1 ) = 0 + n ݁ି௫ݔିଵ݀ݔஶ

= n īQ) īQ QīQ

If n is a positive integer ,

īQ+1 ) = n (n -1) īQ-1 ) īQ = (n-1) īQ-1 )

= n (n -1) (n -2) īQ-2 )

= n (n -1) (n -2) (n -3) (n -««««« ī(1 )

= n (n -1) (n -2) (n -3) (n -««««« ī1) = 1 = n!

īQ+1 ) = n! if n is a positive integer

īQ QīQLQJHQHUDOQLVUDWLRQDOQXPEHU = n! if n is a positive integer

4. ī īQ ሺ୬ାଵሻ ī ሺଵሻൌଵൌλ 5. īଵ

ଶ ) = ξߨ

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221Chapter 10: Beta and Gamma Functions

6. īQ n!

ī(5) = ī(4+1) ------- (ī(n+1) )

= 4! ----- n is positive integer

= 24

ī (ଷ

ଶሻൌȞሺଵ

ଶͳሻ ------- (ī(n+1) )

= ଵ

ଶȞሺଵ

ଶሻ -----( n is rational number )

= ଵ

ଶξߨ ------ īଵ

ଶ ) = ξߨ

ī (ହ

ଶሻ = ī (ଷ

ଶͳሻ

= (ଷ

ଶሻ ī (ଷ

ଶሻ

= (ଷ

ଶሻ ī (ଵ

ଶͳሻ

= (ଷ

ଶሻ (ଵ

ଶሻ ī (ଵ

ଶሻ = (ଷ

ସሻξߨ

ī (ଵଵ

ଶሻ = (ଽ

ଶሻ (

ଶሻሺହ

ଶሻሺଷ

ଶሻሺଵ

ଶሻξߨ

For negative fraction n , we use

īQ ) = ሺ୬ାଵሻ

ī (െହ

ଷሻ = ቀିఱ

యାଵቁ

ିఱ

య = (െଷ

ହሻ ī (െଶ

ଷሻ = (െଷ

ହሻ ቀିమ

యାଵቁ

ିమ

య =

(െଷ

ହሻሺെଷ

ଶሻȞሺଵ

ଷሻ

= ሺଽ

ଵሻȞሺଵ

ଷሻ

10.5 Flow Chart of Gamma Function

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222 APPLIED MATHEMATICSType I – ି Method of Solving: , Put axm=t , then differentiate, check limit points, reduces

the given integral as gamma function, then we can solve by using definition of

gamma function.

Example 1: Evaluate ݔ݁ିଶ௫మ

݀ݔ

Solution : Let I= ݔ݁ିଶ௫మ

݀ݔ --------------------------- (A)

Put 2x2 = t or x2 = t /2

ݔଶൌݐ

ʹെെെെെെെെെെെሺ݅ሻ

ݔൌඨݐ

ʹൌݐଵ

ଶ

ξʹെെെെെെെሺ݅݅ሻ

Differentiating w.r.t. ‘ t’ we get

ௗ௫

ௗ௧ൌଵ

ξଶଵ

ଶݐషభ

మ = ଵ

ଶξଶݐషభ

మ

݀ݔൌͳ

ʹξʹݐିଵ

ଶ݀ݐ

Now limit point from (i) or (ii)

Let x=0 => 0 = ௧

ଶ => t= 0 i.e. x=0, => t= 0

$QG[ ! ௧

ଶ !W x o t 0 ሺሻ becomes

ܫൌනሺݐଵ

ଶ

ξʹሻͳ

ʹξʹݐିଵ

ଶ݁ି௧

݀ݐ

ൌනݐ

ଶ

ሺʹଵଶൗሻ

ͳ

ʹξʹݐିଵ

ଶ݁ି௧

݀ݐ ൌͳʹଶΤͳሺʹሻʹଵଶΤනݐଶൗݐିଵ

ଶ݁ି௧݀ݐ munotes.in

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223Chapter 10: Beta and Gamma FunctionsൌͳʹଶΤͳʹଷଶΤනݐଷ݁ି௧݀ݐ

ൌͳ

ʹሺାଷሻଶΤනݐଷ݁ି௧

݀ݐ

ൌͳ

ʹହනݐସିଵ݁ି௧

݀ݐ

ൌͳ

͵ʹන݁ି௧

ݐସିଵ݀ݐ

Now using definition of gamma function īQ ିିஶ

( n >

0)

( n= 4 , variable is ‘t’ )

ൌͳ

͵ʹන݁െݐ

ͲݐͶെͳ݀ݐൌͳ

͵ʹȞሺͶሻൌͳ

͵ʹǤ͵Ǩൌ

͵ʹൌ͵

ͳ

ൌනݔ݁െʹݔʹ

Ͳ݀ݔൌ͵

ͳ

Example2: Evaluate ൌݔͻ݁െʹݔʹ

Ͳ݀ݔ

Solution: Let I= ݔଽ݁ିଶ௫మ

݀ݔ--------------------------- (A)

Put 2x2 = t or x2 = t /2

ݔଶൌݐ

ʹെെെെെെെെെെെሺ݅ሻ

ݔൌඨݐ

ʹൌݐଵ

ଶ

ξʹ

Differentiating w.r.t. ‘ t’ we get ݀ݔ݀ݐൌͳξʹͳʹݐିଵଶൌͳʹξʹݐିଵଶ munotes.in

## Page 224

224 APPLIED MATHEMATICSݔ݀ൌͳʹξʹݐିଵଶ݀ݐ

Now limit point from (i) Let x=0 => 0 = ௧

ଶ => t= 0 i.e. x=0, => t= 0

$QG[ ! ௧

ଶ !W x o t 0 ሺሻ becomes

ܫൌනሺݐଵ

ଶ

ξʹሻଽͳ

ʹξʹݐିଵ

ଶ݁ି௧

݀ݐ

ൌනͳ

ሺξʹሻଽͳ

ʹξʹݐଽଶൗݐିଵ

ଶ݁ି௧

݀ݐ

ൌͳ

ሺξʹሻଽͳ

ʹξʹනݐଽଶൗݐିଵ

ଶ݁ି௧

݀ݐ

ൌͳ

ʹଽଶΤͳ

ሺʹሻʹଵଶΤනݐଽଶൗݐିଵ

ଶ݁ି௧

݀ݐ

ൌͳ

ʹଽଶΤͳ

ሺʹሻʹଵଶΤනݐସ݁ି௧

݀ݐ

ൌͳ

ሺʹሻʹହනݐସ݁ି௧

݀ݐ

ൌͳ

Ͷනݐହିଵ݁ି௧

݀ݐ

ൌͳ

ͶȞሺͷሻ ሻ

ൌͳ

ͶͶǨൌ͵

ͺൌනݔͻ݁െʹݔʹͲݔ݀ൌ͵ͺ munotes.in

## Page 225

225Chapter 10: Beta and Gamma Functions Example 3: Evaluate

ൌනݔʹ݁െ݇ʹݔʹ

Ͳ݀ݔ

Solution:

ܮ݁ݐܫൌනݔଶ݁ିమ௫మ

݀ݔെെെെെെെെെെെെሺܣሻ

Put h2 x2 = t or x2 = t / k2

ݔଶൌݐ

݇ଶെെെെെെെെെെെሺ݅ሻ

ݔൌඨݐ

݇ଶൌξݐ

݇

Differentiating w.r.t. ‘ t’ we get

݀ݔ

݀ݐൌͳ

݇ͳ

ʹݐିଵଶൗൌͳ

ʹ݇ݐିଵ

ଶ

݀ݔൌͳ

ʹ݇ݐିଵ

ଶ݀ݐ

Now limit point from (i) Let x=0 => 0 = ௧

మ => t= 0 i.e. x=0, => t= 0

$QG[ ! ௧

మ !W x o t 0 ሺሻ becomes

ܫൌනሺξݐ

݇ሻଶͳ

ʹ݇ݐିଵ

ଶ݁ି௧

݀ݐ

ൌනݐ

݇ଶͳ

ʹ݇ݐିଵ

ଶ݁ି௧

݀ݐ ൌͳʹ݇ଷනݐଵ

ଶ݁ି௧݀ݐ munotes.in

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226 APPLIED MATHEMATICSൌͳʹ݇ଷන݁ି௧ݐଷଶିଵ݀ݐ

ൌͳ

ʹ݇ଷȞ൬͵

ʹ൰ሺ ሻ

ൌͳ

ʹ݇ଷͳ

ʹȞ൬ͳ

ʹ൰

ൌͳ

Ͷ݇ଷξɎȞ൬ͳ

ʹ൰ൌξɎǡ

ȞሺͳሻൌȞሺሻǡ

Example 4: Evaluate

Solution:

ܮݐ݁ൌනξݔͶ݁െξݔ

Ͳ݀ݔൌනݔͳͶൗ݁െݔͳʹൗ

Ͳ݀ݔെെെെሺܣሻ

Put x1/2 = t or x = t2 ( Squaring on both sides )

ൌଶെെെെെെെെെെെሺ݅ሻ

Differentiating w.r.t. ‘ t’ we get

݀ݔ

݀ݐൌʹݐ݀ݔൌʹݐ݀ݐ

Now limit point from (i) Let x=0 => t = 0 i.e. x=0, => t= 0

$QG[ !W x o t 0 ሺሻ becomes

ܫൌනሺݐଶሻଵସൗ݁ି௧

ʹݐ݀ݐ

ൌනݐଵଶൗʹݐ݁ି௧

݀ݐ ൌʹනݐଷଶൗ݁ି௧݀ݐ munotes.in

## Page 227

227Chapter 10: Beta and Gamma Functionsൌʹන݁ି௧ݐଷଶൗ݀ݐ

ൌʹන݁ି௧ݐହ

ଶିଵ

݀ݐ

ൌʹȞ൬ͷ

ʹ൰ሺ
ሻ

ൌʹȞ൬ͷ

ʹͳ൰ൌʹ൬͵

ʹ൰Ȟ൬͵

ʹ൰ൌʹ൬͵

ʹ൰൬ͳ

ʹ൰Ȟ൬ͳ

ʹ൰

ൌ͵൬ͳ

ʹ൰ξɎൌ൬͵

ʹ൰ξɎ

܍ܘܡ܂۷۷ൌනஶ

ܽ௫ൌ݁௧

Take log on both sides ܽ௫ൌ݁௧ ֜ݔܽൌݐ݁

ݔൌݐ

ܽ݁ൌͳ

Differentiating w.r.t. ‘t’ we get

݀ݔ

݀ݐൌͳ

ܽ݀ݔൌ݀ݐ

ܽ

- Then checking limit points

- Substitution given integral ( becomes ) reduces to gamma function.

ܕ܉ܠ۳ܘܔ܍ǣൌනݔଷ

͵௫ஶ

݀ݔ

ܗ܁ܔܖܗܑܜܝǣൌනݔଷ

͵௫ஶ

݀ݔ

͵௫ൌ݁௧ǡ

݈݃͵௫ൌ݈݃݁௧ǡ֜ݔ͵ൌݐ݁ ݔൌݐ͵െെെെെെെെሺ݅ሻൌͳ munotes.in

## Page 228

228 APPLIED MATHEMATICSDifferentiating w.r.t. ‘t’ we get ݔ݀ݐ݀ൌͳ͵֜ݔ݀ൌݐ݀͵

Now limits points from (i)

When ݔൌͲ֜ൌͲ ݔൌλ֜ൌλ x o t 0 ൌනݔଷ

͵௫ஶ

ݔ݀ൌනݐ

͵൨

ଷஶ

ͳ

݁௧ݐ݀

͵

ൌනݐଷ

ሺ͵ሻସஶ

݁ି௧ݐ݀ൌනݐସିଵ

ሺ͵ሻସஶ

݁ି௧ݐ݀ൌͳ

ሺ͵ሻସනݐସିଵஶ

݁ି௧ݐ݀

ൌͳ

ሺ͵ሻସȞሺͶሻൌ͵Ǩ

ሺ͵ሻସൌ

ሺ͵ሻସ

܍ܔܘܕ܉ܠ۳ ǣൌනݔସ

Ͷ௫ஶ

ݔ݀

ܗ܁ܔܖܗܑܜܝǣൌනݔସ

Ͷ௫ஶ

ݔ݀

Ͷ௫ൌ݁௧ǡ

݈݃Ͷ௫ൌ݈݃݁௧ǡ֜ݔͶൌݐ݁

ݔൌݐ

Ͷെെെെെെെെሺ݅ሻൌͳ

Differentiating w.r.t. ‘t’ we get

ݔ݀

ݐ݀ൌͳ

Ͷ֜ݔ݀ൌݐ݀

Ͷ

Now limits points from (i)

When ݔൌͲ֜ൌͲ ݔൌλ֜ൌλ x o t 0 munotes.in

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229Chapter 10: Beta and Gamma FunctionsൌනݔସͶ௫ஶ݀ݔൌනݐͶ൨ସஶͳ݁௧݀ݐͶ

ൌනݐସ

ሺͶሻହஶ

݁ି௧݀ݐൌනݐହିଵ

ሺͶሻହஶ

݁ି௧݀ݐൌͳ

ሺͶሻହනݐହିଵஶ

݁ି௧݀ݐ

ൌͳ

ሺͶሻହȞሺͷሻൌͶǨ

ሺͶሻହൌʹͶ

ሺͶሻହ

܍ܘܡ܂۷۷۷ൌනሺͳݔሻ݀ݔනሺെ݈݃ݔሻ݀ݔ

݈݃ͳ

ݔൌݐ݈݃ݔൌെݐݔ

ൌ݁ି௧െെെെሺ݅ሻ

Differentiating w.r.t. ‘t’ we get

݀ݔ

݀ݐൌെ݁ି௧݀ݔൌെ݁ି௧݀ݐ

- Then checking limit points

Now limits points from (i)

When ݔൌͲ֜݁ି௧ൌͲǡൌλ

When ݔൌͳ֜݁ି௧ൌͳǡൌͲ x o 1 t 0

- Substitution given integral ( becomes ) reduces to gamma function.

ܕ܉ܠ۳ܘܔ܍ǣൌනݔ݀ݔ

ට݈݃ͳ

ݔଵ

ܗ܁ܔܖܗܑܜܝǣൌනݔ݀ݔට݈݃ͳݔଵെെെെെെെെെെെሺܣሻ munotes.in

## Page 230

230 APPLIED MATHEMATICSͳݔൌݐͳݔൌ݁௧ݔൌ݁ି௧

ݔൌ݁ି௧െെെെെെെെെሺ݅ሻ

݀ݔ

݀ݐൌെ݁ି௧݀ݔൌെ݁ି௧݀ݐ

Now limits points from (i)

When ݔൌͲ֜݁ି௧ൌͲǡൌ

When ݔൌͳ֜݁ି௧ൌͳǡൌͲ x o 1 t 0 ሺሻ
ൌන݁ି௧ሺെ݁ି௧݀ݐሻ

ξݐൌනെ݁ିଶ௧ݐିଵଶൗ

݀ݐ

ݕܤݏݑ݅݃݊ݎ݁ݎݕݐ݂݅ݎ݃݁ݐ݊ܽݐ݅݊ǡൌන݁ିଶ௧ݐିଵଶൗ

݀ݐൌන݁ିଶ௧ݐିଵమభ

݀ݐ

ܷ݃݊݅ݏන݁ି௬ݕିଵஶ

݀ݕൌȞሺሻ

݇

න݁െʹݐݐെͳʹͳλ

Ͳ݀ݐൌȞ൬ͳ

ʹ൰

ʹͳ

ʹൌξߨ

ξʹൌටߨ

ʹ

܍ܔܘܕ܉ܠ۳ ǣൌන݀ݔ

ඥݔ݈ݔ݃ଵ

ܗ܁ܔܖܗܑܜܝǣൌන݀ݔ

ඥݔ݈ݔ݃ଵ

െെെെെെെെെെെሺܣሻ

ͳ

ݔൌݐͳ

ݔൌ݁௧ݔൌ݁ି௧

ݔൌ݁ି௧െെെെെെെെെሺ݅ሻ

݀ݔ

݀ݐൌെ݁ି௧݀ݔൌെ݁ି௧݀ݐ

Now we check limits points from (i)

When ݔൌͲ֜݁ି௧ൌͲǡൌ munotes.in

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231Chapter 10: Beta and Gamma FunctionsWhen ݔൌͳ֜݁ି௧ൌͳǡൌͲ x o 1 t 0 ሺሻ
ൌනെݐ݀݁ି௧

ξ݁ି௧ݐൌെන݁௧ଶൗݐିଵଶൗ

ݐ݀

݁ି௧

ݕܤ݃݊݅ݏݑݎ݁ݎݕݐ݂ݎ݃݁ݐ݊݅ܽ݊݅ݐǡൌන݁ି௧ଶൗݐିଵଶൗஶ

ݐ݀

ܷ݃݊݅ݏන݁ି௬ݕିଵஶ

݀ݕൌȞሺሻ

݇

න݁ି௧ଶൗݐିଵమభ

ݐ݀൫ܪ݁ݎ݁ǡ݊ൌͳʹൗǡൌͳʹൗ൯֜ൌīቀͳ

ʹቁ

ሺͳʹൗሻଵ

ଶൌξߨ

ටͳʹൗ

ൌξʹߨ

10.6 Beta Function

Beta Function: The definite integral ݔିଵሺͳെݔሻିଵଵ

݀ݔǡ݉Ͳǡ݊Ͳ

is a function of m and n called Beta Function , denoted by B(m,n) ( we read it

as Beta (m,n))

۰ሺܕǡܖሻൌනିሺെሻି

ǡͲǡ݊Ͳ

The Beta function is also called as Euler’s integral of the first kind. Beta

function of negative numbers is not defined.

ǤǤͳሻ൬͵ǡ͵

ʹ൰ൌනݔଶሺͳെݔሻଵଶൗଵ

݀ݔǡ

ʹሻ൬ͷǡͷ

ʹ൰ൌනݐସሺͳെݐሻଷଶൗଵ

ݐ݀

munotes.in

## Page 232

232 APPLIED MATHEMATICS

10.7 Properties of Beta Function :

Ǥሺǡሻൌሺǡሻ

ǣሺǡሻൌݔିଵሺͳെݔሻିଵଵ

݀ݔǡ݉Ͳǡ݊Ͳ

ൌනሺͳെݔሻିଵሺͳെሺͳെݔሻିଵଵ

݀ݔǡ

න݂ሺݔሻ݀ݔൌන݂ሺܽെݔሻ݀ݔ

ǡ݁ݎ݁ܪܽൌͳ

ሺǡሻൌනሺͳെݔሻ݉െͳǤݔ݊െͳͳ

Ͳ݀ݔൌͳනݔ݊െͳሺͳെݔሻ݉െͳͳ

Ͳ݀ݔൌܤሺ݊ǡ݉ሻ

B( m, n ) = B( n, m )

Ǥනݔ݉ሺͳെݔሻ݊ͳ

Ͳ݀ݔൌܤሺ݉ͳǡ݊ͳሻ

Ǥሺǡሻൌʹන݅ݏ݊ʹ݉െͳߠߨȀʹ

Ͳܿݏʹ݊െͳߠ݀ߠ

ǣሺǡሻൌනݔିଵሺͳെݔሻିଵଵ

݀ݔǡݐݑݔൌ݅ݏ݊ଶߠǡ݀ݔ

ൌʹߠܿݏߠ݀ߠ x o 1 ș 0 ʌ ൌන݅ݏ݊ଶିଶߠగȀଶ

ሺͳെ݅ݏ݊ଶߠሻିଵʹ݅ݏ݊ߠݏܿߠ݀ߠ

ሺǡሻൌʹන݅ݏ݊ʹ݉െͳߠߨȀʹ

Ͳܿݏʹ݊െͳߠ݀ߠ

We consider this as a definition of Beta Function. ǡʹെͳൌǡʹെͳൌൌͳʹǡൌͳʹ munotes.in

## Page 233

233Chapter 10: Beta and Gamma Functionsቆͳʹǡͳʹቇൌʹන݊݅ݏߠߨȀʹͲܿݏݍߠ݀ߠ

න݊݅ݏߠߨȀʹ

Ͳܿݏݍߠ݀ߠൌͳ

ʹቆͳ

ʹǡͳ

ʹቇ

Ǥሺǡሻൌනݔ݉െͳ

ሺͳݔሻ݉݊λ

Ͳ݀ݔ

ǣሺǡሻൌනݔିଵሺͳെݔሻିଵଵ

݀ݔǡݐݑݔൌݐ

ͳݐሺ݅Ǥ݁Ǥݔሺͳݐሻ

ൌݐǡݔݔݐൌݐሻ

ݔൌݐെݔݐܴܱݐൌݔ

ͳെݔሺ݈ܲ݁ܽ݁ݏܰ݁ݐݏݑݏܾݐ݅ݐ݅ݑ݊ሻ

ݔൌͲǡݐൌͲ

ͳെͲ݀݊ܽݔൌͳǡݐൌͲ

ͳെͳൌͳ

Ͳൌλ x 0 1 t 0 λ ݀ݔൌሺͳݐሻሺͳሻെݐሺͳሻݐ

ሺͳݐሻଶ݀ݐൌͳ

ሺͳݐሻଶ݀ݐ

ሺǡሻൌනஶ

ݐିଵ

ሺͳݐሻିଵǤ൬ͳെ

ͳ൰ିଵ

Ǥ݀ݐ

ሺͳݐሻଶ

ൌනஶ

ݐିଵ݀ݐ

ሺݐͳሻିଵሺͳݐሻିଵሺͳݐሻଶൌݐିଵ݀ݐ

ሺͳݐሻା

ሺǡሻൌනஶ

ݔିଵ

ሺͳݔሻା݀ݔ

(Note: We consider this result also as another definition of Beta Function)

ሺͳǡͳሻൌනஶ

ݔሺͳെݔሻ݀ݔ Ǥ

ǡሺǡሻൌȞሺሻȞሺሻȞሺሻ munotes.in

## Page 234

234 APPLIED MATHEMATICSǤන݊݅ݏߠߨȀʹͲܿݏݍߠ݀ߠൌͳʹቆͳʹǡͳʹቇൌȞቀʹቁȞሺͳʹሻȞሺʹʹሻ

ൌͲǡൌͲනߨ

ʹ

Ͳ݀ߠൌͳ

ʹȞ൬ͳ

ʹ൰Ȟ൬ͳ

ʹ൰

Ȟሺͳሻ֜ߨ

ʹൌͳ

ʹ൭Ȟቆͳ

ʹቇ൱ʹ

Ȟቆͳ

ʹቇ

ൌξߨ

10.8 Problem based on Beta Function

܍ܘܡ܂۷ܕ܉ܠ۳ܘܔ܍ܛ܊܍ܛ܉܌ܖܗ܍܌ܑܖܑܜܑܖܗܗ۰܉ܜ܍ܝ۴ܖ܋ܖܗܑܜ

ሺǡሻൌනݔିଵሺͳെݔሻିଵଵ

݀ݔ

ܕ܉ܠ۳ܘܔ܍ǣනݔଷ൫ͳെξݔ൯ହଵ

݀ݔ

܁ܖܗܑܜܝܔܗǣൌනݔଷ൫ͳെξݔ൯ହଵ

݀ݔ

ݐݑξݔൌݐǡݔൌݐଶܦ݅݁ݎ݂݂݁݅ݐ݊ܽ݊݅ݐ݃ݓǤݎǤݐǤƍݐƍݓݐ݁݃݁݀ݔ

݀ݐൌʹݐ֜݀ݔ

ൌʹݐ݀ݐ

ݔൌݐଶ

ݔൌͲǡݐଶൌͲ֜ൌͲǡ݀݊ܽݔൌͳǡݐଶൌͳ֜ൌͳ

x 0 1 t 0 1
ൌනሺݐଶሻଷሺͳെݐሻହଵ

ʹݐ݀ݐ ൌනሺݐሻሺͳെݐሻହଵʹݐ݀ݐൌʹනݐሺͳെݐሻହଵ݀ݐ munotes.in

## Page 235

235Chapter 10: Beta and Gamma Functionsൌනݐ଼ିଵሺͳെݐሻିଵଵ݀ݐൌʹܤሺͺǡሻെെെെെെെሺݕܤ݂݅݊݅݁ܦݐ݊݅ሻ

ǡሺǡሻൌȞሺሻȞሺሻ

Ȟሺሻ

ൌȞሺͺሻȞሺሻ

ȞሺͺሻൌʹǤሺǨሻሺͷǨሻ

ሺͳ͵Ǩሻൌͳ

ͷͳͶͺ

܍ܔܘܕ܉ܠ۳ ǣනݔ݀

ξͳെݔଷయଵ

ܗ܁ܔܖܗܑܜܝǣൌනݔ݀

ξͳെݔଷయଵ

ݑݐݔଷൌݐǡݔൌݐଵଷൗ݊݁ݎ݂݂݁݅ܦ ݐܽ݅ݐ݃݊݅ݓǤݎǤݐǤƍݐƍݓݐ݁݃݁ݔ݀

݀ݐൌͳ

͵ݐభ

యషభ

ൌͳ

͵ݐିଶଷൗ

ൌͳ

͵ݐିଶଷൗ݀ݐǡ
ݔൌݐଵଷൗ

ݔൌͲǡ

ݐଵଷൗൌͲ֜ൌͲǡ݀݊ܽݔൌͳǡݐଵଷൗൌͳ֜ൌͳ

x 0 1 t 0 1
ൌනͳ

͵ݐିଶଷൗ݀ݐ

ሺͳെݐሻଵଷൗଵ

ൌͳ

͵නݐିଶଷൗሺͳെݐሻିଵଷൗ݀ݐൌනଵ

ݐభ

యషభሺͳെݐሻమ

యషభ݀ݐଵ

ൌͳ

͵൬ͳ

͵ǡʹ

͵൰െെെെെെݕܤ݂݅݊݅݁ܦݐ݊݅

munotes.in

## Page 236

236 APPLIED MATHEMATICS

ǡሺǡሻൌȞሺሻȞሺሻȞሺሻ

ൌͳ

͵Ȟቀͳ

͵ቁȞቀʹ

͵ቁ

Ȟቀͳ

͵ʹ

͵ቁൌͳ

͵ͳ

͵

Ȟሺͳሻൌͳ

͵Ȟ൬ͳ

͵൰Ȟ൬ʹ

͵൰Ȟሺͳሻൌͳ

ൌͳ

͵Ȟ൬ͳ

͵൰Ȟ൬ͳെͳ

͵൰൏൏ͳǡ݃݊݅ݏܷȞሺሻȞሺͳെሻൌߨ

݊݅ݏߨ

ൌͳ

͵ߨ

ߨ

͵ൌͳ

͵ߨ

ξ͵

ʹൌʹߨ

͵ξ͵

܍ܘܡ܂۷۷െܕ܉ܠ۳ܘܔ܍ܛ۰܍ܛ܉܌ܖܗ

න݊݅ݏߠߨȀʹ

Ͳܿݏݍߠ݀ߠൌͳ

ʹቆͳ

ʹǡͳ

ʹቇൌȞቀ

ʹቁȞ൬ͳ

ʹ൰

Ȟ൬ʹ

ʹ൰

ܕ܉ܠ۳ܘܔ܍ǣ ξܽݐ݊ߠ݀ߠగȀଶ

܁ܔܗܑܜܝܖܗǣൌනξܽݐ݊ߠ݀ߠൌනඨ݊݅ݏߠ

ܿݏߠ݀ߠగȀଶ

గȀଶ

ൌන݊݅ݏଵଶൗߠܿݏିଵଶൗߠ݀ߠగȀଶ

ൌͳʹൗǡൌെͳʹൗන݊݅ݏߠߨ

ʹ

Ͳܿݏݍߠ݀ߠ

ൌͳ

ʹቆͳ

ʹǡͳ

ʹቇ

ൌͳ

ʹቌͳ

ʹͳ

ʹǡെͳ

ʹͳ

ʹቍൌͳ

ʹቆ͵

Ͷǡͳ

Ͷቇ

ൌͳ

ʹȞቀ͵

ͶቁȞ൬ͳ

Ͷ൰

Ȟ൬͵

Ͷͳ

Ͷ൰ൌͳ

ʹȞቀ͵

ͶቁȞ൬ͳ

Ͷ൰

Ȟሺͳሻൌͳ

ʹȞ൬͵

Ͷ൰Ȟቆͳ

ͶቇȞሺͳሻൌͳ

ȞሺሻȞሺͳെሻൌߨ݊݅ݏߨ munotes.in

## Page 237

237Chapter 10: Beta and Gamma FunctionsൌͳʹȞ൬ͳെͳͶ൰Ȟ൬ͳͶ൰ൌͳʹߨߨͶൌͳʹߨͳξʹൌξʹߨʹൌߨξʹ

܍ܜܗۼǣ
නξߠݐܿ݀ߠൌߨ

ξʹగȀଶ

ܔܘܕ܉ܠ۳܍ǣ ξݐܿߠ݀ߠగȀଶ

܁ܔܗܝܜܑܗܖǣൌනξݐ݊ܽߠ݀ߠൌනඨݏܿߠ

ݏ݅݊ߠ݀ߠగȀଶ

గȀଶ

ൌනݏ݅݊ିଵଶൗߠݏܿଵଶൗߠ݀ߠగȀଶ

ൌെͳʹൗǡൌͳʹൗනݏ݅݊ߠߨ

ʹ

Ͳݏܿݍߠ݀ߠ

ൌͳ

ʹቆͳ

ʹǡͳ

ʹቇ

ൌͳ

ʹቌെͳ

ʹͳ

ʹǡͳ

ʹͳ

ʹቍൌͳ

ʹቆͳ

Ͷǡ͵

Ͷቇ

ൌͳ

ʹȞቀ͵

ͶቁȞ൬ͳ

Ͷ൰

Ȟ൬͵

Ͷͳ

Ͷ൰ൌͳ

ʹȞቀ͵

ͶቁȞ൬ͳ

Ͷ൰

Ȟሺͳሻൌͳ

ʹȞ൬͵

Ͷ൰Ȟቆͳ

ͶቇȞሺͳሻൌͳ

ȞሺሻȞሺͳെሻൌߨ

ݏ݅݊ߨ

ൌͳ

ʹȞ൬ͳെͳ

Ͷ൰Ȟ൬ͳ

Ͷ൰ൌͳ

ʹߨ

ߨ

Ͷൌͳ

ʹߨ

ͳ

ξʹൌξʹߨ

ʹൌߨ

ξʹ

ܡ܂܍ܘ۷۷۷െ ሺǡሻൌනݔ݉െͳ

ሺͳݔሻ݉݊λ

Ͳ݀ݔ ܔܘܕ܉ܠ۳܍ǣ۴ܖܑ܌නݔͺെݔͳͶሺͳݔሻʹͶλͲ݀ݔ munotes.in

## Page 238

238 APPLIED MATHEMATICSܖܗܑܜܝܔܗ܁ൌනݔͺെݔͳͶሺͳݔሻʹͶλͲ݀ݔ

ൌනݔͺ

ሺͳݔሻʹͶλ

Ͳ݀ݔെනݔͳͶ

ሺͳݔሻʹͶλ

Ͳ݀ݔ

ൌනݔͻെͳ

ሺͳݔሻͻͳͷλ

Ͳ݀ݔെනݔͳͷെͳ

ሺͳݔሻͳͷͻλ

Ͳ݀ݔ

ൌሺͻǡͳͷሻെܤሺͳͷǡͻሻሺǡሻൌሺǡሻ

ܫൌͲ

ܘܕ܉ܠ۳ܔ܍۴ܖܑ܌නݔͻሺͳെݔͷሻ

ሺͳݔሻʹͷλ

Ͳ݀ݔ

ܖܗܑܜܝܔܗ܁ ൌනݔͻሺͳെݔͷሻ

ሺͳݔሻʹͷλ

Ͳ݀ݔൌනሺݔͻെݔͳͶሻ

ሺͳݔሻʹͷλ

Ͳ݀ݔ

ൌනݔͻ

ሺͳݔሻʹͷλ

Ͳ݀ݔെනݔͳͶ

ሺͳݔሻʹͷλ

Ͳ݀ݔ

ൌනݔͳͲെͳ

ሺͳݔሻͳͲͳͷλ

Ͳ݀ݔെනݔͳͷെͳ

ሺͳݔሻͳͷͳͲλ

Ͳ݀ݔ

ൌሺͳͲǡͳͷሻെܤሺͳͷǡͳͲሻሺǡሻൌሺǡሻ

ܫൌͲ

ܘܕ܉ܠ۳ܔ܍ܚ۾ܗܞ܍ܜܐܜ܉න݀ݔ

ͳݔͶλ

Ͳൌߨ

ʹξʹ

ܖܗܑܜܝܔܗ܁ ൌන݀ݔ

ͳݔͶλ

Ͳൌන݀ݔ

ͳሺݔʹሻʹλ

Ͳ

ݐݑݔଶൌߠǡݔൌξݐ݊ܽߠ݁ݎ݂݂݁݅ܦݐ݊݅ݐܽ݅݊݃ݓǤݎǤݐǤƍݔƍݓݐ݁݃݁

ʹݔ݀ݔ

݀ߠൌݏ݁ܿʹߠʹݔ݀ݔൌݏ݁ܿʹߠ݀ߠ݀ݔൌݏ݁ܿʹߠ݀ߠ

ʹݔൌݏ݁ܿʹߠ݀ߠ

ʹξݐ݊ܽߠ

݀ݔൌݏ݁ܿଶߠ݀ߠʹξݐ݊ܽߠǡ
ݔଶൌߠ munotes.in

## Page 239

239Chapter 10: Beta and Gamma FunctionsݔൌͲǡͲൌߠ֜ߠൌͲǡ݀݊ܽݔൌλǡ֜ߠൌɎʹݐ݊ܽɎʹ

ൌλ

x 0 1 ߠ 0 ߨȀʹ
ǡൌනܿ݁ݏʹߠ݀ߠ

ʹξݐ݊ܽߠ

ͳݐ݊ܽʹߠߨȀʹ

Ͳ

ൌͳ

ʹනߨȀʹ

Ͳܿ݁ݏʹߠ

ξݐ݊ܽߠ݀ߠ

ሺܿ݁ݏʹߠሻൌͳ

ʹනߨȀʹ

Ͳͳ

ξݐ݊ܽߠ݀ߠ

ൌͳ

ʹනඥܿݐߠߨȀʹ

Ͳ݀ߠൌන݅ݏ݊െͳʹൗߠߨ

ʹ

Ͳܿݏͳʹൗߠ݀ߠ

ൌെͳʹൗǡൌͳʹൗන݅ݏ݊ߠߨ

ʹ

Ͳܿݏݍߠ݀ߠ

ൌͳ

ʹቆͳ

ʹǡͳ

ʹቇ

ൌͳ

ʹͳ

ʹቌെͳ

ʹͳ

ʹǡͳ

ʹͳ

ʹቍൌͳ

Ͷቆͳ

Ͷǡ͵

Ͷቇ

ൌͳ

ͶȞቀ͵

ͶቁȞ൬ͳ

Ͷ൰

Ȟ൬͵

Ͷͳ

Ͷ൰ൌͳ

ͶȞቀ͵

ͶቁȞ൬ͳ

Ͷ൰

Ȟሺͳሻൌͳ

ͶȞ൬͵

Ͷ൰Ȟቆͳ

ͶቇȞሺͳሻൌͳ

ȞሺሻȞሺͳെሻൌߨ

݅ݏ݊ߨ

ൌͳ

ͶȞ൬ͳെͳ

Ͷ൰Ȟ൬ͳ

Ͷ൰ൌͳ

Ͷߨ

ߨ

Ͷൌͳ

Ͷߨ

ͳ

ξʹൌͳ

ʹξʹξʹߨ

ͳ

ξʹൌߨ

ʹξʹ ሺܟ܍ۼܘܡ܂܍ሻܕ܉ܠ۳ܘܔ܍ܜ܉ܝܔ܉ܞ۳܍නሺݔെ͵ሻͳʹൗሺͷെݔሻͳʹൗݔ݀ͷ͵ munotes.in

## Page 240

240 APPLIED MATHEMATICSܖܗܑܜܝܔܗ܁ൌනሺݔെ͵ሻͳʹൗሺͷെݔሻͳʹൗ݀ݔͷ͵

ݐݑሺݔെ͵ሻൌሺܷݎ݁݅݉݅ܮݐെݓܮݎ݁݅݉݅ܮݐሻݐ

ሺݔെ͵ሻൌሺͷെ͵ሻݐሺെ͵ሻൌʹݐݔൌሺʹݐ͵ሻ

݀ݔ

݀ݐൌʹǡ݀ݔൌʹ݀ݐ
ݔൌሺʹݐ͵ሻ

ݔൌ͵ǡʹݐൌͲ֜ݐൌͲǡ݊ܽ݀ݔൌͷǡ֜ͷൌʹ͵֜ݐ

ൌͳ x 3 5 t 0 1 ൌනሺʹݐሻͳʹൗሺͷെሺʹݐ͵ሻሻͳʹൗʹ݀ݐͳ

Ͳ

ൌනʹξʹݐଵଶൗሺʹെʹݐሻଵଶൗ݀ݐଵ

ൌʹξʹනݐଵଶൗሺʹሻଵଶൗሺͳെݐሻଵଶൗ݀ݐଵ

ൌʹξʹξʹනݐଵଶൗሺͳെݐሻଵଶൗ݀ݐଵ

ൌͶනݐିଵమయሺͳെݐሻିଵమయ݀ݐଵ

ൌͶܤሺ͵

ʹǡ͵

ʹሻ

ൌͶȞቀ͵

ʹቁȞቀ͵

ʹቁ

Ȟቀ͵

ʹ͵

ʹቁൌͶȞቀ͵

ʹቁȞቀ͵

ʹቁ

Ȟሺ͵ሻൌͶͳ

ʹȞቀͳ

ʹቁͳ

ʹȞቀͳ

ʹቁ

Ȟሺ͵ሻ

ൌͶ

ʹǨͳ

ʹȞሺɎሻͳ

ʹȞሺɎሻൌɎ

ʹ

munotes.in

## Page 241

241Chapter 10: Beta and Gamma Functions

10.9 Duplication Formula of Gamma Functions

ડሺܕሻડ൬ܕ

൰ൌξૈ

ܕିડሺܕሻ

ܗܗܚ۾ǣ ǡͳ

ʹȞቀͳ

ʹቁȞ൬ͳ

ʹ൰

Ȟቀʹ

ʹቁൌන݊݅ݏߠߨ

ʹ

Ͳܿݏݍߠ݀ߠ

ൌͳ

ʹቆͳ

ʹǡͳ

ʹቇ

ൌൌʹെͳǡൌൌʹെͳቆǤǤൌͳ

ʹǡൌͳ

ʹቇ

ͳ

ʹȞሺሻȞሺሻ

Ȟሺʹሻൌන݊݅ݏଶିଵߠగ

ଶ

ܿݏଶିଵߠ݀ߠ

ȞሺሻȞሺሻ

Ȟሺʹሻൌʹ

ʹଶିଵනగ

ଶ

ሺʹ݊݅ݏߠܿݏߠሻଶିଵ݀ߠ

ൌʹ

ʹଶିଵනగ

ଶ

ሺ݊݅ݏʹߠሻଶିଵ݀ߠǡݐݑܲʹߠൌݐǡ݀ߠ

ൌͳ

ʹ݀ݐ ߠ 0 ߨȀʹ t 0 ߨ ൌͳ

ʹଶିଵනగ

ଶ

ሺ݊݅ݏݐሻଶିଵ݀ݐൌͳ

ʹଶିଵʹනగ

ଶ

݊݅ݏଶିଵݐǤ݀ݐሾ݂ሺߨെݐሻ

ൌ݂ሺݐሻሿ ൌʹʹଶିଵනగ

ଶ݊݅ݏଶିଵݐǤܿݏݐǤ݀ݐ munotes.in

## Page 242

242 APPLIED MATHEMATICSൌȞሺሻȞሺሻȞሺʹሻൌʹʹଶିଵͳʹȞቀʹെͳͳʹቁȞቀͲͳʹቁȞቀʹെͳͲʹʹቁൌͳ

ʹଶିଵȞሺሻξɎ

Ȟቀͳ

ʹቁ

ડሺܕሻડቆܕ

ቇൌξૈܕെડ൫ܕ൯

10.10 Exercise

ͳǤනξݔ݁ିξ௫యஶ

݀ݔൌ͵ͳͷ

ͳξߨሺܜܖܑ۶ݔൌଷሻ

ʹǤනݔ݁ିଶ௫మஶ

݀ݔൌ͵

ͳሺܜܖܑ۶ʹݔൌሻ

͵Ǥනݔଶ݁ିమ௫మஶ

݀ݔൌξߨ

Ͷ݄ଷሺܜܖܑ۶݄ଶݔଶൌሻ

ͶǤනඥݕ݁ି௬యஶ

݀ݕൌξߨ

͵ሺܜܖܑ۶ଷൌሻ

ͷǤන݀ݔ

ටݔ݈݃ͳ

ݔଵ

ൌξʹߨ൬ܜܖܑ۶݈݃ͳ

ݔൌ൰

Ǥන݀ݔ

ඥെ݈݃ݔଵ

ൌξߨሺܜܖܑ۶െ݈݃ݔൌݐሻ

Ǥනݔଷଵ

ሺͳെξݔሻହ݀ݔ൫ܜܖܑ۶ξܠൌݐ൯ǣͳ

ͷͳͶͺ

ͺǤන݊

ሺ݊െݔሻ݀ݔሺܜܖܑ۶ݔ

ൌ݊ݐሻǣ݊ାାଵܤሺ݊ͳǡͳሻ munotes.in

## Page 243

243Chapter 10: Beta and Gamma FunctionsͻǤሺǡሻൌනݔିଵ

ሺͳݔሻାஶ݀ݔ

ቆܜܖܑ۶ݔൌͳ

ͳݐ

ሺǡሻቇ

ͳͲǤනݔଶ

ሺͺെݔଷሻଵଷΤ݀ݔൌʹߨ

͵ξ͵ ൬ܜܖܑ۶ݔଷൌݐǡ݁ݏܷȞሺሻȞሺͳെሻൌߨߨǡͲ൏൏ͳ൰

10.12 Summary

In this unit we learn Gamma and Beta Function and its Duplication Formula

Gamma Function

īQ ݁ି௫ݔିଵ݀ݔஶ

( n > 0)

īQ ʹ݁ି௧మݐଶିଵ݀ݐஶ

ī

īQ QīQ

īQ QīQLQJHQHUDOQLVUDWLRQDOQXPEHU

= n! if n is a positive integer

ī īଵ

ଶ ) = ξߨǡīQ Q

Type I – ݁ି௫

݀ݔǡൌௗ௫

ೣஶ

ǡൌ

ሺଵ

௫ଵ

ሻ݀ݔܴܱሺെ݈݃ݔଵ

ሻ݀ݔ

Beta Function

ሺǡሻൌනݔିଵሺͳെݔሻିଵଵ

݀ݔǡ݉Ͳǡ݊Ͳ

ሺǡሻൌሺǡሻ නݔሺͳെݔሻଵ݀ݔൌܤሺ݉ͳǡ݊ͳሻ munotes.in

## Page 244

244 APPLIED MATHEMATICSሺǡሻൌʹන݊݅ݏʹ݉െͳߠߨȀʹͲܿݏʹ݊െͳߠ݀ߠ

ሺǡሻൌනݔ݉െͳ

ሺͳݔሻ݉݊λ

Ͳ݀ݔ

ሺǡሻൌȞሺሻȞሺሻ

Ȟሺሻ

න݊݅ݏߠగ

ଶ

ܿݏߠ݀ߠൌͳ

ʹ൬ͳ

ʹǡͳ

ʹ൰ൌȞቀ

ʹቁȞቀͳ

ʹቁ

Ȟቀʹ

ʹቁ

Duplication Formula

ȞሺሻȞ൬ͳʹ൰ൌξɎʹଶ୫ିଵȞሺʹሻ

10.11 References

1. A Text Book of Applied Mathematics Vol I - P. N. Wartikar and J. N.

Wartikar

2. Applied Mathematics II - P. N. Wartikar and J. N. Wartikar

3. Higher Engineering Mathematics - Dr. B. S. Grewal

munotes.in

## Page 245

245Chapter 1 Differentiation Under the Integral Sign ( Duis ) & Error Functions

245Unit 5

11 DIFFERENTIATION UNDER THE INTEGRAL

SIGN ( DUIS ) & ERROR FUNCTIONS

Unit Structure

11.0 OBJECTIVES

11.1 Introduction

11.2 Rule - I

11.3 Rule - II

11.4 Error Function: -Definition

11.5 Properties of Error Functions

11.6 Differentiation and Integration of Error function

11.7 Exercise

11.8 Summary

11.9 References

11.0 Objectives

After going through this unit, you will be able to:

• Understand the concept of Differential Under the integral sign ( DUIS) and

Error Functions

• Solve the problem based on Leibnitz’s Rule.

• Know the concept of Differentiation and Integration of Error Function.

11.1 Introduction

Not all integrals can be evaluat ed using analytical techniques, such as integration

by substitution, by parts or by partial fractions. People come up with different ways

of solving the integrals and DUIS is one of them. It is an effective technique used

in evaluation of real definite int egrals. When a definite integralܫൌ݂ሺݔǡߙሻ݀ݔǡ

which is to be integrated w.r.t. variable x and contains parameter , by using DUIS. munotes.in

## Page 246

246 APPLIED MATHEMATICS

There are different rules when limits of integral are constants or functions of

parameter ߙ When DUIS technique is used , the definite integral evaluation results

into an ordinary differential equation, the solution of this equation results in the

evaluation of definite integral. The technique is very useful in Laplace Transform.

Error function integral is close to Probability Integral and is used in probability

distribution. Complementary error functions are involved in finding inverse

Laplace transforms of complicated functions.

11.2 Rule I : Integral With Limits ( a,b) as Constants

ሺȽሻൌ݂ሺݔǡߙሻ݀ݔǡ

ǡ

Ƚൌන߲

߲ߙ݂ሺݔǡߙሻ݀ݔ

ܗܗܚ۾ǣ

Ƚൌܕܑܔ

ఋఈ՜۷ሺሺߙߜߙሻെܫሺߙሻ

ߜߙ

ൌܕܑܔ

ఋఈ՜

ߜߙන݂ሺݔǡߙߜߙሻ݀ݔെන݂ሺݔǡߙሻ݀ݔ

ൌܕܑܔ

ఋఈ՜

ߜߙනሾ݂ሺݔǡߙߜߙሻെ݂ሺݔǡߙሻሿ݀ݔ

ൌܕܑܔ

ఋఈ՜නቈ݂ሺݔǡߙߜߙሻȂ݂ሺݔǡߙሻ

ߜߙ

݀ݔ

ൌනܕܑܔ

ఋఈ՜ቈ݂ሺݔǡߙߜߙሻെ݂ሺݔǡߙሻ

ߜߙ

݀ݔ

Ƚൌන߲

߲ߙ݂ሺݔǡߙሻ݀ݔ

ሺݕܾ݁ܦ݂݊݅ݐ݂݅݊݅ݎܽ݅ݐ݈ܽ݀݅ݎ݁ݒ݅ݐܽݒ݁ሻ

ܝ܀ܔ܍െ۷۷۷ሺહሻൌනሺǡࢻሻ

܌۷

܌હൌනࣔ

ࣔࢻሺǡࢻሻ

It may be noted that if integral involves two parameters µ[¶ and ‘ ߙᇱ integration

is to be carried out w.r.t variable ‘ [¶ treating ߙᇱ as constant. Rule (I ) gives method

to differentiate integral w.r.t. parameter ߙ munotes.in

## Page 247

247Chapter 1 Differentiation Under the Integral Sign ( Duis ) & Error Functions܍ܔܘܕ܉ܠ۳ǣන݁ି௫ݔሺͳെ݁ି௫ஶሻ݀ݔሺܽെͳሻ

ܗ܁ܔܖܗܑܜܝǣԄሺሻൌන݁ି௫

ݔሺͳെ݁ି௫ஶ

ሻ݀ݔ

ǤǤǤǡԄ

ൌන߲

߲ܽ݁ି௫

ݔሺͳെ݁ି௫ሻ൨ஶ

݀ݔ

Ԅ

ൌන݁ି௫

ݔሺെ݁ି௫ሻሺെݔሻஶ

݀ݔ

ൌቈ݁ିሺାଵሻ௫

െሺܽͳሻ

ஶ

ൌ݁ିஶ

െሺܽͳሻെ݁

െሺܽͳሻͳͲ݅Ǥ݁Ǥܽ

െͳ

ൌͳ

ሺܽͳሻ

Ԅൌ݀ܽ

ሺܽͳሻ

Ԅሺܽሻൌሺܽͳሻ

ǡܽൌͲԄሺͲሻൌͲ

ǡԄሺͲሻൌන݁ି௫

ݔሺͳെͳሻሺെݔሻஶ

݀ݔൌͲൌͲ

ܪ݁ܿ݊݁ǡԄሺܽሻൌሺܽͳሻ

܍ܔܘܕ܉ܠ۳ නݔെͳ

ݔଵ

݀ݔൌሺͳܽሻǢܽͲ

ܗ܁ܔܖܗܑܜܝǣԄሺሻൌනݔെͳ

ݔଵ

݀ݔ

ǤǤǤǡԄ

ൌන߲

߲ܽݔെͳ

ݔ൨ଵ

݀ݔ ԄൌනͳݔǤݔଵ݈݃ݔǤ݀ݔ munotes.in

## Page 248

248 APPLIED MATHEMATICSԄൌනݔଵ݀ݔൌቈݔାଵܽͳଵܽͲ

.

Ԅൌͳ

ሺܽͳሻ݀ܽ

Ԅሺሻൌሺܽͳሻܥ

ǡܽൌͲԄሺͲሻൌͲ

ǡԄሺͲሻൌනݔെͳ

ݔଵ

݀ݔൌͲǡൌͲ

݁ܿ݊݁ܪǡԄሺܽሻൌሺܽͳሻ

܍ܔܘܕ܉ܠ۳ න
ߣݔ

ݔஶ

ሺ݁ି௫െ݁ି௫ሻ݀ݔ

ൌͳ

ʹቆܾଶߣଶ

ܽଶߣଶቇǢܽͲǡܾͲ

ܗ܁ܔܖܗܑܜܝǣԄሺሻൌන
ߣݔ

ݔஶ

ሺ݁ି௫െ݁ି௫ሻ݀ݔ

ǤǤǤǡԄ

ൌන߲

߲ܽ
ߣݔ

ݔሺ݁ି௫െ݁ି௫ሻ൨ஶ

݀ݔ

Ԅ

ൌන
ߣݔ

ݔሾ݁ି௫ሺെݔሻെͲሿ݀ݔஶ

ൌെන݁ି௫
ߣݔ݀ݔஶ

ൌെ݁ି௫

ܽଶߣଶሺെܽܿݏݔߣߣݏ݅ݔߣ݊ሻ൨

ஶ

ǡන݁௫
ܾݔ݀ݔൌ݁௫

ܽଶܾଶሺܽܿݏݔܾܾݏ݅ݔܾ݊ሻ

Ԅ

ൌെቈͲെ݁

ܽଶߣଶሺെܽͲሻൌെܽ

ܽଶߣଶ

Ԅൌെܽ

ܽଶߣଶ݀ܽ

Ԅሺሻൌെͳʹනʹܽܽଶߣଶ munotes.in

## Page 249

249Chapter 1 Differentiation Under the Integral Sign ( Duis ) & Error FunctionsԄሺሻൌെͳʹሾሺܽଶߣଶሻሿ

ǡܽൌԄሺሻൌെͳ

ʹሾሺܾଶߣଶሻሿ

Ԅሺሻൌනݏܿߣݔ

ݔஶ

ሺି௫െି௫ሻ݀ݔൌͲ

ൌͳ

ʹሺܾଶߣଶሻ

Ԅሺሻൌെͳ

ʹሺܽଶߣଶሻͳ

ʹሺܾଶߣଶሻൌͳ

ʹቆܾଶߣଶ

ܽଶߣଶቇ

܍ܔܘܕ܉ܠ۳ නݐܽ݊ିଵሺܽݔሻ

ݔሺͳݔଶሻஶ

Ǥ݀ݔൌߨ

ʹሺͳܽሻ

ܗ܁ܔܖܗܑܜܝǣԄሺሻൌනݐܽ݊ିଵሺܽݔሻ

ݔሺͳݔଶሻஶ

Ǥ݀ݔ

ǤǤǤǡԄ

ൌන߲

߲ܽቈݐܽ݊ିଵሺܽݔሻ

ݔሺͳݔଶሻஶ

݀ݔ

Ԅ

ൌනஶ

ͳǤሺݔሻ

ሺͳܽଶݔଶሻǤͳ

ݔሺͳݔଶሻ݀ݔ

Ԅ

ൌනஶ

݀ݔ

ሺͳܽଶݔଶሻሺͳݔଶሻ

Ԅ

ൌනஶ

൮ͳ

ሺͳെͳܽଶΤ

ͳܽଶݔଶͳ

ͳെܽଶ

ͳݔଶ൲݀ݔ

Ԅ

ൌ

ͳെܽଶቈන݀ݔ

ͳݔଶஶ

Ȃනܽଶ

ͳܽଶݔଶஶ

݀ݔ

Ԅ

ൌ

ͳെܽଶሾݐܽ݊ିଵݔെܽݐܽ݊ିଵሺܽݔሻሿஶ݀ݔ

ൌͳ

ͳെܽଶቂߨ

ʹെܽǤߨ

ʹቃൌߨ

ʹሺͳെܽሻ

ሺͳെܽሻሺͳܽሻൌߨ

ʹǤͳ

ሺͳܽሻ

ԄൌߨʹǤ݀ܽሺͳܽሻԄሺሻൌߨʹሺͳܽሻܥ munotes.in

## Page 250

250 APPLIED MATHEMATICSǡൌͲԄሺͲሻൌܥ

ԄሺͲሻൌන݊ܽݐିଵሺͲሻ

ݔሺͳݔଶሻ݀ݔൌͲൌͲ

ஶ

Ԅሺሻൌߨ

ʹሺͳܽሻ

܍ܔܘܕ܉ܠ۳ න݀ݔ

ܾܽݏܿݔగ

ǡܽͲǡȁܾȁ൏ܽ݀݊ܽݑ݂݈݈݀݁݀ܿ݁݅ݓ݊݃

න݀ݔ

ሺܾܽݏܿݔሻଶగ

ൌߨܽ

ሺܽଶെܾଶሻଷଶΤ݀݊ܽනݏܿݔ݀ݔ

ሺܾܽݏܿݔሻଶൌగ

െߨܾ

ሺܽଶെܾଶሻଷଶΤ

ܗ܁ܔܖܗܑܜܝǣൌන݀ݔ

ܾܽݏܿݔగ

ܲݐݑݐൌ݊ܽݐݔ

ʹǡ݀ݔൌʹ݀ݐ

ͳݐଶǡݏܿݔൌͳെݐଶ

ͳݐଶ x 0 ʌ t 0 ൌනʹ݀ݐ

ͳݐଶ

ܾܽͳെݐଶ

ͳݐଶஶ

ൌʹන݀ݐ

ܽܽݐଶܾܾݐଶஶ

ൌʹන݀ݐ

ሺܾܽሻሺܽെܾሻݐଶൌʹ

ሺܽെܾሻන݀ݐ

ሺܾܽሻ

ሺܽെܾሻݐଶஶ

ஶ

ൌʹ

ሺܽെܾሻۏێێۍͳ

ටܾܽ

ܽെܾ݊ܽݐିଵቌඨܽെܾ

ܾܽݐቍےۑۑېஶ

ൌʹ

ξܽଶെܾଶሺ݊ܽݐିଵλെ݊ܽݐିଵͲሻൌʹ

ξܽଶെܾଶߨ

ʹ

න݀ݔ

ܾܽݏܿݔగ

ൌߨ

ξܽଶെܾଶെെെെെെെെെെെെെሺͳሻ

Differentiating (1) both sides w.r.t. a

න߲

߲ܽͳ

ܾܽݏܿݔ൨గ

݀ݔൌߨ

ʹሺܽଶെܾଶሻଷଶΤǤʹܽ

නെͳ

ሺܾܽݔݏܿሻଶగ

݀ݔൌെߨܽሺܽଶെܾଶሻଷଶΤ munotes.in

## Page 251

251Chapter 1 Differentiation Under the Integral Sign ( Duis ) & Error Functionsනͳሺܽݔݏܾܿሻଶగ݀ݔൌߨܽሺܽଶെܾଶሻଷଶΤെെെെെെെെെെെെെሺʹሻ

Differentiating (1) both sides w.r.t. b

න߲

߲ܽͳ

ܾܽݔݏܿ൨గ

݀ݔൌߨ

ʹሺܽଶെܾଶሻଷଶΤǤሺെʹܾሻ

නെͳǤݔݏܿ

ሺܽݔݏܾܿሻଶగ

݀ݔൌߨܾ

ሺܽଶെܾଶሻଷଶΤ

݅Ǥ݁Ǥනݔݏܿ

ሺܽݔݏܾܿሻଶగ

݀ݔൌെߨܾ

ሺܽଶെܾଶሻଷଶΤെെെെെെെെെሺ͵ሻ

Hence (1), (2) , (3) are the required results.

11.3 Rule – II Integral With Limits as Functions of the Parameter

: Leibnintz’s Rule

ሺȽሻ

ൌන݂ሺݔǡߙሻ݀ݔǡୠሺఈሻ

ୟሺఈሻ
ߙǡݐ݄݊݁ǡ

Ƚൌනμ

μȽሼሺǡȽሻሽሺǡȽሻୠሺሻ

ୟሺሻ

ȽെሺǡȽሻ

Ƚ

ܗܗܚ۾ǣ
Ƚ
ሺȽሻ

ሺǡȽሻǡ
Ƚǡ

ሺȽሻሺȽሻൌԄሺȽǡǡሻǡǡ

ȽൌμμȽሺሻμμȽμμȽെെെെെെെെെെെെെሺሻ

I

Į

b

a

Į munotes.in

## Page 252

252 APPLIED MATHEMATICSǡሺȽሻൌන݂ሺݔǡߙሻ݀ݔܻܤ݃݊݅ݏݑܴݑ݈݁ܫǡୠሺఈሻୟሺఈሻμμȽሺሻ

ǡ
ǡ

ǡμ

μȽൌනμ

μȽ݂ሺݔǡߙሻ݀ݔୠሺఈሻ

ୟሺఈሻെെെെെെെെെെെെെെെെሺሻ

නሺǡȽሻൌɗሺǡȽሻǡ

ǤǤμ

μሾɗሺǡȽሻሿൌሺǡȽሻെെെെെെെെെെെെെെെെെሺሻ

ԄሺȽǡǡሻൌሺȽሻ

ൌන݂ሺݔǡߙሻ݀ݔൌሾɗሺǡȽሻሿୟሺఈሻୠሺఈሻൌɗሺǡȽሻെɗሺǡȽሻୠሺఈሻ

ୟሺఈሻെെ

െെെെെെെെെെെെെെെെെെെെെെሺܸܫሻ

ǡǡ

μ

μൌμԄ

μൌμ

μɗሺǡȽሻൌሺǡȽሻሺሻെെെെെെെെെሺሻ

μ

μൌμԄ

μൌെμ

μɗሺǡȽሻൌെሺǡȽሻሺሻെെെെെെെሺሻ

ሺሻǡሺሻǡሺሻሺሻ

܍ܔܝ܀۷۷ǣ

Ƚൌ

ȽනሺǡȽሻୠሺሻ

ୟሺሻൌනμ

μሺǡȽሻሺǡȽሻ

ȽെሺǡȽሻ

Ƚୠሺሻ

ୟሺሻ

ܘܕ܉ܠ۳܍ܔ

නሺܽݔሻ݀ݔୟమ

ୟ

ܗ܁ܝܔܖܗܑܜԄሺሻൌනሺሻୟమ

ୟ munotes.in

## Page 253

253Chapter 1 Differentiation Under the Integral Sign ( Duis ) & Error FunctionsԄൌනμμሺሻୟమୟ൜ሺଶሻൠሺǤଶሻെ൜ሺሻൠǤଶ

ൌනͳ

ǤǤୟమ

ୟʹܽǤሺଷሻെʹ

ൌͳ

൨

ୟୟమ

Ǥሺሻെʹ

ൌͳ

ሺଶെሻǤሺሻെʹ

ൌሺെͳሻǤሺሻെʹെെെെെെെെെെെെെെെെሺͳሻ

ԄሺሻൌනሺሻǤͳǤൌሾሺሻǤሿୟୟమെୟమ

ୟනͳ

ǤǤǤୟమ

ୟ

ൌଶଷെଶെሾሿୟୟమൌ͵ଶെʹെሺଶെሻ

Ԅ

ൌ͵ଶǤͳ

െʹെʹǤͳ

െሺʹെͳሻ

Ԅ

ൌെʹെͳെെെെെെെെെെെെെെെሺʹሻ

ሺͳሻሺʹሻ

܍ܔܘܕ܉ܠ۳ ܟܗܐ܁ܜ܉ܐܜԄሺሻൌන

ଶୟΤ

ୟΤ

ܗ܁ܔܝܜܖܗܑԄሺሻ

ൌන

ଶୟΤ

ୟΤ ǡԄᇱሺሻ

Ԅ

ൌනμ

μ൬

൰൜

ቀɎ

ʹቁൠଶୟΤ

ୟΤǤ൬ቀɎ

ʹቁ൰

ቀɎ

ʹቁ

െ൜

ቀɎ

ቁൠǤ൬ቀɎ

ቁ൰

ቌɎ

ቍ

Ԅ

ൌන
ǤǤ

ቀെɎ

ʹଶቁଶୟΤ

ୟΤǤͳ

ሺɎʹሻΤെቀɎ

ଶቁǤͳʹΤ

ሺɎሻΤ munotes.in

## Page 254

254 APPLIED MATHEMATICSൌ൨ୟΤଶୟΤെͳͳʹൌͳቂɎʹെɎቃെͳͳʹ

ൌͳ

െͳ

ʹͳ

ʹെͳ

ൌͲ

ǡԄ

ൌͲԄሺሻ

܍ܔܘܕ܉ܠ۳ ǣ

නିଵ

ୟమ

ܗ܁ܔܖܗܑܜܝԄሺሻൌනିଵ

ୟమ

Ԅᇱሺሻൌනμ

μቀିଵ

ቁୟమ

൜

ሺଶሻൠିଵቆଶ

ቇെ൜

ሺͲሻൠିଵሺͲሻ

ൌනͳ

ͳଶ

ଶቀ

ଶቁୟమ

ʹିଵሺሻൌെන

ଶଶୟమ

ʹିଵሺሻ

ൌെͳ

ʹනʹǤ

ଶଶୟమ

ʹିଵൌെͳ

ʹሾሺଶଶሻሿୟమʹିଵ

ൌെͳ

ʹሾሺଶସሻെଶሿʹିଵ

ൌെͳ

ʹଶሺͳଶሻ

ଶʹିଵ

Ԅᇱሺሻൌെͳ

ʹሺͳଶሻʹିଵെെെെെെെെെെെሺͳሻ

Ԅሺሻൌනିଵቀ

ቁǤͳǤୟమ

munotes.in

## Page 255

255Chapter 1 Differentiation Under the Integral Sign ( Duis ) & Error Functionsൌቂିଵቀቁሺሻቃୟమെනͳͳଶ

ଶǤͳǤǤୟమ

ൌଶିଵെͲെන

ଶଶୟమ

ൌଶିଵെ

ʹሾሺଶଶሻሿୟమ

ൌଶିଵെ

ʹଶሺͳଶሻ

ଶൌଶିଵെ

ʹሺͳଶሻ

Ԅሺሻൌଶିଵെ

ʹሺͳଶሻ

ԄᇱሺሻൌʹିଵଶǤͳ

ͳଶെͳ

ʹሺͳଶሻെ

ʹ൬ʹ

ͳଶ൰

Ԅᇱሺሻൌʹିଵെͳ

ʹሺͳଶሻൌെെെെെെെെെെെെെሺʹሻ

ሺͳሻሺʹሻ

.

܍ܔܘܕ܉ܠ۳ ǣൌනሺሻሺെሻǡଶ

ଶଶൌሺሻ୶

ܗ܁ܔܖܗܑܜܝൌනሺሻሺെሻǡ୶

ǤǤǤǡ

ൌනμ

μሾሺሻሺെሻሿ൜

ሺሻൠሺሻͲെ൜

ሺͲሻൠሺͲሻͲ୶

ൌනሺሻ
ሺെሻሿͲെͲ୶

ǤǤǤǡ

ଶ

ଶൌනμ

μሾሺሻ
ሺെሻሿ

ሺሻ൨ሺሻ
Ͳ୶

െ

ሺͲሻǤሺͲሻǤ
Ͳ

ଶ

ଶൌනሺሻሺെሺെሻሻǤǤǤሺሻെͲ୶

munotes.in

## Page 256

256 APPLIED MATHEMATICSଶଶൌെଶනሺሻሺെሻǤሺሻൌ୶െଶሺሻ

ଶ

ଶଶൌሺሻ

11.4 Error Function: -Definition

ܖܗܑܜܑܖܑ܍۲ ǣ

ʹ

ξɎනି୳మሺሻǤ୶

܍ܚሺܠሻ

ൌ

ξૈන܍ିܝ܌ܝെെെെെെെെെെሺሻܠ

ǡ

Ǥ

۱ܗܔܘܕ܍ܕܜܖ܍܉ܚܡ۳ܚܗܚܚ۴ܖܝ܋ܖܗܑܜǣ

ʹ

ξɎනି୳మஶ

୶
ሺሻǤ

܍ܚ܋ሺܠሻ

ൌ

ξૈන܍ିܝ܌ܝെെെെെെെെെെሺሻஶ

ܠ

ۯ܍ܜ܉ܖܚ܍ܜܔ ܖܗܑܜܑܖܑ܍۲ ܗ۳ܚܗܚܚ۴ܖܝ܋ܖܗܑܜǣሺͳሻǡ

ଶൌǡʹൌൌ

ʹξǢ

u 0 x

t 0 x2

ሺሻൌʹξɎනି୲ʹξൌͳξɎනି୲ିଵଶΤ୶మ୶మ munotes.in

## Page 257

257Chapter 1 Differentiation Under the Integral Sign ( Duis ) & Error Functionsܚ܍ሺܠሻൌξૈන܍ିܜܜିΤ܌ܜെെെെെെെെെെെെെെെሺሻܠ

ሺͳሻሺ͵ሻ ሺሻ

11.5 Properties of Error Functions

Ǥሺλሻൌʹ

ξɎනି୳మെെെെെെെെെെെെሺଶൌሻஶ

ൌʹ

ξɎනି୷ͳ

ʹିଵଶΤൌͳ

ξɎනି୷ିଵଶΤஶ

ஶ

ൌͳ

ξɎξɎൌͳ

ܚ܍ሺλሻൌെെെെെെെെെെሺሻ

ܚ܍ሺλሻൌ

ǤሺͲሻൌʹ

ξɎනି୳మൌͲ

ܚ܍ሺሻൌെെെെെെെെെെሺሻ

ܚ܍ሺሻൌ

Ǥሺሻ
ሺሻൌʹ

ξɎනି୳మනି୳మஶ

୶୶

൩ൌʹ

ξɎනି୳మஶ

൩

ൌሺλሻൌͳ

ܚ܍ሺܠሻ܋ܚ܍ሺܠሻൌെെെെെെെെെሺሻ

ܚ܍ሺܠሻ܋ܚ܍ሺܠሻൌ

Ǥ۳ܚܚܗܚܖܝ۴ܜ܋ܑܗܖܑܛ܉ܖ܌܌ܗܖܗܑܜ܋ܖܝ ܚ܍ሺെܠሻൌെܚ܍ሺܠሻ

۾ܗܗܚǣሺሻൌʹ

ξɎනି୳మ୶

൩െെെെെെെെ
െ

ሺെሻൌʹξɎනି୳మି୶൩ൌെǢൌെ munotes.in

## Page 258

258 APPLIED MATHEMATICSu 0 -x y 0 0 ሺെሻൌʹ

ξɎනି୷మሺെሻ୶

൩ൌʹ

ξɎනି୷మሺെሻ୶

ܚ܍ሺെܠሻൌെܚ܍ሺܠሻെെെെെെെെെሺૠሻ

ܚ܍ሺെܠሻൌെܚ܍ሺܠሻ Ǥ۳ܠܘ܍ܚܖܗܑܛܛܚܗܚ܍ሺܠሻܖܑܛ܍ܑܚ܍ܛ ۾ܗܗܚǣሺሻൌʹ

ξɎනି୳మ୶

ି୲ൌ

ͲǨെଵ

ͳǨଶ

ʹǨെଷ

͵ǨǥǥǥǥǥǤǤൌͳെଶ

ʹǨെଷ

͵ǨǥǥǥǥǥǤǤ ሺሻൌʹ

ξɎනቈͳെଶସ

ʹǨെ

͵Ǩǥǥሺ୶

ൌെଶି୲ሻ

ൌʹ

ξɎቈെଷ

͵ହ

ͳͲെ

Ͷʹǥǥ

୶

ܚ܍ሺܠሻൌʹ

ξɎቈെଷ

͵ହ

ͳͲെ

Ͷʹǥǥെെെെെെെെെሺૡሻ

ܚ܍ሺܠሻൌ

ξૈቈܠെܠ

ܠ

െܠૠ

ǥǥ

ሺሻ
Ǥሺሻ
Ǥ Ǥܜܔۯܚ܍ܕܖܜ܉܍܌ܖܗܑܜܑܖܑ܍ ܗ۱ܚ܉ܜܖ܍ܕ܍ܔܘܕܗ ܡܚܚ܍ܗܚܝܖ܋ܖܗܑܜ

ሺλሻൌͳǡሺλሻൌͳ

ξɎනି୲ିଵଶΤൌͳஶ

ǡͳ

ξɎቐනି୲ିଵଶΤනି୲ିଵଶΤஶ

୶మ୶మ

ቑൌͳ munotes.in

## Page 259

259Chapter 1 Differentiation Under the Integral Sign ( Duis ) & Error FunctionsͳξɎනି୲ିଵଶΤͳξɎනି୲ିଵଶΤஶ୶మ୶మ

ൌͳെെെെെെെെെെെሺૢሻ

ǤǤǤሺͻሻሺሻ
ͳ

ξɎනି୲ିଵଶΤஶ

୶మ
ሺሻǤ

ሺሻͳ

ξɎනି୲ିଵଶΤஶ

୶మെെെെെെെെെെെെെെെെെെ

െሺሻ

܋ܚ܍ሺܠሻ

ξૈන܍ିܜܜିΤ܌ܜஶ

ܠ

ሺͻሻǡǡሺሻ
ሺሻൌͳ

11.6 Differentiation and Integration of Error function

Differentiation of Error function :

ሺሻൌʹ

ξɎනି୳మ୶

ሺሻൌʹ

ξɎනି୳మୟ୶

ǡ

ǤǤǤ
ǤǤǤǤ

ሺሻൌʹ

ξɎනμ

μି୳మ൜

ሺሻൠିୟమ୶మെ൜

ሺͲሻൠିୟ୶

൩

ሺሻൌʹ

ξɎൣͲǤିୟమ୶మെͲ൧ൌʹǤିୟమ୶మ

ξɎ

܌

܌ܠܚ܍ሺ܉ܠሻൌൌ܉Ǥ܍ି܉ܠ

ξૈെെെെെെെെെെെെെെെെെെሺሻ munotes.in

## Page 260

260 APPLIED MATHEMATICS܌ܠ܌ܚ܍ሺ܉ܠሻൌൌ܉Ǥ܍ି܉ܠξૈ

Integration of Error function:

නሺሻ୲

ൌනͳǤሺሻ୲

ሺሻ

ൌሾሺሻǤሿ୲െන

ሺሻǤǤ୲

ൌǤሺሻെͲെනʹǤିୟమ୶మ

ξɎǤǤ୲

ቆ

ሺሻൌൌʹǤିୟమ୶మ

ξɎቇ

ൌǤሺሻͳ

ξɎǤͳ

නିୟమ୶మሺെʹଶሻ୲

ൌǤሺሻͳ

ξɎൣିୟమ୶మ൧୲

ൌǤሺሻͳ

ξɎሺିୟమ୲మെͳሻൌǤሺሻͳ

ξɎିୟమ୲మെͳ

ξɎ

නሺሻ୲

ൌǤሺሻͳ

ξɎିୟమ୲మെͳ

ξɎെെെെെെെെെሺሻ

නܚ܍ሺ܉ܠሻܜ

ܠ܌ൌܜǤܚ܍ሺ܉ܜሻ

܉ξૈ܍ି܉ܜെ

܉ξૈ

۳܉ܠܔܘܕ܍ǣනሺሻන
ሺሻ୲

ൌ୲

ܗ܁ܔܝܜܖܗܑǣනሺሻන
ሺሻ୲

୲

munotes.in

## Page 261

261Chapter 1 Differentiation Under the Integral Sign ( Duis ) & Error Functionsൌනሾሺሻ
ሺሻሿ୲

ൌනሺͳሻǤൌሾሿ୲ൌሼሺሻ
ሺሻൌͳሽ୲

܍ܔܘܕ܉ܠ۳ ǣ
ሺെሻ
ሺሻൌʹ

ܗ܁ܔܖܗܑܜܝሺሻ
ሺሻൌͳǡ
െ

ሺെሻ
ሺെሻൌͳ

ሺെሻ
ሺെሻൌͳ

െሺሻ
ሺെሻൌͳሼሺെሻൌെሺሻሽ

ሺെሻൌͳሺሻ

ሺെሻ
ሺሻൌͳሺሻ + erfc(x)

ሺെሻ
ሺሻൌͳͳൌʹ

܍ܔܘܕ܉ܠ۳ ͳ

ሺሻൌെͳ

ሺሻ

ܗ܁ܔܖܗܑܜܝ
ሺሻൌʹ

ξɎනି୳మஶ

ୟ୶

ሺሻൌʹ

ξɎ൝μ

μන൫ି୳మ൯Ǥ

ሺλሻǤିஶെ

ሺሻǤିୟమ୶మஶ

ୟ୶ൡ

ൌʹ

ξɎ൛ͲͲെିୟమ୶మൟൌെʹିୟమ୶మ

ξɎ

ͳ

ሺሻൌെʹ

ξɎିୟమ୶మെെെെെെെെെെെെെሺͳሻ

ሺሻൌʹ

ξɎ൝μ

μන൫ି୳మ൯Ǥ

ሺሻǤିୟమ୶మെ

ሺͲሻǤୟ୶

ൡ

ൌʹ

ξɎ൛ͲǤିୟమ୶మെͲൟൌʹିୟమ୶మ

ξɎ

െͳ

ሺሻൌെʹ

ξɎିୟమ୶మെെെെെെെെെെെെെെሺʹሻ munotes.in

## Page 262

262 APPLIED MATHEMATICS ሺͳሻሺʹሻǡͳ
ሺሻൌെͳሺሻ

܍ܔܘܕ܉ܠ۳ ǣି୶మൌξ

ଶሾୠ

ୟሺሻെሺሻ ]

ܗ܁ܔܖܗܑܜܝ ሺሻൌʹ

ξɎනି୳మ୶

ൌλǡǡሺλሻൌʹ

ξɎනି୳మஶ

ͳൌʹ

ξɎනି୶మஶ

ሼሺλሻൌͳሽ

ǡ
ǡ

ͳൌʹ

ξɎቐනି୶మୟ

නି୶మୠ

ୟනି୶మஶ

ୠቑ

ͳൌʹ

ξɎනି୶మୟ

ʹ

ξɎනି୶మୠ

ୟʹ

ξɎනି୶మஶ

ୠ

ͳൌሺሻʹ

ξɎනି୶మୠ

ୟ
ሺሻ

ͳെ
ሺሻൌሺሻʹ

ξɎනି୶మୠ

ୟ

ሺሻെሺሻൌʹ

ξɎනି୶మሼሺሻ
ሺሻൌͳሽୠ

ୟ

නି୶మൌξɎ

ʹሾୠ

ୟሺሻെሺሻሿ

܍ܔܘܕ܉ܠ۳ ǣනି୶మିଶୠ୶ൌξɎ

ʹǤୠమሾஶ

ͳെሺሻሿ munotes.in

## Page 263

263Chapter 1 Differentiation Under the Integral Sign ( Duis ) & Error Functionsܖܗܑܜܝܔܗ܁ൌනି୶మିଶୠ୶ஶൌනି୶మିଶୠ୶ିୠమାୠమஶൌୠమනିሺ୶ାୠሻమஶ

ൌǡൌ

x 0

u b

ൌୠమනି୳మൌୠమξɎ

ʹǤʹ

ξɎනି୳మஶ

ୠஶ

ୠ

ൌξɎ

ʹୠమǤ
ሺሻൌξɎ

ʹୠమሾͳെሺሻሿ

ܘܕ܉ܠ۳ܔ܍ǣȽሺሻൌඨʹ

Ɏන୲మ

ିଶሺሻൌȽሾξʹሿ୶

ܖܗܑܜܝܔܗ܁ Ƚሺሻൌඨʹ

Ɏන୲మ

ିଶ୶

Ƚ൫ξʹ൯ൌඨʹ

Ɏන୲మ

ିଶ୶ξଶ

ଶൌʹଶǡʹൌͶǡൌʹ

ൌʹ

ξʹǤൌξʹ

t 0 xξʹ

u 0 x

ൌඨʹ

Ɏනି୳మξʹǤ୶

ൌʹ

ξɎනି୳మξʹǤൌሺሻ୶

munotes.in

## Page 264

264 APPLIED MATHEMATICS

11.7 Exercise

ͳǤනݔെݔ

ݔଵ

݀ݔൌ൬ܽͳ

ܾͳ൰ǢܽͲǡܾͲ

ʹǤ න݊݅ݏݔ

ݔ݀ݔஶ

ൌɎ

ʹǡනͳെ
ܽݔ

ݔଶஶ

݀ݔ

͵Ǥන݁ି௫െ݁ି௫

ݔஶ

݀ݔൌ൬ܾ

ܽ൰ǢܽͲǡܾͲ

ܜܖܑ۶ǣԄᇱሺሻൌെͳ

Ԅሺሻൌെǡൌǡൌ

ͶǤන݁ି௫ݔ

ݔଶஶ

݀ݔൌݐܿିଵܽǤܦ݀݁ݑ݁ܿݐ݄ܽݐනݔ

ݔஶ

݀ݔൌߨ

ʹ

ͷǤනͳെ
ܽݔ

ݔଶஶ

݀ݔൌߨܽ

ʹ

Ǥሺሻൌනሺെሻଶ୶

ୟ

ሺሻଷ

ଷെʹ

ሺሻൌͲ

ܜܖܑ۶ǣ ǡᇱሺሻൌනሺʹሻሺെሻ

ሺሻ୶

ୟǡᇱᇱሺ୶ሻൌන

ሺሻ୶

ୟ

ᇱᇱᇱሺሻൌʹනμ

μ

ሺሻ൜

ൠ

ሺሻെ൜

ൠ

ሺሻ୶

ୟ൩

Ǥ ሺሻൌන୲୶మǡ୲మ

୲

ൌͳ

ʹൣͷଶ୲ఱെ͵୲యെ ሺሻ൧

ͺǤ

Ǥන
ଶଵୟΤ

ξୟ

ൌെනଶǤଶȂͳ

ଶ
ͳ

െͳ

ʹξ
ଶଵୟΤ

ξୟ munotes.in

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265Chapter 1 Differentiation Under the Integral Sign ( Duis ) & Error FunctionsͻǤԄሺሻൌනǡԄୟమୟ

ͳͲǤ

නͳ

ୟమ

ୟ

ͳͳǤ ሺͲሻǡሺλሻǡ
ሺͲሻ

ͳʹǤ

ሺ୬ሻ

ͳ͵Ǥ

൫ξ൯

ͳͶǤනିሺ୶ାୟሻమஶ

ൌξɎ

ʹሾͳെሺሻሿ

ͳͷǤሺሻǡ
ሺሻǡ൫ξ൯ǡ
൫ξ൯Ǥ

11.8 Summary

܍ܔܝ܀െ۷۷۷ሺહሻൌනሺǡࢻሻ

܌۷

܌હൌනࣔ

ࣔࢻሺǡࢻሻ

܍ܔܝ܀െ۷۷ǣሺ۳ۺ۷ۼ۰۷܈܂ᇱܛ܃܀۳ۺሻ

Ƚൌ

ȽනሺǡȽሻୠሺሻ

ୟሺሻ

ൌනμ

μሺǡȽሻሺǡȽሻ

ȽെሺǡȽሻ

Ƚୠሺሻ

ୟሺሻ

۳ܚܗܚܚ۴ܝܑܜ܋ܖܗܖሺሻൌʹ

ξɎනି୳మ୶

۱ܗܘܕ܍ܔܕܜܖ܍܉ܚܡ۳ܚܗܚܚ۴ܝܑܜ܋ܖܗܖ
ሺሻൌʹ

ξɎනି୳మஶ

୶

ۯ܍ܜ܉ܖܚ܍ܜܔ ۲ܖܗܑܜܑܖܑ܍ ܗ۳ܚܗܚܚ۴ܝܑܜ܋ܖܗܖሺሻൌͳ

ξɎනି୲ିଵଶΤ୶మ

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266 APPLIED MATHEMATICSܛ܍ܑܜܚ܍ܘܗܚ۾ܗ۳ܚܗܚܚ܋ܖ۴ܗܑܜܖ

ܚ܍ሺλሻൌ

ܚ܍ሺሻൌ

ܚ܍ሺܠሻ܋ܚ܍ሺܠሻൌ

ܚ܍ሺെܠሻൌെܚ܍ሺܠሻ

܍ܚሺܠሻൌ

ξૈቈܠെܠ

ܠ

െܠૠ

ǥǥ

܋ܚ܍ሺܠሻൌ

ξૈන܍ିܜܜିΤ܌ܜஶ

ܠ

Differentiation of Error function :

܌

܌ܠܚ܍ሺ܉ܠሻൌൌ܉Ǥ܍ି܉ܠ

ξૈ

Integration of Error function:

නܚ܍ሺ܉ܠሻܜ

܌ܠൌܜǤܚ܍ሺ܉ܜሻ

܉ξૈ܍ି܉ܜെ

܉ξૈ

11.9 References

1. A Text Book of Applied Mathematics Vol I - P. N. Wartikar and J. N.

Wartikar

2. Applied Mathematics II - P. N. Wartikar and J. N. Wartikar

3. Higher Engineering Mathematics - Dr. B. S. Grewal

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