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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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 $PîQ 0LQ PQ
d. ȡ ,Q QZKHUH,Q XQLWPDWUL[RIRUGHUQ
e. ,Iȡ $ PDQGȡ % QWKHQȡ $% PLQ PQ
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
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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
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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
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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
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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
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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
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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
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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−ɉ ሻX1 = 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−ɉ ሻX2 = 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
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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
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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−ߣ ሻ| = |ܲିଵ||ሺ−ߣ ሻ|||
= |ܲିଵܲ||ሺ−ߣ ሻ| = | ||ሺ−ߣ ሻ| = |ሺ−ߣ ሻ|
Since |ܤ−ߣܫ| = |ሺ−ߣ ሻ|, 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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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]
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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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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
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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.
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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.
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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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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 ࢋܑࣂ
݁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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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
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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 FRVșLVLQș 3
= (cos3 șíFRVșVLQ2 ș L VLQșFRV2 șíVLQ3 ș
We can equate the real and imaginary coefficients separately,
LHFRVș FRV3 șíFRVșVLQ2 ș FRV3 șíFRVș
VLQș VLQșFRV2 șíVLQ3 ș = VLQșíVLQ3 ș
([,I] FRVșLVLQș VKRZWKDW]n + ଵ
௭ = 2 cos n șDQG]n - ଵ
௭ = 2 isin n ș
Sol: Let z = FRVșLVLQș
%\GH0RLYUH¶VWKHRUHP
zn FRVșLVLQș n FRVQșLVLQQș
ଵ
௭ = z-n FRVQș– LVLQQș
zn + ଵ
௭ = FRVQșLVLQQș FRVQș – 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 = FRVșLVLQș 30 = FRVșLVLQș
Ex 4. Simplify (1+ i)18
Sol: Let 1 +i = r FRVșLVLQș WKHQZHJHW
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
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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 =ࢋܑ࢞ାࢋషܑ࢞
and ݔ ൌࢋܑ࢞ିࢋషܑ࢞
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, ܛܑܖࢠൌࢋܑࢠିࢋషܑࢠ
; ܗ܋ܛࢠ =ࢋܑࢠାࢋషܑࢠ
܉ܜܖࢠ =࢙ࢠ
ࢉ࢙ࢠ =ࢋܑࢠିࢋషܑࢠ
ሺࢋܑࢠାࢋషܑࢠሻ; ܗ܋ܛ܋܍ࢠ =
ܛܑܖࢠ =
ࢋܑࢠିࢋషܑࢠ munotes.in
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49Chapter 2: Complex Numbers܋܍ܛࢠ =܋ܗܛࢠ =ࢋܑࢠାࢋషܑࢠ ; ܋ܜܗࢠ =ܜܖ܉ࢠ =ሺࢋܑࢠାࢋషܑࢠሻࢋܑࢠିࢋషܑࢠ
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
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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
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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
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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 ඥ࢞࢟ + i ܖ܉ܜି࢟
࢞
1. Prove that log (x + iy) = log ඥݔଶݕଶ + i ିଵ௬
௫ + 2nߨ i
Proof: Let z = x + iy munotes.in
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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
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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
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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)
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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 = ξࢇ࢈ (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)]
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61Chapter 2: Complex Numbers
(a - jb) sin pt = ξࢇ࢈ (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.
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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
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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
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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
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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
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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
࢞ାࢉ
Putting x =2
͵ൌሺെͳሻȀሺȀͶሺͳ) + c)
3 = െͳȀʹͺ
c= െͺͷȀ͵
y = ሺିሻ
ሺૠ
ሻܠିૡ
Ans
ப୷
ப୶ = డ
డ௫ሺି
ૠ
࢞ାࢉ = )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 =࢟ࢊȀ࢟െࢇ࢟
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
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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) = െ࢞ࢊȀ࢞
Equation 12
munotes.in
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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(ࢅࢄ)log(1+7(ࢅࢄሻ - 3(܇܆)2 + ሺ࢟࢞ሻሺାૠሺ࢟
࢞ሻିሺ࢟
࢞ሻሻሺሻ = ࢍȁࢄȁࢉ 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
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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
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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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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
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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 = ࢘ᇲି࢘ᇱ
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 ORJ Y ` [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)
ࣔࡹ
ࣔ࢟ =3x2 - 1 ࡺࣔ
࢞ࣔ =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.
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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
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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
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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܍۾ܠൌࢋࡽࡼ࢞ࣔ 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:
ȝࣔ࢟
ࣔ࢞ ȝ0 [ \ ȝ1 [
On the left -hand side of the equation, a particular differential is obtained as
follows:
ࣔࣔ࢞ሺࣆǡ࢟ሻൌࡺࣆሺ࢞ሻ munotes.in
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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 ࣔࡹ
ࣔ࢟ =-1 and ࡺࣔ
࢞ࣔ =1 and the equations are not exact.
In order to make the equations exact the ࣆ should be such that ࣔࡹ
ࣔ࢟ should be equal
to ࡺࣔ
࢞ࣔ .
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. ࣔࡹ
ࣔ࢟ - ࡺࣔ
࢞ࣔ =2xy -4y2
ࡹ( ࣔࡹ
ࣔ࢟ - ࡺࣔ
࢞ࣔ = )ܠܡିܡ
ܠܡିܡ =
࢟
and the integrating factor is ȝ that is dependent on y.
ȝ =
࢟ ,ࣔૄ
ࣔ࢟ =-2࢟,
ૄ( ࣔૄ
ࣔ࢟ = )-
࢟
Upon integrating
-2 ࣔ࢟
࢟ =ln|y| = ȝ =( +-)
࢟
Now the exact equation is got i.e.
ሺଶെʹଷሻȀଶμሺ͵െʹଶሻȀଶμൌሺെʹሻμሺଷ
୷మെ2x) = 0
ࣔࡹ
ࣔ࢟ =-2, ࡺࣔ
࢞ࣔ =-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
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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) = ࢋ࢞ܠ +c or y = (x+1)(c+1/3 ࢋ࢞)
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
= ࢋ࢞
࢟ +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 = - Pࣔ࢞ =y/c =
ࢋିࡼࣔ࢞ 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
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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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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. ࢋష
࢞ࣔ࢞ =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
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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
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86 APPLIED MATHEMATICS
Ans
Example
Solve dy/dx = 3x -4y-2/3x-4y-3
munotes.in
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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.
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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
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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
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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 + ࢋ࢞+k)(y+ ࢋି࢞ +k) = 0 Ans
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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
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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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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
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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 = (
)ࢋିࢉ +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
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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 = ࢉ
+ା
ି 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
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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 = ࢟
ࢇ - 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 Gș WDQ2ș 2
Note that 1+tan2ș VHF2ș
munotes.in
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98 APPLIED MATHEMATICS
Integrating Factor VHF2șGșVHF4ș GșVHF2ș FRV2șGș
Recall that FRVș FRV2șíVR cos2ș FRV2ș
Integrating Factor FRV2șGș Gș
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 VLQșFRVș DUFWDQ[ &
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 1 S 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¶ S ௗ
ௗ௫
= p = p +xௗ
ௗ௫ I¶ S ௗ
ௗ௫
= 0 = xௗ
ௗ௫ I¶ S ௗ
ௗ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
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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
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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: e4x Sí H2ypp2 = 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 = u࢜ࢊ
࢛ࢊ (+࢜ࢊ
࢛ࢊ)2 Ans
Example
(ௗ௬
ௗ௫) + 4x ( ௗ௬
ௗ௫ ) - 4y = 0
Let ௗ௬
ௗ௫ = p then p +4xp -4y = 0
2U\ S[S DQGWKLVLVLQWKH&ODLUDXW¶V(TXDWLRQ
Differentiating with respect to x
S SS¶ [ S S¶
S¶ [ S S¶
Assuming p =c , y = cx +(c2/4)
Eliminating p we have y(x) = -x2 as it satisfies y = (p +4xp)/4 munotes.in
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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[ íS 1/2 munotes.in
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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
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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
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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
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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
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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[ >3 FRVȕ[LVLQȕ[ 4 FRVȕ[ – LVLQȕ[ @
= eĮ[ > 34 FRVȕ[L 3 – 4 VLQȕ[@
y = eĮ[ [c1 FRVȕ[F 2 VLQȕ[@
where, c 1 = P + Q and c 2 = i (P – Q) are arbitrary constants.
y = C eĮ[ FRV ȕ[ș ZKHUH&șDUHDUELWUDU\FRQVWDQWV
using c 1 &FRVșF 2 = – VLQș munotes.in
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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) e ĮLȕ x + (Rx + S) e Į– Lȕ x
= eĮ[ [(P x + Q) eLȕ[ + (Rx + S) e–Lȕ[ ]
= eĮ[ > 3[4 ^FRVȕ[LVLQȕ[` 5[6 ^FRVȕ[ – LVLQȕ[`@
= eĮ[ > 3[45[6 FRVȕ[L (Px + Q – Rx – 6 VLQȕ[@
\ HĮ[> F[F FRVȕ[ F[F VLQȕ[@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
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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 2 FRVȕ[ F 3 x + c 4 VLQȕ[@
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
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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\ ĮLȕ WKHQFRPSOHPHQWDU\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 2 FRVȕ[ F 3 x + c 4 VLQȕ[@
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
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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 ’’ሺD DQGVRRQ
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
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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 ݁௫SURYLGHGI D
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
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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
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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
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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
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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
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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
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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
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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
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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
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124 APPLIED MATHEMATICS
ݐ
݂ሺݏሻǡ
ݏǤ ݂ሺݐሻ
ǡ
ݐ
Ͳǡ݂ҧሺݏሻ
ǡݏ ǡ
Ǥ
݂ሺݐሻ݂ሺݏሻǡࢌሺ࢙ሻൌන܍ିܜܛሾࢌሺ࢚ሻሿࢊ࢚ஶ
െെെെെሺሻ
ࣦሼ݂ሺݐሻሽൌ݂ሺݏሻǡ݂݀݊ܽҧሺݏሻ is called the Laplace
transform of f(t).
ࣦሼܣܨଵሺሻܤܨଶሺሻሽൌܣࣦሼܨଵሺݐሻሽ
ܤࣦሼܨଶሺݐሻሽ
6. 3 Table of Elementary Laplace Transform f(t) f(s) 1 ࢙ ࢋࢇ࢚ ࢙െࢇǡ࢙ܽ
Sin at ࢇ࢙ࢇ
Cos at ࢙࢙െࢇ
sinh at ࢇ࢙െࢇ
cosh at ࢙࢙െࢇ ࢚ࢇ࢙ࢇ࢚ ࢙ሺ࢙ࢇሻ ࢇሺ࢙ࢇ࢚െࢇ࢚ࢉ࢙ࢇ࢚ሻ ሺ࢙ࢇሻ ࢚ ሺሻǨ࢙ା
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
ࢋࢇ࢞ࡱ ǣ݊݅ܨ݀ݐ݄݈݈݁ܽܽܿ݁ݐ݂ݏ݊ܽݎݎ݂݉݅ሻ݁ି௧ݐܽݏܿ
܁ܗܝܔܗܑܜܖࣦሺ
ܽݐሻൌݏ
ݏଶܽଶ
ࣦሼ݁ି௧ݐܽݏܿሽൌݏܾ
ሺݏܾሻଶܽଶ
ࢋࢇ࢞ࡱ ǣ݊݅ܨ݀ݐ݄݈݈݁ܽܽܿ݁ݐ݂ݏ݊ܽݎݎ݂݉݅ሻݐଶ݁ଷ௧
܁ܗܝܔܗܑܜܖࣦሺݐଶሻൌʹǨ
ݏଷ
ࣦሼݐଶ݁ଷ௧ሽൌʹǨ
ሺݏെ͵ሻଶ
ࢋࢇ࢞ࡱ 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
ࢋࢇ࢞ࡱ 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
ࢋࢇ࢞ࡱ 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
ࢋࢉࢋࢎǡखሼࡲሺ࢚ሻሽൌࢋି࢙ࢇࢌതሺ࢙ሻǤ࢝ࢋࢎ࢘ࢋ۴ሺܜሻൌሼ࢚ழࢌሺ࢚ିࢇሻ࢚வ
࢞ࡱࢇࢋǣܨࣦ݀݊݅ሼܨሺݐሻሽ݂ݎ ሺሻൌሼழ௧ழଵሺ࢚ିሻ࢚வଵ
ܖܗܑܜܝܔܗ܁ ݂ሺݐሻൌݐଷǡ
݂ҧሺݏሻൌ͵Ǩ
ݏସ munotes.in
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128 APPLIED MATHEMATICSݕܤ݁ݒܾܽݐ݄݁ݎ݁݉ݏܽܽൌͳǡࣦሼܨሺݐሻሽൌ͵Ǩ݁ି௦ݏସ
܄۷Ǥ۷खሼࢌሺ࢚ሻሽൌࢌതሺܛሻǡܖ܍ܐܜखሼ࢚ࢌሺ࢚ሻሽൌሺെሻࢊ
࢙ࢊࢌതሺ࢙ሻǡࢋ࢘ࢋࢎ࢝
ൌǡǡǥ
ࢀࢎ࢙࢙ࢇࢉࢊࢋ࢛࢚࢚ࢇࢉ࢟࢈࢚
ܚܚܗ۱Ǥ܂܍ܐܛ܍ܚܔܝ۷ܜखሼ࢚ࢌሺ࢚ሻሽൌെࢊ
࢙ࢊࢌതሺ࢙ሻൌെࢌᇱሺ࢙ሻ
࢚ࢋࢎࢊ࢚ࢇ࢚ࢋ࢘ࢋࢌࢌ ࢌ࢚ࢋࢎ࢚࢘ࢌ࢙ࢇ࢘ ࢌࢇࢌ࢚ࢉ࢛ ࢉ࢙ࢊ࢙ࢋ࢘࢘ ࢚
࢚ࢋࢎ࢛࢚࢚ࢇࢉࢌ࢚ࢋࢎࢌ࢚ࢉ࢛ ࢟࢈െ࢚
࢞ࡱࢋࢇૠǣ݅ܨࣦ݊݀ሼܨሺݐሻሽ݂ݎሺ݅ሻݐ
ʹܽ݊݅ݏ݄ܽݐሺ݅݅ሻݐଶ
ܽݐ
܁ܜܝܔܗܑܖܗሺ݅ሻ݂ሺݐሻൌܽݐ
ʹܽǡ݂ሺݏሻൌͳ
ʹͳ
ݏଶെܽଶ
ࣦ൜ݐͳ
ʹܽܽݐൠൌሺെͳሻ݀
݀ݏ൜ͳ
ʹǤͳ
ݏଶെܽଶൠ
ൌሺെͳሻͳ
ʹെʹݏ
ሺݏଶെܽଶሻଶ
ൌݏ
ሺݏଶെܽଶሻଶ
ሺ݅݅ሻ݂ሺݐሻൌ
ܽݐǡࢌതሺܛሻൌݏ
ݏଶܽଶ
ࣦሼݐଶ
ܽݐሽൌሺെͳሻଶ݀ଶ
݀ݏଶቄݏ
ݏଶܽଶቅ
ൌʹݏሺݏଶെ͵ܽଶሻ
ሺݏଶܽଶሻଷ
࢞ࡱࢋࢇૡǣ݅ܨ݀݊ݐ݄݁ܮ݈ܽܽܿ݁ݐݏ݉ݎ݂ݏ݊ܽݎ ݂
ሺ݅ሻݐ
ܽݐሺ݅݅ሻݐଶܽݐሺ݅݅݅ሻݐଷ݁ିଷ௧ሺݒ݅ሻݐ݁ି௧͵ݐ
ࡿ࢚࢛ 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
܄Ǥ۷खሼࢌሺ࢚ሻሽൌࢌതሺܛሻǡܖ܍ܐܜखቊࢌሺ࢚ሻ
࢚ቋ
ൌනࢌതሺ࢙ሻࢊ࢙ஶ
࢙ǡ࢜࢘ࢊࢋࢊܕܑܔ
࢚՜ାࢌሺ࢚ሻ
࢚ࢋ࢙࢚࢙࢞
ࢎࢀ࢙࢙ࢇࢉࢊࢋࢊ࢙࢜࢟࢈࢚
ࡱࢇ࢞ࢋૢǣܨ݀݊݅ݐ݄݁ܮ݈ܽ݁ܿܽݎ݂ݏ݊ܽݎݐ ݉ݏ݂
ሺ݅ሻሺͳെ݁௧ሻ
ݐሺ݅݅ሻ
ݐܽെ
ܾݐ
ݐݐݐܽ
ࡿ࢛࢚ሺ݅ሻܵ݅݁ܿ݊ǡࣦሺͳെ݁௧ሻൌࣦሺͳሻെࣦሺ݁௧ሻൌͳ
ݏെͳ
ݏെͳ
ࣦቆͳെ݁௧
ݐቇൌන൬ͳ
ݏെͳ
ݏെͳ൰݀ݏൌȁݏെሺݏെͳሻȁஶஶ
௦ munotes.in
Page 130
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܄Ǥ۷खሼࢌሺ࢚ሻሽൌࢌതሺܛሻǡܖ܍ܐܜखሼࢌሺࢇ࢚ሻሽ
ൌ
ࢇࢌതቀ࢙
ࢇቁሺࢎࢇࢋࢍࢌࢉ࢙ࢇࢋ࢘ࡼࢋ࢚࢘࢟ሻ
࢞ࡱࢇࢋǣ݂ܫࣦሼ݂ሺݐሻሽൌͺͳʹݏെʹݏଶ
ሺݏଶͶሻଶǡ݂ࣦ݀݊݅ሼ݂ሺʹݐሻሽ
ܝܔܗ܁ܜܗܑܖǣ ǡ
ࣦሼ݂ሺʹݐሻሽൌͳ
ʹە۔ۓͺͳʹቀݏ
ʹቁെʹቀݏ
ʹቁଶ
൬ቀݏ
ʹቁଶ
Ͷ൰ଶۙۘۗ
ൌͶሺͳͳʹݏെݏଶሻ
ሺݏଶͳሻଶ
۷۷܄Ǥ܉ܚ܂ܗܛܖܕܚܗ۳ܚܗܚܚ۴ܖܝ܋ܖܗܑܜ
ܹ݁݇݊ݓ݁ݎ݂൫ξݔ൯ൌʹ
ξߨන݁ି௧మݐ݀ξ௫
munotes.in
Page 131
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
खቐනࢌሺ࢚െ࢛ሻࢌሺ࢛ሻࢊ࢛࢚
ቑൌࢌሺ࢙ሻࢌሺ࢙ሻൌࢌതതതሺ࢙ሻࢌതതതሺ࢙ሻ
ൌखනࢌሺ࢛ሻࢌሺ࢚െ࢛ሻࢊ࢛࢚
݄ܶݏ݅ݐ݄݁݁ݎ݉ݏ݅ݑ݂݁ݏݑ݈ݐ݂݅݀݊݊݅ݒ݁ݏݎ݁ݐ݂ݏ݊ܽݎݎ݉ܽݐ݅݊Ǥ
ࢋࢇ࢞ࡱ ǣܸ݂݅ݎ݁ݕݐ݄݁݊݅ݐݑ݈ݒ݊ܿ ݐ݄݁݁ݎ݉ݎ݂݄݁ݎ݂݂݅ܽݑ݊ܿݐ݅ݏ݊
݂ଵሺݐሻൌݐǡ݂ଶሺݐሻൌ݁௧
ࡿ࢚࢛ǣ݂ଵഥሺݏሻൌͳ
ݏଶǡ݂ଶഥሺݏሻൌͳ
ݏെܽ
݂ଵഥሺݏሻ݂ଶഥሺݏሻൌͳ
ݏଶሺݏെܽሻ
ܰݓǡන݂ଵሺݑሻ݂ଶሺݐെݑሻݑ݀௧
ൌනݑǤ݁ሺ௧ି௨ሻݑ݀௧
ൌെݑ
ܽ݁ሺ௧ି௨ሻെͳ
ܽଶ݁ሺ௧ି௨ሻ൨
௧
ൌͳ
ܽଶሾ݁௧െܽݐെͳሿ खቐන݂ଵሺݑሻ݂ଶሺݐെݑሻݑ݀௧ቑൌख൜ͳܽଶሾ݁௧െܽݐെͳሿൠ munotes.in
Page 132
132 APPLIED MATHEMATICSൌͳܽଶͳݏെܽെܽݏଶെͳݏ൨
ൌͳ
ݏଶሺݏെܽሻൌ݂ଵഥሺݏሻ݂ଶഥሺݏሻ
ǤǤࢋࢉࢇࢇࡸ ࢘ࢀࢇ࢘ࢌ࢙ ࢌࢇࡵ࢚ࢋࢍࢇ࢘
ݕܤ݂݁݀݊݅ݐ݅݊݅ǡ
खቐන݂ሺݑሻ݀ݑ௧
ቑൌන݁ି௦௧න݂ሺݑሻ݀ݑ௧
ஶ
ݐ݀െെെെെെെെെሺ݅ሻ
ݏ݊݅ܿ݁݀
ݐ݀න݂ሺݑሻ݀ݑ௧
ൌ݂ሺݐሻǡ
ݓ݁݃ݐܾ݁ݕ݃݁ݐ݊݅ݎ݃݊݅ݐܾܽݕݏݐݎܽǡݐ݄݁ݑݏ݁ݎ݈ݐሺ݅ሻܽݏ
ࣦቐන݂ሺݑሻ݀ݑ௧
ቑൌെͳ
ݏ݁ି௦௧න݂ሺݑሻ݀ݑ௧
ஶ
ͳ
ݏන݁ି௦௧݂ሺݐሻݐ݀ஶ
ൌͳ
ݏ݂ҧሺݏሻ
݄ܶݑݏǡखቐනࢌሺ࢛ሻࢊ࢛࢚
ቑൌ
࢙ࢌതሺ࢙ሻ
݅Ǥ݁Ǥ݂݊݅ݐܿ݊ݑݏ݅݃݁ݐ݊݅ݎ݀݁ݐܽ݊ݒݎ݁ሺͲǡݐሻݐ݄݁ݎ݂ݏ݊ܽݎݐ ݂݉݃݊݅ݐ݄݁݃݁ݐ݊݅ݎ݈ܽݏ݅ ݀݁݊݅ܽݐܾ ܾݕ݅݀ݒ݊݅݀݅݃ݐ݄݁ݎ݂ݏ݊ܽݎݐ ݂݉ݐ݄݂݁݊݅ݐܿ݊ݑܾݕݏǤ
ࡾࡻ
ܫ݂ࣦሺݐሻൌ݂ҧሺݏሻǡݐ݄ࣦ݊݁ቐන݂ሺݑሻ݀ݑ௧
ቑൌͳ
ݏ݂ҧሺݏሻǡ
ܮݐ݁ሺݐሻൌන݂ሺݑሻ݀ݑ௧
ǡݐ݄݊݁ᇱሺݐሻൌ݂ሺݐሻܽ݀݊ሺͲሻൌͲ
ࣦሼᇱሺݐሻሽൌݏഥሺݏሻെሺͲሻ
ܱݎഥሺݏሻൌͳ
ݏࣦሼᇱሺݐሻሽ݅Ǥ݁Ǥࣦቐන݂ሺݑሻ݀ݑ௧
ቑൌͳ
ݏ݂ҧሺݏሻ munotes.in
Page 133
133Chapter 6: The Laplace Transformࢋࢇ࢞ࡱǣݕ݂݅ݎܸࣦ݁ቐනݑଶ݁ି௨݀ݑ௧ቑൌͳݏखሼݐଶ݁ି௧ሽ
ࡿ࢚࢛ǣනݑଶ݁ି௨݀ݑ௧
ൌሾെሺݑଶʹݑʹሻ݁ି௨ሿ࢚
ൌʹെሺݐଶʹݐʹሻ݁ି௧
ࣦቐනݑଶ݁ି௨݀ݑ௧
ቑ
ൌखሼʹെሺݐଶʹݐʹሻ݁ି௧ሽെെെെെെെെെെሺሻ
ࣦሺݐଶ݁ି௧ሻൌሺെͳሻଶ݀ଶ
݀ݏଶ൬ͳ
ݏͳ൰ൌʹ
ሺݏͳሻଷ
ࣦሺʹݐ݁ି௧ሻൌʹǤሺെͳሻ݀
݀ݏ൬ͳ
ݏͳ൰ൌʹ
ሺݏͳሻଶ
ሺሻǡ
ࣦቐනݑଶ݁ି௨݀ݑ௧
ቑൌʹ
ݏെʹ
ሺݏͳሻଷʹ
ሺݏͳሻଶʹ
ሺݏͳሻଵ൨
ൌʹ
ݏሺݏͳሻଷൌͳ
ݏࣦሼݐଶ݁ି௧ሽ
ࢋࢇ࢞ࡱ ǣ݈ܽݒܧݑܽ݁ݐݐ݄݂݈݈݁ݓ݅݃݊ǣ
ሺ݅ሻනݐ݁ିଷ௧ݐ݀ݐሺ݅݅ሻஶ
න݉ݐ
ݐ݀ݐஶ
ሺ݅݅݅ሻන݁௧൬
ݐܽെ
ܾݐ
ݐ൰݀ݐஶ
ሺݒ݅ሻࣦቐනି୲ݐ
ݐ݀ݐ௧
ቑ
ࡿ࢚࢛ሺ݅ሻනݐ݁ିଷ௧ݐ݀ݐൌනݐ݁ି௦௧ሺݐݐሻ݀ݐݓ݄݁ݎ݁ݏൌ͵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
ࡾࡻܾݕ݁݀݅ݐ݂݅݊݅݊න݁ି௦௧ݐ݉
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
ǤǤࢋࢉࢇࢇࡸ ࢘ࢀࢇ࢘ࢌ࢙ ࢌࡰ࢘ࢋ࢜ࢇ࢚࢜ࢋ
ሺሻ
ൌͲ
ሺǤǤ
՜ͲሻǤ
ࣦሾ݂ሺݐሻሿൌ݂ҧሺሻሺሻ
ቂ݅Ǥ݁Ǥ
՜ஶ݁ି௦݂ሺ݉ሻൌͲǡݎ݂ݏቃǡݐ݄݁݊खሼࢌᇱሺ࢚ሻሽൌ࢙ࢌതሺ࢙ሻെࢌሺሻ
ܹ݄݁ݎ݂݁ሺͲሻ݅ݏݐ݄݁ݑ݈ܽݒ݂݂݁ሺݐሻݐܽݐൌͲǤ
खሼࢌᇱሺ࢚ሻሽൌ࢙ࢌതሺ࢙ሻെࢌሺሻെെെെെെെെെെെሺࡵሻ
࢘ࢇ࢘࢟െ
ܫ݂ࣦሼ݂ሺݐሻሽൌ݂ҧሺݏሻݐ݄ࣦ݁݊ሼ݂ᇱԢሺݐሻሽൌݏଶ݂ҧሺݏሻെݏ݂ሺͲሻെ݂ᇱሺͲሻ
ܮݐ݁ܨሺݐሻൌ݂ᇱሺݐሻݐ݄݁݊
ࣦሼ݂ᇱԢሺݐሻሽൌࣦሼܨᇱሺݐሻሽ
ൌݏࣦሼܨሺݐሻሽെܨሺͲሻെെെെെെݕܾሺࡵሻ
ൌݏࣦሼ݂ᇱሺݐሻሽെ݂ᇱሺͲሻ
ൌݏൣ݂ҧሺݏሻെ݂ሺͲሻ൧െ݂ᇱሺͲሻെെെെെെݕܾሺࡵሻ
ൌݏଶ݂ҧሺݏሻെݏ݂ሺͲሻȂ݂ᇱሺͲሻെെെെെെݕܾሺࡵሻ
ܤݕ݃݊݅ݏݑ݉ݐ݄݈ܽ݁ܽܿ݅ݐܽ݉݊݅ݐܿݑ݀݅݊ǡݓ݁ܿ݊ܽݏ݄ݓݐ݄ݐܽ
6.5 Additional Problems
࢞ࡱࢋࢇǣܨ݀݊݅ݐ݄݈݈݁݁ܿܽܽݐ݉ݎ݂ݏ݊ܽݎ ݂݄݂݁ܿܽݐ݄݁݃݊݅ݓ݈݈݂ ݏ݊݅ݐܿ݊ݑ݂ ሺ݅ሻ
ݐ
ʹݐሺ݅݅ሻݐଶെ͵ݐͷሺ݅݅݅ሻݐଶݐܽሺ݅ݒሻ݁ସ௧ܿݏ݄ͷݐ
ࡿ࢛࢚ሺ݅ሻࣦሼ
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
ࢋࢇ࢞ࡱ ǣܨࣦ݀݊݅ሼ݂ሺݐሻሽǡ݂݂݅ሺݐሻൌሼǡழఈୡ୭ୱሺ௧ିఈሻǡ௧வఈ
ࡿ࢚࢛ݕܤ݀݊݅ݐ݂݅݊݅݁ ǡ
ࣦሼ݂ሺݐሻሽൌන݁ି௦௧݂ሺݐሻ݀ݐஶ
ൌන݁ି௦௧ሺͲሻ݀ݐන݁ି௦௧
ሺݐെߙሻ݀ݐஶ
ఈఈ
ൌන݁ି௦ሺ௨ାఈሻ
ݑ݀ݑሾݓ݄݁ݎ݁ሺݑൌݐെߙሻሿஶ
ൌ݁ିఈ௦න݁ି௦௨
ݑ݀ݑஶ
ൌ݁ିఈࣦሼ
ݑሽൌ݁ି௦ఈݏݏଶͳ munotes.in
Page 137
137Chapter 6: The Laplace Transformࢋࢇ࢞ࡱǣࢊࡲࢎ࢚ࢋࢇࢇࢉࢋ࢚࢘ࢇࢌ࢙࢘ࢌࢎࢉࢇࢋࢌࢎ࢚ࢋ
ࢌ࢝ࢍࢌ࢛࢙࢚ࢉ
ሺ݅ሻݐହଶΤሺ݅݅ሻ݁ିଷ௧ݐିଵଶΤሺ݅݅݅ሻξݐ
ࡿ࢚࢛ 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
ࢋࢇ࢞ࡱ ͳǣܩ݊݁ݒࣦ݅ቐʹඨݐ
ߨቑൌͳ
ݏଷଶൗݏ݄ݓݐ݄ܽݐࣦ൜ͳ
ξߨݐൠൌͳ
ξݏ munotes.in
Page 138
138 APPLIED MATHEMATICS࢚࢛ࡿݐ݁ܮ݂ሺݐሻൌʹඨݐߨ
ܨᇱሺݐሻൌʹ
ξߨͳ
ʹݐିଵଶൗൌͳ
ξߨݐ
ࣦሺ݂ᇱሺݐሻሻൌࣦͳ
ξߨ
ൌݏ݂ҧሺݏሻെ݂ሺͲሻെെെെܽܮǤݎܶܽǤ݊ܦݎ݁ݒ݅ݐܽݒ݅ݏ݁
ൌݏࣦቐʹඨݐ
ߨቑൌݏǤͳ
ݏଷଶൗୀͳ
ξݏ
ࢇ࢞ࡱࢋͳͺǣܧܽݒݑ݈݁ݐܽනݐ݁ିଷ௧ݐ݀ݐஶ
࢚࢛ࡿ 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
ͶǤ ࣦ݂ሺݐሻൌݏଶെݏͳ
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
ࢌሺ࢙ሻൌන܍ିܜܛሾࢌሺ࢚ሻሿࢊ࢚ஶ
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܍ܚܕ ࣦሼ݂ሺݐሻሽൌ݂ҧሺݏሻǡࣦሼ݁ି௧݂ሺݐሻሽ
ൌ݂ҧሺݏܽሻ
܁܍܋ܗ܌ܖܐ܁ܑܑܜܖ܂܍ܐܗ܍ܚܕ ࣦሼ݂ሺݐሻሽൌ݂ҧሺݏሻ ሺሻ
ൌሼ௧ழሺ௧ିሻ௧வࣦሼܨሺݐሻሽൌ݁ି௦݂ҧሺݏሻ
ࣦሼ݂ሺݐሻሽൌ݂ҧሺሻǡࣦሼݐ݂ሺݐሻሽൌሺെͳሻ݀ݏ݂݀ҧሺݏሻǡݓ݄݁ݎ݁݊ൌͳǡʹǡ͵ǥ munotes.in
Page 141
141Chapter 6: The Laplace Transform࢙ࢎࢀ࢙ࢊࢋࢇࢉ࢛࢚ࢇࢉ࢚࢟࢈࢚
ࣦሼ݂ሺݐሻሽൌ݂ҧሺሻǡࣦቊ݂ሺݐሻ
ݐቋൌන݂ҧሺݏሻ݀ݏஶ
௦ǡݎ݅ݒ݀݁݀
௧՜ା݂ሺݐሻ
ݐ݁ݔݏݐݏ݅
࢙ࢎࢀ࢙ࢊࢋࢇࢉࢊ࢙࢜࢟࢈࢚
ࣦሼ݂ሺݐሻሽൌ݂ҧሺሻǡࣦሼ݂ሺܽݐሻሽ
ൌͳ
݂ܽҧቀݏ
ܽቁሺࢇࢎࢍࢋࢌࢋࢇࢉ࢙࢘ࡼࢋ࢚࢘࢟ሻ
ܕܚܗܛܖ܉ܚ܂ ܗ۳ܚܗܚܚܝ۴ܖܑܜ܋ܖܗࣦ൛݂ݎ݁൫ξݐ൯ൟൌͳ
ݏξݏെͳ
܍ܐ܂ܖܗ܋ܞܗܔܖܗܑܜܝܕ܍ܚܗ܍ܐ܂ :
ࣦቐන݂ଵሺݐെݑሻ݂ଶሺݑሻ݀ݑ௧
ቑൌ݂ଵሺݏሻ݂ଶሺݏሻൌ݂ଵഥሺݏሻ݂ଶഥሺݏሻ
ൌࣦන݂ଵሺݑሻ݂ଶሺݐെݑሻ݀ݑ௧
ࡸࢋࢉࢇࢇࢀࢇ࢙࢘࢘ࢌࢌࢇࡵࢇ࢘ࢍࢋ࢚ ǣࣦቐන݂ሺݑሻ݀ݑ௧
ቑൌͳ
ݏ݂ҧሺݏሻ
ࡸࢋࢉࢇࢇࢀࢇ࢙࢘࢘ࢌࢌࡰࢋ࢚࢜ࢇ࢜࢘ࢋ ࣦ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
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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
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143Chapter 7: Inverse Laplace Transform
Definition: ࣦሼ݂ሺݐሻሽൌ݂ሺݏሻǡ݂ሺݐሻ is called the inverse Laplace Transform
of ݂ҧሺݏሻ.
खି൛ࢌതሺ࢙ሻൟൌࢌሺ࢚ሻ ࣦି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
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144 APPLIED MATHEMATICSࢋࢇ࢞ࡱǣ ሺ݅ሻͳݏͶሺ݅݅ሻʹݏݏଶͶ
ܖܗܑܜܝܔܗ܁ ሺ݅ሻͳ
ݏͶǡࣦିଵͳ
ݏͶ൨ൌ݁ିସ௧ǡࡾࡻ݂ሺݐሻൌ݁ିସ௧
ሺ݅݅ሻʹݏ
ݏଶͶൌʹݏ
ݏଶͶ
ݏଶͶൌʹݏ
ݏଶͶ͵ʹ
ݏଶͶ
ࣦሺ
ʹݐሻൌݏ
ݏଶͶǡࣦሺʹݐሻൌʹ
ݏଶͶ
ࣦିଵʹݏ
ݏଶͶ൨ൌʹࣦିଵቄݏ
ݏଶͶቅ͵ࣦିଵ൜ʹ
ݏଶͶൠ
ൌʹ
ʹݐ͵ʹݐ
ࢋࢇ࢞ࡱ 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܍
ሺࡵሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻǡݐ݄ࣦ݊݁ିଵ൛݂ҧሺݏെܽሻൟൌ݁௧݂ሺݐሻൌ݁௧ࣦିଵ൛݂ҧሺݏሻൟ
ሺࡵࡵሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻ݂ܽ݊݀ሺͲሻൌͲǡݐ݄ࣦ݊݁ିଵ൛ݏ݂ҧሺݏሻൟൌ݀
ݐ݀ሼ݂ሺݐሻሽൌ݂ᇱሺݐሻ
݅Ǥ݁Ǥ ݂ҧሺݏሻ ݏǡ
݂ሺݐሻ
ǡࣦିଵ൛ݏ݂ҧሺݏሻൟൌൌ݀
ݐ݀ሼ݂ሺݐሻሽǡ ݂ሺͲሻൌ݂ᇱሺͲሻൌڮ
ൌ݂ିଵሺͲሻൌͲ
ࣦሼݐ݂ሺݐሻሽൌሺെͳሻ݀
݀ݏ݂ҧሺݏሻൌሺെͳሻ݂ҧሺሻሺݏሻ
ࣦିଵ൛݂ሺሻሺݏሻൟൌሺെͳሻݐ݂ሺݐሻ
ሺࡵࡵࡵሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻݐ݄ࣦ݊݁ିଵቊ݂ҧሺݏሻ
ݏቋൌන݂ሺݐሻݐ݀௧
݈ܣݏࣦିଵቊ݂ҧሺݏሻ
ݏଶቋൌන௧
ቐන݂ሺݐሻݐ݀௧
ቑݐ݀
ࣦିଵቊ݂ҧሺݏሻ
ݏଶቋ
ൌන௧
ቐනቌන݂ሺݐሻ௧
ݐ݀ቍ௧
ݐ݀ቑݐ݀Ǥ
ሺࢂࡵሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻݐ݂ሺݐሻൌࣦିଵ൜Ȃ݀
ݏ݀ൣ݂ҧሺݏሻ൧ൠǡ
ࣦሺݐ݂ሺݐሻሻൌȂ݀
ݏ݀ൣ݂ҧሺݏሻ൧
ሺࢂሻࣦቆ݂ሺݐሻ
ݐቇൌන݂ҧሺݏሻݏ݀ǡ݂ሺݐሻ݂ሺݏሻǡஶ
௦
න݂ҧሺݏሻ
ஶ
௦ munotes.in
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146 APPLIED MATHEMATICSࢋࢇ࢞ࡱǣ݀݊݅ܨ݂ሺݐሻǡ݂݂݅ҧሺݏሻൌݏݏଶʹݏͷ
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
ࢋࢇ࢞ࡱ 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
ࢋࢇ࢞ࡱ 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
ሺࡵࡵሻܫ݂ࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻ݂݀݊ܽሺͲሻൌͲǡݐ݄݁݊खି൛࢙ࢌതሺ࢙ሻൟൌࢊ
࢚ࢊሼࢌሺ࢚ሻሽൌࢌᇱሺ࢚ሻ
݅Ǥ݁Ǥ ݂ҧሺݏሻݏǡ
݂ሺݐሻ
ܫ݊݃݁݊݁ݎ݈ܽǡࣦିଵ൛ݏ݂ҧሺݏሻൟൌൌ݀
ݐ݀ሼ݂ሺݐሻሽǡ
ݎݒ݂݀݁݀݅ሺͲሻൌ݂ᇱሺͲሻൌڮൌ݂ିଵሺͲሻൌͲ
ܵ݉݅ݐ݁݉ݏ݈݁ܽ݊݃ݓݐ݄݅ݐ݄ܾ݁ܽ݁ݒݎ݁ݑݏݐ݈ݓ݁݁ݎݍ݁ݎ݅ݑݐ݁ݏݑ݂݃݊݅ݓ݈݈
ࣦሼݐ݂ሺݐሻሽൌሺെͳሻ݀
݀ݏ݂ҧሺݏሻൌሺെͳሻ݂ҧሺሻሺݏሻݓ݄݄ܾܿ݅݊ܽܿ݁ݔ݁ݎ݁ݏݏ݁݀ܽݏ
खି൛ࢌሺሻሺ࢙ሻൟൌሺെሻ࢚ࢌሺ࢚ሻ
ࢋࢇ࢞ࡱ ǣ
ǣ
ሺሻݏଶ
ሺݏଶܽଶሻଶሺ݅݅ሻݏଶ
ሺݏܽሻଷ
ሺሻ݈݃ቆͳܽଶ
ݏଶቇሺ݅ݒሻݐ݊ܽିଵ൬ʹ
ݏଶ൰
ܝܔܗ܁ܜܑܗܖሺ݅ሻ݂ҧሺሻൌͳ
ݏଶܽଶࣦିଵ൬ͳ
ݏଶܽଶ൰ൌͳ
ܽܽݐൌ݂ሺݐሻ
ܰݓǡ݂ҧᇱሺሻൌെʹݏ
ሺݏଶܽଶሻଶǡܷ݃݊݅ݏखି൛ࢌሺሻሺ࢙ሻൟ
ൌሺെሻ࢚ࢌሺ࢚ሻݓ݁݃ݐ݁ munotes.in
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151Chapter 7: Inverse Laplace Transformࣦିଵ൜െʹݏሺݏଶܽଶሻଶൠൌሺെͳሻሺଵሻݐଵͳܽܽݐ
ࣦିଵ൜ݏ
ሺݏଶܽଶሻൠൌͳ
ʹܽܽݐ
࢝ࡺǡࢍ࢙ࢁखି൛࢙ࢌതሺ࢙ሻൟൌࢊ
࢚ࢊሼࢌሺ࢚ሻሽǡ࢝ࢋ࢜ࢇࢎࢋ
ࣦିଵቊݏଶ
ሺݏଶܽଶሻଶቋൌࣦିଵቊݏǤݏଶ
ሺݏଶܽଶሻଶቋ
ൌ݀
݀ݐ൜ͳ
ʹܽܽݐൠൌͳ
ʹܽሺܽ
ܽݐሻ
ሺ݅݅ሻݏଶ
ሺݏܽሻଷ
ݓܰǡ݂ҧሺሻൌͳ
ሺݏܽሻǡ݂ሺݐሻൌ݁ି௧
ݓܰǡ݂ҧᇱሺሻൌͳ
ሺݏܽሻଶǡ݂ҧᇱᇱሺሻൌʹ
ሺݏܽሻଷ
ܷ݅ݏ݃݊खି൛ࢌሺሻሺ࢙ሻൟൌሺെሻ࢚ࢌሺ࢚ሻ݁ݓݐ݁݃
ࣦିଵ൜ʹ
ሺݏܽሻଶൠൌሺെͳሻଶݐଶ݁ି௧
ࣦିଵ൜ͳ
ሺݏܽሻଷൠൌͳ
ʹݐଶ݁ି௧
࢝ࡺǡࢍ࢙ࢁखି൛࢙ࢌതሺ࢙ሻൟൌࢊ
࢚ࢊሼࢌሺ࢚ሻሽǡ࢝ࢋ࢜ࢇࢎࢋ
ࣦିଵቊݏଶ
ሺݏܽሻଷቋൌ݀ଶ
݀ݐଶ൜ͳ
ʹݐଶ݁ି௧ൠൌͳ
ʹሾܽଶݐଶെͶܽݐʹሿ݁ି௧
ሺሻ݂ሺݏሻൌ݈݃ቆͳܽଶ
ݏଶቇൌሺݏଶܽଶሻെʹݏ
݂ഥᇱሺݏሻൌʹݏ
ݏଶܽଶെʹ
ݏൌܨሺݏሻ
ܷ݅ݏ݃݊खି൛ࢌሺሻሺ࢙ሻൟൌሺെሻ࢚ࢌሺ࢚ሻ݁ݓݐ݁݃
ࣦି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
۷खି۴തሺ࢙ሻൌࡲሺ࢚ሻǡखି۵ഥሺ࢙ሻൌࡳሺ࢚ሻࢊࢇࢌഥሺ࢙ሻൌࡲഥሺ࢙ሻכࡳഥሺ࢙ሻࢋࢎ࢚
൛खିࢌതሺ࢙ሻൟൌखିሼࡲഥሺ࢙ሻכࡳഥሺ࢙ሻሽൌනࡲሺ࢚െ࢛ሻࡳሺ࢛ሻ࢛ࢊ࢚
ǣ
खି൬
࢙൰ൌࢊࢇखିࢌതሺ࢙ሻൌࢌሺ࢚ሻ
ܮݐ݁ܨഥሺݏሻൌͳ
ܩҧሺݏሻൌ݂ഥሺݏሻǡ݄݁݊݁ܿ݉ݎ݂ܾܽ݁ݒݏ݁ݎ݈ݑݐݓ݁݃ݐ݁ǡ खିቊࢌሺ࢙ሻ࢙ቋൌනǤࢌሺ࢛ሻ࢛ࢊ࢚ࡺࢋ࢚ǣܨሺݐሻ݊ܽ݀ܩሺݐሻ݁ݎܽ݅ܿݎ݁ݐ݄ܾ݈݊ܽ݁݃݊ܽ݁munotes.in
Page 153
153Chapter 7: Inverse Laplace Transform
ࢋࢇ࢞ࡱǣ݊݅ܽݐܾܱݐ݄݁݊݅ݒ݁ݏݎ݁ܮ݈ܽܽܿ݁ݏ݊ܽݎݐ݂ݎ݂݉ݐ݄݈݈݂݁ݓ݊݅݃ǣ
ሺሻ
࢙ሺ࢙ሻሺሻ࢙
ሺ࢙ࢇሻ
ࡿ࢚࢛ǣሺ݅ሻͳ
ݏଶሺݏͳሻଶൌͳ
ݏଶͳ
ሺݏͳሻଶ
ࣦିଵ൜ͳ
ݏଶൠൌݐൌܨሺݐሻࣦ݊ܽ݀ିଵ൜ͳ
ሺݏͳሻଶൠൌݐ݁ି௧ൌܩሺݐሻ
ܷ݃݊݅ݏǡ൛खିࢌതሺ࢙ሻൟൌखିሼࡲഥሺ࢙ሻכࡳഥሺ࢙ሻሽൌනࡲሺ࢚െ࢛ሻࡳሺ࢛ሻࢊ࢛࢚
ࣦିଵ൜ͳ
ݏଶሺݏͳሻଶൠൌනሺݐെݑሻݑ݁ି௨ݑ݀௧
ൌሾെሺݑݐെݑଶሻ݁ି௨െሺݐെʹݑሻ݁ି௨െሺെʹሻ݁ି௨ሿ௧
ൌݐ݁ି௧ʹ݁ି௧ݐെʹ
खିቊࢌሺ࢙ሻ
࢙ቋൌනǤࢌሺ࢛ሻࢊ࢛࢚
ǡݑ݃݊݅ݏ݈݀݁ݐܽ݁݁ݎ ݕǡࣦିଵ൜ͳ
ሺݏͳሻଶൠ
ൌݐ݁ି௧ൌ݂ሺݐሻ
ࣦିଵ൜ͳ
ݏଶͳ
ሺݏͳሻଶൠ
ൌනݑ݁ି௨ݑ݀ൌሾെݑ݁ି௨െ݁ି௨ሿ௧ൌͳെሺݐͳሻ݁ି௧௧
ࣦିଵ൜ͳ
ݏଶሺݏͳሻଶൠൌࣦିଵ൜ͳ
ݏଶͳ
ሺݏͳሻଶൠ
ൌනሾͳെሺݑͳሻ݁ି௨ሿݑ࢚݀
ൌሾݑሺݑͳሻ݁ି௨݁ି௨ሿ௧ൌݐ݁ି௧ʹ݁ି௧ݐെʹ
ሺሻݏ
ሺݏଶܽଶሻଶൌͳ
ݏଶܽଶǤݏ
ݏଶܽଶ
ࣦିଵቄݏ
ݏଶܽଶቅൌݏܿݐܽൌܨሺݐሻ munotes.in
Page 154
154 APPLIED MATHEMATICSࣦିଵ൜ͳݏଶܽଶൠൌݐܽܽൌܩሺݐሻ
۷खି۴തሺ࢙ሻൌࡲሺ࢚ሻǡखି۵ഥሺ࢙ሻൌࡳሺ࢚ሻࢊࢇࢌഥሺ࢙ሻ
ൌࡲഥሺ࢙ሻכࡳഥሺ࢙ሻࢋࢎ࢚
൛खିࢌതሺ࢙ሻൟൌखିሼࡲഥሺ࢙ሻכࡳഥሺ࢙ሻሽ
ൌනࡲሺ࢚െ࢛ሻࡳሺ࢛ሻ࢛ࢊ࢚
ࣦିଵ൜ݏ
ሺݏଶܽଶሻଶൠൌࣦିଵ൜ͳ
ሺݏଶܽଶሻǤݏ
ሺݏଶܽଶሻൠ
ൌන
ܽሺݐെݑሻܽݑ
ܽ݀ݑ௧
ൌͳ
ʹܽනሾݐܽሺʹܽݑെݐܽሻሿ݀ݑ௧
ൌͳ
ʹܽݑݐܽെͳ
ʹܽ
ሺʹܽݑെݐܽሻ൨
௧
ൌͳ
ʹܽݐݐܽ
࢞ࡱࢇࢋૠǣ ǣ
ሺ݅ሻͳ
ሺݏെʹሻସሺݏ͵ሻ
ࡿ࢚࢛ǣሺ݅ሻࣦିଵ൜ͳ
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
ݏଶͶࡾࡻݔҧൌݏܽ
ሺݏଶͻሻݏ
ሺݏଶͶሻሺݏଶͻሻ
ࡾࡻݔҧൌݏܽ
ሺݏଶͻሻݏ
ሺݏଶͶሻሺݏଶͻሻ
ࡾࡻݔҧൌܽ
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 I W
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 ሺ ƍ
ሻ
Ǥ
Ǥܖ܃ܑܜ܁܍ܜܘܝ۴ܖ܋ܑܜܗܖሺ܍܌ܑܛܑܞ܉܍۶ᇱܛܖ܃ܑܜܝ۴ܖ܋ܑܜܗܖሻ
ݑሺݐെܽሻǣ
ݑሺݐെܽሻൌቄͲ൏ܽͳ ǡܽǤ ܪሺݐെܽሻǤ
Fig (Unit Step Function )
ሻܚ܂ܛܖ܉ܗܕܚܗܜܑܖܝܖܗܑܜ܋ܖܝ Ǥ
ࣦሼݑሺݐെܽሻሽൌන݁ି௦௧ݑሺݐെܽሻ݀ݐൌஶ
ൌන݁ି௦௧ǤͲ݀ݐන݁ି௦௧Ǥͳ݀ݐൌஶ
Ͳቤ݁ି௦௧
െݏቤ
ஶ
ࣦሼݑሺݐെܽሻሽൌ݁ି௦
െݏ
݂ሺݐሻݑሺݐെܽሻൌ൜Ͳݎ݂ݐ൏ܽ
݂ሺݐሻݎ݂ݐܽ
݂ሺݐെܽሻǤݑሺݐെܽሻ ݂ሺݐሻ
ࢇ
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
ߜᖡሺݐെܽሻൌͳᖡΤǡࢇ࢚ࢇᖡ
ൌͲǡݐ݄݅ݓݎ݁ݏ݁
ᖡ՜ͲǤ
ᖡ՜Ͳǡ
Ǥ
ߜሺݐെܽሻǣ ߜሺݐെܽሻൌλݐൌܽǢൌͲݐ്ܽǡ
munotes.in
Page 169
169Chapter 7: Inverse Laplace Transform
නߜሺݐെܽሻஶ
݀ݐൌͳሺܽͲሻ
ǡݓ
ݔൌܽ
ݓᖡ Τ ݔൌܽݔൌܽᖡǤ
ݓሺݔሻൌݓᖡΤǡࢇ൏࢞൏ࢇᖡ
ൌͲǡݐ݄ݎ݁ݓ݁ݏ݅
݅Ǥ݁Ǥݓሺݔሻൌݓߜሺݔܽሻ
Ǥܕܚܗܛܖ܉ܚ܂ ܗܝܖܜܑܕܑܘܝܔܛ܍ሺ۲܉ܚܑ܋۲܉ܜܔ܍ሻܖܗܑܜ܋ܖܝ Ǥ
݂ሺݐሻ
ݐݐൌܽǡ
න݂ሺݐሻஶ
ߜᖡሺݐെܽሻǤ݀ݐൌන݂ሺݐሻୟାᖡ
ᖡǤ݀ݐ
ൌሺܽᖡെܽሻ݂ሺߟሻͳ
ᖡൌ݂ሺߟሻǡݓ݄݁ݎ݁ܽ൏ߟ൏ܽᖡǡ
ᖡ՜ǡࢋ࢝ࢍࢋ࢚න݂ሺݐሻஶ
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
Ǥ ݅
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
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171Chapter 7: Inverse Laplace Transform
݅ൌͲǡ
ݐൌͲǤ
ܧȀܮ
7.5 Exercise
ͳǤ
ǣ
ሺ݅ሻͳ
ሺݏെͳሻହሺ࢙ͳ
ʹͶݐସ݁ି௧ሻ
ሺ݅݅ሻͶݏͳͷ
ͳݏଶെʹͷ൬࢙ͳ
Ͷܿݏ݄ͷ
Ͷݐ͵
Ͷݏ݄݅݊ͷ
Ͷݐ൰
ሺ݅݅݅ሻ͵ሺݏଶെͳሻଶ
ʹݏହ൬࢙͵
ʹെ͵
ʹݐଶͳ
ͳݐସ൰
ሺݒ݅ሻͳ
ݏଷଶΤቌ࢙ʹඨݐ
ߨቍ
ሺݒሻͳ
ξʹݏ͵൬࢙ͳ
ξʹߨݐ݁ିଷ௧ଶൗ൰
ʹǤ
ǣ
ሺ݅ሻͶݏͳʹ
ݏଶͺݏͳሺ࢙Ͷ݁ିସ௧ሺͳെݐሻሻ
ሺ݅݅ሻ͵ݏ
ݏଶെʹݏെ͵ሺ࢙Ͷ݁ଷ௧െ݁ି௧ሻ
ሺ݅݅݅ሻݏଶͳ
ݏଷ͵ݏଶʹݏ൬࢙ͳ
ʹെʹ݁ି௧ͷ
ʹ݁ିଶ௧൰
ሺݒ݅ሻݏʹͻ
ሺݏͶሻሺݏଶͻሻ൬࢙݁ିସ௧െܿݏ͵ݐͷ
͵ݏ݅݊͵ݐ൰
ሺݒሻݏʹ
ݏଷሺݏെͳሻଶሺ࢙ሺ͵ݐെͺሻ݁௧ݐଶͷݐͺሻ
ሺ݅ݒሻͳ
ݏଷሺݏଶͳሻ൬࢙ͳ
ʹݐଶܿݏݐെͳ൰
ሺ݅݅ݒሻݏଶെܽଶ
ሺݏଶܽଶሻଶሺ࢙ݐܿݏܽݐሻ
ሺ݅݅݅ݒሻݏ
ሺݏଶͳሻሺݏଶͶሻ൬࢙ͳ
͵ሺܿݏݐെܿݏʹݐሻ൰
ሺݔ݅ሻͳ
ሺݏଶܽଶሻଶ൬࢙ͳ
ʹܽଷሺݏ݅݊ܽݐെܽݐܿݏܽݐሻ൰ ሺݔሻݏݏସͶܽସ൬࢙ͳʹܽଶሺݏ݅݊ܽݐݏ݄݅݊ܽݐ൰ munotes.in
Page 172
172 APPLIED MATHEMATICS͵Ǥ
݉ݎ݂ݏ݊ܽݎݐ݂݄݂݁ܽܿݐ݄݂݈݈݁݅ݓ݊݃
ሺ݅ሻܽݏሺݏെܽሻሺܛܖۯୟ୲െͳሻ
ሺ݅݅ሻͳ
ݏሺݏଶܽଶሻ൬࢙ͳ
ܽଶሺͳെݏܿݐܽሻ൰
ሺ݅݅݅ሻͳ
ݏξݏͶ൬࢙ͳ
ʹ൫ʹξݐ൯൰
ሺ݅ݒሻͳ
ݏ݈݃ሺݏ͵ሻ
ሺݏʹሻቌ࢙න݁ିଶ௫െ݁ିଷ௫
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
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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 .
ࢋ࢙࢘ࢋ࢜ࡵ खି൛ࢌതሺ࢙ሻൟൌࢌሺ࢚ሻ ࣦିଵͳݏ൨ൌͳ ࣦିଵͳݏെܽ൨ൌ݁௧ ࣦିଵͳݏ൨ൌݐିଵሺ݊െͳሻǨǡ݊ൌͳǡʹǡ͵ǤǤ ࣦିଵͳሺݏെܽሻ൨ൌ݁௧ݐିଵሺ݊െͳሻǨ ࣦିଵͳݏଶܽଶ൨ൌͳܽܵ݊݅ܽݐ ࣦିଵቂݏݏଶܽଶቃൌݏܿܽݐ ࣦିଵͳݏଶെܽଶ൨ൌ݄ܵ݊݅ܽݐ ࣦିଵቂݏݏଶെܽଶቃൌݏ݄ܿܽݐ ࣦିଵͳሺݏെܽሻଶܾଶ൨ൌͳܾ݁௧ܵ݊݅ݐܾ ࣦିଵݏെܽሺݏെܽሻଶܾଶ൨ൌ݁௧ܥݏݐܾ ࣦିଵݏሺݏଶܽଶሻଶ൨ൌͳʹܽݐܵ݊݅ܽݐ ࣦିଵͳሺݏଶܽଶሻଶ൨ൌͳʹܽଷሺܵ݊݅ܽݐെܽݐݏܿܽݐሻ
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174 APPLIED MATHEMATICSܖܑܜܑܐ܁܂ܐܕ܍ܚܗ܍ǣ
ሺࡵሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻǡݐ݄ࣦ݊݁ିଵ൛݂ҧሺݏെܽሻൟൌ݁௧݂ሺݐሻൌ݁௧ࣦିଵ൛݂ҧሺݏሻൟ
ሺࡵࡵሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻ݂ܽ݊݀ሺͲሻൌͲǡݐ݄ࣦ݊݁ିଵ൛ݏ݂ҧሺݏሻൟൌ݀
ݐ݀ሼ݂ሺݐሻሽ
ൌ݂ᇱሺݐሻ
ሺࡵࡵࡵሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻݐ݄ࣦ݊݁ିଵቊ݂ҧሺݏሻ
ݏቋൌන݂ሺݐሻݐ݀௧
ሺࢂࡵሻ݂ܫࣦିଵ൛݂ҧሺݏሻൟൌ݂ሺݐሻݐ݄݊݁ݐ݂ሺݐሻൌࣦିଵ൜Ȃ݀
ݏ݀ൣ݂ҧሺݏሻ൧ൠ
ሺࢂሻࣦቆ݂ሺݐሻ
ݐቇൌන݂ҧሺݏሻݏ݀ஶ
௦
܉۾ܑܜܚܔ܉܉ܚ܋ܖܗܑܜۻܗܐܜ܍ܛ܌
݂ҧሺݏሻ
ܨഥሺݏሻ
ܩҧሺݏሻ
ܨഥሺݏሻܩҧሺݏሻ
ܣ
ሺܽݏܾሻǡܣ
ሺܽݏଶܾݏܾሻሺݎൌͳǡʹǡǥሻ
ǡࣦିଵ൛݂ҧሺݏሻൟ
ࣦିଵ൛ݏ݂ҧሺݏሻൟൌൌ݀
ݐ݀ሼ݂ሺݐሻሽǡݎ݅ݒ݂݀݁݀ሺͲሻൌ݂ᇱሺͲሻൌڮൌ݂ିଵሺͲሻൌͲ
ࣦሼݐ݂ሺݐሻሽൌሺെͳሻ݀
݀ݏ݂ҧሺݏሻൌሺെͳሻ݂ҧሺሻሺݏሻݓ݄݄ܾ݅ܿ݊ܽܿ݁ݔ݁ݎ݁ݏݏ݁݀ܽݏ
खି൛ࢌሺሻሺ࢙ሻൟൌሺെሻ࢚ࢌሺ࢚ሻ
܃܍ܛܗ۱ܖܗܞܔܗܝܖܗܑܜ܂ܐܕ܍ܚܗ܍
۷खି۴തሺ࢙ሻൌࡲሺ࢚ሻǡखି۵ഥሺ࢙ሻൌࡳሺ࢚ሻࢇࢊࢌഥሺ࢙ሻൌࡲഥሺ࢙ሻכࡳഥሺ࢙ሻࢋࢎ࢚
൛खିࢌതሺ࢙ሻൟൌखିሼࡲഥሺ࢙ሻכࡳഥሺ࢙ሻሽൌනࡲሺ࢚െ࢛ሻࡳሺ࢛ሻ࢛ࢊ࢚
ǣ
खି൬
࢙൰ൌࢇࢊखିࢌതሺ࢙ሻൌࢌሺ࢚ሻ
खିቊࢌሺ࢙ሻ
࢙ቋൌනǤࢌሺ࢛ሻ࢛ࢊ࢚
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
ǡܽǤ ܪሺݐെܽሻǤ
۲܋܉ܚܑെ܍܌ܔ܉ܜ۴ܖܗܑܜ܋ܖܝሺ܃ܜܑܖܕ۷܍ܛܔܝܘ۴ܖܗܑܜ܋ܖܝሻ
܃ܜܑܖ܍ܛܔܝܘܕܑ ሺ۲܋܉ܚܑ۲܉ܜܔ܍ሻ
ߜᖡሺݐെܽሻൌͳᖡΤǡࢇ࢚ࢇᖡ
ൌ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
Where the integral signs are written in order of i ntegration taken from the right, munotes.in
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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
නݔ݀න݂ሺݔǡݕሻ݀ݕ௬మሺ௫ሻ௬భሺ௫ሻ࢘න݀ݕන݂ሺݔǡݕሻݔ݀௫మሺ௬ሻ௫భሺ௬ሻ 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 FRQVW ZKLFKDUHVWUDLJKW - 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
munotes.in
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190 APPLIED MATHEMATICS ൌන݀Ʌగସିగ
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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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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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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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
munotes.in
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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
ݔݕݖͳൌͳ munotes.in
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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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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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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
ൌͳʹනݎଶ݀Ʌఉఈ munotes.in
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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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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
݀ munotes.in
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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.
[The second term vanishes as the integrand is an odd function] munotes.in
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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
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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
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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
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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
munotes.in
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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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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 ī Q ࢋି࢞࢞ିࢊ࢞ஶ ( 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. ī Q ࢋି࢞࢞ିࢊ࢞ஶ
Proof: ī Q ݁ି௫ݔିଵ݀ݔஶ
Put x = t2 , dx = 2t dt
= ݁ି௧మݐଶିଶʹݐ݀ݐஶ
= ʹ݁ି௧మݐଶିଵ݀ݐஶ
ī Q ࢋି࢚࢚ିࢊ࢚ஶ ---------------------------------- (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 ī Q ݁ି௫ݔିଵ݀ݔஶ
put n =1
ī 1) = ݁ି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 :
ī Q Qī Q
Proof: By definition ī Q ݁ି௫ݔିଵ݀ݔஶ
Replace n by n+1
ī Q+1 ) = ݁ି௫ݔ݀ݔஶ
Now, integrating by parts
ī Q+1 ) = ሾݔሺെ݁௫ሻሿஶെ݊ݔିଵሺ݁ି௫ሻ݀ݔஶ
Now,
௫՜ஶ௫
ೣൌͲǡͲǡ௫
ೣൌͲݔൌͲቂ௫
ೣቃ
ஶ
= 0
ī Q+1 ) = 0 + n ݁ି௫ݔିଵ݀ݔஶ
= n ī Q) ī Q Qī Q
If n is a positive integer ,
ī Q+1 ) = n (n -1) ī Q-1 ) ī Q = (n-1) ī Q-1 )
= n (n -1) (n -2) ī Q-2 )
= n (n -1) (n -2) (n -3) (n - ««««« ī(1 )
= n (n -1) (n -2) (n -3) (n - ««««« ī 1) = 1 = n!
ī Q+1 ) = n! if n is a positive integer
ī Q Qī Q LQJHQHUDOQLVUDWLRQDOQXPEHU = n! if n is a positive integer
4. ī ī Q ሺ୬ାଵሻ ī ሺଵሻൌଵൌλ 5. ī ଵ
ଶ ) = ξߨ
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221Chapter 10: Beta and Gamma Functions
6. ī Q 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 ------ ī ଵ
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
ī Q ) = 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 – ࢋି࢞ࢇࢊ࢞ 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 ī Q ࢋି࢞࢞ିࢊ࢞ஶ
( 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
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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͵
ͺ ൌනݔͻ݁െʹݔʹͲݔ݀ൌ͵ͺ munotes.in
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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:
ܮݐ݁ ൌනξݔͶ݁െξݔ
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
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227Chapter 10: Beta and Gamma Functionsൌʹන݁ି௧ݐଷଶൗ݀ݐ
ൌʹන݁ି௧ݐହ
ଶିଵ
݀ݐ
ൌʹȞ൬ͷ
ʹ൰ሺ
ሻ
ൌʹȞ൬ͷ
ʹͳ൰ൌʹ൬͵
ʹ൰Ȟ൬͵
ʹ൰ൌʹ൬͵
ʹ൰൬ͳ
ʹ൰Ȟ൬ͳ
ʹ൰
ൌ͵൬ͳ
ʹ൰ξɎൌ൬͵
ʹ൰ξɎ
܍ܘܡ܂۷۷ൌන࢞ࢊࢇ࢞ஶ
ܽ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܍ǣ ൌනݔଷ
͵௫ஶ
݀ݔ
ܗ܁ܔܖܗܑܜܝǣ ൌනݔଷ
͵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 ǣ ൌනݔସ
Ͷ௫ஶ
ݔ݀
ܗ܁ܔܖܗܑܜܝǣ ൌනݔସ
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
Page 229
229Chapter 10: Beta and Gamma Functions ൌනݔସͶ௫ஶ݀ݔൌනݐͶ൨ସஶͳ݁௧݀ݐͶ
ൌනݐସ
ሺͶሻହஶ
݁ି௧݀ݐൌනݐହିଵ
ሺͶሻହஶ
݁ି௧݀ݐൌͳ
ሺͶሻହනݐହିଵஶ
݁ି௧݀ݐ
ൌͳ
ሺͶሻହȞሺͷሻൌͶǨ
ሺͶሻହൌʹͶ
ሺͶሻହ
܍ܘܡ܂۷۷۷ൌනሺͳݔሻ݀ݔࡾࡻනሺെ݈݃ݔሻ݀ݔ
݈݃ͳ
ݔൌݐࡾࡻ݈݃ݔൌെݐࡾࡻݔ
ൌ݁ି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܍ǣ ൌනݔ݀ݔ
ට݈݃ͳ
ݔଵ
ܗ܁ܔܖܗܑܜܝǣ ൌනݔ݀ݔට݈݃ͳݔଵെെെെെെെെെെെሺܣሻ munotes.in
Page 230
230 APPLIED MATHEMATICSͳݔൌݐࡾࡻͳݔൌ݁௧ࡾࡻݔൌ݁ି௧
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 ǣ ൌන݀ݔ
ඥݔ݈ݔ݃ଵ
ܗ܁ܔܖܗܑܜܝǣ ൌන݀ݔ
ඥݔ݈ݔ݃ଵ
െെെെെെെെെെെሺܣሻ
ͳ
ݔൌݐࡾࡻͳ
ݔൌ݁௧ࡾࡻݔൌ݁ି௧
ݔൌ݁ି௧െെെെെെെെെሺ݅ሻ
݀ݔ
݀ݐൌെ݁ି௧݀ݔൌെ݁ି௧݀ݐ
Now we check limits points from (i)
When ݔൌͲ֜݁ି௧ൌͲǡൌ munotes.in
Page 231
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))
۰ሺܕǡܖሻൌන࢞ିሺെ࢞ሻି
ࢊ࢞ǡͲǡ݊Ͳ
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
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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
܁ܖܗܑܜܝܔܗǣ ൌනݔଷ൫ͳെξݔ൯ହଵ
݀ݔ
ݐݑξݔൌݐǡݔൌݐଶܦ݅݁ݎ݂݂݁݅ݐ݊ܽ݊݅ݐ݃ݓǤݎǤݐǤƍݐƍݓݐ݁݃݁݀ݔ
݀ݐൌʹݐ֜݀ݔ
ൌʹݐ݀ݐ
ݔൌݐଶ
ݔൌͲǡݐଶൌͲ֜ൌͲǡ݀݊ܽݔൌͳǡݐଶൌͳ֜ൌͳ
x 0 1 t 0 1
ൌනሺݐଶሻଷሺͳെݐሻହଵ
ʹݐ݀ݐ ൌනሺݐሻሺͳെݐሻହଵʹݐ݀ݐൌʹනݐሺͳെݐሻହଵ݀ݐ munotes.in
Page 235
235Chapter 10: Beta and Gamma Functions ൌනݐ଼ିଵሺͳെݐሻିଵଵ݀ݐൌʹܤሺͺǡሻെെെെെെെሺݕܤ݂݅݊݅݁ܦݐ݊݅ሻ
ǡሺǡሻൌȞሺሻȞሺሻ
Ȟሺሻ
ൌȞሺͺሻȞሺሻ
ȞሺͺሻൌʹǤሺǨሻሺͷǨሻ
ሺͳ͵Ǩሻൌͳ
ͷͳͶͺ
܍ܔܘܕ܉ܠ۳ ǣනݔ݀
ξͳെݔଷయଵ
ܗ܁ܔܖܗܑܜܝǣ ൌනݔ݀
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
܁ܔܗܑܜܝܖܗǣ ൌනξܽݐ݊ߠ݀ߠൌනඨ݊݅ݏߠ
ܿ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 Functions ൌͳʹȞ൬ͳെͳͶ൰Ȟ൬ͳͶ൰ൌͳʹߨߨͶൌͳʹߨͳξʹൌξʹߨʹൌߨξʹ
܍ܜܗۼǣ
නξߠݐܿ݀ߠൌߨ
ξʹగȀଶ
ܔܘܕ܉ܠ۳܍ǣ ξݐܿߠ݀ߠగȀଶ
܁ܔܗܝܜܑܗܖǣ ൌනξݐ݊ܽߠ݀ߠൌනඨݏܿߠ
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
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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
ሺͳെξݔሻହ݀ݔ൫ܜܖܑ۶ξܠൌݐ൯࢙ǣͳ
ͷͳͶͺ
ͺǤන݊
ሺ݊െݔሻ݀ݔሺܜܖܑ۶ݔ
ൌ݊ݐሻ࢙ǣ݊ାାଵܤሺ݊ͳǡͳሻ 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
ī Q ݁ି௫ݔିଵ݀ݔஶ
( n > 0)
ī Q ʹ݁ି௧మݐଶିଵ݀ݐஶ
ī
ī Q Qī Q
ī Q Qī Q LQJHQHUDOQLVUDWLRQDOQXPEHU
= n! if n is a positive integer
ī ī ଵ
ଶ ) = ξߨǡī Q Q
Type I – ݁ି௫
݀ݔǡ ൌௗ௫
ೣஶ
ǡ ൌ
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܍െ۷۷۷ሺહሻൌනࢌሺ࢞ǡࢻሻࢊ࢞ࢋࢎ࢚࢈
ࢇ܌۷
܌હൌනࣔ
ࣔࢻࢌሺ࢞ǡࢻሻࢊ࢞࢈
ࢇ
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܁ܔܖܗܑܜܝǣ ൌන݀ݔ
ܾܽ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
ሺȽሻ ሺȽሻൌԄሺȽǡǡሻǡǡ
Ƚൌμ μȽሺ ሻμ μȽμ μȽെെെെെെെെെെെെെሺ ሻ
I
Į
b
a
Į munotes.in
Page 252
252 APPLIED MATHEMATICSǡ ሺȽሻൌන݂ሺݔǡߙሻ݀ݔܻܤ݃݊݅ݏݑܴݑ݈݁ܫǡୠሺఈሻୟሺఈሻμ μȽሺ ሻ
ǡ
ǡ
ǡμ
μȽൌනμ
μȽ݂ሺݔǡߙሻ݀ݔୠሺఈሻ
ୟሺఈሻെെെെെെെെെെെെെെെെሺ ሻ
නሺǡȽሻൌɗሺǡȽሻǡ
ǤǤμ
μሾɗሺǡȽሻሿൌሺǡȽሻെെെെെെെെെെെെെെെെെሺ ሻ
ԄሺȽǡǡሻൌ ሺȽሻ
ൌන݂ሺݔǡߙሻ݀ݔൌሾɗሺǡȽሻሿୟሺఈሻୠሺఈሻൌɗሺǡȽሻെɗሺǡȽሻୠሺఈሻ
ୟሺఈሻെെ
െെെെെെെെെെെെെെെെെെെെെെሺܸܫሻ
ǡǡ
μ
μൌμԄ
μൌμ
μɗሺǡȽሻൌሺǡȽሻሺ ሻെെെെെെെെെሺሻ
μ
μൌμԄ
μൌെμ
μɗሺǡȽሻൌെሺǡȽሻሺ ሻെെെെെെെሺ ሻ
ሺ ሻǡሺሻǡሺ ሻሺ ሻ
܍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 ǣ ൌනሺሻሺെሻǡଶ
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܋ܖܗܑܜǣ ሺͳሻǡ
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
Ǥ ǤǤሺͻሻሺሻ
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
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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
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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
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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
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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܍ǣ Ƚሺሻൌඨʹ
Ɏන୲మ
ିଶሺሻൌȽሾξʹሿ୶
ܖܗܑܜܝܔܗ܁ Ƚሺሻൌඨʹ
Ɏන୲మ
ିଶ୶
Ƚ൫ξʹ൯ൌඨʹ
Ɏන୲మ
ିଶ୶ξଶ
ଶൌʹଶǡʹൌͶǡൌʹ
ൌʹ
ξʹǤൌξʹ
t 0 xξʹ
u 0 x
ൌඨʹ
Ɏනି୳మξʹǤ୶
ൌʹ
ξɎනି୳మξʹǤൌሺሻ୶
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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
Ǥ ሺሻൌනሺെሻଶ୶
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 FunctionsͻǤ ԄሺሻൌනǡԄୟమୟ
ͳͲǤ
නͳ
ୟమ
ୟ
ͳͳǤ ሺͲሻǡሺλሻǡ
ሺͲሻ
ͳʹǤ
ሺ୬ሻ
ͳ͵Ǥ
൫ξ൯
ͳͶǤනିሺ୶ାୟሻమஶ
ൌξɎ
ʹሾͳെሺሻሿ
ͳͷǤሺሻǡ
ሺሻǡ൫ξ൯ǡ
൫ξ൯Ǥ
11.8 Summary
܍ܔܝ܀െ۷۷۷ሺહሻൌනࢌሺ࢞ǡࢻሻࢊ࢞ࢋࢎ࢚࢈
ࢇ܌۷
܌હൌනࣔ
ࣔࢻࢌሺ࢞ǡࢻሻࢊ࢞࢈
ࢇ
܍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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