Case Three: Complex Roots
Chapter One Hundred Twenty-One
Syllabus topic Module 2, "2.3 Linear Differential Equations with Constant Coefficients"
Pages 269 to 270 of 303
In one line
A complex pair a plus or minus ib gives e to the ax times (A cos bx plus B sin bx), and Euler's formula is what turns the complex answer real.
The case
The auxiliary equation has a negative discriminant, so its roots are a conjugate pair.
m = a + i b
and its conjugate a minus ib. They always come in pairs because the coefficients are real, which is the fact the conjugate chapter of Module 1 established.
The answer
y = e^(a x)(A cos(b x) + B sin(b x))
The real part of the root becomes the exponential, and the imaginary part becomes the frequency.
Where it comes from: Module 1 paying for itself
The auxiliary equation's roots are complex, so the two solutions are e to the (a plus ib)x and e to the (a minus ib)x. The complementary function is a combination of them.
Split each exponential: e to the ax times e to the ibx. And by Euler's formula, which the chapter on the exponential form of a complex number proved:
e^(i b x) = cos(b x) + i sin(b x)
e^(-i b x) = cos(b x) - i sin(b x)
So a combination P e to the (a+ib)x plus Q e to the (a-ib)x becomes e to the ax times ((P plus Q)cos bx plus i(P minus Q)sin bx).
Naming A for (P plus Q) and B for i(P minus Q), which is legitimate because P and Q are arbitrary, gives the real answer.
The arbitrary constants absorbed the i. That is the whole of the argument, and it is why a real differential equation with complex roots still has a real solution.
And the two combinations that do it are precisely the two from Module 1: adding Euler's formula to its conjugate gives twice the cosine, and subtracting gives twice i times the sine.
(e^(i b x) + e^(-i b x))/2 = cos(b x)
(e^(i b x) - e^(-i b x))/(2i) = sin(b x)
Worked: a pair with no real part
Solve y'' plus 9y = 0.
The auxiliary equation is m squared plus 9 = 0, so m is plus or minus 3i. Here a is 0 and b is 3.
d2y/dx2 + 9y = 0
y = C1 cos(3x) + C2 sin(3x)
With a = 0 there is no exponential, so the solution oscillates for ever without growing or decaying. That is simple harmonic motion, and this is its equation.
Worked: a decaying oscillation
Solve y'' plus 2y' plus 5y = 0.
The auxiliary equation is m squared plus 2m plus 5 = 0, whose discriminant is 4 minus 20, that is minus 16. So the roots are minus 1 plus or minus 2i, and a is minus 1 with b equal to 2.
Case Three: Complex Roots
d2y/dx2 + 2 dy/dx + 5y = 0
y = C1 e^(-x) cos(2x) + C2 e^(-x) sin(2x)
A decaying oscillation: it swings, and each swing is smaller. That is underdamping, and it is what a plucked guitar string does.
Worked: a growing oscillation
Solve y'' minus 2y' plus 5y = 0.
The roots are 1 plus or minus 2i, so a is plus 1.
d2y/dx2 - 2 dy/dx + 5y = 0
y = C1 e^x cos(2x) + C2 e^x sin(2x)
An oscillation whose amplitude grows without limit. In an engineered system that is a fault, and the sign of a is what tells you: a negative means stable, a positive means unstable, and zero means it oscillates for ever.
The alternative form, and why it is worth knowing
The answer can equally be written as a single sine or cosine with a phase.
A cos(b x) + B sin(b x) = sqrt(A^2 + B^2) cos(b x - atan(B/A))
The identity is written for a positive A; for a negative A the same expression holds with pi added to the phase, because the arctangent only knows the ratio and not the quadrant, which is exactly the trap the argument chapter of Module 1 warned about.
That form separates the amplitude, which is the square root of A squared plus B squared, from the phase, which is the arctangent of B over A. Both are what a physical measurement actually gives you, so a question about amplitude or phase wants this form.
And notice where it comes from: the modulus and the argument of the complex number A plus iB, which is the polar form of Module 1 in a new costume.
Fitting initial conditions
Solve y'' plus 4y = 0 with y(0) = 3 and y prime of 0 = 2.
The general solution is C1 cos 2x plus C2 sin 2x. At x = 0 it gives C1 = 3. Its derivative is minus 2C1 sin 2x plus 2C2 cos 2x, which at 0 gives 2C2 = 2, so C2 = 1.
d2y/dx2 + 4y = 0
y = 3 cos(2x) + sin(2x)
Check yourself
d2y/dx2 + y = 0
y = C1 cos(x) + C2 sin(x)
d2y/dx2 + 6 dy/dx + 13y = 0
y = C1 e^(-3x) cos(2x) + C2 e^(-3x) sin(2x)
d2y/dx2 + 16y = 0
y = C1 cos(4x) + C2 sin(4x)
For the second, the discriminant is 36 minus 52, that is minus 16, so the roots are minus 3 plus or minus 2i: a is minus 3 and b is 2.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.