Differential Equations – Complex Roots

In order to achieve complex roots, we have to look at the differential equation:
Ay” + By’ + Cy = 0

Then we look at the roots of the characteristic equation:
Ar² + Br + C = 0

After solving the characteristic equation the form of the complex roots of r1 and r2 should be:
λ ± μi

We refer back to the characteristic equation, we then assume that all the solution to the differential equation will be:
y(t) = e^(rt)

By plugging in our two roots into the general formula of the solution, we get:
y1(t) = e^(λ + μi)t
y2(t) = e^(λ – μi)t

Since these two functions are still in complex form, and we started the differential equation with real numbers. It would best if our solution is also real numbers. In order to transform the complex solution into a real solution, we need to use the Euler’s Formula.
e^(iƟ) = cosƟ + isinƟ
Another form of the Euler’s Formula is:
e^(-iƟ) = cos(-Ɵ) + isin(-Ɵ)
e^(-iƟ) = cosƟ – isin(Ɵ)

Now we split up both of the solutions into two parts one with real exponent and one with an imaginary exponent. Afterwards, we then applied Euler’s Formula.
y1(t) = e^(λt) e^(μit) = e^(λt) [cos(μt) + isin(μt)]
y2(t) = e^(λt) e^(-μit) = e^(λt) [cos(μt) – isin(μt)]

After doing all of that work, we are still left with a part of a complex solution. But we are able to get rid of the complex part. We can rewrite this solution as:
y(t) = c1y1(t) + c2y2(t)
y1(t) + y2(t) = 2e^(λt) cos(μt)
Simply it further:
u(t) = ½y1(t) + ½y2(t) = e^(λt) cos(μt)
Therefore:
c1 = c2 = ½

We can get the second part of the solution by subtracting the two original solutions:
y1(t) – y2(t) = 2ie^(λt) sin(μt)

At a glance, it still look like a complex solution, but looking at the two constants c1 and c2 we can result in a real solution by dividing it by 2i.
c1 = ½i
c2 = -½i

The second solution would be:
v(t) = ½i y1(t) – ½i y2(t) = e^(λt) sin(μt)

All of the work just to achieve the two real solutions:
u(t) = e^(λt) cos(μt)
v(t) = e^(λt) sin(μt)

These two real solutions is also a general solution. So if the roots of the characteristic equation result in r1,2 = λ ± μi, the general solution to this would be:
y(t) = c1 e^(λt) cos(μt) + c2 e^(λt) sin(μt)

EXAMPLE
y” – 10y’ + 29y = 0, y(0) = 1, y'(0) = 3

The characteristic equation would be:
r² – 10r + 29 = 0

The roots would be:
5 ± 2i

General Solution:
y(t) = e^(5t)[c1 cos(2t) + c2 sin(2t)]
y'(t) = 5e^(5t)[cos(2t) + c2 sin(2t)] + e^(5t)[-2 sin(2t) + 2c2 cos(2t)]

We use the initial values to find the constants.
y(0) = 1
We substitute all the t’s with 0’s and set it to equal 1.
y(0) = e^(5(0))[c1 cos(2(0)) + c2 sin(2(0))] = 1
1[c1(1) + c2(0)] = 1
“c1 = 1

y'(0) = 3
We substitute all the t’s with 0’s and set it to equal 3.
y'(0) = 5e^(5(0))[cos(2(0)) + c2 sin(2(0))] + e^(5(0))[-2 sin(2(0)) + 2c2 cos(2(0))] = 3
5[1 + 0] + 1[0 + 2c2] = 3
5 + 2c2 = 3
c2 = -1

Particular Solution:
y = e^(5t)[cos(2t) – sin(2t)]

If you still do not understand this study guide you may go visit these websites or you can personally ask me, they go into more detail solving the differential equation of complex root solutions with characteristic equation. Thank you.

Videos from Khan Academy
Part I

Part II

Part III