Complex Values in Second Order Differential Equations

Question

Recently I've learned that second order linear homogeneous differential equations can be solved by assuming the function to be something like this. $$Ay''+By'+Cy=0$$ $$y = e^{St} $$ $$AS^2+BS+C=0$$ When encountering underdamped systems, the value of S would be imaginary, leaving you with Euler's identity. $$e^{i\alpha}=\cos(\alpha t)+i\sin(\alpha t)$$

When solving for the fundamental solutions our professor disregarded the imaginary coefficient and claimed that the fundamental solutions are the imaginary component and the real component. How is this so??

Answer 1

For a linear equation, the sum of two solutions is a solution. If $A,B,C$ are real, then the complex conjugate of a solution is a solution. The real part of a complex function $f\left(x\right)$ is $\left(f\left(x\right)+f^{*}\left(x\right)\right)/2$ and the imaginary part is $\left(f\left(x\right)-f^{*}\left(x\right)\right)/2i$. So, given a solution, it's complex conjugate is a solution (when $A,B,C$ are real) and so the real and imaginary parts are solutions.

Answer 2

I am puzzled by your saying that "our professor disregarded the imaginary coefficient". What "imaginary coefficient" do you mean? What you are talking about requires that your differential equation has all real coefficients so that a solution must be real. The characteristic equation is $AS^2+ BS+ C= 0$ which, since it has real coefficients, if it has complex solution, must have two complex conjugate solutions, S= a+ bi and S= a- bi. The general solution to the differential equation is $y(t)= Pe^{(a+ bi)t}+ Qe^{(a- bi)t}= Pe^{at}e^{bit}+ Qe^{at}e^{-bit}$. An obvious thing to do is factor out $e^{at}$ to get $y(t)= e^{at}(Pe^{bit}+ Qe^{-bit})$. Now use Euler's identity: $y(t)= e^{at}(P(cos(bt)+ isin(bt)+ Q(cos(bt)+ isin(bt))= e^{at}((P+ Q)cos(bt)+ i(P+ Q)sin(bt))$.

We can write that as $y(t)= e^{at)(Acos(by)+ Bsin(bt))$ where I have NOT "ignored" any thing- I have include "i" in the coefficient B. I can reasonably expect that, since this is to be a solution of a differential equation with real coefficients with initial or boundary constraints that are real numbers, those coefficients, A and B, are real numbers.

Answer 3

This transformation is functional only for linear differential equations with constant coefficients.

Note that if $i \alpha $ is solution $-i \alpha$ is also solution because the characteristic polynomial has real coefficients.

Now if the solution is

$y = C_1e^{i \alpha t}+C_2 e^{-i \alpha t}$

This is equivalent to

$ y = D_1 \sin(\alpha t)+D_2\cos(\alpha t) $

with $D_1, D_2$ real constants, because solving for initial conditions

$ y(0) = C_1+C_2 = y_0\\ y'(0)= i\alpha C_1-i\alpha C_2 = y'_0 $

Solving for $C_1,C_2$ gives

$$ C_1 = \frac{\alpha y_0-i y'_0}{2\alpha}\\ C_2 = \frac{\alpha y_0+i y'_0}{2\alpha} $$

and finally

$$ y = y_0 \cos(\alpha t)+\frac{y'_0}{\alpha}\sin(\alpha t) $$

hence $D_1 = \frac{y'_0}{\alpha}$ and $D_2 = y_0$