First off - this isn't a homework question, just something I got wrong on a midterm and can't quite figure out.
The question:
Given that $e^t cos(2t)$ is a solution of the constant coefficients differential equation $y''' + ay''-3y'+by=0$, find a and b.
Here's what I've done so far:
Since $e^t cos(2t)$ is a solution, I know $e^t sin(2t)$ must also be a solution. This means that $(1+2i)$ and $(1-2i)$ must be roots of the characteristic equation. We clearly aren't done though, as if we use $ \lambda $ for $y'$, our characteristic equation would be:
$$ ( \lambda - ((1+2i))( \lambda - ((1-2i))$$ or, expanded,
$$ \lambda^2 - 2\lambda -5 $$ or
$$y''-2y'-5y.$$
I know I need another simple root (I know now that it's $\lambda - 4$, giving a,b=2,20) but I can't for the life of me figure out how to find that root. I've tried setting $y''' + ay''-3y'+by= ( \lambda - ((1+2i))( \lambda - ((1-2i))(\lambda - x)$ and solving for x algebraically, but I keep getting stuck with too many variables.
What am I missing?
Thanks in advance!
Instead of worrying about the characteristic equation, you can substitute the two solutions you know into the ODE and solve the resulting system of two equations with the two unknowns $a$ and $b$.
For instance,
$$y=e^t\cos(2t)\implies\begin{cases}y'=e^t(\cos(2t)-2\sin(2t))\\ y''=e^t(-3\cos(2t)-4\sin(2t))\\y'''=e^t(-11\cos(2t)+2\sin(2t))\end{cases}$$
$$y'''+ay''-3y'+by=0\implies (-3a+b-14)\cos(2t)-4(a-2)\sin(2t)=0$$
Similarly, $y=e^t\sin(2t)$ gives
$$(-3a+b-14)\sin(2t)+4(a-2)\cos(2t)=0$$
Now just solve for the coefficients.
If you want to continue with your method, you know the characteristic polynomial has roots $1\pm2i$, so $(\lambda-1-2i)(\lambda-1+2i)=\lambda^2-2\lambda+5$ divides it exactly and leaves no remainder. We have
$$\frac{\lambda^3+a\lambda^2-3\lambda+b}{\lambda^2-2\lambda+5}=\lambda+a+2+\frac{(2a-4)\lambda-5a+b-10}{\lambda^2-2\lambda+5}$$
For the remainder to be $0$, you need
$$\begin{cases}2a-4=0\\-5a+b-10=0\end{cases}$$