Consider the following DE.
$$ y''+4y'+\lambda y, \space\space\space\space\space y'(0)=y'(\pi)=0$$
Is $\lambda$=8 an eigenvalue? If so, what is the corresponding eigenfunction?
I did the following, skipping a few steps:
$y(x)=Ae^{-2x}cos(2x)+Be^{-2x}sin(2x)$
$y'(x)=-2Ae^{-2x}cos(2x)-2Ae^{-2x}sin(2x)-2Be^{-2x}sin(2x)+2Be^{-2x}cos(2x)$
$y'(0)\Rightarrow A=B$
Updating function
$y(x)=A(e^{-2x}cos(2x)+e^{-2x}sin(2x))$
$y'(x)=-4Ae^{-2x}sin(2x)$
$y'(\pi)= -4Ae^{-2\pi}sin(2\pi)=0$
Since $sin(2\pi)=0$ this is a non-trivial solution.
Taking A=1, would the corresponding eigenfunction simply be:
$y(x)=e^{-2x}cos(2x)+e^{-2x}sin(2x)$ $\space$ ?
I think I got the answer but just want to make sure I'm not doing anything wrong.
Thanks!