Sturm Liouville Eigenvalue Problem for boundary conditions I have the following boundary value problem that I am struggling to solve.
$$y''-\lambda y=0 , y(1)=y(2)=0$$
I am working through the different cases for $\lambda$ but am unsure how to solve the last case:
$\lambda =0,$ $$y=ax+b$$ $$0=a+b=2a+b$$ this gives trivial solution $a=b=0$
$\lambda = \mu ^2>0$, $$0=a\sinh(\mu)+b\cosh(\mu)=a\sinh(2\mu)+b\cosh(2\mu)$$ this also gives the trivial solution $a=b=0$
$\lambda = -\mu ^2<0$, $$0=a\sin(\mu)+b\cos(\mu)=a\sin(2\mu)+b\cos(2\mu)$$ I know this solution has to be the non-trivial one but am unsure of a method to find out the general equation for $y$ and $\lambda$?

For the last case where $\lambda =-\mu^2$, you are correct in that $$0 = a \sin\mu + b \cos \mu =  a \sin2\mu + b \cos2 \mu, $$ which can be written as a linear system
$$ \begin{bmatrix}
\sin \mu & \cos \mu \\
\sin 2\mu & \cos 2\mu
\end{bmatrix} \begin{bmatrix} a \\
b 
\end{bmatrix} = \begin{bmatrix} 0 \\
0 
\end{bmatrix}.$$
In order for this system not to have the trivial solution, the determinant of the matrix must be zero:
$$ \sin \mu \cos 2\mu - \cos \mu \sin 2 \mu = \sin(\mu - 2\mu) = -\sin \mu= 0.$$
This condition will give you what $\mu$ is equal to, and there will be countably infinite values of $\mu$. The solution $y$ can be written as a superposition of all of these (eigen)functions, each scaled up to an arbitrary constant that cannot be solved for.
