Non zero solutions for $y''+(1+x)y=ay,0<x<1,y(0)=y(1)=0$

Question

For which $a$ does the differential equation $$y''+(1+x)y=ay,0<x<1,y(0)=y(1)=0$$ have a non zero solution?

Answer 1

Plotting $y(1)$ (for initial values $y(0)=0$, $y'(0)=1$) in dependence of $a$ gives the following picture:

From the table underlying the graph one finds suitable $a$ close to $-8.375$, $-37.981$, $-87.328$, $-156.41$, $-245.24$.

Try to argue why there are no solutions for $a>2$.

For large negative $a$ the $x$ is a small perturbations to $(-a+1)$ in $$ y''+(-a+1+x)y=0. $$ Disregarding $x-0.5$ gives the standard harmonic oscillator $$y''+\omega^2y=0$$ with solutions satisfying the boundary conditions $$y(x)=\sin(n\pi x),$$ which means that $-a+1.5\approx\omega^2= (n\pi)^2$. The first values are \begin{array}{l|l}n&a\approx1.5-(n\pi)^2\\\hline 1 & -8.36960440109 \\ 2 & -37.9784176044 \\ 3 & -87.3264396098 \\ 4 & -156.413670417 \\ 5 & -245.240110027 \\ 6 & -353.805758439 \\ 7 & -482.110615653 \\ 8 & -630.15468167 \\ 9 & -797.937956488 \\ \end{array} which is close enough to the numerical results for a first approximation.