The given problem is
$$\displaystyle -\frac{d^2y}{dx^{2}}+(1+x)y=\lambda y, \qquad x\in(0,1), \qquad y(0) = y(1) = 0$$
This equation can be transformed into a Sturm-Liouville problem of the form
$$-\frac{d}{dx} \left[\frac{d}{dx}\right]y+(1+x)y=\lambda y\ $$
I do not get how to proceed, as I only know the way of solving equations of the kind
$$\displaystyle \frac{d^2y}{dx^2}+\lambda y=0$$
Please help.
Thank you.
On
This Schroedinger equation can be in Airy Eq. form as $$\frac{d^2 \psi}{dy^2}-y \psi(y)=0,~~~ y=1-\lambda+x$$ See https://en.wikipedia.org/wiki/Airy_function This equation has two linearly independent solutions as $Ai(y), Bi(y)$ So $$\psi(x)=C_1~ Ai(y(x) + C_2 ~Bi(y(x))$$ by using $\psi(0)=0$, we get $$C_1 ~ Ai(1-\lambda) + C_2 ~ Bi (1-\lambda)= 0 \implies \frac{C_1}{C_2}= -\frac{Bi(1-\lambda)}{Ai(1-\lambda)}.~~~~(1)$$ Next by using $\psi(1)=0$ we get $$C_1 ~ Ai(2-\lambda) + C_2 ~ Bi (2-\lambda)= 0 \implies \frac{C_1}{C_2}= -\frac{Bi(2-\lambda)}{Ai(2-\lambda)}.~~~~(2)$$ By (1) and (2) we get the eigenvalue formula for $\lambda$ to calculate the allowed discrete values of $\lambda$ as $$f(\lambda)=[Ai(2-\lambda) Bi( 1-\lambda)-Ai (1-\lambda) Bi(2-\lambda)]=0.$$ For the solutions of $f(\lambda)=0$, see the curve $f(\lambda)$ below.. The first three eigenvalues of $\lambda$ are 11.36, 40.97, 90.32.
$f(\lambda) $ vs
Just an idea, use the eigenfunctions of the last equation (a Fourier expansion) which are
$$ \phi_n(x)=\sin(n \pi x) $$
ie
$$ y(x)=\sum_{k=1}^\infty a_k \phi_k(x) $$ Write y'', y and x in terms of $\phi_n$ and get a system for the coefficients. Look for non-zero solutions. You may have to rely on numerical methods in case there are no analytic solutions.