The problem is the following $$\begin{cases}-y''+(1+x)y= λy\\ y(0)=y(1)=0\\ x\in(0,1)\end{cases}$$ For which value if $\lambda$ the problem has nonzero solutions?
The answers given are
- for all $\lambda<0$
- for all $\lambda\in[0,1]$.
- for some $\lambda\in(2,\infty)$
- for a countable number of $\lambda$'s
Just a partial answer: Multiply your equation with $y$ and integrate. Then you obtain $$-\int_0^1 y'' ydx + \int_0^1(1+x)y^2dx = \lambda \int_0^1y^2dx$$ Now use integration by parts on the first integral to get $$-[y'y]_0^1 + \int_0^1(y')^2dx + \int_0^1(1+x)y^2dx = \lambda \int_0^1y^2dx,$$ which simplifies by the boundary values to $$\int_0^1(y')^2dx + \int_0^1(1+x)y^2dx = \lambda \int_0^1y^2dx.$$ $x\in[0,1]$ further yields $$\int_0^1y^2\, dx\leq \lambda \int_0^1 y^2dx,$$ which exclude option 1) and 2). For $\lambda=1$ you need go through the argument again and proceed by contradiction.
Existence can be done in a more general context. Look up Gilbarg/Trudinger's book on elliptic partial differential equations of second order. It should be Theorem 8.37. This is a reference to the book: https://www.springer.com/de/book/9783540411604