What are the $\lambda$ values such that $$\; y''-(\frac{1}{4}+\frac{\lambda}{x}).y=0 \;;\; 0<x<\infty$$ $$ y(0)=0 \; ; \; \lim_{x\rightarrow\infty}y(x)=0$$ have a non trivial solution.
Trying to use the Frobenius method, I found that $\; (s^2+s-\lambda).a_1=0\;$ and the recurrence relation $\; a_k=\frac{-1}{k.(k-1)}.(\frac{1}{4}.a_{k-2}+\lambda.a_{k-1})\;$. These results that I got are correct ? How can I get the answer by these results ?
One hint the professor gives is that $y(x)=\exp(-x/2).v(x)$.