Find the first positive eigenvalue $\lambda$ of the boundary value problem over $x\in [0,\frac{1}{2}]$. $$y''-\lambda y'+\frac{2\lambda-1}{x}y=0, \quad y(0)=y(\tfrac{1}{2})=0.$$
My approach: I have tried to use Frobenius Theorem because $x=0$ is a regular-singular point and also the indicial equation implies that the eigenfunction (non-trivial solution) will not a similar form of a Bessel function.
I have managed to use self-adjoint properties but the differential operator of the left hand side turns out to be non self-adjoint.
When writing $y(x) = x w(z)$ with $z= \lambda x$, the differential equation is transformed into $$ z w''(z)+ (2-z) w'(z) -(\lambda^{-1} -1) w(z) =0$$ which is Kummer's equation. The regular solution to this equation (fulfilling $y(0)=0$) is $$ w(z) = {}_1 F_1(\lambda^{-1} -1; 2; z).$$ The first positive eigenvalue, corresponds to the first zero of the function $$ f(\lambda) = w(\lambda/2).$$
Numerics shows that this is situated at $$\lambda \approx 4.60571.$$