# Differential equation, eigenvalues and eigenfunctions

How does one find all the permissible values of $b$ for $-{d\over dx}(-e^{ax}y')-ae^{ax}y=be^{ax}y$ with boundary conditions $y(0)=y(1)=0$? I assume we have a discrete set of $\{b_n\}$ where they can be regarded as eigenvalues? After that how does one find the corresponding $\{y_n\}$? I am sure we substitute the $\{b_n\}$ into the equation, but then I still don't know how this equation is solved. Please help!

Thanks.

Update: Perhaps it is easier to find the permissible $b$'s if we write the equation in the form $y''+ay'+(a+b)y=0$?

Update 2: OK, so the general solution is $y(x)=A\exp[\frac{-a+\sqrt{a^2+4(a+b)}}{2}]+B\exp[\frac{-a-\sqrt{a^2+4(a+b)}}{2}]$ And then the BC's mean that $A+B=0$ and $\exp[\frac{-a+\sqrt{a^2+4(a+b)}}{2}]-\exp[\frac{-a-\sqrt{a^2+4(a+b)}}{2}]=0$, therefore we need $\sqrt{a^2+4(a+b)}=0$ i.e. $b={-1\over 4}(a^2+4a)$? Am I right? Is this the only permissible $b$?

Final update: This problem has been resolved. Thanks anyway!

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