Finding the eigenvalues and eigenfunction (tricky) I'm given $$X"- vX' +X \lambda=0$$ (v is a constant)
I have worked x' to be:
X'(x) = $$\frac{1}{2} B v e^{\frac{v x}{2}} \sin \left(\frac{1}{2} x \sqrt{v^2-4 \beta
   ^2}\right)+\frac{1}{2} B \sqrt{v^2-4 \beta ^2} e^{\frac{v x}{2}} \cos
   \left(\frac{1}{2} x \sqrt{v^2-4 \beta ^2}\right)$$
My BCs are: $$X(0)=X'(L)=0$$
I need to show that the eigenvalue
$$\lambda _{n} = \frac{v^{2}}{4} + \frac{(2n-1)^{2}\pi^{2}}{4L^{2}}$$
I believe I'm close. I've already applied to BC X(0) so that I arrived at X'(x). I have tried using X'(L) on the above X'(x) but it got really messy. 
I shall add that after applying X'(L), I arrived at some B[Sin(...)+Cos(...)]=0 which doesn't possible to solve for an eigenvalue.
Please help.
 A: The polynomial $z^{2}-vz+\lambda = (z-v/2)^{2}+(\lambda-v^{2}/4)=0$ has solutions
$$
      X(x)=e^{vx/2}\left[A\sin(\sqrt{\lambda-\frac{v^{2}}{4}}x)+B\cos(\sqrt{\lambda-\frac{v^{2}}{4}}x)\right].
$$
The unique $X$ for which $X(0)=0$, $X'(0)=1$ is
$$
     X_{\lambda}(x)=\frac{1}{\sqrt{\lambda-\frac{v^{2}}{4}}}
                  e^{vx/2}\sin(\sqrt{\lambda-\frac{v^{2}}{4}}x).
$$
The advantage of using fixed conditions is that special cases work out through limits. For example, as $\lambda\rightarrow\frac{v}{2}$, the above converges to
$$
             X_{v/2}(x) = e^{vx/2}x.
$$
The eigenvalues $\lambda$ are solutions of
$$
       0=X_{\lambda}'(L) = \left\{\frac{\frac{v}{2}}{\sqrt{\lambda-\frac{v^{2}}{4}}}
    \sin(\sqrt{\lambda-\frac{v^{2}}{4}} L)+\cos(\sqrt{\lambda-\frac{v^{2}}{4}}L)\right\}e^{vL/2}
$$
The limiting form for $\lambda=v/2$ is
$$
    \frac{vL}{2}+1=0,
$$
which happens iff $v  = -2/L$. Otherwise, the equation is
$$
        \tan(\sqrt{\lambda-\frac{v^{2}}{4}}L)=-\frac{2}{v}\sqrt{\lambda-\frac{v^{2}}{4}},
$$
which is solved by finding solutions of
$$
           \tan(\mu) = -\frac{2}{vL}\mu
$$
and then setting
$$
    \lambda = \frac{\mu^{2}}{L^{2}}+\frac{v^{2}}{4}.
$$
Just plotting $y=\tan(\mu)$ against $y=-\frac{2}{vL}\mu$ shows you that you will get something asymptotic to what you want, but not exactly what you want.
