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.
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.