Solve S.L problem $$ y'' + \lambda y = 0 , ~~0 < x<1$$
$$ y'(0) = 0, y'(1) - y(1) = 0.$$
I have shown that there are no eigenvalues corresponding to $\lambda=0$.
For $\lambda < 0$, I let $\lambda = - k^2, k^2>0.$
Then solving the ODE gives 
$$ y = A\cosh(kx)+B\sinh(kx) $$
and applying Boundary conditions gives $B=0$ and $\tanh(k)=\frac{1}{k}$.
Hence there is one solution for $k^2 > 0$ where $\frac{1}{\sqrt{\lambda_0}}=\tanh(\sqrt{\lambda_0})$.
Then for $\lambda > 0$ I get 
$$ -\frac{1}{\lambda_n} = tan(\sqrt{\lambda_n})$$
Is what I have correct? Because I want to give a closed form solution for $\lambda$ to help solve a PDE and I am not sure how to derive one.
 A: Any non-zero solution of your eigenvalue equation, once scaled properly, will satisfy
$$
               y''+\lambda y = 0,\;\; y(0)=1,\; y'(0)=0,
$$
which has unique solution
$$
                   y(x) = \cos(\sqrt{\lambda}x).
$$
This function $y$ is a solution of the full equation iff $\lambda$ satisfies
$$
       0=y'(1)-y(1)=-\sqrt{\lambda}\sin(\sqrt{\lambda})-\cos(\sqrt{\lambda}).
$$
The eigenvalue equation is a power series equation in $\lambda$, which is always true of regular Sturm-Liouville problems. That's why eigenvalues of regular problems do not cluster in the finite plane. And, as with all such Sturm-Liouville equations, the eigenvalues $\lambda$ must be real. $\lambda=0$ is not an eigenvalue.
The solutions $\lambda > 0$ are obtained by finding the positive real $\mu$ such that $\tan(\mu)=-1/\mu$ and letting $\lambda=\sqrt{\mu}$ be the positive square root. Any solution $\lambda < 0$ is obtained by finding the solutions $\lambda=-\mu^2$ of the equation $\tanh(\mu)=\frac{1}{\mu}$. You correctly derived these equations. The equation for $\lambda > 0$ is transcendental; there is not closed form for these solutions. I believe the same is true for the equation where $\lambda < 0$.
