Finding eigenfunctions and eigenvalues from a differential equation Consider the differential equation 
$$X''(x)+\lambda X=0$$ 
on $0 \leq x \leq 1$with boundary conditions 
$$X'(0)+X(0)=0  \ \ \ \ \text{and}  \ \ \ \ X(1)=0.$$
I have a few problems here that I think I figured out, but I would appreciate another look or some hints as to what I can fix.  Or, maybe I am totally wrong! 
$\textbf{My first goal:}$

Find an eigenfunction associated with eigenvalue $\lambda=0.$

An eigenvalue $\lambda =0$ would mean that $X''(x)=0$.  This means that the solution takes the form 
$$X(x)=Ax+B.$$
Since $X'(0)=A$ and $X(0)=B$, 
$$X'(0)+X(0)=0 \iff A=-B.$$
Therefore an eigenfunction that works would be $\boxed{X_{0}(x)=-2x+2}.$
$\textbf{My second goal:}$

Find an expression for all eigenvalues $\lambda = \beta ^2>0.$

This one requires a little more work.  The solution to $X'(x)+\beta ^2 X=0$ takes the form 
$$X(x)=A\cos\beta x +B\sin \beta x.$$
Taking derivatives, one easily finds that $X'(0)=B\beta$ and $X(0)=A.$ Thus we obtain 
$$B\beta + A=0 \implies \beta= \frac{-A}{B}.$$
Finally, this gives $\boxed{\beta=\frac{A^2}{B^2}}$
What are your thoughts?  Thanks in advance!
Edit: I just realized that my last problem did not necessarily satisfy $X(1)=0.$
 A: Since you want $X(1)=0$, you cat set $X'(1)$ to any non-zero constant ($0$ leads to $X\equiv 0$.) So I'd pick a value of $1$. Then
$$
   X_{\lambda}(x) = \frac{\sin(\sqrt{\lambda}(x-1))}{\sqrt{\lambda}}
$$
Choosing a constant value for $X'(1)$ forces the limiting case as $\lambda\rightarrow 0$ to be the correct solution for $\lambda=0$ as well,  which is
$$
    X_{0} =  x-1.
$$
You can check that this is an eigenfunction. So $\lambda=0$ is an eigenvalue, with corresponding eigenfucntion $x-1$.
The general eigenvalue equation becomes
$$
        X_{\lambda}(0)+X_{\lambda}'(0)=0 \\
      -\frac{\sin(\sqrt{\lambda})}{\sqrt{\lambda}}+\cos(\sqrt{\lambda})=0
$$
The limit at $\lambda\rightarrow 0$ is $0$. So $\lambda_0=0$ is an eigenvalue with $X_{0}=x-1$. For $\lambda\ne 0$, the solutions are zeros of
$$
      \tan(\sqrt{\lambda})=\sqrt{\lambda}.
$$
This is a transcendental equation. You can plot the graphs of $y=\tan(x)$ and $y=x$, and check the intersections of the graphs for $x \ge 0$. The negative values of $\sqrt{\lambda}$ can be ignored because they lead to duplicate values of $\lambda$.
The non-negative solutions are ordered as follows:
$$
       \sqrt{\lambda_0} = 0 < \sqrt{\lambda_1} < \frac{\pi}{2} < \sqrt{\lambda_2} < \frac{3\pi}{2} < \sqrt{\lambda_3} < \frac{5\pi}{2} < \cdots
$$
Asymptotically, $\sqrt{\lambda_n} \approx \frac{(2n-1)\pi}{2}$ or $\lambda_n \approx \frac{(2n-1)^2\pi^2}{4}$ as $n\rightarrow\infty$.
