Vrftsjtryk

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
$$X(x)=Acosbeta x +Bsin beta x.$$
Taking derivatives, one easily finds that $X'(0)=Bbeta$ and $X(0)=A.$ Thus we obtain
$$Bbeta + A=0 implies beta= frac{-A}{B}.$$
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 $X(1)=0.$

Since you want $X(1)=0$, you cat set $X'(1)$ to any non-zero constant ($0$ leads to $Xequiv 0$.) 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 $lambdarightarrow 0$ 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 $lambdarightarrow 0$ is $0$. So $lambda_0=0$ is an eigenvalue with $X_{0}=x-1$. For $lambdane 0$, 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:
$$
sqrt{lambda_0} = 0 < sqrt{lambda_1} < frac{pi}{2} < sqrt{lambda_2} < frac{3pi}{2} < sqrt{lambda_3} < frac{5pi}{2} < cdots
$$
Asymptotically, $sqrt{lambda_n} approx frac{(2n-1)pi}{2}$ or $lambda_n approx frac{(2n-1)^2pi^2}{4}$ as $nrightarrowinfty$.