A question about solving a Sturm–Liouville Problem

Question

I am to solve the following SL problem: $$X''+\lambda X=0$$ Given $$X'(a)=X'(b)=0$$ The general soultion is obviously $$X(x)=A\cos(\sqrt\lambda x)+B\sin(\sqrt\lambda x)$$ But I cannot manage to solve these system of equations other than finding the eigenvalues $\lambda=(\pi n/(b-a))^2$ which I have obtained by adding the two equations. Also, tried to define an auxiliary function $Y(x)=X(a+x)$ but to no avail.

Answer 1

First, note that you also have a solution for an eigenvalue of 0 in this case, which is the constant solution X(x)=A.

Now, for positive eigenvalues, take the derivative, and you get $X'(x)=-A\sqrt \lambda cos(\sqrt \lambda x)+B\sqrt \lambda sin(\sqrt \lambda x)$ From here, to solve for A and B, you put in your initial values. Typically one of the values is 0 so you get one of $A,B$ is 0, and that'll lead the other one to be of some form of $n\pi$ or $(2n+1)\pi /2$ usually.

Now, if we don't have 0 bounds, I'm experimenting here....but still plugging in $X'(x)=0$ and moving the -A term to the other side, we get $A\sqrt \lambda cos(\sqrt \lambda a)=B\sqrt \lambda sin(\sqrt \lambda a)$, and the exact same equation with $b$.Dividing through, this simplifies to $\frac A B=tan(\sqrt \lambda a)$, and the same for $b$. So,we need to find a $\lambda$ so that $tan(\sqrt \lambda a)=tan(\sqrt \lambda b)=\frac A B$.

Going from there, I'm not quite sure