Sturm-Liouville Problem: Finding eigenvalues and eigenfunctions 
I am trying to find the eigenvalues and eigenfunctions of the
  following Sturm-Liouville problem:
$$  \begin{cases}
 -u''=\lambda u, \ &x\in (a, b), \\  u(a)=u(b)=0, \ &b>a.\\  \end{cases}   $$

I already checked the eigenvalues for when $\lambda<0$ and $\lambda=0$ and got that the only there are only trivial results for those two cases.
However, for $\lambda>0$, found a general solution of $$u_{gen}(x)=c_1\sin((\sqrt\lambda)x) + c_2\cos((\sqrt\lambda)x)$$ and then plugging in initial conditions to get $$u(a)=c_1\sin((\sqrt\lambda)a) + c_2\cos((\sqrt\lambda)a)$$ and $$u(b)=c_1\sin((\sqrt\lambda)b) + c_2\cos((\sqrt\lambda)b)$$
Then I put these values into a matrix, took the determinant and ended up with
$$\sin(\sqrt\lambda a)\cos(\sqrt\lambda b)-\sin(\sqrt\lambda b)\cos(\sqrt\lambda a)=0$$
which eventually simplified to $$\tan(\sqrt\lambda a)=\tan(\sqrt\lambda b)$$
which led me to an eigenvalue of 
$$\lambda_n = \left( 
     \frac{n\pi
     }{
       (b-a)
     }
\right)^2$$
Here is where I am getting stuck. I am not completely sure how to find the eigenfunctions for this problem, and I feel like the answer is already in my grasp, I just have no idea what it is. In examples we did in class, the initial conditions worked out so that in order to not have a trivial solution, one of the c values was being multiplied by a sin function, meaning the sin function had to be zero. Can anyone help me figure out what the eigenfunction is?
 A: Oooh, you could have made it so much easier on yourself by starting from
$$u_{gen}=c_1\sin\left(\sqrt{\lambda}(x-a)\right)+c_2\cos\left(\sqrt{\lambda}(x-a)\right)$$
Then you have
$$u(a)=c_2=0$$
And
$$u(b)=c_1\sin\left(\sqrt{\lambda}(b-a)\right)=0$$
So for a nontrivial solution
$$\sqrt{\lambda}(b-a)=n\pi$$
So
$$\lambda=\frac{n^2\pi^2}{(b-a)^2}$$
And then
$$u_n(x)=c_1\sin\left(\frac{n\pi(x-a)}{b-a}\right)$$
If we want to normalize to unity, let
$$\int_a^bc_1^2\sin^2\left(\frac{n\pi(x-a)}{b-a}\right)dx=\frac12c_1^2(b-a)=1$$
So we get the normalized eigenfunctions
$$u_n(x)=\sqrt{\frac2{b-a}}\sin\left(\frac{n\pi(x-a)}{b-a}\right)$$
A: Any non-trivial solution $u$ of $-u''=\lambda u$ with $u(a)=0$ will be a non-constant multiple of the solution where $u(a)=0,u'(a)=1$. That solution is
$$
        \varphi_{\lambda}(x) = \frac{\sin(\sqrt{\lambda}(x-a))}{\sqrt{\lambda}}.
$$
The limiting form where $\lambda=0$ is also correct. In this case $\varphi_{0}(x)=x-a$. So, the solutions $u_{\lambda}$ where $u_{\lambda}(a)=0=u_{\lambda}(b)$ are the solutions $\varphi_{\lambda}$ for which $\lambda$ satisifes the algebraic equation
$$
            \frac{\sin(\sqrt{\lambda}(b-a))}{\sqrt{\lambda}}=0.
$$
$\lambda=0$ is not a solution because the limiting equation where $\lambda=0$ is $b-a=0$, which is not valid. So, the solutions $\lambda$ are
$$
             \sqrt{\lambda}(b-a)=n\pi \\
                \lambda = \frac{n^2\pi^2}{(b-a)^2},\;\; n=1,2,3,\cdots.
$$
These are the eigenvalues.
