A Sturm-Liousville Problem

Find the eigenvalues and corresponding eigenfunctions of the Sturm-Liousville problem
$$-y^{\prime\prime }=\lambda y, \: \: y'(0)=0=y'(2).$$

Here's my attempt:

The general solution to $-y'' = \lambda y$ is $A \cos(\sqrt{\lambda} t) + B \sin(\sqrt{\lambda}t)$. Applying the first boundary condition, we get

$$
\begin{align*}
y'(0) &= \left(-\sqrt{\lambda}A\sin(\sqrt{\lambda}t) + \sqrt{\lambda} B \cos(\sqrt{\lambda}t) \right)_{t=0}
\\
&= 0 + \sqrt{\lambda} B
\\
&= 0
\end{align*}
$$

If we allowed $\lambda = 0$, then this would result in $y(t) = A$, and applying the second boundary condition would require $A=0$, so $\lambda = 0$ generates the trivial solution. Thus, $B = 0$, and

$$y(t) = A \cos(\sqrt{\lambda}t).$$

Now, applying the next boundary condition, we get

$$
\begin{align*}
y'(2) = -\sqrt{\lambda} A \sin(2 \sqrt{\lambda}) &= 0
\\
\Rightarrow \sin(2\sqrt{\lambda}) &= 0.
\end{align*}
$$

Sine is zero when the argument is an odd-integer multiple of $\pi$, so

$$
\begin{align*}
2 \sqrt{\lambda} &= n\pi
\\
\Rightarrow \sqrt{\lambda} &= \frac{n\pi}{2};
\end{align*}
$$

so our eigenvalues are $\lambda_n = (n\pi)^2/4$ and the eigenfunctions are

$$
y_n = A \cos\left(\frac{n\pi}{2} \, t\right).
$$

@cromer: Shouldn't $2\sqrt{\lambda}=\frac{(2n+1)\pi}{2}$ turn into
$$\sqrt{\lambda}=\frac{(2n+1)\pi}{4}?$$
Thus, the eigenvalues would be $\lambda_n=\frac{(2n+1)^2\pi^2}{16}$. Just a minor detail; otherwise, everything looks great!

Thanks! My inability to divide properly has been corrected.

why did you use $y(2)=0$ instead of $y'(2)=0$?

Thanks! I apparently did not read carefully enough. I have again made a correction to my work.