There's a nice technique you can use in such problems. Because you're only interested in solutions that are not identically $0$, you can pick one of the endpoints such as $-4$ and add another condition $y(-4)=0$, $y'(-4)=1$. You can assume this because $y'(-4)\ne 0$--otherwise, $y\equiv 0$ due to the uniqueness of solutions of $y''+\lambda y=0$ with $y(-4)=y'(-4)=0$.
The solution of $y''+\lambda y=0$ with $y(-4)=0$, $y'(-4)=1$ is
$$
y_{\lambda}(x) = \frac{\sin(\sqrt{\lambda}(x+4))}{\sqrt{\lambda}}.
$$
The other nice part of doing things this way is that special cases are handled by limiting cases. This is because you are guaranteed that two fixed endpoint conditions will result in a solution that has an everywhere convergent power series expansion in $\lambda$. That's a general theorem of ODEs. So,
$$
y_{0}(x) = \lim_{\lambda\rightarrow 0}y_{\lambda}(x)=x+4.
$$
You can check--but you don't have to--that this is solution where $\lambda=0$.
Any non-zero solution of $y''+\lambda y=0$ with $y(-4)=0$, $y(4)=0$ must be a non-zero multiple of $y_{\lambda}(x)$. Therefore, the permissible eigenvalues $\lambda$ are the solutions of
$$
\frac{\sin(\sqrt{\lambda}8)}{\sqrt{\lambda}}=0.
$$
Checking separately, $y_0$ does not work. The only $\lambda$ that work are
$$
\sqrt{\lambda}8 = \pm \pi,\pm 2\pi,\pm 4\pi,\cdots, \\
\lambda = \frac{n^2\pi^2}{64},\;\; n=1,2,3,\cdots.
$$
