All the SL problems I have seen before had $ \lambda$ in them, so finding the eigenvalues meant finding the values of $ \lambda$. However I don't know what I am supposed to do in the following:
Find the eigenvalues & normalised eigenvectors for $y''+4y'+(4+n^2 \pi^2)y=0,$ with conditions $y(0)=y(1)=0$, where $n$ is an integer.
I suppose I should put it in the SL usual form but I don't know how to deal with the absense of $ \lambda$.
On
You can write this as $$ -y''-4y'-4y = n^2\pi^2 y,\;\;\; y(0)=y(1)=0. $$ If you multiply by $e^{4x}$, then the equation can be written as $$ -(e^{2x}y)'' =n^2\pi^2(e^{2x}y). $$
So, letting $w=e^{2x}y$ gives $$ -w''=n^2\pi^2w,\;\; w(0)=w(1)=0. $$
The solutions are $w_n(x)=\sin(n\pi x)$ or $y_n(x)=e^{-2x}\sin(n\pi x)$,
Giving an operator $\mathcal{D}(\cdot)$ their eigenvalues are all $\lambda$ such that
$$ \mathcal{D}(y)=\lambda y $$
solving
$$ y''+4y'=\lambda y $$
we have
$$ y = c_1 e^{(-2-\sqrt{\lambda+4})x}+ c_2e^{(-2+\sqrt{\lambda+4})x} $$
here $\lambda = -(4+n^2\pi^2)$ and then
$$ y = c_1 e^{(-2-i n \pi)x}+ c_2e^{(-2+i n\pi)x}\Rightarrow y = \left(C_1 \cos(n\pi x)+C_2\sin(n\pi x)\right)e^{-2x} $$
so due to the contour conditions
$$ y(0) = C_1 = 0\\ y(1) = C_1(-1)^ne^{-2} = 0 $$
then
$$ y_n = C_2\sin(n\pi x)e^{-2x} $$
are the eigenfunctions associated to the boundary conditions.