Eigenvalues and eigenfunctions of the Fredholm integral equation of the second kind I'm trying to solve the following question:
Find the eigenfunctions and eigenvalues of the following integral operator:
\begin{equation}
Ku(x) = \int_{0}^{\pi}(\sin(x)\sin(y)+\alpha\cos(x)\cos(y))u(y) dy.
\end{equation}
My attempted solution is as follows:
Let $(\lambda, u)$ be such an eigenpair. Then,
\begin{align}
\lambda u(x) & = \int_{0}^{\pi}(\sin(x)\sin(y)+\alpha\cos(x)\cos(y))u(y)dy \\
& = \sin(x)\int_{0}^{\pi}\sin(y)u(y)dy + \alpha \cos(x)\int_{0}^{\pi}\sin(y)u(y)dy \\
\end{align}
We convert the above equation into a differential equation. Differentiating twice, we get,
\begin{equation}
\lambda u^{''}(x) = - u(x).
\end{equation}
As $\lambda = 0$ is clearly not an eigenvalue (if not, then the only possible eigenfunction is $u = 0$ almost everywhere), we have the general solution is,
\begin{equation}
u(x) = A\sin\bigg(\frac{x}{\sqrt{\lambda}}\bigg) + B\cos\bigg(\frac{x}{\sqrt{\lambda}}\bigg).
\end{equation}
We need two boundary conditions. From the eigenvalue equation above, we have,
\begin{align}
u(0) & = \frac{\alpha}{\lambda} \int_{0}^{\pi}\sin(y)u(y)dy \\
u(\pi) & = \frac{-\alpha}{\lambda} \int_{0}^{\pi}\sin(y)u(y)dy \\
\end{align}
However, I don't how to proceed from here. The boundary conditions above are functions of $u$.
Please provide relevant hints. This is a homework problem, and I would like to finish it on my own.
 A: There is a error, you have that 
\begin{align}
u(0) & = \frac{\alpha}{\lambda} \int_{0}^{\pi}\cos(y)u(y)dy \\
u(\pi) & = \frac{-\alpha}{\lambda} \int_{0}^{\pi}\cos(y)u(y)dy \\
\end{align}
So $u(0)=-u(\pi)$. Another condition can be 
\begin{align}
u(\frac{\pi}{2}) & = \frac{\alpha}{\lambda} \int_{0}^{\pi}\sin(y)u(y)dy \\
u(\frac{3\pi}{2}) & = \frac{-\alpha}{\lambda} \int_{0}^{\pi}\sin(y)u(y)dy \\
\end{align}
So $u(\frac{\pi}{2})=-u(\frac{3\pi}{2})$
But 
$u(0)=Asin(\frac{0}{\sqrt{\lambda}})+Bcos(\frac{0}{\sqrt{\lambda}})=B$
and 
$u(\pi)= Asin(\frac{\pi}{\sqrt{\lambda}})+Bcos(\frac{\pi}{\sqrt{\lambda}})$
So you have that 
$Asin(\frac{\pi}{\sqrt{\lambda}}) =-B(1+cos(\frac{\pi}{\sqrt{\lambda}}))$
To the other hand 
$u(\frac{\pi}{2})=Asin(\frac{\pi}{2\sqrt{\lambda}})+Bcos(\frac{\pi}{2\sqrt{\lambda}}) $
while 
$u(\frac{3\pi}{2})= Asin(\frac{3\pi}{2\sqrt{\lambda}}) +Bcos(\frac{3\pi}{2\sqrt{\lambda}}) $
so 
$A(sin(\frac{\pi}{2\sqrt{\lambda}})+sin(\frac{3\pi}{2\sqrt{\lambda}}))=-B(cos(\frac{\pi}{2\sqrt{\lambda}}) + cos(3\frac{\pi}{2\sqrt{\lambda}}) $
