I am asked to find the equivalent Sturm-Liouville problem for this integral operator equation and then proceed to finding its eigenfunctions and eigenvalues:
$$\Psi (x) = \int _0 ^x s \left( \frac{\kappa}{1-\kappa} x - 1\right) \Phi(s) ds + \int _x ^1 x \left( \frac{k}{1-k} s - 1 \right) \Phi(s) ds$$
So, differentiating twice, i get: $$\Psi '' (x) = \Phi(x)$$
and now (where i'm stuck) i want to find two boundary conditions for $\Psi(x)$. The first one, $\Psi(0)=0$ is trivial but as for the second, i am supposed to find $\Psi'(1)=\kappa \Psi(1)$. Although, i keep getting $\Psi'(1) = \frac{\kappa}{2\kappa -1} \Psi(1)$. I know i've probably done something stupid but can't figure out exactly what. Any help is appreciated.
Your kernel is $$ K(s,x)= s\left(\frac{\kappa}{1-\kappa}x-1\right) $$ and $$ \Psi(x)=\int_{0}^{x}K(s,x)\Phi(s)ds+\int_{x}^{1}K(x,s)\Phi(s)ds \\ = \left(\frac{\kappa}{1-\kappa}x-1\right)\int_{0}^{x}s\Phi(s)ds+x\int_{x}^{1}\left(\frac{\kappa}{1-\kappa}s-1\right)\Phi(s)ds. $$
And, cancellation of half of the terms leads to $$ \Psi'(x)=\frac{\kappa}{1-\kappa}\int_{0}^{x}s\Phi(s)ds+\int_{x}^{1}\left(\frac{\kappa}{1-\kappa}s-1\right)\Phi(s)ds. $$ Therefore,
$$ \Psi(1)=\left(\frac{\kappa}{1-\kappa}-1\right)\int_{0}^{1}s\Phi(s)ds =\frac{2\kappa-1}{1-\kappa}\int_{0}^{1}s\Phi(s)ds \\ \Psi'(1)=\frac{\kappa}{1-\kappa}\int_{0}^{1}s\Phi(s)ds = \frac{\kappa}{2\kappa-1}\Psi(1). $$ I'd say YOU are right.