I have the following equation : $$ \phi(x) = \frac{x}{2} + \frac{\pi^2}{4}\int_{0}^{1}K(x,t)\phi(t)dt $$
where $$ K(x,t)= \left\{ \begin{array}{ll} \frac{x(2-t)}{2} & \mbox{if } 0 \leq x \leq t \\ \frac{t(2-x)}{2} & \mbox{if } t \leq x \leq1 \end{array} \right. $$
I tried differentiating twice to try and find the eigenfucntions of the homogenic equation but the ode is too hard to solve and I can't get a sturm liueville condition properly because phi on 0 is 0 but I there is not second condition.
** Edit : The question was made much simpler when a small change was made; probably due to a mistake in the exam that this question was taken out of.
If the correct Kernel is what I mentioned in the comment above - given that you mentioned about symmetry - then differenting twice we obtain $$\phi''(x)+\frac{\pi^2}{4}\phi(x)=0.$$Now this is standard, knowing that $\phi(0)=0$ and $\phi'(1)+\phi(1)=\frac12$.