Let $u\in C^2[0,1]$ satisfy for some $\lambda \neq 0$ and $a\neq 0$ $$u(x)+\frac{\lambda}{2}\int_0^1 |x-s|u(s)ds=ax+b$$ Then $u$ also satisfies
$(a)$ $\frac{d^2u}{dx^2}+\lambda u=0$
$(b)$ $\frac{d^2u}{dx^2}-\lambda u=0$
$(c)$ $\displaystyle{\frac{du}{dx}-\frac{\lambda}{2}\int_0^1 \frac{x-s}{|x-s|}u(s)ds=a}$
$(d)$ $\displaystyle{\frac{du}{dx}+\frac{\lambda}{2}\int_0^1 \frac{x-s}{|x-s|}u(s)ds=a}$
This is an example of an Fredholm equation with symmetric kernel. We know that its solutions can be obtained using the normalized eigenfunctions $\phi_m(x)$ corresponding to eigenvalues $\lambda_m$ as $$u(x)=ax+b-\frac{\lambda}{2}\int_0^1 \bigg(\sum_m \frac{\phi_m(x)\phi_m(t)}{\lambda_m-\lambda}\bigg)(at+b)dt$$
But I find it difficult to find this solution and then again applying the solution in the above options to see whether the equations are satisfied. I cannot apply Leibnitz rule either for this problem as limits of integration are constant. Is there any alternative method for this? Thanks in advance.