I have the following system
$$\frac{d}{dx}\left(a(x)\frac{du}{dx}\right)=f, \text{ for } x \in (0,1)$$ with boundary conditions $u_x(0)=0$ and $u(1)=0$. For $a(x)>0$, and $b(x)=\frac{1}{a(x)}$, I am required to obtain an integral equation of the form $u(x)=\int_0^1 k(x,y)b(y)dy$. I have no idea about f
It seems that using the fundamental theorem of calculus two times to integrate the given equation yields something confusing, which doesn't match what is required to be obtained.
I am not able to get past this, after the first step of integration:
$$a(x)\frac{du}{dx}=\int_0^xf(t)dt $$
Any ideas?