I'm trying to solve the following Fredholm integral equation of the 2nd kind:
$$u(x)=f(x)+\lambda\int_{0}^{1}x^4y^2u(y)dy, 0<x<1$$ where $f(x)$ is given, $u(x)$ is the unknown function.
This is what I've done so far:
$u(x)=f(x)+ax^4$
$\Rightarrow a=\lambda\int_{0}^{1}y^2u(y)dy=\lambda\int_{0}^{1}y^2[f(y)+ay^4]dy=\lambda\int_{0}^{1}y^2f(y)dy+a\lambda\int_{0}^{1}y^4dy=\lambda\int_{0}^{1}y^2f(y)dy+\frac{a\lambda}{5}$
$\Rightarrow a=\frac{\lambda}{1-\frac{\lambda}{3}}\int_{0}^{1}y^2f(y)dy$
$\Rightarrow u(x)=f(x)+x^4\frac{\lambda}{1-\frac{\lambda}{3}}\int_{0}^{1}y^2f(y)dy$
I'm not really sure where to go from here or even if what I have done so far is correct. If it's correct, how do proceed from here? If it's wrong, what have I done wrong/what should I be doing instead?
$$u(x)=f(x)+x^4\lambda\int_{0}^{1}y^2u(y)dy\tag 1$$ $$u'(x)=f'(x)+4x^3\lambda\int_{0}^{1}y^2u(y)dy=f'(x)+\frac{4}{x}\big(u(x)-f(x)\big)$$ $$u'(x)-\frac{4}{x}u(x)=f'(x)-\frac{4}{x}f(x) \tag 2$$ $f'(x)-\frac{4}{x}f(x)$ is a known function.
Eq.$(2)$ is a first order linear ODE. I suppose that you can take it from here.