solution of the integral equation

Question

let $y(x)$ be the solution of the integral equation

$$ y(x) = x-\int _0 ^x xt^2y(t)dt, x>0$$

Then value of $y(\sqrt 2)?$

Answer 1

We have that $$\frac{y(x)}{x} = 1-\int _0 ^x t^2y(t)dt$$ and after taking the derivative we obtain $$y'(x)x-y(x)=-x^4y(x)$$ that is, for $y(x)\not=0$, $$\frac{y'(x)}{y(x)}=\frac{1}{x} -x^3$$ By solving the separable differential equation we get $$y(x)=Cxe^{-x^4/4}\quad \text{with $C\in \mathbb{R}$.}$$ Now go back to the integral equation, find the constant $C$ and determine the value of $y(\sqrt 2)$.