$y(x)=x-\int_0^x xt^2y(t)dt, x>0.$ , $y(\sqrt 2)=?$

Question

Given $$y(x)=x-\int_0^x xt^2y(t)dt, x>0.$$ How to find the value of $y(\sqrt 2)$? Solving using resolvent kernel is lengthy procedure.

Answer 1

Well if you write $g(x)=y(x)/x$ you get

$$g(x)=1-\int_0^x t^3g(t)dt \;\;\;\Longrightarrow \;\;\;g'(x)=-x^3g(x)$$

$$ [\ln g(x)]'= {g'(x)\over g(x)} = -x^3 \;\;\;\Longrightarrow \;\;\; \ln g(x) = -{x^4\over 4} + c$$

So $g(x) = A\cdot e^{-x^4\over 4}$ and thus $y(\sqrt{2}) = A\cdot e^{-1}$. Now you have to find this $A=e^c$.

Edit: How to calculate $A$? Bring back to the integral equation $g(x)= A\cdot e^{-x^4\over 4}$:

$$ A\cdot e^{-x^4\over 4} = 1- A\int_0^x t^3e^{-t^4\over 4}dt $$ It is easy to calculate this integral and you get $A=1$.