Volterra integral equation of second type solve using resolvent kernel

Solve the integral equation

$$ y(t)= f(t) + \lambda \int_{0}^{t} (t-s) y(s) ds $$

where $f$ is continuous using the method of finding the resolvent kernel and Newmann series.

Here it is what I did:

$ K_1 (t,s) \equiv K(t,s) =t-s$

$ K_2 (t,s) = \int_{s}^{t} K(t, \xi) K_1 (\xi ,s) d \xi= \frac{1}{2} (t+s)^2(t-s)-ts(t-s) +\frac{1}{3} (s^3 -t^3) $

From here and on the calculations are too difficult.

Is there any trick?

Any help?

Thank's in advance!

P.S Is there another way to solve it (without using this method) ?

edit: I didn't made any proccess. Some help?

Solution 1:

A related problem. I am answering your question about the other way to solve the problem. The other technique is to use the Laplace transform technique. Taking the Laplace transform of both sides gives

$$ Y(s)= F(s)+\lambda L(x*y(x))=F(s)+\lambda L(x)L(y(x))=F(s)+\frac{\Gamma(2)\lambda}{s^2}Y(s). $$

Simplifying the above gives

$$ Y(s)= \frac{s^2F(s)}{s^2-\lambda}. $$

Taking the inverse Laplace transform yields the solution

$$ y(x)=f \left( x \right) +\sqrt {\lambda}\int _{0}^{x}\!f \left( t \right) \sinh \left( \sqrt {\lambda} \left( x-{t} \right) \right) {d{t}} .$$

Notes: We used the facts

i)$$ L(\delta(x)+\sqrt {\lambda}\sinh \left( \sqrt { \lambda}x \right) ) = \frac{s^2}{s^2-\lambda}.$$

ii) The Laplace transform of the convolution equals the product of the Laplace.