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?
 A: 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.

A: No need to expand the integrand. Linear change of variables mapping $[s,t]$ to $[0,1]$ reduces integrals for $K_n$ to beta-function. It will be easy to calculate several first  ones, guess the formula for $K_n$ and prove it by induction.
Edit
Making change of variables $\xi=s+y(t-s)$ we have
$$
\int_s^t(t-\xi)(\xi-s)\,d\xi=
(t-s)^3\int_0^1(1-y)y\,dy=
(t-s)^3B(2,2).
$$
