Volterra equation $y(x)=\cos(x)-x-2+ \int_0^x(t-x)y(t)dt$

Question

Solve the following Volterra integral equation:

$$y(x)=\cos(x)-x-2+ \int_0^x(t-x)y(t)dt.$$

My attempt so far:

By iterated kernel method we get \begin{align} K_n(x,t) &=\frac{(-1)^{n-1}}{(2n-1)!}(t-x)^{2n-1}, \qquad t\le x,\\\\ K_n(x,t) &=0, \qquad x\le t \end{align}

Then it could be solved in a couple of ways:

1. $$y(x)=f(x)+\sum_{n=1}^\infty \lambda ^n \int_a^b K_n(x,t)f(t)dt$$
2. $$y(x)=f(x)+\lambda \int_a^b R(x,t,\lambda)f(t)dt$$ where the resolvent is $$R(x,t,\lambda)= \sum_{n=1}^\infty \lambda ^ {n-1} K_n(x,t).$$

In any case, I don't get anything as simple as it could be solvable easily:

I tried to calculate the resolvent:

$$R(x,t,\lambda)=\sum_{n=1}^\infty \lambda ^ {n-1} K_n(x,t)=\sum_{n=1}^\infty \lambda ^ {n-1} \frac{(-1)^{n-1}}{(2n-1)!}(t-x)^{2n-1}=\sum_{k=0}^\infty \lambda ^k \frac{(-1)^k}{(2k+1)!}(t-x)^{2k+1} = \text{[by wolframalpha. What's the trick?]} =\frac{\sin(\sqrt{\lambda} (t-x))}{\sqrt{\lambda}}.$$ I see that it's going to be a really messy integration further since we have to calculate

$$y(x)=f(x)+\lambda \int_a^b R(x,t,\lambda)f(t)dt \\ = \cos(x)-x-2 + \sqrt{\lambda} \int_t^x \sin \left(\sqrt{\lambda} \left(t-x\right)\right)(\cos(t)-t-2)dt$$

Is there an easier way to solve this problem or is it going to be so problematic?

Answer 1

Just after I posted it, I found the way to solve it...

Just differentiate the whole equation and we get $$y'(x)= - \sin(x)-1-\int_0^x y(t)dt.$$ Differentiate it once again and we get $$y''(x)=- \cos (x)-y(x) \\ y''(x)+y(x)=- \cos (x)$$ with initial conditions $$y(0)=y'(0)=-1.$$

Now it's an easy PDE problem.