Proving particular solution in differential equation 
Given the second-order ordinary differential equation: 
  $$
{y}''+y=f(x)
$$
  prove that:
  $$
y_p(x)=\int_{0}^{x}f(u)\sin(x-u)du
$$
  is the particular solution of the equation.

I know this is homework but I've been trying to solve it for the past few days and I can't. I even asked my teacher for help but he doesn't answer.
Thanks.

Hint. We assume our $f$ is nice enough to be allowed to use the Leibniz rule,
$$
\frac{d}{dx} \left (\int_{0}^{b(x)}f(x,u)\,du \right) = f\big(x,b(x)\big)\cdot \frac{d}{dx} b(x) + \int_{0}^{b(x)}\frac{\partial}{\partial x} f(x,u) \,du
$$ giving here
$$
\begin{align}
y'_P(x)&=\frac{d}{dx} \left (\int_{0}^{x}f(u)\sin(x-u)\,du \right) 
\\\\&= 0+ \int_{0}^{x}f(u) \cos(x-u) \,dt
\end{align}
$$ then differentiate once more using the same tool and see what happens.
