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.

Variation of constants tells you to find $y_p$ in the form
$$
y_p(x)=c_1(x)\cos(x)+c_2(x)\sin(x)
$$
where the coefficient functions satisfy the differential equation system
\begin{align}
c_1'(x)\cos x+c_2'(x)\sin x &= 0\\
-c_1'(x)\sin x+c_2'(x)\cos x &= f(x)
\end{align}
which leads to
$$
c_1'(x)=-f(x)\sin x,\ \ c_2'(x)=f(x)\cos(x) 
$$
and thus
$$
y_p(x)=-\int_0^xf(u)\sin u\,du\,\cos x+\int_0^xf(u)\cos u\,du\,\sin x
=\int_0^xf(u)\sin(x-u)\,du.
$$
