How to find the particular solution of $y''+y=f(x)$

Question

I have to solved the above equation using the method of variation of parameters. So please try to answer proof using the method of variation of parameters.

Answer 1

Assuming you know how to find the general solution (here $y_g$ is the the general solution, $y_p$ the particular and $W(y_1,y_2)(x)$ the Wronskian)$$y_g(x)=c_1\cos(x)+c_2\sin(x)\implies y_1(x)=\cos(x),\quad y_2(x)=\sin(x)$$ $$W(y_1,y_2)=y_1(x)y_2'(x)-y_1'(x)y_2(x)=\cos^2(x)+\sin^2(x)=1$$ so $$y_p(x)=y_2(x)\int\frac{y_1(x)f(x)}{W(y_1,y_2)}\,dx-y_1(x)\int\frac{y_2(x)f(x)}{W(y_1,y_2)}\,dx=\\ =\sin(x)\int\cos(x)f(x)\,dx-\cos(x)\int\sin(x)f(x)\,dx.$$