Show that method of variation of parameters applied to the equation $y'' + y = f(x)$ leads to a particular solution $y(x) = \int_0^x f(t) \cdot \sin (x-t) dt$. I tried like this Wronskian of $y_1$ and $y_2$ is $1$. Two solutions of corresponding homogeneous DE are $\sin(x)$ and $\cos(x)$.Then put these values into formula used in variation of parameters.Bit i did not get answer.of this form.
In fact this just involves some elementary functions simplifications.
By variation of parameters, you can get the particular solution is $y_p=\sin x\int_0^xf(x)\cos x~dx-\cos x\int_0^xf(x)\sin x~dx$
Note that the particular solution can also rewrite as
$y_p=\sin x\int_0^xf(t)\cos t~dt-\cos x\int_0^xf(t)\sin t~dt$
$y_p=\int_0^xf(t)(\sin x\cos t-\cos x\sin t)~dt$