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.
[Math] general solution using variation of parameters
ordinary differential equations
Best Answer
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
$y_p=\sin x\int_0^xf(t)\cos t~dt-\cos x\int_0^xf(t)\sin t~dt$
$y_p=\int_0^xf(t)(\sin x\cos t-\cos x\sin t)~dt$
$y_p=\int_0^xf(t)\sin(x-t)~dt$