Past exam of the mathematical methods I course of the University of Oxford / 2017 / 7 / b / Solution

Using the complex forcing method for a real differential equation, $\mathrm{Im}[(4x+2i)e^{ix}]=4x\sin x+2\cos x$, a suitable complex forcing is $f(x)=(4x+2i)e^{ix}$. Solve $u^{\prime \prime }+2u=f$ and take its imaginary part. Writing $u=p(x)e^{ix}$ reduces the equation toA linear polynomial $p=ax+b$ gives $a=4$ and $b=-6i$. ThusFor a direct check, $(4x\sin x{)}^{\prime \prime }+2(4x\sin x)=8\cos x+4x\sin x$, while $(-6\cos x{)}^{\prime \prime }+2(-6\cos x)=-6\cos x$. Their sum is the required forcing. Adding $A\cos (\sqrt{2}x)+B\sin (\sqrt{2}x)$ gives the full real solution if needed.