 Solution

The homogeneous linear differential equation $y^{\prime \prime }+y=0$ has general solution $C_1\cos x+C_2\sin x$. For a particular solution, use variation of parameters with $y_1=\cos x$, $y_2=\sin x$ and Wronskian one:The double-angle formulas and integration by parts giveSubstitution simplifies to $-x^2\cos x/4+x\sin x/4+\cos x/8$. Absorb the last term into the homogeneous linear differential equation solution. ThereforeIndeed, applying $d^2/d{x}^2+1$ to $x\sin x$ gives $2\cos x$, and applying it to $x^2\cos x$ gives $2\cos x-4x\sin x$; their difference divided by four is the required $x\sin x$.