$x\sin x$ is a solution of $n$th order linear differential equation. Find minimum $n$ Let $y(x)=x\sin x$ be one of solution of $n^{th}$ order linear differential equation with constant coefficients. Find the minimum value of $n$.
I have no idea where how to approach this problem. In book the answer is $4$.
 A: Let's think about the characteristic equation.  For $\sin x$ (or $\cos x$) to be a solution, there must be a pair of conjugate complex roots. For $x\sin x$ (or $x\cos x$) to be a solution, the root must be repeated. 
Hence a conjugate pair of solutions of the characteristic equation is repeated and there is a minimum of $2 \times 2 = 4$ roots. In other words, the equation is at least $4^{th}$ order.
Explicitly, the roots would be $\pm i$; the minimal characteristic equation up a non-zero multiple is $(\lambda-i)^2(\lambda+i)^2 = (\lambda^2 + 1)^2 = \lambda^4 + 2\lambda^2 + 1$. This corresponds to the homogeneous ODE
$y'''' + 2y'' + y = 0$.
A: It doesn't say "homegeneous"?  Probably an oversight.  But since it does not, we can get order $1$:
$$
y' = \sin x + x \cos x
$$
... or maybe you can allow order zero ...
$$
y = x \sin x
$$
A: so you are looking for a homogeneous linear differential equation whose indicial equation is $(r^2+1)^2 = 0$ so that you have repeated roots $i, i, -i, -i.$ this corresponds to the differential equation $$\dfrac{d^4y}{dx^4} + 2\dfrac{d^2y}{dx^2} + y = 0 $$
