[Math] Construct a Second Order ODE given the fundamental solutions

Question

I need to construct a second order linear differential equation for which $\{ \sin (x), x \sin (x) \}$ is the set of fundamental solutions.
I am completely lost on this problem and have been trying different approaches for days now. First, I tried simple variations of $ y'' + y = 0$ but I couldn't come up with anything successful. Then I tried using the identity $$ \dfrac{dW}{dx} + PW = 0 $$ and solving for $~P~$, knowing that the Wronskian $~W~$ of the solution set is $ \sin^2 (x) $. I got something like $ P = -\cot(x) $ so I tried variations of $$ y'' -\cot(x)y' + y = 0 $$ but again nothing worked.
So I started looking through my book over and over to see what types of ODE's could produce such a solution set, and from what I can tell, there are no constant-coefficient ODE's that can possibly produce this solution set. The only way I learned to solve variable coefficient second order equations was by Cauchy-Euler's method, but I don't see how that can output such a solution set either. I've never come across a second order ODE that has the solution $ \sin(x) $ without $ \cos(x) $.
Any ideas?

Answer 1

In fact we can solve best with this approach:

$\because$ the ODE whose having the general solution $u=C_1+C_2x$ is $u''=0$

$\therefore$ the ODE whose having the general solution $y=C_1\sin x+C_2x\sin x$ is

$(y\csc x)''=0$

$y''\csc x-2y'\csc x\cot x+y(\csc^2x+\cot^2x)\csc x=0$

$y''\sin^2x-2y'\sin x\cos x+y(1+\cos^2x)=0$

$\dfrac{1-\cos2x}{2}y''-y'\sin2x+\dfrac{3+\cos2x}{2}y=0$

$(\cos2x-1)y''+2y'\sin2x-(\cos2x+3)y=0$