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?

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    $\begingroup$ $y=x\sin x$ can not solve a second order homogenous linear ODE because $y(0)=y'(0)=0$. $\endgroup$ – user138530 Oct 12 '15 at 1:24
  • $\begingroup$ I'm not sure I understand what you mean, you're saying that {$\sin(x), x\sin(x)$} must be non-homogeneous? If so, don't non-homogeneous equations normally have more than two terms, because the homogeneous solution is part of it? $\endgroup$ – Har Wiltz Oct 12 '15 at 1:36
  • $\begingroup$ I'm saying $y=x\sin x$ does not solve any ODE of the form $y''+a(x)y'+b(x)y=0$ since this would contradict uniqueness. $\endgroup$ – user138530 Oct 12 '15 at 1:40
  • $\begingroup$ Ok, so you're saying that it's impossible to construct this ODE? I had never considered that. Your reasoning looks pretty solid though. Thanks. $\endgroup$ – Har Wiltz Oct 12 '15 at 14:20
  • $\begingroup$ Just realized I forgot to include the interval, $ I = (o, \pi) $. So 0 isn't really part of the interval. According to the Wronskian of the set, it should be possible to construct an ODE since $ W = \sin^2 (x) \neq 0$ on $ I $ $\endgroup$ – Har Wiltz Oct 12 '15 at 14:24

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

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

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

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




The general solution is

$$y(x)=(C_1+C_2x)\sin x.$$

We can get rid of the two constants simply with

$$\color{green}{\left(\frac{y(x)}{\sin x}\right)''=0},$$ which is the desired ODE.

Optionally, to obtain a linear form, we develop $$\left(\frac{y}{\sin x}\right)''=\left(\frac{y'\sin x-y\cos x}{\sin^2x}\right)'$$ and the numerator is

$$(y''\sin x+y\sin x)\sin^2x-2(y'\sin x-y\cos x)\sin x\cos x.$$

After simplification, we get an homogeneous equation,

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


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