4
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ $y=x\sin x$ 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
1
$\begingroup$

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

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

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

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

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

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

$\endgroup$
1
$\begingroup$

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,

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.