Find a 3rd order linear homogeneous differential equation with constant coefficients whose solution is $y=x\sin(x)$
Here is my attemot so far
$$y = x\sin x\\y'=x\cos x+\sin x\\y'' = -x\sin x + 2\cos x\\y''' = -x\cos x – 3\sin x$$
I am stuck here. Any help ?
Best Answer
You have computed $y$, $y'$, $y''$, and $y'''$ for the given function $y(x):=x\sin x$. The first thing to try is to introduce undetermined coefficients $a_0$,$a_1$,$a_2$, $a_3$, and to check whether it is possible to attain $$a_0y(x)+a_1y'(x)+a_2y''(x)+a_3y'''(x)\equiv0$$ with a suitable choice of the $a_k$. Now the $y^{(k)}$ contain terms of the form $\cos x$, $\sin x$, $x\cos x$, $x\sin x$, so that we will obtain a system of four linear equations in the four unknowns $a_k$. If this system has a nontrivial solution $(a_0,a_1,a_2,a_3)$ we can present $$a_0y+a_1y'+a_2y''+a_3y'''=0$$ as a solution to the given problem. Doing the calculations you'll find out that the only solution is the trivial one: $a_0=a_1=a_2=a_3=0$. This means that there is no third-order linear homogeneous ODE having $x\mapsto x\sin x$ as solution.
This has a reason: The solution space of a linear homogeneous ODE is translation invariant. Therefore, if $x\mapsto \sin x$ is a solution of such an ODE then so is $x\mapsto\cos x$, and as a consequence the functions $x\mapsto e^{ix}$ and $x\mapsto e^{-ix}$ both are solutions of this ODE. Therefore both $i$ and $-i$ are roots of the characteristic polynomial associated to this ODE. Similarly: If $x\mapsto x\sin x$ is a solution of such an ODE then necessarily both $i$ and $-i$ have to be at least double roots of the characteristic polynomial. The simplest polynomial (even allowing complex coefficients) of this kind is $$p(\lambda):=(\lambda-i)^2(\lambda+i)^2=\lambda^4+2\lambda^2+1\ .$$ It follows that the simplest linear homogeneous ODE having $x\mapsto x\sin x$ as a solution is $$y^{(4)}+2y''+y=0\ .$$