What is the solution to $xy''+2y'+xy=0$ where $y(0),y(\infty)<\infty$

Question

I was hoping someone might be able to explain to me how I find the family of solutions to the following:

$$xy''+2y'+xy=0$$

where $y=y(x)$ and $y$ is bounded at both $0$ and infinity. I have been attempting to recast it as

$$(xy)''+xy = 0$$

and transforming it via $z=xy$, however my solutions aren't making any sense! It's been about 5 years since I've worked with an equation like this so I apologize for the triviality of this problem.

My workings:

If one makes the transformation $z=xy$ we get

$$z''+z=0$$

with $z(0)=0\times y(0)=0$, which gives

$$z=A\sin(x)+B\cos(x)$$

$B=0$ by $z(0)=0$. At this point I was expecting to find a way to create an infinite family of solutions, I can if I restrict x to a finite domain, however I wish to solve over an infinite one.

Answer 1

$$xy''+2y'+xy=0\implies x^2y''+2xy'+x^2y=0$$

Without loss of generality, we assume that $y(x)=x^af(x)$ for some unknown function $f(x)$. Thus, we have

\begin{align} y'(x) &= x^af'(x)+ax^{a-1}f(x)\\ y''(x) &= x^af''(x)+2ax^{a-1}f'(x)+a(a-1)x^{a-2}f(x) \end{align}

Plugging this into our differential equation, we have

$$x^2\left(x^af''(x)+2ax^{a-1}f'(x)+a(a-1)x^{a-2}f(x)\right)\\+2x\left(x^af'(x)+ax^{a-1}f(x)\right)+x^{a+2}f(x) = 0$$

Multiplying out and sorting by powers of $x$, we have

$$x^{a+2}\left(f''(x)+f(x)\right)+2x^{a+1}(a+1)f(x)+ax^a(a+1)f(x)=0$$

If $a=-1$, the last two terms reduce $0$ and we are left with

$$x^{a+2}\left(f''(x)+f(x)\right)=0\implies f''(x)+f(x)=0\implies f(x)=A\sin(x)+B\cos(x)$$

Since $a=-1$, this gives us our general solution:

$$y(x)=\frac{A\sin(x)+B\cos(x)}{x}$$

This solution has a singularity at $x=0$ if $B\neq0$, therefore $B=0$. This gives you the solution

$$y(x)=\frac{A\sin(x)}{x}$$

on an infinite domain.