What is the solution on the interval (-3, 3) for this ODE? I took Differential Equations a few years ago and am reading back through my old book in a little more detail, especially with regard to intervals of validity for solutions. I found one of the example problems which finds a solution only for the intervals (-$\infty$, -3) and (3, $\infty$) using the standard integrating factor method. One of the section problems is to go back and find the solution on the interval (-3, 3). This is the equation: $$(x^2 - 9)\frac{dy}{dx}+xy=0$$
I've banged my head against the wall for a little bit before I remembered to check the Existence and Uniqueness theorem for IVP's to see if a solution exists on the interval: $$f(x,y)=-\frac{x}{x^2-9}y$$ and $$\frac{\partial f}{\partial y}=-\frac{x}{x^2-9}$$
Both of which are defined everywhere but x=-3 and x=3 so I believe a unique solution must exist on all of the intervals I mentioned above. I follow the books solution fine for the example but in this case I don't see any way to move forward on (-3,3) without bringing in complex numbers which I'm not very confident with. Further plugging the equation into Symbolab gives a plot with no solution on this interval. Usually I'm missing something simple but could someone enlighten me as to what it is?
 A: I imagine the book solves the equation for the intervals $(-\infty,-3)$ and $(3,\infty)$ by setting up
$$ y' + \frac{x}{x^2-9}y = 0 $$
Then multiplying both sides of the equation by the integrating factor $\sqrt{x^2-9}$. This is where the original domain restriction appears. But we will not use this integrating factor! Integrating factors are not unique, and using different integrating factors may give us solutions for different intervals. Let's find the integrating factor we need for $(-3,3)$. The general form of the integrating factor of this ODE is
$$ \exp\biggl({\int\frac{x}{x^2-9}}\biggr) = \exp\biggl(\frac{\ln\lvert x^2-9\rvert}{2}+C\biggr) = C\cdot\sqrt{\lvert x^2-9\rvert}$$
Where $C$ is nonzero. Since $\lvert -a \rvert = \lvert a \rvert$, a valid integrating factor is
$$\sqrt{9-x^2}$$ You will notice this is defined on the required interval $(-3,3)$. From here, apply the integrating factor method as normal to reach your desired solution.
This is one of my first answers on this site, so I would appreciate any formatting and/or etiquette tips on how to improve my responses. I hope this was helpful!
