How to solve $xy'=3y-6x^2$ using integrating factors?

Question

I am trying to solve $xy'=3y-6x^2$ using integrating factors. I am facing 2 issues when doing so. In order to find $P(x)$ to be used in $e^{\int P(x) dx}$, I am dividing by $x$ which, it seems, restricts the domain for which the solution holds. Also, when I do find that $P(x)$ is $-3 \over x$ from dividing everything by $x$, the integrating factor becomes $|x|^{-3}$, and after multiplying the DE by this and trying to integrate both sides, I am uncertain how to deal with the absolute value. I would appreciate help on these 2 matters.

Answer 1

The absolute value is sort of a non-issue. If you ignore it and proceed as you have stated, you will divide the given equation by $x^4$ to get $$x^{-3}y'-3x^{-4}y=-6x^{-2}$$ which can be written $$(x^{-3}y)'=-6x^{-2}$$ and hence $$x^{-3}y=6x^{-1}+C\quad\hbox{so}\quad y=6x^2+Cx^3\ .\tag{$*$}$$ The thing is, there is no actual need to calculate an integrating factor - you could just pretend someone came up and whispered in your ear, "Hint: what happens if you divide by $x^4$?". Since you did not integrate to get a log, there is no need to worry about the absolute value.

The $x=0$ issue is more of a problem. In fact, though it is easy to check that $(*)$ is a solution for any $C$, there are other solutions too. Here is one: $$y=\cases{6x^2+x^3&if $x\ge0$\cr 6x^2-x^3&if $x<0$.\cr}$$ I leave it to you to check that $y'$ exists for all $x$ (be careful when $x=0$) and satisfies the equation. What we should really do for an equation like this is to specify that $x>0$, or $x<0$ if you prefer, but not both.

This is related to the issue of putting an absolute value in a log integral, which is not strictly correct for a similar reason. I have explained it in some detail here.

As a final comment, it is quite likely that your teachers expect you to get the answer $(*)$ without asking awkward questions. Congratulations on asking them!!!