1st order differential linear equation, question on absolute value

Question

I'm trying to find the general solution to this equation: $$x \frac{dy}{dx}+3(y+x^2)=\frac{\sin(x)}{x} $$ Standard form puts it like this: $$\frac{dy}{dx}+\frac{3}{x}y=\frac{\sin(x)-3x^3}{x^2} $$ To determine the integrating factor I did $e^{\int{3/x}\,dx}$ and got $e^{\ln{\lvert x\rvert}^3}$.

Does this not simplify to $\lvert x\rvert^3$? In all the online calculators I've used, they've ignored the absolute value? The problem would be much easier if that was the case but I'm not convinced.

I wouldn't know how to integrate the following with the absolute value: $$ \int{\frac{\lvert x\rvert^3}{x^2}\cdot (\sin(x)- 3x^3)\,dx}$$ I'd appreciate any help.

Answer 1

Hint

You can get rid of the problem if you start using $y=\frac z {x^3}$ which makes the equations to become $$z'=x \sin(x)-3x^4$$

Answer 2

Your equation is singular at $x=0$, so that the domain for the continuous ODE is split at this line. Consequently you can use different integrating factors for both parts of the domain, as the solutions there are independent from each other. If you chose $|x|^3$ for $x>0$ and $-|x|^3$ for $x<0$, then you could also chose $x^3$ uniformly, or just multiply the original equation with $x^2$.