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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.

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    $\begingroup$ For the sake of continuity of $\ln$ function, I prefer to assume $x>0$. $\endgroup$ – Sujit Bhattacharyya Mar 4 at 5:46
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    $\begingroup$ The general solution is $C|x|^3$, but for an integrating factor you only need one solution, so $x^3$ will do. Try it out for $-x^3$, you'll get the same result regardless. $\endgroup$ – Dylan Mar 4 at 7:43
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Hint

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

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