Differential Equation $x^3y' + (2-3x^2)y = x^3$ Solve the following differential equation on the biggest possible interval $$x^3y' + (2-3x^2)y = x^3,  \space \space \space\space (x>0).$$
First I write it in the form $$y' = 1 - \frac{2-3x^2}{x^3}y$$but I'm not sure how to proceed from there. The things I've learned so far (separation of variables, variation of constants) don't seem applicable here.
Can anyone please share a hint?
 A: Hint: Write $y= z+\frac{x^3}{2}$ and derive
$$
\frac{z'}{z} = \frac{3x^2-2}{x^3},
$$
and then integrate.
A: To arrive to Mohammad Khosravi's good answer, suppose that a particular solution is $y=kx^3$ and plug this expression in the differential equation. You will arrive to $(1-2k)x^3=0$. Then, ...
A: I will give you a fairly general method to solve 1st-order linear ODEs. Let:
$$y' + p \, y = f, \quad y = y(x),$$ where $p$ and $f$ are functions of $x$. Assume you are interested in solutions satisfying:
$$\frac{d}{dx}\left( I \, y \right) = I \, f, \tag{1}$$  so we have $I y' + I' y = I f$, which is equivalent to (if $I \neq 0$):
$$ y' + \frac{I'}{I} \, y = f. $$ If you compare this result with the first result you obtain a differential equation for $I$ which is straightforward to solve:
$$I'/I = p, \quad I = e^{\int p \, dx}.$$ 
$I$ is called integrating factor. Once $I$ is known, you can solve for $y$ from eq (1).
I'm sure you can take it from here.
Cheers!
A: Great, thanks for the help. This is how I've tried it now:
$$ I = e^{\int p \, dx} = e^{-x^{-2}} x^{-3},$$
thus, $$\frac{d}{dx}\left( I \, y \right) = I \, f$$ $$\iff  I \, y  = \int I \, f \,dx $$ $$\iff y =  I^{-1} \int I \, f \,dx = x^3 e^{x^{-2}} \int e^{-x^{-2}} x^{-3}  \,dx = \frac{1}{2} e^{-x^{-2}} e^{x^{-2}}x^3 =  \frac{x^3}{2}.$$
One more thing: How do I now determine the maximal interval for which y solves the differential equation?
