Solve $x^3+3y-xy^\prime=0$ 
Solve $x^3+3y-xy^\prime=0$

Divided by $x$ to get $x^2+3\frac{y}{x}-y^\prime=0$
Let $v=\frac{y}{x}$ then $y^\prime=v^\prime x+v$
then $x^2+3v=v^\prime x+v$
then $v^\prime=x+2\frac{v}{x}$
Used integrating factor $e^{\int -2/x dx}=e^{-x^2}$
then $e^{-x^2}v=\int e^{-x^2}\cdot x dx$
But when I solve this $e^{-x^2}v=\frac{-1}{2}e^{-x^2}$
$v=\frac{-1}{2}$
But this isn't the right answer.


Divided by $x$ to get $x^2+3\frac{y}{x}-y^\prime=0$

From this step, multiply by negative one to form
$$y'-\frac{3}{x}y=x^2$$
therefore the integrating factor is 
$$\mu(x)=\text{exp}\left(-{\int {\frac{3}{x}}dx}\right)=x^{-3}$$
hence the equation can be written as
$$\frac{d}{dx}\left(\frac{y}{x^3}\right)=\frac{1}{x}$$
Can you finish?
