Solving $(y+x^4y^2)dx+xdy=0$ How to solve (in terms of $y$) $(y+x^4y^2)dx+xdy=0$.
I know I'm supposed to multiply by an integrating factor to turn this equation into an exact equation.
In the previous exercise I proved that in $Mdx+Ndy$ the functions:


*

*$\frac{1}{N}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)$

*$\frac{1}{M}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)$

*If $M=yf(xy)$ and $N=xg(xy)$ then $\frac{1}{xM-yN}$ is an integrating factor


I've tried using 1. and 2., but the calculations are horrible and do not seem to work. I don't think that's the way to go.
About 3. I don't know what $f$ and $g$ I should choose.
how can I go about solving this differential equation?
this isn't homework.
 A: $(y+x^4y^2)dx+xdy=0$
Subtract $x^4 y^2$ from both sides, yielding:
$y + x \frac{dy}{dx} = -x^4 y^2$
Divide both sides by $-xy^2$, yielding:
$\displaystyle -\frac{1}{xy} - \frac{\frac{dy}{dx}}{y^2} = x^3$
Let $\displaystyle v(x) = \frac{1}{y}$, which yields:
$\displaystyle \frac{dv}{dx} - \frac{v}{x} = x^3$
Choose an integrating factor $\displaystyle \mu = \text{exp}\left(\int -\frac{1}{x} dx\right) = \frac{1}{x}$, so
$\displaystyle \frac{\frac{dv}{dx}}{x} - \frac{v}{x^2} = x^2$
Substitute $\displaystyle -\frac{1}{x^2} = \frac{d}{dx}\frac{1}{x}$, yielding:
$\displaystyle \frac{d}{dx}(\frac{v}{x}) = x^2$
Integrate both sides, yielding:
$\displaystyle v = x(c_1 + \frac{x^3}{3})$
Solve for the original $y$, yielding:
$$\displaystyle y = \frac{3}{x^4 + cx}$$
Note: you should be able to do it the way you suggested, so give that another go.
A: An integrating factor, as you can check (!), is
$$f(x,y) = \frac1{xy(x^4y-3)}\,.$$
Where does this come from, you ask? By noting that the $1$-parameter group on $\mathbb R^2-\{0\}$
$$\{(x,y) \rightsquigarrow (e^t x, e^{-4t}y)\}$$
leaves the differential equation invariant. I was told years ago that it was exactly for the purposes of finding such obscure integrating factors that Sophus Lie "invented" Lie groups.
One then integrates and gets 
$$\frac{3-x^4y}{xy} = C,$$
which is, indeed, $y=\dfrac3{x^4+Cx}$, as @Amzoti obtained.
