Solving the differential equation $(xy^{4}+y)dx - xdy = 0$ The differential equation is : 
$(xy^{4}+y)dx - xdy = 0$
I am trying to simplify the equation to the form $\dfrac{fdg-gdf}{f^{2}}$ so that I can reduce it to $d(\dfrac{g}{f})$ but I am unable to do it.
Any ideas are appreciated.
 A: If you instead let $y(x)=x u(x)$ your differential equation takes the form
$$
\bigl(x(xu)^4+xu)\bigr)\,dx-x(x\,du+u\,dx)=0
$$
that is
$$
x^5u^4\,dx=x^2\,du.
$$
I'm sure you can solve this differential equation.
A: You can do it ......
$$(xy^{4}+y)dx - xdy = 0$$
$$(xy^{4}+y)= xy' $$
$$xy^{4}= -(y-xy') $$
Here you have what you wanted
$$xy^{2}= -\left(\frac {y-xy' }{y^2}\right )$$
$$xy^{2}= -\left(\frac {x }{y}\right)' \implies \int x^{3}dx= -\int\left(\frac xy \right )^2d\left(\frac {x }{y}\right)$$
You can conitnue but I prefer to substitute $\frac xy=z$
$$xy^{2}= -\left(\frac {x }{y}\right)' \implies \frac {x^{3}} {z^2}= -z'$$
Which is easy to integrate 
$$\int x^{3}dx= -\int z^2 dz \implies \frac {x^4}4+K=- \frac {z^3}3 \implies  y^3(\frac {x^4}4+K)=- \frac {x^3}3 $$
$$\boxed {y^3(x)=- \frac {4x^3}{({3x^4}+K)}} $$
A: Rewrite the equation like this
$$ \frac{dy}{dx} - \frac{1}{x}y = y^4 $$
This is a Bernoulli equation. Divide through by $y^4$ 
$$ y^{-4}\frac{dy}{dx} - \frac{1}{x}y^{-3} = 1$$
Make the substitution $z = y^{-3}$ to obtain
$$ \frac{dz}{dx} + \frac{3}{x}z = -3 $$
This is now just a regular linear first-order equation. The integrating factor is $\mu = e^{3\ln x} = x^3$, you can do the rest
