Solving a differential equation with natural log I am given:
$x\dfrac{dy}{dx}=\dfrac{1}{y^3}$
After separating and integrating, I have:
$y^4/4=\ln x+C$
I am supposed to solve this equation, but I'm stuck here.  Should I solve explicitly so I can keep $C$?
EDIT:
A solution I came up with last night was:
$y=(4\ln x+C)^{1/4}$

Try differentiating to see if you got the correct solution!
You can compute
$$
\frac{dy}{dx} = \frac{1}{4}\left(4 \ln x + C\right)^{-3/4}\left(\frac{4}{x}\right) = \frac{1}{x}\left(4 \ln x + C\right)^{-3/4}
$$
so 
$$
x\frac{dy}{dx} = \left(4 \ln x + C\right)^{-3/4}.
$$
Is this equal to $\dfrac{1}{y^{3}}$?
