Answer
$y= \sqrt \frac{2x-y^4}{C}$
Work Step by Step
We are given
$ydx+(2x+y^4)dy=0$
Here, $M(x,y)=y$
$N(x,y)=2x+y^4$
We have:
$\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{N}=\frac{-3}{y}$
The integrating factor is:
$I(x)=e^{\int f(x)dx}=e^{\int \frac{-3}{y}dy}=e^{-3\ln y}=\frac{1}{y^3}$
Multiplying equation (1) by $I(x)=\frac{1}{y^3}$
$\frac{1}{y^3}(ydx+(2x+y^4)dy)=0$
$y^{-2}dx+(2xy^{-3}+y)dy=0$
Then:
$\frac{\partial \phi}{\partial x}=y^{-2}$
Integrating both sides:
$\phi=xy^{-2}+f(y)$
We have:
$\frac{\partial phi}{\partial y}=-2xy^{-3}+f'(y)=-2xy^{-3}-y$
Since $f'(y)=0 \rightarrow f(y)=-\frac{y^2}{2}+C$
C is a constant integration
The solution is:
$\phi (x,y)=xy^{-2}-\frac{y^2}{2}+C$
or
$y= \sqrt \frac{2x-y^4}{C}$
