Solve the equation `ydx+(x-y^2)dy=0`

Answer

`x=y^(3)/3+C/y`

Explanation

`ydx+(x-y^(2))dy=0`
or `ydx=(y^(2)-x)dy`
or `(dx)/(dy)=(y^(2)-x)/(y)=y-x/y`
or `(dx)/(dy) + x/y=y`
This is a linear differential equation of the form
`(dy)/(dx) + Px=Q` where `P=1/y` and `Q=y`
Now, `I.F.= e^(int(Pdy))=e^(1/ydy)=e^(logy)+C`
or `xy=inty^(2)dy+C`
or `xy=y^(3)/3+C`
or `x=y^(3)/3+C/y`