Solving for dy/dx: x/(x2y + y3)

[checkmatechamp]

Homework Statement

dy/dx = x/(x²y + y³)

Homework Equations

The Attempt at a Solution

The middle term is what's throwing me off. I can't put it in terms of y = vx (Dividing by x² means that the y term gets screwed up. (It would turn it into yv²). Out of curiosity, I tried flat-out substituting y = vx (and turning dy/dx into x*dv/dx + v), but that didn't work either. It's obviously not exact, and I don't see how it could be made linear, so I'm not sure where to begin.

 

[Saitama]

Homework Statement

dy/dx = x/(x²y + y³)

Homework Equations

The Attempt at a Solution

The middle term is what's throwing me off. I can't put it in terms of y = vx (Dividing by x² means that the y term gets screwed up. (It would turn it into yv²). Out of curiosity, I tried flat-out substituting y = vx (and turning dy/dx into x*dv/dx + v), but that didn't work either. It's obviously not exact, and I don't see how it could be made linear, so I'm not sure where to begin.

Doesn't look like there's an elementary solution to this: http://www.wolframalpha.com/input/?i=dy/dx=x/(x^2y+y^3)

 

[Quesadilla]

The answer will involve the Lambert W function. Is this something you are familiar with? Basically, one arrives at something like $z + \ln z = c$ (or $c = ze^z$) and needs to solve for $z$.

Start by substituting $u = y^2$ and then do another substitution. This should give you a separable equation which, after you solve it, gives you something like $v - \ln v = g(x)$. This would be where the Lambert W function comes into play.

Edit: If you don't like Lambert's W function, you can find a function $h$ such that $x = h(y)$.