I've been stuck trying to solve this system:
$$\ \frac{\partial u}{\partial x} = \frac{-2xy^2}{u} + 3y $$ $$\ \frac{\partial u}{\partial y} = \frac{-2x^2y}{u} + 3x $$
Which must satisfy $ \ u(0,0) = 1$ and the solution lies on $u(x,y) = z$
I've started by choosing vector fields $ \ X = \ \left(1,0,\dfrac{\partial u}{\partial x}\right)$ and $ \ Y = \ \left(0,1,\dfrac{\partial u}{\partial y}\right)$ which by the Frobenius theorem must satisfy $\left[X,Y\right] =0$ where the brackets indicate the jacobi brackets.
Then to work out the flow of the X vector field:
$ \dfrac{dx}{dt} = 1$ => $ x = t $
$ \dfrac{dy}{dt} = 0$ => $ y = 0 $
$ \dfrac{dz}{dt} = \dfrac{-2xy^2}{z} + 3y$ where at $t=0, z= 1$
Now I am stuck integrating this last equation. I believe there is a simple trick such as making a substitution. Don't worry, this isn't homework