Finding a particular solution to a partial differential equation.

Question

I am given the equation: \begin{equation}2 y \frac{\partial u}{\partial x}-x \frac{\partial u}{\partial y}=2 x y^{3}-x^{3} y \end{equation} And told that from this, it is possible to obtain: $$\frac{\partial u}{\partial x}=x y^{2}, \quad \frac{\partial u}{\partial y}=x^{2} y$$ Can someone explain how these last equations are obtained from the first?

Answer 1

$$2 y \frac{\partial u}{\partial x}-x \frac{\partial u}{\partial y}=2 x y^{3}-x^{3} y$$ System of characteristic ODEs (Charpit-Lagrange) : $$\frac{dx}{2y}=\frac{dy}{-x}=\frac{du}{2xy^3-x^3y}$$ A first characteristic equation from solving $\frac{dx}{2y}=\frac{dy}{-x}$ : $$x^2+2y^2=c_1$$ A second characteristic equation from : $$\frac{dx}{2y}=\frac{dy}{-x}=\frac{du}{2xy^3-x^3y}=\frac{xy^2dx+x^2ydy-du}{xy^2(2y)+x^2y(-x)-(2xy^3-x^3y)}=\frac{xy^2dx+x^2ydy-du}{0}$$ $$xy^2dx+x^2ydy-du=0\quad\implies\quad \frac{\partial u}{\partial x}=xy^2\quad\text{and}\quad \frac{\partial u}{\partial y}=x^2y$$ $$u-\frac12(x^2y^2)=c_2$$ General solution of the PDE on implicit form $c_2=F(c_1)$ : $$u(x,y)=\frac12 x^2y^2+F(x^2+2y^2)$$ with arbitrary function $F$ , to be determined according to some boundary condition.