Solving : $x^2 z_x + y^2 z_y = 2xy$

Question

Exercise :

Find the general integral and compute three different solutions for the PDE : $$x^2 z_x + y^2 z_y = 2xy$$

Attempt :

The general integral is given by a function $F \in C^1$ : $F(u_1,u_2) = 0$, where $u_1$ and $u_2$ are the integral curves, calculated by the differential problem : $$\frac{\mathrm{d}x}{x^2} = \frac{\mathrm{d}y}{y^2} = \frac{\mathrm{d}z}{2xy}$$ But then I am at loss on how to calculate the general solutions asked.

Also, for $u_1$ and $u_2$ :

$$\frac{\mathrm{d}x}{x^2} = \frac{\mathrm{d}y}{y^2} \implies u_1 = \frac{y-x}{xy}$$

but I am also unable to grasp a calculation for $u_2$.

Any help and explanation about the general solutions and $u_2$ will be greatly appreciated as this is a new subject I am getting in.

Answer 1

Starting from your result $$\frac{\mathrm{d}x}{x^2} = \frac{\mathrm{d}y}{y^2} \implies c_1 = \frac{y-x}{xy}$$
You can deduce y as a function of x $$\implies y=\frac x {1-c_1x}$$

$$ \frac{\mathrm{d}x}{x^2} = \frac{\mathrm{d}z}{2xy}$$ $$\implies \frac{2y\mathrm{d}x}{x} = dz$$ $$\implies \frac{2\mathrm{d}x}{{1-c_1x}} = dz$$ $$\implies \int \frac{\mathrm{d}x}{{x-\frac 1 {c_1}}} = -\frac {c_1}2\int dz$$

Which is easy to integrate...

Answer 2

The hint "finding three different general solutions" it could mean the solutions of $$ ds=\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{2xy} $$ as functions of $s$. As $x(s)=\frac{x_0}{1-sx_0}$, $y(s)=\frac{y_0}{1-sy_0}$ one gets $$ \frac{dz}{ds}=\frac{2x_0y_0}{(1-sx_0)(1-sy_0)}=\frac{2x_0y_0}{x_0-y_0}\left(\frac{x_0}{1-sx_0}-\frac{y_0}{1-sy_0}\right) $$ so that $$ z(s)=z_0+\frac{2x_0y_0}{x_0-y_0}\left(\ln(1-sy_0)-\ln(1-sx_0)\right) $$