I.V.P. : $ x^2 z_x + y^2 z_y = z^2 \; ; \; z=1 $ on the initial curve $C : y=2x $

Question

Exercise :

Solve the Initial Value Problem (I.V.P.) $$\begin{cases} x^2 z_x + y^2 z_y = z^2 \\[.5em] z=1 \text{ on } C =\{(x,y): y=2x \} \end{cases}$$ where $C$ is the initial curve of the problem.

Attempt :

First of all, we yield the differential problem :

$$\frac{\mathrm{d}x}{x^2} = \frac{\mathrm{d}y}{y^2} = \frac{\mathrm{d}z}{z^2}$$

We solve :

$$\frac{\mathrm{d}x}{x^2} = \frac{\mathrm{d}y}{y^2} \implies u_1 = \frac{1}{x} - \frac{1}{y} = \frac{y-x}{xy} $$

$$\frac{\mathrm{d}x}{x^2} - \frac{\mathrm{d}y}{y^2} = \frac{\mathrm{d}z}{z^2} \implies u_2 = \frac{y-x}{xy} -\frac{1}{z} $$

The parameters of the initial curve as a position vector, are : $$C : r_C(t) = (x,2x,1) \; x \in \mathbb R$$

And then we yield the system : $$\begin{cases} Z_1 = \frac{1}{x} \\ Z_2 = \frac{1}{x} - 1 \end{cases} \Rightarrow Z_2 = Z_1 - 1 \implies z_2(x,y,z) = z_1(x,y,z) - 1$$ which finally implies that $z=1$ is a solution but also $z_0 = z(x,y) = z(x,2x) = 1$.

Question : Is my approach correct ? Have I left something out of my solution ?

Answer 1

Your second step seems wrong, by the Lagrange equations $\dfrac{dx}{x^2}-\dfrac{dy}{y^2}=0$, not $=\dfrac{dz}{z^2}$.

You should get something like $u_2=z^{-1}-x^{-1}$, so that then with $u_2=\phi(u_1)$ and inserting the initial conditions one gets $$ 1-x^{-1}=\phi(x^{-1}-(2x)^{-1})\implies \phi(u)=1-2u $$ and then in general $$ z^{-1}=x^{-1}+1-2(x^{-1}-y^{-1})\implies z=\frac{xy}{2x-y+xy}. $$