Exercise :
Consider the equation $$z z_x + y z_y = x$$ and the initial curve $$C : x=t, y=t \; ; \; t >0$$ Decide whether there is a unique solution, no solution or infinitely many solutions in a neighborhood of the point $(1,1)$, for the initial data $z=2t$.
Attempt :
We yield the Lagrange equations :
$$\frac{\mathrm{d}x}{z} = \frac{\mathrm{d}y}{y} = \frac{\mathrm{d}z}{x}$$
and, easily, we can calculate two linearly independent integral curves :
$$\frac{\mathrm{d}x}{z} = \frac{\mathrm{d}y}{y} \implies z_1 = y(x-z) $$
$$\frac{\mathrm{d}x}{z} = \frac{\mathrm{d}z}{x} \implies z_2 = \frac{1}{2}(z^2-x^2)$$
Thus, the general solution is given by an expression of a $C^1$ differentiable function $F$, such that :
$$F(z_1,z_2) = 0 \Rightarrow F\bigg(y(x-z), \frac{1}{2}(z^2-x^2)\bigg) = 0 $$
Taking into account the parameters of the initial curve $C$, we yield :
$$C : \mathbf{r}_C(t) = (t,t,2t) \; ; \; t > 0$$ $$\frac{\mathrm{d}\mathbf{r}_C(t)}{\mathrm{d}t} = (1,1,2) \; ; \; t > 0$$
Since :
$$\frac{\mathrm{d}\mathbf{r}_C(t)}{\mathrm{d}t} = \frac{1}{t} \mathbf{r}_C(t)$$
there exists a proportionality function-constant $μ(t) = \frac{1}{t}$ such that $\frac{\mathrm{d}\mathbf{r}_C(t)}{\mathrm{d}t} = μ(t) \mathbf{r}_C(t)$, which means that there exist infinitely many solutions.
Alternativelly, substituting the parameters into $F$, we get :
$$F\bigg(-t^2, \frac{3}{2}t^2\bigg) = z(t,t) = 2t = 0$$
But that would only hold for $t=0$, thus for $F(0,0) = 0$ and there can be infinitely many $C^1$ differentiable functions $F$ such that this case holds, thus we have infinitely many solutions again.
Questions : Is my approach correct ? Is there a more standard way of approaching the existence and uniqueness of a PDE ? Maybe more strictly and formally ? Am I missing something ?
Let $c_1=y(z-x)$ and $c_2=z^2-x^2$ be the constants of the characteristic curves. None are constant on the initial data, indeed there $c_1=t^2$ and $c_2=3t^2$ so that $c_2=3c_1$ and inserting that relation backwards $$ z^2-x^2=3y(z-x)\iff z=x\lor z=3y-x $$ Both are solutions of the PDE, $z(x,y)=x$ and $z(x,y)=3y-x$, but the first does not satisfy the initial conditions except at $t=0$, which is not enough.