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 $z=t$.
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 :
$$C : \mathbf{r}_C(t) = (t,t,t) \; ; \; t > 0$$ $$\frac{\mathrm{d}\mathbf{r}_C(t)}{\mathrm{d}t} = (1,1,1) \; ; \; t > 0$$
From now on, how would one continue to make a conclusion about the existence and uniqueness of the IVP ? Any tip or hint or thorough solution will be greatly appreciated