Basic First-order quasi-linear PDEs question, involving characteristics I would be incrediabely gratful if someone could go through step by step and explain how to do this question, as i'm rather stuck -and the lecture notes have a lot to be desired!
'Find $z(x,y)$ explicitly if
$$-yz_x+xz_y=\frac{xz}{\sqrt{x^2+y^2}}$$
and $z(x,0)=1$ for $x \geq 1$
So i'm trying to go about solving the characteristics of this, the equations
$$\frac{dx}{ds}=-y$$
$$\frac{dy}{ds}=x$$
$$\frac{dz}{ds}=\frac{xz}{\sqrt{x^2+y^2}}$$
Then use the solution along the curve $l(t)=(t,0,1)$ for $t\geq 1$ to find the coefficients of the solutions to the equation in terms of t. We would then have the solution space in terms of parameters $t,s$ correct? Then we try to eliminate these to get the space in terms of $x,y,z$. I can't seem to solve the equations however.
Any help is greatly appreciated.

You've got the idea right.
The characterics starting on the positive $x$ axis (at $(x,y)=(t,0)$ to be precise) when $s=0$ are $(x(s),y(s))=(t \cos s,t \sin s)$. Then the equation for $z$ reduces to $dz/ds=z \cos s$ (independent of $t$, as it turns out). Separation of variables gives
$$
\int_{1}^{z(s)} \frac{dz}{z} = \int_0^s \cos s \, ds.
$$
(I'm being a little sloppy with the notation here, using the same name for the dummy variable as in the upper limit of integration.)
Integrating this, we get $\ln z(s)=\sin s$, hence
$$
z(x,y) = \exp\frac{y}{\sqrt{x^2+y^2}}.
$$
