Find the solution of the Cauchy problem for the pde Find the solution of the Cauchy problem for the pde 
$x\dfrac{\partial z }{\partial x}+y\dfrac{\partial z}{\partial y}=z$
on $D=\{(x,y,z):x^2+y^2\neq 0,z>0\}$ with the initial condition $x^2+y^2=1,z=1$.
By Lagrange's Equations 
$\dfrac{dx}{x}=\dfrac{dy}{y}=\dfrac{dz}{z}$ 
If I solve the above I get ;$\dfrac{x}{y}=c_1,\dfrac{y}{z}=c_2,\dfrac{x}{z}=c_3$
But I am unable to use the boundary conditions.How to do this?

The set of characteristic equations is :
$$\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{z}$$
With the condition $z=1$ on the curve $x^2+y^2=1$
So, it should be easy to satisfy this condition if this curve would be a characteristic curve. So, it is of interest to make appear the terme $x^2+y^2$ in a combination of the characteristic equations :
$$\frac{dx}{x}=\frac{dy}{y}=\frac{xdx}{x^2}=\frac{ydy}{y^2}=\frac{xdx+ydy}{x^2+y^2}$$
This uses the well known identity : $\frac{A}{B}=\frac{C}{D}=\frac{A+C}{B+D}$
$$\frac{xdx+ydy}{x^2+y^2}=\frac{dz}{z}$$
$$\frac{1}{2}\frac{d(x^2+y^2)}{x^2+y^2}=\frac{dz}{z}$$
$$z=C\:\sqrt{x^2+y^2}$$
with the condition $1=C\sqrt{1}\quad\to\quad C=1\quad$. So the solution is :
$$z=\sqrt{x^2+y^2}$$
NOTE:
A more extended application of the method of characteristics should be to express the general solution of the PDE.
As seen above, a first characteristic equation is : 
$$\frac{z}{\sqrt{x^2+y^2}}=C_1$$
As already found by learnmore, another characteristic curve is $\frac{x}{y}=C_2$
Thus, an implicit form of the general solution is the equation :
$$\Phi\left(\frac{z}{\sqrt{x^2+y^2}} \:,\: \frac{x}{y}\right)=0$$
where $\Phi$ is any differentiable function of two variables.
Solving it for the the first variable gives the explicit form of the general solution :
$$\frac{z}{\sqrt{x^2+y^2}} =F\left( \frac{x}{y}\right)$$
where $F$ is any differentiable function.
The condition $z=1 \:,\: x^2+y^2=1 \quad\to\quad \frac{1}{\sqrt{1}} =F\left( \frac{x}{y}\right)$ 
This implies that $F$ is a constant function  $F=1 \quad\to\quad \frac{z}{\sqrt{x^2+y^2}} =1$
$$z=\sqrt{x^2+y^2}$$
Alternatively, one could easily solve the PDE with the change to polar coordinates.
