Find solution of PDE $x \frac {\partial z} {\partial x} +y \frac {\partial z} {\partial y}=z$

Question

Problem:Find solution of Cauchy problem for the first order PDE $x \frac {\partial z} {\partial x} +y \frac {\partial z} {\partial y}=z$,on $ D= {(x,y,z): x^2 +y^2 \neq0,z>0} $ with initial condition $x^2+y^2=1,z=1$

Solution:Using Lagrange's solution we get,

$\phi \left ( x \over y \right )={y \over z} $ where $\phi$ is some arbitrary function

But I don't know "How to do this with above conditions"

Answer 1

This is Euler's equation for homogeneous functions of degree $k=1$. Hence we know that $z( \lambda x, \lambda y)=\lambda z(x,y)$ for all $\lambda>0$. Since $z \equiv 1$ on the unit circle, we find that $$z(x,y)=z \left( \sqrt{x^2+y^2}\frac{(x,y)}{\sqrt{x^2+y^2}} \right)=\sqrt{x^2+y^2} \;z \left( \frac{(x,y)}{\sqrt{x^2+y^2}} \right)=\sqrt{x^2+y^2}.$$