How to solve this first order nonlinear PDE? Given 
$u=x{{u}_{x}}+y{{u}_{y}}+\frac{1}{2}\left( u_{x}^{2}+u_{y}^{2} \right)$  , find a solution with $u\left( x,0 \right)=\frac{1}{2}\left( 1-{{x}^{2}} \right)$ .  
Not confortrable with my solution as follows.  Please help.

Standard Charpit's method leads to    
$\frac{dp}{0}=\frac{dq}{0}=\frac{du}{xp+{{p}^{2}}+qy+{{q}^{2}}}=\frac{dx}{x+p}=\frac{dy}{y+q}.$     
$\Rightarrow p=a,q=b\Rightarrow u=ax+by+\frac{1}{2}\left( {{a}^{2}}+{{b}^{2}} \right)$  
$u\left( x,0 \right)=\frac{1}{2}\left( 1-{{x}^{2}} \right)=ax+0+\frac{1}{2}\left( {{a}^{2}}+{{b}^{2}} \right).$   
$\Rightarrow {{b}^{2}}=\frac{1}{2}\left( 1-{{x}^{2}}-{{a}^{2}} \right)-ax.$   
$\Rightarrow u=\frac{1}{2}\left( 1-{{x}^{2}} \right)+y\sqrt{\frac{1}{2}\left( 1-{{x}^{2}}-{{a}^{2}} \right)-ax}\text{   where }a\text{ is an arbitary constant}.$ 

Solution alternative (for Clairaut's form)  :  
$dz=pdx+qdy$  
$z=ax+by+c\text{ }.$   
$u\left( x,0 \right)=\frac{1}{2}\left( 1-{{x}^{2}} \right)=ax+0+c\text{ }\Rightarrow \text{ }c=\frac{1}{2}\left( 1-{{x}^{2}} \right)-ax\text{ }.$  
$\Rightarrow u=\frac{1}{2}\left( 1-{{x}^{2}} \right)+by.$ 

Hint It is in Clairaut's form. Can you see?

 $u=ax+by+\frac{1}{2}(a^2+b^2)$
then use the initial condition to eliminate $b$

