Doubt in solving PDE. Given that 
$yu_x+xu_y=xy$ , $x\geqslant0$, $y\geqslant0$ with $u(0,y)=e^{-y^2}$ for $y>0$ , and $u(x,0)=e^{-x^2}$ for $x>0$.
My doubt is that how to use initial values in this case?  
The answer given is 
$ 
\left\{
\begin{aligned} \frac12y^2 + e^{-(x^2-y^2) } \quad for  \quad  x>y\\ 
 \frac12 x^2 + e^{-(y^2-x^2) } \quad   for  \quad  x<y
\end{aligned}
\right.$
Please guide me.
I got the solution.Let me answer it.
 A: $yu_x+xu_y=xy$ , $x\geqslant0$, $y\geqslant0$ with $u(0,y)=e^{-y^2}$ for $y>0$ , and $u(x,0)=e^{-x^2}$ for $x>0$.
Dividing it by $xy$, we get,
${u_x\over x}$+ ${u_y\over y}$=1
${dx\over dt} = { 1\over x}$ with $x(0)=0$
$\implies$${ x^2\over 2}=t$
${dy\over dt}={1\over y}$ with $y(0)=y_0$ $\implies$ ${ y^2\over 2}=y_0+t$
$\implies$ ${ y^2\over 2}-t=y_0 $ $\implies$ ${ y^2\over 2}- { x^2\over 2}=y_0$
${ du\over dt}=1$ with $u(0)=f(y_0)$ $\implies$ $u=t+f(y_0)$
$\implies$ $u= { x^2\over 2}+f(   -{  x^2\over 2}$ +${ y^2\over 2} )$
Now $u(x,0)={ x^2\over 2}+ f(   -{  x^2\over 2})$  
$\implies$ $e^{-x^2}= { x^2\over 2}+ f(   -{  x^2\over 2})$    
put     $-{  x^2\over 2}=t$ ,we get $e^{2t}-t=f(t)$ 
$\implies$ $u={ x^2\over 2} +e^{2(-{  x^2\over 2}  + { y^2\over 2})}    -{ x^2\over 2}+{ y^2\over 2}$ 
$\implies$ $u=   { y^2\over 2}+ e^{  -({  x^2 }  - { y^2 })}     $
Also $u(0,y)=e^{-y^2}=f(    {  y^2\over 2}) $
Put ${y^2\over2}=t$, we get  $e^{-2t}=  f(   t )$
$\implies$ $u={x^2\over 2}+e^{  -2(-{   x^2\over2 }  + { y^2 \over 2})} $ 
$\implies$ $u={x^2\over 2}+e^{    ( {   x^2  }  - { y^2  })} $
Or $u={x^2\over 2}+e^{      -({   y^2  }  - { x^2  })} $
I still have a doubt. Why are $x>y$ and $y>x$   written in the final answer?  
