How can I solve these pde's? Three different problem I got:
1.. $xu_x+2x^2u_y-u=x^2e^x$ and $u(x,x^2+x)=xe^x+x^2$
2.. $yu_{xx}+(x+y)u_{xy}+xu_{yy}=0, \quad x\neq  y$ 
3.. $(y+xu)u_x+(x+yu)u_y=u^2-1$
Couldnt even start. Could you at least give me a hint? thanks.
edit1: 
at 2.  from $B^2-AC=\frac {(x-y)^2}4>0 $ so the equation is     hyperbolic. so we can write its cononical forms. is it true?
edit2: for the first one :
by taking $\eta=x^2-y$ and $\xi=x$ I found $u(x,y)=xe^x+x(-x+1-e^{-x})$
for the second one : 
I took $\xi=\frac{x^2-y^2}{2}$ and $\eta=x-y$  and found $u(x,y)=\frac {f^*(\frac{x^2-y^2}{2})}{x-y}+g(x-y)$
edit 3:----------------------
1-) $\frac ba=\frac {2x^2}{x}=2x=\frac {dy}{dx}$
$x^2-y=c$,  $\eta =x^2-y$ and $\xi =x$
$u_x=w_{\xi}+w_{\eta}2x$  and $ u_y=-w_{\eta}$
$\xi(w_{\xi}+w_{\eta}2\xi)-2\xi ^2w_{\xi}-w=\xi^2e^\xi $
$w_{\xi}-\frac w \xi=e^\xi$ multiplying by $\frac 1\xi$
$\frac{d(w\frac 1\xi)}{d\xi}=e^\xi$
so we got $w=\xi e^\xi+g(\eta)$
$u(x,y)=xe^x+g(x^2-y)$
and $u(x,x^2+x)=xe^x+x^2 =xe^x+g(x^2-x^2-x)$
$g(-x)=x^2$, $g(x)=x^2$ so the  $u(x,y)=xe^x+x^2$
//--------------------------------------------
3)$\frac {\frac{dx}{dt}}{y+ux}=\frac {\frac{dy}{dt}}{x+yu}=\frac {\frac{du}{dt}}{u^2-1}$
$\frac {\frac{dx}{dt}-\frac{dy}{dt}}{y+ux-x-yu}=\frac {\frac{du}{dt}}{u^2-1}$
$\frac {\frac{dx}{dt}-\frac{dy}{dt}}{(1-u)(y-x)}=\frac {\frac{du}{dt}}{u^2-1}$
$\frac {\frac{d(x-y)}{dt}}{y-x}=\frac {\frac{du}{dt}}{u+1}$
$ln(u+1)+ln(x-y)=c_1$------------------- (1) we got one independent solution
we need another
$\frac {\frac{dx}{dt}+\frac{dy}{dt}}{y+ux+x+yu}=\frac {\frac{du}{dt}}{u^2-1}$
$\frac {\frac{dx}{dt}+\frac{dy}{dt}}{(1+u)(x+y)}=\frac {\frac{du}{dt}}{u^2-1}$
$\frac {\frac{d(x+y)}{dt}}{x+y}=\frac {\frac{du}{dt}}{u-1}$
$ln(u-1)-ln(x+y)=c_2$----------------------(2)
from (1) and (2) which form is $h(x,y,u)=c_1$ and
$j(x,y,u)=c_2$
general solution would be in $j(x,y,u)=F(h(x,y,u))$ Where F is an arbitrary.
so $ln(u-1)-ln(x+y)=F(ln(u+1)+ln(x-y))$

1..
$xu_x+2x^2u_y-u=x^2e^x$
$u_x+2xu_y=\dfrac{u}{x}+xe^x$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$
$\dfrac{dy}{dt}=2x=2t$ , letting $y(0)=y_0$ , we have $y=t^2+y_0=x^2+y_0$
$\dfrac{du}{dt}=\dfrac{u}{x}+xe^x=\dfrac{u}{t}+te^t$ , we have $u=te^t+tf(y_0)=xe^x+xf(y-x^2)$
$u(x,x^2+x)=xe^x+x^2$ :
$xe^x+xf(x)=xe^x+x^2$
$f(x)=x$
$\therefore u(x,y)=xe^x+x(y-x^2)=xe^x-x^3+xy$
2.. Solving $yu_{xx}+(x+y)u_{xy}+xu_{yy}=0$
3.. :How to find the general solution of $(y+ux)u_x+(x+yu)u_y=u^2-1$?
