I wanted to solve following PDE
$$u_x+2xu_y=2xu,$$ $u(x,0)=x^2$ for $x\geq 0$ and $u(0,y)=y^2$ for $y\geq 0$
I had find its solution as follows
for $x\geq 0$ $u(x,y)=(x^2-y)e^y$
for $y\geq 0$ $u(x,y)=(x^2-y)^2e^{x^2}$
But solution is same but domain they given is different
Solution is given in book is
$u=(x^2-y)e^y$ for $x\geq 0 ,y\leq x^2$
$u(x,y)=(x^2-y)^2e^{x^2}$ for $x\geq 0 ,y\geq x^2$
I do not know how to obtain domain of solution
The method of characteristics gives
$\frac{\text d}{\text d s}x = 1$, letting $x(0) = x_0$, we know $x = s+x_0$;
$\frac{\text d}{\text d s}y = 2x = 2(s+x_0)$, letting $y(0) = y_0$, we know $y = s^2+2x_0 s+ y_0$;
$\frac{\text d}{\text d s}u = 2xu = 2(s+x_0)u$, letting $u(0) = u_0$, we know $u = u_0 e^{s^2 + 2x_0s}$;
For characteristics starting at $y_0=0$, $x_0>0$, we deduce from the expression of characteristics that $x_0 = x-s$ and $s = x-\sqrt{x^2-y} \geq 0$. Hence, $$ u(x,y) = (x^2 - y) e^{y} $$ for $y = x^2 - {x_0}^2 < x^2$ and $x > 0$.
For characteristics starting at $x_0=0$, $y_0>0$, we deduce from the expression of characteristics that $s = x$ and $y_0 = y-x^2$. Hence, $$ u(x,y) = {(x^2 - y)}^2 e^{x^2} $$ for $y = x^2+ y_0 > x^2$ and $x > 0$.
The boundary between the two zones is the characteristic line starting at the origin $(x_0, y_0)=0$, which equation is $y=x^2$.