The PDE is
$x\frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = u+1$
With $u(x,y) = x^2$ on $y = x^2$
The characteristic equations are
$\frac{dx}{x} = \frac{dy}{y} = \frac{du}{u+1}$
Integrating the characteristic equations I get that $C_1 = 0$ and that $C_2 = x$ does this mean the solution is $(0,x)$?
$$x\frac{\partial u}{\partial x} + y \frac{ \partial u}{\partial y} = 1 + u $$ On the characteristic curves : $$\frac{dx}{x}=\frac{dy}{y}=\frac{du}{u+1}$$ A first characteristic equation comes from $\frac{dx}{x}=\frac{dy}{y}$ : $$\frac{y}{x}=c_1$$ A second characteristic equation comes from $\frac{dx}{x}=\frac{du}{u+1}$ : $$\frac{u+1}{x}=c_2$$ The general solution of the PDE expressed on implicit form $c_2=F(c_1)$ is : $$\frac{u+1}{x}=F\left(\frac{y}{x}\right)$$ $F$ is an arbitrary function. $$u(x,y)=-1+xF\left(\frac{y}{x}\right)$$ The function $F$ has to be determined according to the condition $u(x,x^2)=x^2$
$x^2=-1+xF\left(\frac{x^2}{x}\right)=1+xF(x)$ for any value of $x$. This determines the function $F(X)$ :
$$F(X)=\frac{X^2+1}{X}=X+\frac{1}{X}$$ We put $F(X)$ into the above general solution where $X=\frac{y}{x}$ :
$F\left(\frac{y}{x}\right)=\frac{y}{x}+\frac{x}{y}$
$u(x,y)=-1+x\left(\frac{y}{x}+\frac{x}{y}\right)$ $$u(x,y)=-1+y+\frac{x^2}{y}$$