I am going through an example in my textbook, which solves the boundary value problem
$$u_{x} + xu_{y} = x^{2}$$ such that $u(0,y) = y$.
The author finds the characteristics $\frac{dy}{dx} = x \implies y(x) = \frac{x^{2}}{2} + y_{0}$. Along the curves $(x, y(x))$, we have the ODE:
$$\frac{d}{dx} u(x,y(x)) = x^{2}$$
whose solution is $u(x,y(x)) = u(0, y_{0}) + \frac{x^{3}}{3}$
My problem: why does the solution to the ODE $\frac{d}{dx} u(x,y(x)) = x^{2}$ have integration constant given by $u(0,y_{0})$?
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$
$\dfrac{dy}{dt}=x=t$ , letting $y(0)=y_0$ , we have $y=\dfrac{t^2}{2}+y_0=\dfrac{x^2}{2}+y_0$
$\dfrac{du}{dt}=x^2=t^2$ , letting $u(0)=f(y_0)$ , we have $u(x,y)=\dfrac{t^3}{3}+f(y_0)=\dfrac{x^3}{3}+f\left(y-\dfrac{x^2}{2}\right)$
$u(0,y)=y$ :
$f(y)=y$
$\therefore u(x,y)=\dfrac{x^3}{3}-\dfrac{x^2}{2}+y$