Solving quasilinear p.d.e. with method of characterstics

I am currently working on solving the p.d.e.

$$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 1$$

with the initial condition

$$u = 0 \text{ on } x + y = 0$$

using the method of characteristics. Thus, so far, I have obtained a system of characteristic equations

$$\begin{cases} \frac{dx}{dt} = 1 \implies x = t + x_0\\ \frac{dy}{dt} = 1 \implies y = t + y_0\\ \frac{du}{dt} = 1 \implies u = t + u_0 \end{cases}$$ However, I'm unsure how to proceed to encorporate the initial conditions or solve for $$u(x, y)$$ explicitly. Any help is appreciated.

....

• Eliminate $t$ from each of the equations i.e $$x-x_{0} = y-y_{0} = u-u_{0}$$ which implies $y = x + c_{1}$ and $u = x + f(c_{1}) = x + f(y-x)$. Now apply your boundary data. – Mattos Mar 10 at 1:36
• How do you deduce $u = x + f(c_1)$? All that seems clear to me is $u = x - x_0 + u_0$. – Scriniary Mar 10 at 1:59
• Because the implicit solution to the PDE problem is given by $$\phi(c_{1}, c_{2}) = 0 \implies \phi(y-x, u-x) = 0$$ which, in a different form, is $$u-x = f(y-x)$$ i.e $c_{2} = f(c_{1})$. – Mattos Mar 10 at 2:52