From Essential Partial Differential Equations, by Griffiths, Dold, and Silvester:
Example 4.2 (Half-plane problem) Solve the PDE $pux + quy = f(x, y)$ (with $p$ and $q$ constant) in the domain $\alpha x + \beta y > 0$ given that $u = g(x)$ on the line $\mathcal{l} : \alpha x + \beta y = 0$, where $q > 0$ and $\beta > 0$.
As in the previous example, the characteristic equations (4.4) are readily solved to give
$$x = pt + C_1, y = qt + C_2$$
where $C_1$ and $C_2$ are constant along any characteristic. The assumption $q > 0$ means that $y$ increases along a characteristic as $t$ increases -- thus the parameter $t$ may be viewed as a time-like variable. This is illustrated in Fig. 4.1 where the arrows on the characteristics indicate increasing t.
In order to determine the solution at a point $P(x, y)$, we trace the characteristic through this point backwards in $t$ until it intersects the line $\mathcal{l}$. We shall suppose that this occurs at the point $Q$ having coordinates $x = \beta s$, $y = − \alpha s$ when $t = 0$. Then, from (4.8), we find $C_1 = \beta s$, $C_2 = − \alpha s$. Here $s$ is a parameter that plays a role similar to that of $k$ in the previous example. It varies along the line l and each choice of s selects a different characteristic:
$$x = pt + \beta s, y = qt − \alpha s$$
Next we consider the last characteristic equation
$$\frac{du}{dt} = f(x(s, t), y(s, t))$$
Integrating gives
$$u(s, t) = u(s, 0) + \int_0^t f(x(s, t'), y(s, t')) \ dt'$$
Here $u(s, 0)$ is a constant along any characteristic but varies from one characteristic to another.
My question is, how did the author get
$$u(s, t) = u(s, 0) + \int_0^t f(x(s, t'), y(s, t')) \ dt'$$
by integrating?
Where did the $u(s, 0)$ term come from? And how was $\int_0^t f(x(s, t'), y(s, t')) \ dt'$ calculated by integration? For instance, I have never seen something written like $dt'$?
Thank you for any explanation you could offer.