I have the given problem :
A string is at rest and at time t=0 it is exposed to a constant force-distribution perpendicular from the longitude of the string. This force distribution remains for all $t>0$. Determine the signal of the string $u=u(x,t)$ if it is fastened ($u=0$ at the ends) and has the length $L$.
The original question is "How do I solve this using d'Alembert equation? \begin{equation} u(x,t)=\frac{1}{2}\big(f(x+Ct)+f(x-Ct)\big)+\frac{1}{2C}\int_{x-ct}^{x+ct}g(t)dt \end{equation} ?"
UPDATE:
We prepare the nonhomogeneous PDE:
\begin{equation} u_{xx}-u_{tt}=F(x,t) \\ u(0,t)=0 \ \ u_t(0,t)=0 \ \ \ \ \ \ t>0, \ 0<x<L\\ u(x,0)=F(x,t)=const \end{equation}
So we use the extended d'Alemberts formula, as implied by Hans Lundmark:
\begin{equation} u(x,t)=\frac{1}{2}\big(f(x+Ct)+f(x-Ct)\big)+\frac{1}{2C}\int_{x-ct}^{x+ct}g(t)dt+\frac{1}{2C}\int_0^t\int_{x+c(t-s)}^{x-c(t-s)}F(x,t)d\xi ds \end{equation}
since the general form of the PDE problem is:
\begin{equation} C^2u_{xx}-u_{tt}=F(x,t) \\ u(0,t)=f(x) \ \ u_t(0,t)=g(x) \ \ \ \ \ \ t>0, \ 0<x<L\\ u(x,0)=F(x,t)=const \ \ \ \ ,(set\ const=D) \end{equation}
then we have, $C=1$, $f(x)=0$, g(x)=0 and F(x,t)=D. So we have:
\begin{equation} u(x,t)=\frac{1}{2}\big(0\big)+\frac{1}{2}\int_{x-ct}^{x+ct}0ds+\frac{1}{2}\int_0^t\int_{x+1(t-s)}^{x-1(t-s)}Dd\xi ds \end{equation}
So we only need solve
\begin{equation} u(x,t)=\frac{D}{2}\int_0^t\int_{x+(t-s)}^{x-(t-s)}d\xi ds\\ u(x,t)=\frac{D}{2}\int_0^t 2t-2sds\\ u(x,t)=\frac{D}{2}\bigg[t^2-2st\bigg]_0^t\\ u(x,t)=\frac{D}{2}\bigg[t^2-2t^2\bigg] \end{equation}
and obtain:
\begin{equation} u(x,t)=-\frac{D t^2}{2} \end{equation}
When I test this, I get that:
\begin{equation} \frac{\partial^2}{\partial x^2}\bigg[-\frac{D t^2}{2}\bigg]-\frac{\partial^2}{\partial t^2}\bigg[-\frac{D t^2}{2}\bigg]=D\\ D=D \end{equation}
So the problem is fixed. A big thanks to Hans Lundmark.
I haven't done this in many years so this is not a full solution but hopefully you'll be able to finish. You boundary condition for $t=0$ should be $u(x,0)=0, u_t(x,0)=0$ as the string is at rest at the beginning.
We start with expressing $F(x,t)$ as a Fourier sine series because it's equal to zero at the ends. $F(x,t)=\sum_\limits{n=1}^{\infty}F_n(t)\sin \frac{n\pi x}{L}$. Because $F$ is constant, $F_n(t)$ should be constants. Let's find them:
$F_n=\frac{2}{L}\int_\limits{0}^{L}C\sin\frac{n\pi x}{L}dx=\frac{2C}{\pi n}\left(1+(-1)^{n+1}\right)$
We will look for a general solution in the following form: $u(x,t)=\sum_\limits{n} a_n(t)\sin \frac{n\pi x}{L}$
Thus, $$\sum_na''(t)\sin \frac{n\pi x}{L}+\alpha\sum_na_n(t)\frac{n^2\pi^2}{L^2}\sin \frac{n\pi x}{L}=\sum_n F_n\sin \frac{n\pi x}{L}$$ From here we obtain ODE for $a_n$: $$a''_n(t)+\frac{\alpha n^2 \pi^2}{L^2}a_n(t)=F_n$$ We also know that $a_n(0)=0, a'_n(t)=0$. This should be enough information to solve the problem.