I'm trying to prove a unique solution for this PDE
\begin{align} u_{tt} (x,t)+(d+c) u_{xt} (x,t)+cdu_{xx} (x,t) &= F(x,t), \quad x \in \mathbb{R}, t > 0 \\ u(x,0) &= f(x) \\ u_t(x,0) &= g(x) \end{align}
I know the solution is
$$u(x, t) = \frac{cf(x - dt) - df(x - ct)}{c-d} -\frac{1}{cd}\int_{x-ct}^{x-dt}g(s)ds + \frac{1}{c-d}\int_{0}^{t}\int_{x-c(t-\tau)}^{x-d(t-\tau)}F(s,\tau)dsd\tau$$
What I'm thinking is to assume that $v(x,t)$ is also a solution, and some how plug in $w(x,t)=u(x,t)-v(x,t)$. Because I know that the homogeneous equation
$$u_{tt} (x,t)+(d+c) u_{xt} (x,t)+cdu_{xx} (x,t)=0$$
(already proved), I can say that
$$w_{tt} (x,t)+(d+c) w_{xt} (x,t)+cdw_{xx} (x,t)=0$$
is the unique solution for the homogeneous problem, and therefore $u(x,t)=v(x,t)$.
If I'm correct, what is the formal way to write that?