Rewriting your equation a little, we have
$$ \partial_t C - Q\partial_V C = r. $$
This can be written as
$$ DC\begin{pmatrix} 1\\ -Q\end{pmatrix} = r $$
That is the directional derivative in direction $(1,-Q)^t$ is $r$. So let us consider $C$ in this direction, define $f \colon [0,\infty) \to \mathbb R$ by $f(t) := C(t, V_0 - tQ)$, we have
\begin{align*}
f'(t) &= \partial_t C(t, V_0 - tQ) - Q\partial_V C(t, V_0 - tQ)\\
&= r
\end{align*}
for any $t \ge 0$. Hence, by integrating
\begin{align*}
f(t) &= f(0) + \int_0^t f'(s)\, ds\\
&= C(0, V_0) + rt
\end{align*}
So, the solution is given by
$$ C(t, V) = C(t, V + tQ - tQ) = C(0, V+tQ) + rt $$
for some initial function $C(0,\cdot)$.