Consider the ODE:
$$\frac{dy}{dt} = Ay + B$$
where $A$ and $B$ are constant coefficients. This has the solution:
$$y(t) = c_1 e^{A t} - \frac{B}{A}$$
Suppose knowing this solution, I want to find the solution for arbitrary values of $A$ and $B$. It is usually as simple as substituting the values into the solution. This method does not appear to work for the case $A=0$.
But it is not the case that a solution does not exist for the differential equation when $A=0$. The solution is $y(t) = Bt + c_2$.
I can't see a way to obtain this solution from the solution of the general ODE. Is there any way to do so without solving the special case of the ODE separately?
You can obtain your solution as a limit if you take into account an initial condition $y(t_0)= y_0 = c_1e^{At_0} - \frac{B}{A}$. Then your solution becomes $$ y(t) = (y_0 + \frac{B}{A})e^{A(t-t_0)} - \frac{B}{A} = y_0e^{A(t-t_0)} + \frac{B(e^{(A(t-t_0))}-1)}{A} $$ Taking limits $$ \lim_{A\to 0} y(t) = y_0 + B(t-t_0) $$ which is a solution to the differential equation when $A=0$ with initial condition $y(t_0) = y_0$
More generally, given an initial value problem $\dot x = f(x,t,\lambda), x(t_0) = x_0$, where $\lambda$ is a parameter, if $f$ is smooth enough, then $\lim_{\lambda \to \lambda_0} x(\lambda, t) = x(\lambda_0, t) \ \forall t$