Two state exponential growth model
In general

Analytic solutions only exist for special cases. Therefore a good way of approaching these problems is to first act like they don’t and be happy if you can see how to derive one. It’s also more straightforward than wild guessing.

Your problem

Notation:


$P_\text{w}$ denotes the original population and $g_\text{w}$ its growth rate.
$P_\text{m}$ denotes the mutant population and $g_\text{m}$ its growth rate.
$r$ denotes the mutation rate.


The differential equations that describe your problem are:
$$
\dot{P}_\text{w} = g_\text{w} P_\text{w} - rP_\text{w},\\
\dot{P}_\text{m} = g_\text{w} P_\text{m} + rP_\text{w}.
$$
This system of differential equations is linear as it can be written in the form $\dot{\vec{P}} = A·\vec{P}$ with $\vec{P} = (P_\text{w},P_\text{m})$ and:
$$A = \pmatrix{g_\text{w}-r &amp; 0 \\ r &amp; g_\text{m}}.$$
Linear systems in turn can be solved by finding the eigenvalues and eigenvectors of the matrix $A$, which are:


$\vec{v}_1 = (0,\frac{r}{g_\text{w}-g_\text{m}-r})$ for the eigenvalue $λ_1 = g_\text{m}$.
$\vec{v}_2 = \left( 1, \frac{r}{g_\text{w}-g_\text{m}-r} \right)$ for the eigenvalue $λ_2 = g_\text{w}-r$.


This means that the solution can be written as:
$$\vec{P} = c_1 e^{λ_1 t} \vec{v}_1 + c_2 e^{λ_2 t} \vec{v}_2,$$
where $c_1$ and $c_2$ are constants determined by the initial condition:
$\vec{P}(0) = c_1 \vec{v}_1 + c_2 \vec{v}_2.$
If you assume that you have no mutants at the beginning and you initial wild-type population is $1$, you have $c_1=-1$ and $c_2=1$.
With this you get:
$$
P_\text{w} = e^{ (g_\text{w}-r) t},\\
P_\text{m} = \frac{r}{g_\text{w}-g_\text{m}-r} \Big(e^{(g_\text{w}-r) t} -e^{g_\text{m} t} \Big).
$$