Here is simple decay equation – its initial conditions are being derived: $$ m(t) = Ce^{-kt}\\ m_0 = C^{-kt_0}\text{, which gives}\\ C = m_0e^{kt_0}\\ \text{After inserting $C$ to first equation}\\ m(t)=m_0e^{-k(t-t_0)} $$
I can obtain the same by not skipping any constant during integrations. From my observations, many authors skip constant when integrating left side in the following:
$$ m'(t) = -km(t)\\ \frac{dm}{dt} = -km\\ \frac{dm}{m} = -kdt\\ -\frac{dm}{mk} = dt\\ -\frac{1}{k}\int\frac{dm}{m} = \int dt\\ -\frac{1}{k}(\log|m| + C_1) = t+C_2\\ \log|m| + C_1 = -k(t+C_2)\\ \log|m| = -C_1 -k(t+C_2)\\ m = e^{-C_1}e^{-k(t+C_2)}\\ \text{I assume that $e^{-C_1}$ is $m_0$, and $(t+C_2)$ is $(t-t0)$:}\\ m = m_0e^{-k(t-t_0)}\\ $$
I didn't skip $C_1$ and this way was able to give form of the equation that includes initial condition $m_0$. Why do authors often skip $C_1$ and then derive $m_0$ in separate approaches, like the first one I included? Is my derivation correct? I quite arbitrarily assumed what's mass and what's time.
Don't think there is a good answer to this question, its just ease of notation. Your $C_1 = -log|m_0|$, where $m_0$ is the value m attains when $t = t_0$. If you want to be very explicit, you could write: \begin{align} -\frac{dm}{mk} &= dt\\ -\frac{1}{k}\int\frac{dm}{m} &= \int dt\\ -\frac{1}{k}(\log|m| - \log|m_0|) &= t - t_0\\ ... \end{align} This, to me, looks clearer than cramming everything into one constant $C = kt_0+log|m_0|$, but it can get messy.