I have the following equation for the evolution of the membrane potential ($V$) of a neuron:
$$ \frac{dV}{dt} = [-g_L(V-V_{rest}) + I_{syn}(t) + I_0] / C. $$
According to Equation 2.13 of this paper, the solution of this equation (for $V$) is
$$ V = V(0)e^{-t/\tau}+\frac{I_0}{g_L}(1 - e^{-t/\tau})+V_{rest}(1 - e^{-t/\tau}) + \int^t_0e^{-(t'-t)/\tau}\frac{I_{syn}(t')}{C}dt', $$
where $\tau=\frac{C}{g_L}$.
I'm having trouble getting to this solution.
We have
\begin{align} \frac{dV}{dt} &= [-g_L(V-V_{rest}) + I_{syn}(t) + I_0] / C\\ &= \frac{-g_L(V-V_{rest})}{C} + \frac{I_{syn}(t)}{C} + \frac{I_0}{C}\\ &= \frac{-(V-V_{rest})}{\tau} + \frac{I_{syn}(t)}{C} + \frac{I_0}{g_L}\frac{1}{\tau}\\ &=V_{rest}\frac{1}{\tau} - V\frac{1}{\tau} + \frac{I_{syn}(t)}{C} + \frac{I_0}{g_L}\frac{1}{\tau}. \end{align}
The Laplace transform of $\frac{dV}{dt}$ is
$$ \mathcal{L}\bigg\{\frac{dV}{dt}\bigg\} = -V(0)+sV_L(s), $$
where $V_L(s)$ is the Laplace transform of $V$.
Hence,
\begin{align} -V(0)+sV_L(s)&=\mathcal{L}\bigg\{V_{rest}\frac{1}{\tau} - V\frac{1}{\tau} + \frac{I_{syn}(t)}{C} + \frac{I_0}{g_L}\frac{1}{\tau}\bigg\}. \end{align}
Because the Laplace transform of the sum of two functions is the sum of the Laplace transforms of the functions taken separately, we have
\begin{align} sV_L(s)&=V(0)+\mathcal{L}\bigg\{V_{rest}\frac{1}{\tau}\bigg\} - \mathcal{L}\bigg\{V\frac{1}{\tau}\bigg\} + \mathcal{L}\bigg\{\frac{I_{syn}(t)}{C}\bigg\} + \mathcal{L}\bigg\{\frac{I_0}{g_L}\frac{1}{\tau}\bigg\}. \end{align}
Applying a table of commonly encountered Laplace transforms to this gives us
\begin{align} sV_L(s)+V_L(s)\frac{1}{\tau}&=V(0)+V_{rest}\frac{1}{\tau}\frac{1}{s} + \mathcal{L}\bigg\{\frac{I_{syn}(t)}{C}\bigg\} + \frac{I_0}{g_L}\frac{1}{\tau}\frac{1}{s}\\ V_L(s)\bigg(s+\frac{1}{\tau}\bigg)&=V(0)+V_{rest}\frac{1}{\tau}\frac{1}{s} + \mathcal{L}\bigg\{\frac{I_{syn}(t)}{C}\bigg\} + \frac{I_0}{g_L}\frac{1}{\tau}\frac{1}{s}\\ V_L(s)&=\bigg[V(0)+V_{rest}\frac{1}{\tau}\frac{1}{s} + \mathcal{L}\bigg\{\frac{I_{syn}(t)}{C}\bigg\} + \frac{I_0}{g_L}\frac{1}{\tau}\frac{1}{s}\bigg]\frac{1}{s+\frac{1}{\tau}}\\ &=V(0)\frac{1}{s+\frac{1}{\tau}}+V_{rest}\frac{1}{\tau}\frac{1}{s}\frac{1}{s+\frac{1}{\tau}} + \mathcal{L}\bigg\{\frac{I_{syn}(t)}{C}\bigg\}\frac{1}{s+\frac{1}{\tau}} + \frac{I_0}{g_L}\frac{1}{\tau}\frac{1}{s}\frac{1}{s+\frac{1}{\tau}}. \end{align}
Taking the inverse Laplace transform of this (again using a table) gives us
\begin{align} V&=V(0)e^{-t/\tau}+V_{rest}\frac{1}{\tau}\int^t_0e^{-(t'-t)/\tau}dt'+ \int^t_0e^{-(t'-t)/\tau}\frac{I_{syn}(t')}{C}dt' + \frac{I_0}{g_L}\frac{1}{\tau}\int^t_0e^{-(t'-t)/\tau}dt'. \end{align}
Because
\begin{align} \int^t_0e^{-(t'-t)/\tau}dt'&=-\tau\bigg[e^{-(t'-t)/\tau}\bigg]^t_0\\ &=-\tau\bigg(e^{-(t-t)/\tau}-e^{t/\tau}\bigg)\\ &=\tau\bigg(e^{(t-t)/\tau}-e^{-t/\tau}\bigg)\\ &=\tau\bigg(1-e^{-t/\tau}\bigg)\\ \end{align}
we have
\begin{align} V&=V(0)e^{-t/\tau}+V_{rest}\bigg(1-e^{-t/\tau}\bigg)+ \int^t_0e^{-(t'-t)/\tau}\frac{I_{syn}(t')}{C}dt' + \frac{I_0}{g_L}\bigg(1-e^{-t/\tau}\bigg). \end{align}