Answer
\begin{equation}
x(t)=\frac{g m}{c}\left(t+\frac{m}{c} e^{-\frac{c}{m} t}\right)-\frac{g m^{2}}{c^{2}}
\end{equation}
Work Step by Step
\begin{equation}
\begin{array}{l}{\text { Substituting }(1.10) \text { yields }} \\ {\qquad \frac{d x}{d t}=\frac{g m}{c}\left(1-e^{-\frac{c}{m} t}\right)} \\ {\text { The solution of this differential equation can be obtained by simply integrating both sides. }}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\int d x=\int \frac{g m}{c}\left(1-e^{-\frac{c}{m} t}\right) d t} \\ {x(t)=\frac{g m}{c}\left(t+\frac{m}{c} e^{-\frac{c}{m} t}\right)+C} \\ {x(0)=0 \Rightarrow C=-\frac{g m^{2}}{c^{2}}} \\ {\Rightarrow x(t)=\frac{g m}{c}\left(t+\frac{m}{c} e^{-\frac{c}{m} t}\right)-\frac{g m^{2}}{c^{2}}}\end{array}
\end{equation}
