(This exercise is in reference to Example 2 on page 314 of Apostol, though I’ve added some details to hopefully make it self-contained.)
The problem of finding the speed 
of a falling body of mass 
in a resisting medium with resistance 
under the influence of the Earth’s gravitational attraction 
is governed by the differential equation
![\[ v' + \frac{k}{m} v = g. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9b47159fc4dc2c5d8a97663c467859fb_l3.png "Rendered by QuickLaTeX.com")
If 
is the distance the body has fallen at time 
then 
. Use the chain rule to write
![\[ \frac{dv}{dt} = \frac{ds}{dt} \cdot \frac{dv}{ds} = v \frac{dv}{ds}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4aeb7e1f38bf4f2026dcc7aafd6c4d9c_l3.png "Rendered by QuickLaTeX.com")
Use this to show that the first-order differential equation for a falling body in a resisting medium can be rewritten as
![\[ \frac{ds}{dv} = \frac{bv}{c-v}, \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-85e27339d5973fb0de87d7e67778a48b_l3.png "Rendered by QuickLaTeX.com")
where
![\[ b = \frac{m}{k}, \qquad c = \frac{gm}{k}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-bf25cd31e1624acfc257505e4cd7be7e_l3.png "Rendered by QuickLaTeX.com")
Express 
in terms of by integrating this differential equation and check this result against the results derived in Example 2 which tell us that
![\[ s = \frac{mg}{k} t + \frac{gm^2}{k^2} \left( e^{-\frac{kt}{m}} - 1 \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5c24fdedd1316fd100662fab75d26c21_l3.png "Rendered by QuickLaTeX.com")
Considering as a function of and applying the chain rule, we have
![\[ \frac{dv}{dt} = \frac{ds}{dt} \cdot \frac{dv}{ds} = v \cdot \frac{dv}{ds}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a8b342a9f50b278f9a19648e00352024_l3.png "Rendered by QuickLaTeX.com")
Thus,
![\begin{align*} v' + \frac{k}{m} v = g && \implies && \frac{m}{k} v' + v &= \frac{mg}{k} \\[9pt] && \implies && bv' + b &= c \\[9pt] && \implies && v' &= \frac{c-v}{b} \\[9pt] && \implies && \frac{dv}{dt} &= \frac{c-v}{b} \\[9pt] && \implies && v \cdot \frac{dv}{ds} &= \frac{c-v}{b} \\[9pt] && \implies && \frac{ds}{dv} &= \frac{bv}{c-v}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-75cf15ee27c8e3bb8ac1b99e37f1479c_l3.png "Rendered by QuickLaTeX.com")
Then, integrating we have
![\begin{align*} s &= \int \frac{bv}{c-v} \, dv \\[9pt] &= -bv - bc \log |c-v| + K, \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-06420732863cd624d0072e124d8392b1_l3.png "Rendered by QuickLaTeX.com")
for some constant 
. Since 
when 
we have that 
. Furthermore we know,
![\[ v = \frac{mg}{k} \left( 1 - e^{-\frac{kt}{m}} \right) = c \left( 1 - e^{-\frac{t}{b}} \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8f3ee1398fb98db683c4d39020e82742_l3.png "Rendered by QuickLaTeX.com")
Therefore,
![\begin{align*} s &= -bc \left( 1 - e^{-\frac{t}{b}} \right) - bc \log \left| -ce^{-\frac{t}{b}} \right| + bc \log |c| \\[9pt] &= (bc)e^{-\frac{t}{b}} - bc - bc \log |c| - bc \log e^{-\frac{t}{b}} + bc \log |c| \\[9pt] &= (bc) e^{-\frac{t}{b}} - bc + ct \\[9pt] &= \frac{gm^2}{k^2} e^{-\frac{kt}{m}} - \frac{gm^2}{k^2} + \frac{mg}{k} t \\[9pt] &= \frac{mg}{k} t + \frac{gm^2}{k^2} \left( e^{-\frac{kt}{m}} - 1 \right). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9280fd3fb4232a87377c39011ac22376_l3.png "Rendered by QuickLaTeX.com")
This is the result we expected from Example 2.