I have a problem question in here.
The differential equation for an object falling under the influence of
gravity near the surface of the earth with air resistance comparable
to velocity $v$ is$$\frac{\mathrm{d}v}{\mathrm{d}t} = -g – av$$with
$g = 32 \text{ feet per second per second}$ is the gravitation
accleration and $a > 0$ is shear coefficient. Show that:(a) $v(t) = (v_0 – v_\infty)e^{-at} + v_\infty$, with $v_0 = v(0)$ and
$v_\infty = -\frac{g}{a} = \underset{t \to \infty}{\lim} v(t)$; and(b) If $y(t)$ is the height, then$$y(t) = y_0 + tv_\infty + \frac{1}{a}(v_0 – v_\infty)(1 – e^{-at})$$
So, I already proved the part of (a)
$\begin{align} \frac{\mathrm{d}v}{\mathrm{d}t} &= -g – av \\ ve^{at} &= \int -ge^{at} \ \mathrm{d}t \\ ve^{at} &= -\frac{g}{a}e^{at} + C \\ \text{Let } C = v_0 + \frac{g}{a} \\ ve^{at} &= -\frac{g}{a}e^{at} + \left(v_0 + \frac{g}{a}\right) \\ v &= (v_0 – v_\infty)e^{-at} + v_\infty \ \blacksquare \text{ Q.E.D.}\end{align}$
For part of (b)
$\begin{align} v(t) &= \frac{\mathrm{d}y}{\mathrm{d}t} \\ \int v(t) \ \mathrm{d}t &= \int \left(v_0 + \frac{g}{a}\right)e^{-at} – \frac{g}{a} \\ y(t) &= y_0 + tv_\infty + \frac{1}{a}(v_0 – v_\infty)e^{-at}\end{align}$
And here is the problem. It should be $y(t) = y_0 + tv_\infty + \frac{1}{a}\left(v_0 – v_\infty\right)(1-e^{-at})$. I tried to fix my mistake here but I even didn't had any hint.
Oh, and please correct me if my works had mistake
Best Answer
You didn't apply limits to the exponential term.$$y(t)-y_0=v_\infty[x]_0^t+\frac1{\color{red}-a}(v_0-v_\infty)[e^{-ax}]_0^t$$which will give you the required result.