# Homework Help: A Simple(?) Differential Equation

1. Jul 5, 2014

### blintaro

This is question 1.4 of Chapter 8 of Mary L. Boas's Mathematical Methods in the Physical Sciences, Edition 2. I'm using it as a substitute for my ordinary differential equations class since my textbook has apparently been lost somewhere in the mail.

1. The problem statement, all variables and given/known data
Find the distance which an object moves in time t if it starts from rest and has an acceleration d^2/dt^2 = ge^(-kt). Show that for a small t the result is approximately (x = 1/2(gt^2)) and for very large t, the velocity dx/dt is approximately constant. (This problem corresponds roughly to the motion of a parachutist.)

2. Relevant equations
The solution manual gives the solution as: x=k^(-1)gt + k^(-2)g(e^(-kt)-1)

3. The attempt at a solution
I integrated both sides of d^2/dt^2 = ge^(-kt) twice with respect to t, ending up with x = k^(-2)ge^(-kt) + v_0(t) + x_0.
However substituting t=0 into the equation gives x = g/(k^2) + x_0, not (1/2)gt^2, so, obviously on the wrong track.

I put the process in an attached image below since it might be easier to read that than my terrible equation formatting. Thanks!

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2. Jul 5, 2014

### LCKurtz

Please post your equations instead of images of them as per forum rules. Your integrations are not correct. For example, if $x''(t) = ge^{-kt}$, then $x'(t) = -\frac g k e^{-kt} + C$. When you put $t=0$ in both sides you do not get $C = v_0$.

3. Jul 5, 2014

### Zondrina

Here are the equations in a latex format:

$\vec a = ge^{-kt}$
$\vec v = - \frac{g}{k} e^{-kt} + c_1$
$\vec r = \frac{g}{k^2} e^{-kt} + c_1t + c_2$

Using the initial condition $r(0) = 0$, you should get something useful.

$- \frac{g}{k^2} = c_2$

The initial condition that $v(0) = 0$ also gives you some more information.

$\frac{g}{k} = c_1$

Then:

$\vec r = \frac{g}{k^2} e^{-kt} + \frac{g}{k} t - \frac{g}{k^2}$

4. Jul 5, 2014

### blintaro

Oh, I see! So that's how to deal with the c's that wind up after integration. Excellent, thank you very much!