# Thread: Help with Differential Eqn

1. ## Help with Differential Eqn

I've been tearing my hair out for the past two hours trying to figure out why I can't do a differential equation.

Here is the problem and the solution:

The bottom equation is supposed to be solution, but I have no idea how they got that.

Here is my work trying to solve it (sorry about legibility, had to use a mouse):

The (kvo+g) seems to just pop out of nowhere, and I have no idea where that comes from. Can anyone help?

EDIT: to be clear, the Vo is just the constant of integration

2. Originally Posted by bigallah
I've been tearing my hair out for the past two hours trying to figure out why I can't do a differential equation.

Here is the problem and the solution:

The bottom equation is supposed to be solution, but I have no idea how they got that.

Here is my work trying to solve it (sorry about legibility, had to use a mouse):

The (kvo+g) seems to just pop out of nowhere, and I have no idea where that comes from. Can anyone help?

EDIT: to be clear, the Vo is just the constant of integration
Seeing the question in its entirety would be helpful; you've given us just a part of the question and I'm sure the beef of the information is contained in the part you haven't shown us.

So please post the whole question.

P.S.: This is more suitable for the differential equations subforum, not the calculus subforum. So post future DE questions in the Differential Equations subforum. Consider this a friendly warning.

3. Problem is a part of 2-9)b) that can be found here:

Instructor's Solutions Manual - Marion, Thornton - Classical Dynamics of Particles and Systems, 5th Ed!!!!!!!!!!

I posted in the Calc section since this DiffEq is supposed to be stupid simple -- I remember learning how to solve 1st order ODEs in Calc rather than DiffEQ. You can keep it in whatever section you want, I just need an answer.

4. Your solution is $\frac{C}{k}e^{-kt}- \frac{g}{k}$
and the given solution is $\frac{kv_0+ g}{k}e^{-kt}- \frac{g}{k}$
so the only difference is that your constant of integration, C, is replace by their $kv_0+ g$.

I have not looked at the entire problem but I imagine there is some other, initial, condition that gives $C= kv_0+ g$

5. $\displaystyle m\,\frac{dv}{dt} = -mg - kmv$

$\displaystyle \frac{dv}{dt} = -g - kv$

$\displaystyle \frac{dv}{dt} + kv = -g$.

This is first order linear, so use an Integrating Factor $\displaystyle = e^{\int{k\,dt}} = e^{kt}$.

Multiplying both sides by the Integrating Factor gives

$\displaystyle e^{kt}\,\frac{dv}{dt} + k\,e^{kt}v = -g\,e^{kt}$

$\displaystyle \frac{d}{dt}\left(e^{kt}v\right) = -g\,e^{kt}$

$\displaystyle e^{kt}v = \int{-g\,e^{kt}\,dt}$

$\displaystyle e^{kt}v = -\frac{g\,e^{kt}}{k} + C$

$\displaystyle v = -\frac{g}{k} + C\,e^{-kt}$.

They would have used some initial or boundary conditions to find $\displaystyle C$.

6. Originally Posted by HallsofIvy
Your solution is $\frac{C}{k}e^{-kt}- \frac{g}{k}$
and the given solution is $\frac{kv_0+ g}{k}e^{-kt}- \frac{g}{k}$
so the only difference is that your constant of integration, C, is replace by their $kv_0+ g$.

I have not looked at the entire problem but I imagine there is some other, initial, condition that gives $C= kv_0+ g$
Wow, I see it now. Thanks. If you take the $\frac{kv_0+ g}{k}e^{-kt}- \frac{g}{k}$ equation and give it the conditions v(t=0)=vo then it works out to what I want. Lol, I guess I just made a bad assumption when trying to follow the solution.

Many thanks to everyone who helped out!