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Old Dec 10th 2013, 08:33 AM   #1
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Resistive Force Calculus Help

I have the differentialequation dv/dt = g - kv^n and need to solve it with the initial condition that v(o) = v.inital.

I tried seperating the variables so i get dv = g-kv^n dt and then 1/(g-kv^n)dv = dt.

Then I wanted to integrate both sides but I am not sure how or if this is even correct. Please help!
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Old Dec 10th 2013, 04:37 PM   #2
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I don't believe there is a single, closed-form solution for this integral that works for all values of n. If you know that n = 0, or n=1, or n=2 it's pretty straight-forward, but not for higher values of n.
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Old Dec 10th 2013, 07:12 PM   #3
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Originally Posted by Jokerpoker View Post
I have the differentialequation dv/dt = g - kv^n and need to solve it with the initial condition that v(o) = v.inital.

I tried seperating the variables so i get dv = g-kv^n dt and then 1/(g-kv^n)dv = dt.

Then I wanted to integrate both sides but I am not sure how or if this is even correct. Please help!
The solution for general n is a hypergeometric function. Could you tell us where this problem came from? Maybe that will help.

-Dan
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Old Dec 11th 2013, 12:50 AM   #4
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It is just from my physics text book, where it is related to newtons second law, and the mass cancels out on both sides.

So how would you solve it (if possible where n is just n, and if not then for example n = 2)?
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Old Dec 11th 2013, 05:46 AM   #5
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Originally Posted by Jokerpoker View Post
It is just from my physics text book, where it is related to newtons second law, and the mass cancels out on both sides.

So how would you solve it (if possible where n is just n, and if not then for example n = 2)?


Let v = sqrt(g/k) * u. Then dv = sqrt(g/k) du. The integral becomes:


So


Which you can finish from here. (The solution for v can be put into the form of an inverse hyperbolic tangent for a "cleaner" solution, but I think the log format is easier to grasp.)

-Dan
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Old Dec 11th 2013, 07:44 AM   #6
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Im not really sure how you arrive at v = sqrt(g/k)*u?

Im assuming you are using u to mean v.initial? - How can you just replace v with u?

- Thanks by the way, this is really helpful.
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Old Dec 11th 2013, 09:10 AM   #7
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Originally Posted by Jokerpoker View Post
Im not really sure how you arrive at v = sqrt(g/k)*u?

Im assuming you are using u to mean v.initial? - How can you just replace v with u?

- Thanks by the way, this is really helpful.
It's not a speed it's a change of variables. As far as v = sqrt(g/k) u, this substitution gives the integrand as 1/(1 - u^2), which is easier to integrate than when the constants were in with it.

-Dan
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Old Dec 11th 2013, 09:51 AM   #8
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Okay then Im not sure how you arrive at the substitue u. If I was to substitute something with u, I would say u = what ever I want to substitute?
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Old Dec 11th 2013, 05:02 PM   #9
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Originally Posted by Jokerpoker View Post
Okay then Im not sure how you arrive at the substitue u. If I was to substitute something with u, I would say u = what ever I want to substitute?
Okay then, let u = sqrt{k/g} v. I'm not sure why this is a sticking point. Would you be more comfortable with x = sqrt{k/g} v? Or is it that you are having trouble with why I'm choosing that substitution?

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Old Dec 11th 2013, 11:56 PM   #10
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Yeah, Im definitely having a hard time seeing why/how you chose that substitution?
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