Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum 
Dec 10th 2013, 08:33 AM

#1  Junior Member
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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 = gkv^n dt and then 1/(gkv^n)dv = dt.
Then I wanted to integrate both sides but I am not sure how or if this is even correct. Please help!

 
Dec 10th 2013, 04:37 PM

#2  Physics Team
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I don't believe there is a single, closedform 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 straightforward, but not for higher values of n.

 
Dec 10th 2013, 07:12 PM

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Originally Posted by Jokerpoker 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 = gkv^n dt and then 1/(gkv^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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Dec 11th 2013, 12:50 AM

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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)?

 
Dec 11th 2013, 05:46 AM

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Originally Posted by Jokerpoker 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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Dec 11th 2013, 07:44 AM

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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.

 
Dec 11th 2013, 09:10 AM

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Originally Posted by Jokerpoker 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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Dec 11th 2013, 09:51 AM

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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?

 
Dec 11th 2013, 05:02 PM

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Originally Posted by Jokerpoker 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?
Dan
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Dec 11th 2013, 11:56 PM

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Yeah, Im definitely having a hard time seeing why/how you chose that substitution?

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