differential equations help Watch

Dado Prso
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Hi

Solve dv/dt = g - (kv^2)/m

i tried taking the (kv^2/m) term to the otherside so i could use the integrating factor but then i realised the v was squared and then questioned whether i could do this or not.

any help appreciated, thanks.
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7589200
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Since you've got no t term, use separation of variable...
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Dado Prso
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(Original post by Vazzyb)
Since you've got no t term, use separation of variable...
how are the variables separable?
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7589200
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by solve you mean get v in terms of t

dv/dt = g - (kv^2)/m

then

integral of 1/(g - (k/m)v^2) dv = integral of 1 dt + C

Then you can use partial fractions to integrate
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Dado Prso
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(Original post by Vazzyb)
by solve you mean get v in terms of t

dv/dt = g - (kv^2)/m

then

integral of 1/(g - (k/m)v^2) dv = integral of 1 dt + C

Then you can use partial fractions to integrate
sorry i'm not sure you can do that, perhaps i've not wrote it out correctly.

\frac{dv}{dt} = g - \frac{kv^2}{m}
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7589200
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why can you not do that lol
if you have

dy/dx = 3y^2, you can turn that to 1/(3y^2) dy = 1 dx + C

if you had

dy/dx = 9.8 + 3y^2, you can still turn that into 1/(9.8 + 3y^2) = 1 dx + C

when you've got

dy/dx = y + e^x for example, there's no way you can separate completely y and x


I dunno i may be wrong, may be someone else should help
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ukdragon37
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(Original post by Dado Prso)
sorry i'm not sure you can do that, perhaps i've not wrote it out correctly.

\frac{dv}{dt} = g - \frac{kv^2}{m}
As Vazzyb says, rearranging to \dfrac{m dv}{mg - kv^2} = dt is a way.
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anshul95
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(Original post by Vazzyb)
by solve you mean get v in terms of t

dv/dt = g - (kv^2)/m

then

integral of 1/(g - (k/m)v^2) dv = integral of 1 dt + C

Then you can use partial fractions to integrate
partial fractions will not work here. A hyperbolic substitution will work. You can also do it via trig susbtitution.
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anshul95
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(Original post by Vazzyb)
why can you not do that lol
if you have

dy/dx = 3y^2, you can turn that to 1/(3y^2) dy = 1 dx + C

if you had

dy/dx = 9.8 + 3y^2, you can still turn that into 1/(9.8 + 3y^2) = 1 dx + C

when you've got

dy/dx = y + e^x for example, there's no way you can separate completely y and x


I dunno i may be wrong, may be someone else should help
yes for that you would need integrating factors.
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Dado Prso
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(Original post by ukdragon37)
As Vazzyb says, rearranging to \dfrac{m dv}{mg - kv^2} = dt is a way.
how would i go about integrating that?

is it possible to use partial fractions given that i don't know the values of m,g and k?
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ukdragon37
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(Original post by Dado Prso)
how would i go about integrating that?

is it possible to use partial fractions given that i don't know the values of m,g and k?
Assume m, g and k are constants then you can integrate thus:

Let A^2 = \dfrac{mg}{k}

\displaystyle \int \dfrac{mdv}{mg-kv^2} = \dfrac{m}{k} \displaystyle \int \dfrac{dv}{A^2 - v^2} = \dfrac{m}{Ak} \displaystyle\int \dfrac{Adv}{A^2 - v^2} = \dfrac{m}{Ak} \tanh^{-1} \left(\dfrac{v}{A} \right) + C

Dunno whether they taught you the standard derivative of hyperbolic functions and their inverses....

It looks like terminal velocity you are doing, or I could be totally barking up the wrong tree?
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7589200
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(Original post by anshul95)
yes for that you would need integrating factors.
No you wouldn't , read the thread...
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7589200
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(Original post by anshul95)
partial fractions will not work here. A hyperbolic substitution will work. You can also do it via trig susbtitution.
Inverse tanh is faster, yes

But partial fractions would work...you'd just get two lns which would combine into one, and essentially, that's what Inv tanh is
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Farhan.Hanif93
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(Original post by anshul95)
partial fractions will not work here.
Why not? Of course it will!
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anshul95
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(Original post by ukdragon37)
Assume m, g and k are constants then you can integrate thus:

Let A^2 = \dfrac{mg}{k}

\displaystyle \int \dfrac{mdv}{mg-kv^2} = \dfrac{m}{k} \displaystyle \int \dfrac{dv}{A^2 - v^2} = \dfrac{m}{Ak} \displaystyle\int \dfrac{Adv}{A^2 - v^2} = \dfrac{m}{Ak} \tanh^{-1} \left(\dfrac{v}{A} \right) + C

Dunno whether they taught you the standard derivative of hyperbolic functions and their inverses....

It looks like terminal velocity you are doing, or I could be totally barking up the wrong tree?
that is terminal velocity stuff. This version assumes that the resistive force is proportional to the velocity squared.
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anshul95
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(Original post by Vazzyb)
Inverse tanh is faster, yes

But partial fractions would work...you'd just get two lns which would combine into one, and essentially, that's what Inv tanh is
ah yes you can-I missed that. It would be quite weird to do it that way. For the integrating factors I meant the example which you gave that wasn't separable.
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7589200
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(Original post by anshul95)
ah yes you can-I missed that. It would be quite weird to do it that way. For the integrating factors I meant the example which you gave that wasn't separable.
ah right lol yeah
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anshul95
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(Original post by Vazzyb)
ah right lol yeah
just realised your name sounds like Jazzy B - if you know who he is
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7589200
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(Original post by anshul95)
just realised your name sounds like Jazzy B - if you know who he is
lol i didn't, but i googled him
it seems his name was based on me:cool:
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Dado Prso
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(Original post by ukdragon37)
Assume m, g and k are constants then you can integrate thus:

Let A^2 = \dfrac{mg}{k}

\displaystyle \int \dfrac{mdv}{mg-kv^2} = \dfrac{m}{k} \displaystyle \int \dfrac{dv}{A^2 - v^2} = \dfrac{m}{Ak} \displaystyle\int \dfrac{Adv}{A^2 - v^2} = \dfrac{m}{Ak} \tanh^{-1} \left(\dfrac{v}{A} \right) + C

Dunno whether they taught you the standard derivative of hyperbolic functions and their inverses....

It looks like terminal velocity you are doing, or I could be totally barking up the wrong tree?
yeah, we have been taught the derivative of hyperbolic functions and their inverses. i just couldn't recognise it. cheers for the help! any chance of seeing how its done by partial fractions?

and yes, it is some terminal velocity stuff.
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