differential equations help

Since you've got no t term, use separation of variable...

Reply 2

Original post by Vazzyb

Since you've got no t term, use separation of variable...

how are the variables separable?

Reply 3

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

Reply 4

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}

Reply 5

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

Reply 6

ukdragon37

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.

Reply 7

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.

Reply 8

Original post by Vazzyb

yes for that you would need integrating factors.

Reply 9

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?

Reply 10

ukdragon37

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?

(edited 13 years ago)

Reply 11

Original post by anshul95

yes for that you would need integrating factors.

No you wouldn't , read the thread...

Reply 12

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

Reply 13

Farhan.Hanif93

Original post by anshul95

partial fractions will not work here.

Why not? Of course it will! :wink:

Reply 14

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.

Reply 15

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.

(edited 13 years ago)

Reply 16

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

Reply 17

Original post by Vazzyb

ah right lol yeah

just realised your name sounds like Jazzy B - if you know who he is :smile:

Reply 18

Original post by anshul95

just realised your name sounds like Jazzy B - if you know who he is :smile:

lol i didn't, but i googled him
it seems his name was based on me :cool:

Reply 19