Question

hi. i'm from germany and i have a problem with this function: mv' = mg - cv know i saw your solution but i don't understand your step to get this form: e^(k/m t) v = {S} e^(k/m t) g dt = mg/k e^(k/m t) + C v = mg/k + C e^(k/m)t link: http://openstudy.com/updates/4f425378e4b0803c63534640 i tried so often but didn't get the same solution. i do this: mv' = mg - cv mv' + cv = mg v' + (c/m)v = g v' = g - (c/m)v knwo i say (dv/dt) = g - (c/m)v from know on i cannot go on. how did you made this? thx for help.

Answer 1

ok.
mv' = mg - cv

and i should get the v(t) function.
i do not know where my fault is. i am looking in my math book and ask some friends. we don't know the answer.
please help. :)

Answer 2

mv' = mg - cv\[m\frac{dv}{dt} + cv = mg\]\[\frac{dv}{dt} + \frac{c}{m}v = g\]
This is of the form
\[\frac{dy}{dx} + Py = Q\]
where P and Q are functions of x
Consider multiplying \[e^{\int\limits Pdx}\] throughout the equation and note that
\[d(y.e^{\int\limits Pdx}) = e^{\int\limits Pdx}dy + y.e^{\int\limits Pdx}.Pdx\]

Thus, \[LHS = \frac{d(y.e^{\int\limits Pdx})}{dx}\]
\[RHS = Q.e^{\int\limits Pdx}\]
\[So, y.e^{\int\limits Pdx} = \int\limits Q.e^{\int\limits Pdx}dx + c\]

Here, P = c/m, Q = g

Try from here and I'l be glad to help if you get stuck further on

Answer 3

thx alot. know i have to look. it looks very strange to me. :/ do you have a book were i can find your descriptions? i am using papula and it doesn't really helps.

Answer 4

Ok .. this method is extremely horrible ... I was terribly blind.
This is simply
\[\frac{dv}{dt} = g - \frac{c}{m}v\]
\[\frac{dv}{dt} = -\frac{c}{m}(v -\frac{mg}{c})\]
\[\frac{dv}{v-\frac{mg}{c}} = -\frac{c}{m}dt\]
\[\log(v - \frac{mg}{c}) = -ct/m + K\]
\[v = \frac{mg}{c} + K.e^{-ct/m}\]
K is the constant which can be determined by initial conditions

Answer 5

wow. this look very simple. is this the main function. also the function v(t).
in germany we say "allgemeine Lösung v(t)".

Answer 6

i see my fault.
i know that v' ist the (dv/dt) but i always tried to get rid of the m.
but how do you get the last form. if you have a log or ln you use the exp function. where get the K the e−ct/m? when i use exp there would be stand "e(−ct/m+K)" or not?

Answer 7

e^(-ct/m + k) = e^(-ct/m)*e^k
I have just written e^k as another constant K

Answer 8

ah. know i know. the K could be written as K or as e^k. it's a constant and it doesn't care, how a constant is written. our teacher said something in this direction. thanks a lot. know i realize all. this is a really simple thing.

thx alot.
UMMD

Answer 9

A medal for FoolAroundMath for converting a complicated thing to a thing of understanding.