Question

An irritating differential equation constant

Answer 1

\[m\frac{dv}{dt}=mg-kv\]

Answer 2

v is evidently\[v=e^{pt}+c\], but how do you prove what c is (even though it's evident that it's 0)?

Answer 3

why do you say it has to be zero?
it depends on the initial velocity

Answer 4

Sorry, I meant to say that at t=0, v=0

Answer 5

I've got as far as \[p=g-\frac{k}{m}(1+c)\]

Answer 6

e^0 = 1 so c would need to be -1

Answer 7

if\[v(0)=0\]then just plug that in to your answer\[0=e^0+c\implies c=-1\]

Answer 8

so yeah, it ain't zero

Answer 9

I don't really get what p is

Answer 10

A constant dependent on the initial conditions

Answer 11

that constant is not dependent on anything other than m and k...

Answer 12

Yes and yes, you're right (earlier) in that c=-1, so p=g

Answer 13

\[v'(t)=g-\frac kmv(t)\]this is a linear DE with integrating factor\[\large\mu(t)=e^{\int-\frac kmdt}=e^{-\frac{kt}m}\]

Answer 14

Integrating factor? Is that the same thing as my 'p' in the example?

Answer 15

no, it's how you solve a linear DE like this one
if you haven't heard of it you should read up on it

http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

Answer 16

first rearrange this into a way such that the derivative has coefficient 1 and the constant g is on the other side\[v'+\frac kmv=g\]I messed up a bit, the integrating factor is\[\mu(t)=e^{\int\frac kmdt}=e^{\frac{kt}m}\]the trick is to multiply everything by the integrating factor and we get\[\large v'e^{\frac{kt}m}+\frac kmve^{\frac{kt}m}=ge^{\frac{kt}m}\]\[\large (ve^{\frac{kt}m})'=ge^{\frac{kt}m}\]now integrate both sides...

Answer 17

\[\large\int d(ve^{\frac kmt})=\int ge^{\frac kmt}dt\]\[\large ve^{\frac kmt}=\int ge^{\frac kmt}dt\]\[\large ve^{\frac kmt}=\frac{gm}ke^{\frac kmt}+c\]\[\large v=\frac{gm}k+ce^{-\frac kmt}\]

Answer 18

Excellent, thanks. Does the link explain why this works?

Answer 19

yeah it has a derivation of the integrating factor, but I can never get it into my mind to derive it very easily and usually just remember the formula blindly :p

Answer 20

It's always horrible to do that, but often unavoidable.
Anyway- thanks a lot for that!

Answer 21

very welcome :)