Question

@Kainui

Answer 1

Alright, I'm given \[F = - \alpha e^{\beta v}\] and I have to find v(t), my set up is right, just want to check the math :P
F = ma <3

\[- \alpha e^{\beta v} = ma \implies \frac{ dv }{ dt } = \frac{ -\alpha e^{\beta v} }{ m }\]

Answer 2

So far so good. =)

Answer 3

\[\frac{ dv }{ e^{\beta v} } = - \frac{ \alpha }{ m } dt \implies \int\limits \frac{ dv }{ e^{bv} } = \int\limits - \frac{ \alpha }{ m } dt\]

\[\int\limits e^{-\beta v} dv = - \frac{ \alpha }{ m } \int\limits dt \implies - \frac{ 1 }{ \beta} e^{-\beta v} = - \frac{ \alpha }{ m }t +C\]

Looking at my initial conditions t = 0, v = v0, now I'll solve for C.

\[C = - \frac{ 1 }{ \beta } e^{- \beta v_0}\]

\[- \frac{ 1 }{ \beta } e^{-\beta v} - \frac{ 1 }{ \beta } e^{-\beta vo} = - \frac{ \alpha }{ m }t \implies - \frac{ e^{-\beta v}+e^{-\beta v_0} }{ ? } = - \frac{ \alpha }{ m } t\]

\[e^{-\beta v}+ e^{- \beta v_0} = \frac{ \alpha }{ m } \beta t\]

\[e^{-\beta v} = \frac{ \alpha }{ m } \beta t - e^{-\beta v_0}\]

\[- \beta v = \ln \left( \frac{ \alpha }{ m } \beta t - e ^{-\beta v_0}\right)\]

\[v(t) = - \frac{ 1 }{ \beta } \ln ( \frac{ \alpha }{ m } \beta t - e^{-\beta v_0})\]

Answer 4

Oh that should be -(stuff)/beta don't know why it didn't show up

Answer 5

So far I think I found one error right after you solved for C you plugged it in on the wrong side so the sign is wrong on that term.

Answer 6

Oh crap!

Answer 7

I thought there was something fishy going on haha

Answer 8

Yes, that's it haha!

Answer 9

I am still looking though, not done yet =P

Answer 10

Yeah done, aside from that I think it checks out to be perfectly right

Answer 11

Haha, alright, so quick question, when dealing with separable equations and placing constant when you integrate it doesn't matter where you put the constant right, like it would be alright if I put +C on the left side, rather on the right?

Answer 12

\[\int\limits\limits e^{-\beta v} dv = - \frac{ \alpha }{ m } \int\limits\limits dt \implies - \frac{ 1 }{ \beta} e^{-\beta v} +C = - \frac{ \alpha }{ m }t \] like I could do that rather than have constant on the right

Answer 13

Yeah that's perfectly fine cause when you work out initial conditions, here's an example:

y+C=x
y=x+K

Let's say at x = 0 we have M

M+C=0
M=-C
y-M=x

other equation

M=0+K
y=x+M

yay math

Answer 14

Got ya, thanks man! :P

Answer 15

Yeah no problem any time, now you're getting into fun stuff. =D

Actually one last question I forgot to ask, what is alpha and beta? Like, for instance, do they depend on time or are they constants?

Answer 16

constants

Answer 17

Oh then you're good.

Answer 18

Yup, thanks again :P