A seperation of variables intergration

[thomas49th]

Homework Statement

\frac{dV}{dt} = 20 - kV

By solving this show that

V = A + Be^{-kt}

Homework Equations

Well I am guessing there is a ln coming into play somewhere during the intergration
if the diff the bottom = the top then you get a ln(bottom)

The Attempt at a Solution

seperate the variables

\int{\frac{1}{20-kV}dv} = \int{1dt}

i don't know how to form the ln part out of the LHS, cus you can't just take a constant out here?

Thanks :)

 

[gabbagabbahey]

Hint: what is \frac{d}{dx} \ln(20-kV)?

 

[Mark44]

Let w = 20 - kV. Then dw = -k dV

 

[thomas49th]

ahh take a constant of -1/k out

so -1/k ln (20-kV) = t
multiply through by -k to give

ln(20-kV) = -tk
then e it man
gives you
20-kV = e^(-tk)
V = 20/k -e^(-tk)/k
but the stupid mark scheme says
V = 20/k -20e^(-tk)/k

who is right?

Thanks :)

 

[gabbagabbahey]

Remember, your integrals are indefinite integrals; so you need to include a constant of integration:

\Rightarrow \frac{-1}{k} \ln(20-kV) = t +C

 

[thomas49th]

Ooops missed that.
however how does that effect the co-efficient of e being 20?

Thanks :)

 

[gabbagabbahey]

Well, what do you get when you solve it with the constant C?

 

[thomas49th]

using v=0 and t=0
(as I am told the container is empty)
so that means c = -1/k ln 20

which means i can combine the ln s!

cheers :)