A seperation of variables intergration

  • Thread starter thomas49th
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  • #1
thomas49th
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Homework Statement



[tex]\frac{dV}{dt} = 20 - kV[/tex]

By solving this show that

[tex]V = A + Be^{-kt}[/tex]

Homework Equations



Well im 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

[tex]\int{\frac{1}{20-kV}dv} = \int{1dt}[/tex]

i dont know how to form the ln part out of the LHS, cus you cant just take a constant out here?

Thanks :)
 

Answers and Replies

  • #2
gabbagabbahey
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Hint: what is [tex]\frac{d}{dx} \ln(20-kV)[/tex]?
 
  • #3
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Let w = 20 - kV. Then dw = -k dV
 
  • #4
thomas49th
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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 :)
 
  • #5
gabbagabbahey
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Remember, your integrals are indefinite integrals; so you need to include a constant of integration:

[tex]\Rightarrow \frac{-1}{k} \ln(20-kV) = t +C[/tex]
 
  • #6
thomas49th
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Ooops missed that.
however how does that effect the co-efficient of e being 20?

Thanks :)
 
  • #7
gabbagabbahey
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Well, what do you get when you solve it with the constant C?
 
  • #8
thomas49th
655
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using v=0 and t=0
(as im told the container is empty)
so that means c = -1/k ln 20

which means i can combine the ln s!!!

cheers :)
 

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