Skar02
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#1
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#1
A little stuck with trying to solve this differential equation, dx/dt = kx(x-1)
The question asks me to solve it expressing x in terms of k and t.
I've got to this far:
dx = kx (1-x) dt
1/x(1-x) dx = k dt
Put integral signs in and then integrate the left side using partial fractions to get:
ln(x) - ln(x-1)
Putting it all together:
ln(x)- ln(x-1) = 1/2 k^2 +c

What would I do know? The question asks to express x in terms of k and t?
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Skar02
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#2
Report Thread starter 48 minutes ago
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Please, any help would be great
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Zay07
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#3
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on the right hand side of the equation integrate with respect to t not with respect to k (it says dt not dk)
subtract logs then raise both sides to the power of e, make x the subject
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Zay07
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#4
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i forgot to add +c on the right hand side haha rip
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Muttley79
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#5
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(Original post by Zay07)
on the right hand side of the equation integrate with respect to t not with respect to k (it says dt not dk)
subtract logs then raise both sides to the power of e, make x the subject
Remove the attachment - you are breaking forum rules
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Muttley79
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(Original post by Zay07)
sorry, look at this instead
REMOVE - read the rules please
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Skar02
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#7
Report Thread starter 27 minutes ago
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(Original post by Zay07)
sorry, look at this instead
Thank you, that helps a lot .
So to finally solve it, I would have to plug in t=0 and x=0.2.
That would be:
0.2 = 1/1-A
A=-4.
Is that correct?
Last edited by Skar02; 25 minutes ago
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Zay07
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#8
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ok my bad

(Original post by Muttley79)
REMOVE - read the rules please
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Zay07
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#9
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(Original post by Skar02)
Thank you, that helps a lot .
So to finally solve it, I would have to plug in t=0 and x=0.2.
That would be:
0.2 = 1/1-A
A=-4.
Is that correct?
i dont think so, multiply both sides by (1-A) and rearrange
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Zay07
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#10
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wait.. -4 is correct
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