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A Level Maths

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?
Reply 1
Avatar for Skar02
Skar02
OP
14
Please, any help would be great :smile:
Reply 2
Avatar for Zay07
Zay07
9
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
Reply 3
Avatar for Zay07
Zay07
9
i forgot to add +c on the right hand side haha rip
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 :smile:
Original post by Zay07
sorry, look at this instead

REMOVE - read the rules please :smile:
Reply 6
Avatar for Skar02
Skar02
OP
14
Original post by Zay07
sorry, look at this instead

Thank you, that helps a lot :smile:.
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?
(edited 3 years ago)
Reply 7
Avatar for Zay07
Zay07
9
ok my bad

Original post by Muttley79
REMOVE - read the rules please :smile:
Reply 8
Avatar for Zay07
Zay07
9
Original post by Skar02
Thank you, that helps a lot :smile:.
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
Reply 9
Avatar for Zay07
Zay07
9
wait.. -4 is correct

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