C4 differential equations help
Given that k is a constant, find the solution of the differential equation:
dy/dt +ky = 2k
for which y=3 and t=0.
So far I have the following:
dy/dt = k(2-y)
1/(2-y) dy = kdt
∫1/(2-y) = ∫k dt
-ln|2-3| = kt + c
-ln(1) = 0
c=0
-ln|2-y| = kt
-2 + y = e^kt
y= 2 + e^kt.
However, the answer is y= 2 + e^-kt. I don't understand why the k is negative.
Any help would be really appreciated.
Thanks
dy/dt +ky = 2k
for which y=3 and t=0.
So far I have the following:
dy/dt = k(2-y)
1/(2-y) dy = kdt
∫1/(2-y) = ∫k dt
-ln|2-3| = kt + c
-ln(1) = 0
c=0
-ln|2-y| = kt
-2 + y = e^kt
y= 2 + e^kt.
However, the answer is y= 2 + e^-kt. I don't understand why the k is negative.
Any help would be really appreciated.
Thanks
Original post by mr_brightside_
Given that k is a constant, find the solution of the differential equation:
dy/dt +ky = 2k
for which y=3 and t=0.
So far I have the following:
dy/dt = k(2-y)
1/(2-y) dy = kdt
∫1/(2-y) = ∫k dt
-ln|2-3| = kt + c
-ln(1) = 0
c=0
-ln|2-y| = kt
-2 + y = e^kt
y= 2 + e^kt.
However, the answer is y= 2 + e^-kt. I don't understand why the k is negative.
Any help would be really appreciated.
Thanks
dy/dt +ky = 2k
for which y=3 and t=0.
So far I have the following:
dy/dt = k(2-y)
1/(2-y) dy = kdt
∫1/(2-y) = ∫k dt
-ln|2-3| = kt + c
-ln(1) = 0
c=0
-ln|2-y| = kt
-2 + y = e^kt
y= 2 + e^kt.
However, the answer is y= 2 + e^-kt. I don't understand why the k is negative.
Any help would be really appreciated.
Thanks
I think your mistake came from the bolded line.
Original post by keromedic
I think your mistake came from the bolded line.
Should it have been ln|2=y| = -kt ???
Original post by ETRC
from -ln|2-y| = kt
you always take whatever you have outside the ln bit to the other side
so you get
ln|2-y| = -kt
then you will get your answer
you always take whatever you have outside the ln bit to the other side
so you get
ln|2-y| = -kt
then you will get your answer
Oh I see. Thank you so much!
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