Question

can anyone help me get started on this?
dh/dt= 0.016(2.5-h)
Solve the differential equation to find h in terms of t.

Answer 1

dh/(2.5-h)=0.016 dt
now integrating, -log(2.5-h)=0.016 t+c

Answer 2

Is this enough, ersp?

Answer 3

i think so my brains kinda died doing this work XD thanks :)

Answer 4

ok :)

Answer 5

actually i dont get where the minus log comes from when integrating

Answer 6

Comes from the -h...

Answer 7

\[\int\limits_{}{}\frac{dh}{2.5-h}=-\int\limits_{}{}\frac{(2.5-h)'dh}{2.5-h}=-\log (2.5-h)+c\]

Answer 8

I got bumped. Is this explanation okay?

Answer 9

i think so cheers

Answer 10

so what method is this?

Answer 11

Method? To solve the differential equation?

Answer 12

The d.e. is separable.

Answer 13

Or, you can just take the reciprocal of both sides and integrate directly, but it's the same thing.

Answer 14

thanks makes a bit more sense to me now, my mind is really failing me today!

Answer 15

np ;)

Answer 16

shouldnt both sides be logged?

Answer 17

No. It goes like this:\[\frac{dh}{dt}=0.016(2.5-h) \rightarrow \frac{dh}{2.5-h}=0.016dt \rightarrow \int\limits_{}{}\frac{dh}{2.5-h}=\int\limits_{}{}0.016dt\]\[\rightarrow -\log (2.5-h) = 0.016t + c \rightarrow \log (2.5-h)=-0.016t+c_1 \rightarrow 2.5-h=c_2e^{-0.016t}\]\[\rightarrow h=2.5-ce^{-0.016t}\]

Answer 18

I added subscripts as I went along, but we just lump constants into constants, so normally, the subscripts are left off...which is what I did in the end.

Answer 19

ah that makes sense thanks very much, this work ive been throwing my head in to a wall for the last few days!

Answer 20

lol, it happens :p

Answer 21

Thanks for fanning me, ersp :D

Answer 22

after that much help it seems logical! XD

Answer 23

ah it doesnt seem to fit for the next part of the question, at max h is 2.5, how long does it take to get to this point, so i have to rearrange it to make t the subject but the values dont seem to fit at all!