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.

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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.

Mathematics
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dh/(2.5-h)=0.016 dt now integrating, -log(2.5-h)=0.016 t+c
Is this enough, ersp?
i think so my brains kinda died doing this work XD thanks :)

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Other answers:

ok :)
actually i dont get where the minus log comes from when integrating
Comes from the -h...
\[\int\limits_{}{}\frac{dh}{2.5-h}=-\int\limits_{}{}\frac{(2.5-h)'dh}{2.5-h}=-\log (2.5-h)+c\]
I got bumped. Is this explanation okay?
i think so cheers
so what method is this?
Method? To solve the differential equation?
The d.e. is separable.
Or, you can just take the reciprocal of both sides and integrate directly, but it's the same thing.
thanks makes a bit more sense to me now, my mind is really failing me today!
np ;)
shouldnt both sides be logged?
No. It goes like this:\[\frac{dh}{dt}=0.016(2.5-h) \rightarrow \frac{dh}{2.5-h}=0.016dt \rightarrow \int\limits_{}{}\frac{dh}{2.5-h}=\int\limits_{}{}0.016dt\]\[\rightarrow -\log (2.5-h) = 0.016t + c \rightarrow \log (2.5-h)=-0.016t+c_1 \rightarrow 2.5-h=c_2e^{-0.016t}\]\[\rightarrow h=2.5-ce^{-0.016t}\]
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.
ah that makes sense thanks very much, this work ive been throwing my head in to a wall for the last few days!
lol, it happens :p
Thanks for fanning me, ersp :D
after that much help it seems logical! XD
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!

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