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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dh/(2.5h)=0.016 dt now integrating, log(2.5h)=0.016 t+c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this enough, ersp?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think so my brains kinda died doing this work XD thanks :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually i dont get where the minus log comes from when integrating

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}{}\frac{dh}{2.5h}=\int\limits_{}{}\frac{(2.5h)'dh}{2.5h}=\log (2.5h)+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got bumped. Is this explanation okay?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so what method is this?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Method? To solve the differential equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The d.e. is separable.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Or, you can just take the reciprocal of both sides and integrate directly, but it's the same thing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks makes a bit more sense to me now, my mind is really failing me today!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0shouldnt both sides be logged?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No. It goes like this:\[\frac{dh}{dt}=0.016(2.5h) \rightarrow \frac{dh}{2.5h}=0.016dt \rightarrow \int\limits_{}{}\frac{dh}{2.5h}=\int\limits_{}{}0.016dt\]\[\rightarrow \log (2.5h) = 0.016t + c \rightarrow \log (2.5h)=0.016t+c_1 \rightarrow 2.5h=c_2e^{0.016t}\]\[\rightarrow h=2.5ce^{0.016t}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah that makes sense thanks very much, this work ive been throwing my head in to a wall for the last few days!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks for fanning me, ersp :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0after that much help it seems logical! XD

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah 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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