anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
dh/(2.5-h)=0.016 dt now integrating, -log(2.5-h)=0.016 t+c
anonymous
  • anonymous
Is this enough, ersp?
anonymous
  • anonymous
i think so my brains kinda died doing this work XD thanks :)

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