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November 01, 2025, 03:45:39 pm
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Author Topic: Differential Equations  (Read 11030 times)  Share 

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d0minicz

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Re: Differential Equations
« Reply #45 on: July 13, 2009, 05:30:37 pm »
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lols i have TI-84
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d0minicz

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Re: Differential Equations
« Reply #46 on: July 13, 2009, 07:15:20 pm »
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How would you do it by hand?
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TrueTears

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Re: Differential Equations
« Reply #47 on: July 13, 2009, 07:17:11 pm »
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You're gonna have to do linear approximation 50 times.

GL mate.
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d0minicz

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Re: Differential Equations
« Reply #48 on: July 13, 2009, 07:22:52 pm »
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wow this is gonna be fun
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/0

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Re: Differential Equations
« Reply #49 on: July 13, 2009, 08:18:22 pm »
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You can input:



Where

(i think)
« Last Edit: July 13, 2009, 08:29:03 pm by /0 »
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dcc

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Re: Differential Equations
« Reply #50 on: July 13, 2009, 08:41:24 pm »
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Quote
Use Euler's method with to find the approximate value of at if given that .

Euler's method:


In this case, that means , which is the answer you SEEK.

note: Perhaps I have used a poor choice for the index of summation, but the final sum is correct, if you ignore the multiple uses of n throughout.
« Last Edit: July 13, 2009, 08:44:39 pm by dcc »
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QuantumJG

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Re: Differential Equations
« Reply #51 on: July 17, 2009, 02:34:15 pm »
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Quote from: d0minicz on May 15, 2009, 09:01:35 pm
A city with the population P, at time t years after a certain date, has a population which increases at a rate proportional to the population at that time.
a)i) Set up a differential equation to describe the situation.
ii) Solve to obtain a general solution.
b) If the initial population was 1000 and after two years the population had risen to 1100:
i) find the population after five years
ii) sketch a graph of P against t

need workings for part b)
thanks =]

a) part ii) is actually P = Ae^kt, A is an constant

b) if P(0) = 1000 and P(2) = 1100

.: A = 1000

1100 = 1000e^2k

k = 0.5ln(1.1)

i) P(5) = 1000e^2.5ln(1.1)

= 1000*1.269

= 1269

ii) this is pretty much self done! You have an exponential graph going through (0,1000), (2,1100) and (5,1269)
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kamil9876

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Re: Differential Equations
« Reply #52 on: July 17, 2009, 05:21:43 pm »
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Quote from: QuantumJG on July 17, 2009, 02:34:15 pm
Quote from: d0minicz on May 15, 2009, 09:01:35 pm
A city with the population P, at time t years after a certain date, has a population which increases at a rate proportional to the population at that time.
a)i) Set up a differential equation to describe the situation.
ii) Solve to obtain a general solution.
b) If the initial population was 1000 and after two years the population had risen to 1100:
i) find the population after five years
ii) sketch a graph of P against t

need workings for part b)
thanks =]

a) part ii) is actually P = Ae^kt, A is an constant

b) if P(0) = 1000 and P(2) = 1100

.: A = 1000

1100 = 1000e^2k

k = 0.5ln(1.1)

i) P(5) = 1000e^2.5ln(1.1)

= 1000*1.269

= 1269

ii) this is pretty much self done! You have an exponential graph going through (0,1000), (2,1100) and (5,1269)


If you havn't already noticed, latest posts are found at the bottom of the last page, not the top of the first. Furthermore, notice that this topic hasn't had a reply in 4 days and ussually all hw problems are solved at most 30min after the question is being posted.
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Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

TrueTears

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Re: Differential Equations
« Reply #53 on: July 17, 2009, 05:23:09 pm »
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kamil let him have his fun, he think he's a pr0
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TonyHem

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Re: Differential Equations
« Reply #54 on: July 17, 2009, 05:42:43 pm »
0
He did only join 3 days ago. Probably just didn't realise that they were old Q's.
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zzdfa

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Re: Differential Equations
« Reply #55 on: July 17, 2009, 05:52:27 pm »
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Quote from: TrueTears on July 17, 2009, 05:23:09 pm
kamil let him have his fun, he think he's a pr0
he's just new&enthusiastic :)
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