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population growth using ODE's |
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| Apr3-10, 11:53 AM | #18 |
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Recognitions:
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population growth using ODE's
Yes that is correct. Now replace all Ps and dPs in your original integral with u and du.
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| Apr3-10, 12:10 PM | #19 |
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does the equation become
just for the subtitution [itex] \frac{-1}{ln(1000-P)} [/itex] |
| Apr3-10, 01:09 PM | #20 |
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Recognitions:
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No. It appears that you do not know how to integrate by substitution. Looking at the problem you have presented you ought to have followed a course at some point that taught you how to do it. I suggest reviewing the basics of integration before continuing with this problem.
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| Apr3-10, 11:17 PM | #21 |
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[itex] \frac{1}{5} lnP - \frac{1}{5} ln(1000-P)= t+C [/itex]
[itex] lnP - ln(1000-P) = 5t+5c [/itex] using log rules [itex] ln \frac{P}{1000-P} = 5t+5c [/itex] multiplying by e [itex] \frac{P}{1000-P}=Ae^{5t} , A=e^{5c} [/itex] [itex] P = (1000-P)(Ae^{5t}) [/itex] multiplying it out [itex] P=1000Ae^{5t} - PAe^{5t} [/itex] [itex] P + PAe^{5t} = 1000Ae^{5t} [/itex] [itex] Ae^{5t}= 1000Ae^{5t}/P [/itex] [itex] P = 1000 [/itex] but how do i get P(t) |
| Apr4-10, 03:34 AM | #22 |
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turns out i just had to use rearranging to find my P(t)
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| growth, ode, population |
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