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Population growth using ODE's

by cheddacheeze
Tags: growth, ode, population
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cheddacheeze
#19
Apr3-10, 12:10 PM
P: 42
does the equation become
just for the subtitution

[itex] \frac{-1}{ln(1000-P)} [/itex]
Cyosis
#20
Apr3-10, 01:09 PM
HW Helper
P: 1,495
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.
cheddacheeze
#21
Apr3-10, 11:17 PM
P: 42
[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)
cheddacheeze
#22
Apr4-10, 03:34 AM
P: 42
turns out i just had to use rearranging to find my P(t)


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