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Population growth using ODE's

  • #1

Homework Statement


Consider a population whose number we denote by [tex] P [/tex] . Suppose that [tex]b[/tex] is the average number of births per capita per year and[tex]d[/tex] is the average number of deaths per capita per year. Then the rate of change of the population is given by the following differential equation,

[tex]dP/dt = bP-dP [/tex] where t is in years

[tex]d = 4 + P/200, b = 9 and P(0) = 200[/tex]

solve for [tex]P(t)[/tex]
(Hint: you might find the partial fraction in part (c) useful in determining your solution.)

Homework Equations



part (c) being:

[tex] 200/P(1000-P) = A/P + B/1000-P [/tex]
A=1/5, B=1/5

The Attempt at a Solution


[tex]dP/dt = 9P-(4+P/200)[/tex]
[tex]dP/dt = 9P-4-P/200 [/tex]
[tex]dP=(9P-4-P/200)dt[/tex]
[tex]dp/(9P-4-P/200)=dt[/tex]

im thinking the partial fraction comes somewhere in between my 3rd and 4th line but i cant seem to think what factor and exactly where....
 

Answers and Replies

  • #2
Cyosis
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From your data [itex] bP-dP=9P-(4+P/200)P[/itex]. You're missing that last P in your attempt.
 
  • #3
From your data [itex] bP-dP=9P-(4+P/200)P[/itex]. You're missing that last P in your attempt.
youre right, going from there
[tex]dP/dt=9P-(4+P/200)P[/tex]
[tex]5P-(P^2)/200[/tex]
[tex]P(5-P/200)[/tex]
[tex]dP/dt=P(1000-P/200)[/tex]

inversing i get
[tex]dt/dP=200/P(1000-P)=1/5P+1/5000-5P[/tex]
[tex]200/P(1000-P)=1/5P+1/5000-5P[/tex]
is from A=1/5, B=1/5 using partial fractions
[tex](1/5P+1/5000-5P)dP=200/dt[/tex]

im confused on what to do with [tex]200/dt[/tex]
how do you integrate that?
[tex]/int(1/5P+1/5000-5P)dP[/tex]
is [tex]1/5lnP+ 1/5 ln 5-5P[/tex] ?
 
  • #4
Cyosis
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You may want to use parenthesis to make it readable.

You have a differential equation of the form dt/dP=f(P) in equation line (6) therefore f(P)dP=dt and not 1/dt.

Integrate by using a substitution for the second fraction.
 
  • #5
basically the equation becomes
[tex]\int (1/5) (1/P + 1/1000-P) = \int 200dt [/tex]
or you cant take the common factor of 1/5?
and how do you use parenthesis, still kind of new to these forums
 
  • #6
Cyosis
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Yes you can take the factor 1/5 out. Parenthesis are (). For example 1/1000-P as you've written it down is 0.001-P, but you mean [itex]1/(1000-P)=\frac{1}{1000-P}[/itex]. Where does the 200 come from? The rest looks good.
 
  • #7
Yes you can take the factor 1/5 out. Parenthesis are (). For example 1/1000-P as you've written it down is 0.001-P, but you mean [itex]1/(1000-P)=\frac{1}{1000-P}[/itex]. Where does the 200 come from? The rest looks good.
doesnt the 200 come from

[tex]dP/dt=P(1000-P)/200[/tex]

then i inversed

[tex]dt/dP = 200/(P(1000-P))[/tex]

and how do you make it so that it looks like a fraction? which tool was that
 
  • #8
Cyosis
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To create a fraction a/b you can write \frac{a}{b}.

You have the equation [itex]\frac{dt}{dP}=\frac{1}{5P}+\frac{1}{5000-5P}[/itex]. There is no 200 in there.
 
  • #9
ohhhhhh right i forgot i turned [itex] \frac{200}{P(1000-P)} [/itex]
into [itex] \frac{1}{5P}+\frac{1}{5000-5P} [/itex]
so its possible to do
[itex] \frac{1}{5} \int \frac{1}{P} + \frac{1}{1000-P} [/itex] ?
 
Last edited:
  • #10
Cyosis
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Don't forget to put [tex] or [itex] brackets around your code. Yes you can now do the integral through a substitution.
 
  • #11
cant i just do

[itex] \frac{1}{5} \int \frac{1}{P} + \frac{1}{5} \int \frac{1}{1000-P} [/itex]

turning into:
[itex] \frac{1}{5} lnP + \frac{1}{5} ln(1000-P) [/itex]

forgetting my basics...
 
Last edited:
  • #12
Cyosis
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What you have written down now is no longer readable. Put \ in front of all your fractions. Anyhow you can check your answer by taking the derivative of your answer. Do it and you will see one term is correct and one is not.
 
  • #13
ok it seems the term

[itex] \int \frac{1}{1000-P} [/itex]

is giving me trouble, what would be the substitution
 
Last edited:
  • #14
Cyosis
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Can you please fix your latex code before proceeding with making a new post. Post 9,11 and 13 need fixing. What kind of substitution do you think would help?
 
  • #15
letting u = 1000-P
du = -1?
so confused
 
  • #16
Cyosis
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Looking good now. Don't forget the dP in your integration though. The substitution u=1000-P is the correct one. However what is du=-1.... (something is missing here)?
 
  • #17
i have totally forgot how to do integration using substitution,
you mean du=-1dP
will have to read up on it...
 
  • #18
Cyosis
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Yes that is correct. Now replace all Ps and dPs in your original integral with u and du.
 
  • #19
does the equation become
just for the subtitution

[itex] \frac{-1}{ln(1000-P)} [/itex]
 
  • #20
Cyosis
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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.
 
  • #21
[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)
 
Last edited:
  • #22
turns out i just had to use rearranging to find my P(t)
 

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