Population growth using ODE's

  • #1
cheddacheeze
42
0

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
Homework Helper
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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
cheddacheeze
42
0
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
cheddacheeze
42
0
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
Homework Helper
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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
cheddacheeze
42
0
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
Homework Helper
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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
cheddacheeze
42
0
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
Homework Helper
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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
cheddacheeze
42
0
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
cheddacheeze
42
0
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
cheddacheeze
42
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letting u = 1000-P
du = -1?
so confused
 
  • #16
Cyosis
Homework Helper
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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
cheddacheeze
42
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i have totally forgot how to do integration using substitution,
you mean du=-1dP
will have to read up on it...
 
  • #18
Cyosis
Homework Helper
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Yes that is correct. Now replace all Ps and dPs in your original integral with u and du.
 
  • #19
cheddacheeze
42
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does the equation become
just for the subtitution

[itex] \frac{-1}{ln(1000-P)} [/itex]
 
  • #20
Cyosis
Homework Helper
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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
cheddacheeze
42
0
[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
cheddacheeze
42
0
turns out i just had to use rearranging to find my P(t)
 

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