My Math Forum Population Model

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 November 20th, 2017, 04:43 AM #1 Newbie   Joined: Jan 2014 Posts: 17 Thanks: 0 Population Model I want to show that $p=10000\exp(\frac{t}{3}\ln\frac{23}{20})$ is the unique solution to the population growth problem ($t$ is in years): $\displaystyle dp/dt=kp, p(0)=10000, p(3)=11500$ In this case $\displaystyle t$ is restricted to nonnegative values (because $t$ is in years) and this restriction made it difficult to show that the solution is the unique solution. How to explain in words the solution is the unique solution by using the following existence and uniqueness theorem? Theorem: Let the functions $\displaystyle f$ and $\displaystyle âˆ‚f /âˆ‚p$ be continuous in some rectangle $\displaystyle Î± < t < Î²$, $\displaystyle Î³ < p < Î´$ containing the point $\displaystyle (t_0, p_0)$. Then, in some interval $\displaystyle t_0 âˆ’ h < t < t_0 + h, h>0$, contained in $\displaystyle Î± < t < Î²$, there is a unique solution $\displaystyle p = Ï†(t)$ of the initial value problem $\displaystyle p' = f (t, p), y(t_0) = p_0$. Last edited by woo; November 20th, 2017 at 04:49 AM.
 November 20th, 2017, 08:56 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,544 Thanks: 1476 By using $e^{-kt}$ as an integrating factor, the equation's solution is $p = 10000e^{kt}$ in order that $p(0) = 10000$. In order that $p(3) = 11500$, the value of $k$ must be $\frac13\ln\left(\frac{23}{20}\right)$.
November 20th, 2017, 09:19 AM   #3
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Quote:
 Originally Posted by skipjack By using $e^{-kt}$ as an integrating factor, the equation's solution is $p = 10000e^{kt}$ in order that $p(0) = 10000$. In order that $p(3) = 11500$, the value of $k$ must be $\frac13\ln\left(\frac{23}{20}\right)$.
@skipjack

I want to show that the solution is a UNIQUE solution.

 November 20th, 2017, 02:57 PM #4 Global Moderator   Joined: Dec 2006 Posts: 18,544 Thanks: 1476 If integrating the equation (after applying the integrating factor) leads to only one solution, that solution must be unique.

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