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 April 21st, 2014, 05:41 PM #1 Member   Joined: Jun 2012 From: San Antonio, TX Posts: 84 Thanks: 3 Math Focus: Differential Equations, Mathematical Modeling, and Dynamical Systems Diff Eqs and the spread of rumors Relevant Info: Let $y(t)$ be the number of people who have heard the rumor at time t and assume that everyone who has heard the rumor passes it to r others in unit time. Thus, from time t to time (t+h) the rumor is passed on hry(t) times, but a fraction of the people have heard it already, thus there are only $hry(t) \left(\frac{K-y(t)}{K}\right)$ people who hear the rumor for the first time. Make and expression for $y(t+h)-y(t)$, divide by h, and take the limit as $h \rightarrow 0$ to obtain a differential equation satisfied by y(t). So the model I came up with is the logistic d.e. $y'(t)=ry(t) \left(1-\frac{y(t)}{K}\right)$. My Problem: At 9 AM one person in a village of 100 inhabitants starts a rumor. Suppose that everyone who hears the rumor tells one other person per hour. Using the model of exercise 8, determine how long it will take until half the village has heard the rumor. I'm a bit rusty with differential equations, but this sounds like an IVP, where $y(t=9AM)=y(0)=1$. $\int\frac{1}{y\left(1-\frac{y}{K} \right)}dy=\int r dt$. Now, using partial fractions.... $\int\left(\frac{1}{y}+\frac{1/K}{1-\frac{y}{K}}\right)dy=\int rdt \Rightarrow ln \left( \frac{y}{1-\frac{y}{K}}\right)=rt + C \Rightarrow \frac{y}{1-\frac{y}{K}}=C_2 e^{rt} \Rightarrow y(t)=\frac{C_2 e^{rt}}{1+\frac{C_2}{K} e^{rt}}$ Using $y(0)=1$: $C_2=\frac{100}{99}$. From my problem r should equal 1 and K=100. So by solving $y(t)=50$ for t should give me my answer which should be 2 hours, but my calculations keep giving me t=ln(99). Sorry for the length of the post. I just didn't want to leave any information out.
 April 21st, 2014, 07:50 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra I agree with you. $$C_2=\frac{k}{k-1}$$ It's fewer than 16 seconds short of 2 hours though. For what it's worth, in this instance I'd leave the solution of the integral as an expression for $t$ in terms of $y$. Thanks from MadSoulz
April 22nd, 2014, 06:42 AM   #3
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Quote:
 Originally Posted by v8archie It's fewer than 16 seconds short of 2 hours though.
My answer is?

 April 22nd, 2014, 09:09 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra Yes, $\log{99}$ hours is fewer than 16 seconds short of 2 hours.
 April 25th, 2014, 05:47 AM #5 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,161 Thanks: 734 Math Focus: Physics, mathematical modelling, numerical and computational solutions I get $\displaystyle \ln\left(99\right)$ = 4.6 hours also. Perhaps the person who originally set the question accidentally pressed $\displaystyle \log\left(99\right)$, or perhaps we have both made a mistake. Let me know if you resolve this. Thanks from MadSoulz
April 26th, 2014, 03:23 PM   #6
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Quote:
 Originally Posted by Benit13 I get $\displaystyle \ln\left(99\right)$ = 4.6 hours also. Perhaps the person who originally set the question accidentally pressed $\displaystyle \log\left(99\right)$, or perhaps we have both made a mistake. Let me know if you resolve this.
Yeah, I think the answer my book has is a mistake. I believe the rate is too low for the rumor to spread to 50 people in two hours. I'll be able to ask my professor next week.

May 21st, 2014, 04:27 PM   #7
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Quote:
 Originally Posted by Benit13 I get $\displaystyle \ln\left(99\right)$ = 4.6 hours also. Perhaps the person who originally set the question accidentally pressed $\displaystyle \log\left(99\right)$, or perhaps we have both made a mistake. Let me know if you resolve this.
Hey Benit13. Sorry for replying a month later, but our work is correct. The correct time is 4.6 hours, just like we got.

Last edited by MadSoulz; May 21st, 2014 at 04:31 PM.

 May 21st, 2014, 05:37 PM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra Ah! My calculator (probably Google) uses $$\log_{10}{x} \equiv \log{x} \\ \log_e{x} \equiv \ln{x}$$ Thanks from MadSoulz
May 21st, 2014, 05:44 PM   #9
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Quote:
 Originally Posted by v8archie Ah! My calculator (probably Google) uses $$\log_{10}{x} \equiv \log{x} \\ \log_e{x} \equiv \ln{x}$$
I think the same thing is what happened with the author of my textbook.

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