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Gossip and Rumors
One model for the spread of a rumor is that the rate of spread is
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<blockquote data-quote="Croony" data-source="post: 2418612" data-attributes="member: 830521"><p>proportional to the product of the fraction? y of the population who have heard the rumor and the fraction who have not heard the rumor.</p><p>(a) Write a differential equation that is satisfied by y. (Use k for the constant of proportionality.) </p><p>My Ans: dy/dt=ky(1-y) //which is right</p><p>(b) Solve the differential equation. Assume y(0) = c., y=?</p><p> My Ans: y=e^(kt)+y-1 //I got this part wrong</p><p> A small town has 1300 inhabitants. At 8 AM, 100 people have heard a rumor. By noon half the town has heard it. At what time will 90% of the population have heard the rumor?</p><p>Not sure how to do it Kindly show the steps on how to do it.</p><p></p><p></p><p> Thank you in advance for trying</p></blockquote><p></p>
[QUOTE="Croony, post: 2418612, member: 830521"] proportional to the product of the fraction? y of the population who have heard the rumor and the fraction who have not heard the rumor. (a) Write a differential equation that is satisfied by y. (Use k for the constant of proportionality.) My Ans: dy/dt=ky(1-y) //which is right (b) Solve the differential equation. Assume y(0) = c., y=? My Ans: y=e^(kt)+y-1 //I got this part wrong A small town has 1300 inhabitants. At 8 AM, 100 people have heard a rumor. By noon half the town has heard it. At what time will 90% of the population have heard the rumor? Not sure how to do it Kindly show the steps on how to do it. Thank you in advance for trying [/QUOTE]
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