[Math] DiffEQ and rumor spreading

calculusordinary differential equations

One model for the spread of a rumor is that the rate of spread is proportional to the product of the fraction ($y$) of people who have heard the rumor and the fraction who have not heard the rumor.
A) Write a DE involving $y$ to fit the description above.
B) Factor to resemble the logistic DE, then write the solution function.
C) A school has 2000 student. At 8am, 80 students have heard the rumor. By noon half the school has heard it. At what time will 90% of the students have heard the rumor?

A) $\frac{dy}{dt}$ = $ky(1-y)$

B)My question is for this part: Do I just need it in the form of $ln|| = kt+c$?
If so, I need some help. Please let me know if you would like to see more work below.

I got this into the form of partial fractions and have it up until $ln|y| – ln|u| = kt+c$.
However, after I exponentiate that, I'm not sure what to do next, the negative $ln$ confuses me..

Any help is much appreciated! Thank you in advance.
I think I can probably do part C once I figure the rest out.

Best Answer

the equation $$\frac{dx}{dt} = kx(1-\frac{x}{M})$$ is called the logistic equation. you can verify that $$x = \dfrac{M}{1 + Ce^{-kt}} $$ is a solution of $(1).$ the equation has two parameters $M$ and $k$. the significance of $M,$ called the carrying capacity, is the limit of $x$ as $t \to \infty.$ and of $k$ is the relative growth rate when $x$ is small.

to fit you problem: we have $M = 2000, x(0) = 80, x(4) = 1000, x(T) = 1800.$ you should now be able to find first $k, C$ and then $T$.


$\bf edit:$

finding $k$ and $C.$

$80 = \dfrac{*2000*}{1+C} $ gives $C = 24$

$1000 = \dfrac{2000}{1 + 24e^{-4k}}$ gives $k = 0.7945$

and $1800 = \dfrac{2000}{1+24e^{-kT}}$ gives $T = \frac1k\ln(9*24)$ = 6.765

i hope i have not made too many errors in the computations.