According to the given statement the answer will be [tex]y=\frac{A e^{k t}}{1+A e^{k t}}[/tex].
An element of a whole is a fraction. The number is represented mathematically as a quotient, where the numerator and denominator are split. Both of these are integers in a simple fraction. A fraction appears in the numerator as well as denominator of a complicated fraction. Every numerator of a correct fraction is smaller than the denominator.
[tex]y= $ fraction of people who heared rumour\\$(1-y)=$ fraction who have not heard the rumor.\\Rate of rumour spread $\propto y(1-y)$[/tex]
[tex]\begin{aligned}& \frac{d y}{d t}=k y(1-y) \\& \int \frac{d y}{y(1-y)}=k \cdot \int d t \\& \ln \left|\frac{y}{1-y}\right|=k t+C \\& \frac{y}{1-y}=e^{k t+C}=e^C e^{k t}=A e^{k t} \\& y=\left(A e^{k t}\right)(1-y) \\& y=\left(A e^{k t}\right)-\left(A e^{k t}\right) y \\& y\left(1+A e^{k t}\right)=A e^{k t} \\& y=\frac{A e^{k t}}{1+A e^{k t}}\end{aligned}[/tex]
y(0)=y0 is the beginning condition.
[tex]\begin{aligned}& y(0)=y_0=\frac{A e^{k * 0}}{1+A e^{k * 0}}=\frac{A}{1+A} \\& y_0(1+A)=A \\& y_0+y_0 A=A \\& A\left(1-y_0\right)=y_0 \\& A=\frac{y_0}{1-y_0}\end{aligned}[/tex]
[tex]y(t)=\frac{\left(\frac{y_0}{1-y_0}\right) e^{k t}}{1+\left(\frac{y_0}{1-y_0}\right) e^{k t}}[/tex]
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The complete question is-
One model for the spread of a rumor is that the rate of spread is 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.) dy dt = Correct: Your answer is correct. (b) Solve the differential equation. (Let y(0) = y0.) (c) A small town has 4500 inhabitants. At 8 AM, 360 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? (Do not round k in your calculation. Round your final answer to one decimal place.) hours after 8 AM