Question

(a) Find the differential equation for given data. The spread of a rumor is proportional to the product of the population who has heard the rumor and the population who have not heard the rumor.

(b) Solve the differential equation:\(\frac{{dy}}{{dt}} = ky(1 - y)\)

(c) Find the time at which rumor will spread across\(90\% \)of the population.

Short answer

(a) The differential equation is \(\frac{{dy}}{{dt}} = ky(1 - y)\).

(b) The solution to the differential equation is \(y(t) = \frac{{{y_0}}}{{{y_0} + \left( {1 - {y_0}} \right){e^{ - kt}}}}\).

(c) The rumor will spread across \(90\% \) of the population at \(3:36{\rm{PM}}\).

Step-by-step solution

Substitute the proportionality constant \(k\)for the differential equation.

Let the population is \(1\).

Consider\(\,y\) be the rumor heard by the population and \(y - 1\) not heard the rumor.

So that the rumor spread is

\(\frac{{dy}}{{dt}} \propto y(1 - y)\)

Substitute the proportionality constant\(k\)

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

Therefore, the differential equation is \(\frac{{dy}}{{dt}} = ky(1 - y)\).

Solution for the differential equation.

The solution of the differential equation is,

\(P(t) = \frac{M}{{1 + A_e^{ - kt}}}\) where \(A = \frac{{M - {P_0}}}{{{P_0}}}\).

Consider the logistic equation for the solution of the differential equation.

The differential equation is,

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

Consider the logistic equation,

\(\frac{{dP}}{{dt}} = kP\left( {1 - \frac{P}{M}} \right)\) …… (2)

Compare the equation (1) and (2),

\(P = y\)and\(M = 1\)

Substitute\(P = y\)and\(M = 1\)in using the result,

\(\begin{aligned}(t) &= \frac{1}{{1 + \left( {\frac{{1 - {y_0}}}{{{y_0}}}} \right){e^{ - kt}}}}\\ &= \frac{{{y_0}}}{{{y_0} + \left( {1 - {y_0}} \right){e^{ - kt}}}}\end{aligned}\)

Therefore, the solution of the differential equation is \(y(t) = \frac{{{y_0}}}{{{y_0} + \left( {1 - {y_0}} \right){e^{ - kt}}}}\).

Use   from (b) for calculation.

Population \( = 1000\)

At\(8AM\)the number of people who heard the rumor\( = 80\)

At\(12PM\)the number of people who heard the rumor\( = 500\)

At the initial stage\(80\)people heard the rumor at time zero.

So that\(y(0) = \frac{{80}}{{1000}} = 0.08\)

In the afternoon, people heard the rumor is\(500\). That means after\(4\)hours.

So that\(y(4) = \frac{{500}}{{1000}} = 0.5\).

From part (b)

…… (3)

Simplify further,

\(\begin{aligned}{e^{ - 4k}} &= \frac{{0.08}}{{0.92}}\\{e^{ - 4k}} = 0.0869\end{aligned}\)

Take natural logarithm at both sides,

\(\begin{aligned} - 4k &= \ln (0.087)\\ - 4k &= - 2.442\\k &= \frac{{2.442}}{4}\\k &= 0.611\end{aligned}\)

Substitute\(k\)value in equation (3),

\(\,y(t) = \frac{{0.08}}{{0.08 + (0.92){e^{ - 0.611t}}}}\) …… (4)

Obtain the\(90\% \)of heard population,

Consider\({\rm{y(t) = 90 \% }}\)

From equation (4),

\(\begin{aligned}\frac{{90}}{{100}} &= \frac{{0.08}}{{0.08 + {{(0.92)}_{{e^{ - 0.611t}}}}}}\\0.08 + (0.92){e^{ - 0.611t}} &= \frac{{0.08}}{{0.9}}\\(0.92){e^{ - 0.611t}} &= 0.0888 - 0.08\\{e^{ - 0.611t}} &= \frac{{0.0088}}{{0.92}}\end{aligned}\)

Simplify further,

\(\begin{aligned}{e^{ - 0.611t}} &= \frac{{0.000}}{{0.92}}\\{e^{ - 0.611t}} &= 0.0095\end{aligned}\)

Take natural logarithm on both sides,

\(\begin{aligned}{l} - 0.6107t &= \ln (0.0095)\\ - 0.6107t = - 4.6564\\t = \frac{{4.6564}}{{0.6107}}\\t \approx 7.62\end{aligned}\)

Thus, the\(90\% \)heard the rumor in\(7.6\)hours.

\(90\% \)heard the rumor time in an initial stage to complete\(90\% \)heard.

That means\(8AM\)to after\(7.6\)hours.

\(7.47\)hours written as\(7\)hours\(0.6 \times 60\)minutes. So that\(7\)hours\(36\)minutes.

Therefore, in \(90\% \)of population is heard the rumor time is \(3:36{\rm{PM}}\).