A rumor spreads through a school. Let y(t)y(t) be the fraction of the population that has heard the rumor at time tt and assume that the rate at which the rumor spreads is proportional to the product of the fraction yy of the population that has heard the rumor and the fraction 1−y1−y that has not yet heard the rumor.

Answer:Step-by-step explanation:Rate the rumor spread so proportional to y∙(1 - y). So the rate of change of y equals this rate. Therefore   dy/dt = k∙y∙(1 - y) That differential is a separable first order equation. Separation of variables yields: (1/(y∙(1 - y)) dy = k dt Hence, ∫ (1/(y∙(1 - y)) dy = ∫ k dt Partial fraction expansion for the left hand side: 1/(y∙(1 - y) = (A/y) + (B/(1 - y)) Residue method gives A = 1/(1 - 0) = 1 B = 1/1 = 1 ∫ (1/(y∙(1 - y)) dy = ∫ k dt <=> ∫ (1/y) + (1/(1 - y)) dy = ∫ k dt <=> ln(y) - ln(1 - y) = k∙t + C (C is the constant of integration) Since ln(a) - ln(b) = ln(a/b) you can rewrite   ln( y/(1 - y) )= k∙t + C To evaluate C apply the initial value y(t=0) = 10% = 0.1 => ln( 0.1/(1 - 0.1) )= k∙0 + C => C = ln(0.1/0.9) = ln(1/9) So   ln( y/(1 - y) ) = k∙t + ln(1/9) To evaluate k, solve this equation for k and apply the given value y = 0.4 at t = 2 days ln( y/(1 - y) ) = k∙t + ln(1/9) <=> k = ( ln( y/(1 - y) ) - ln(1/9) ) / t = ( ln( (y/(1 - y) ) / (1/9) ) / k∙t = ( ln( 9∙y/(1 - y) ) / t = ( ln( 9∙0.4/(1 - 0.4) ) / 2 d = (1/2)∙ln( 6 ) d⁻¹ ≈ 0.8959 d⁻¹