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
