a. Consider the following models: (i.) A linear model. Suppose that P satisfies:

Question

a. Consider the following models:

(i.) A linear model. Suppose that P satisfies: dP/dt = 30.2. with P(0) = 95.

Solve the differential equation and use your solution to predict the number of

cases in Hong Kong by June 12 (t = 87). Answer is 2,722 cases reported

(ii.) An exponential model. Suppose that P satisfies: (1/P)(dP/dt) = 0.12 with P(0) = 95.

Solve the differential equation and use your solution to predict the number of

cases in Hong Kong by June 12 (t = 87). Answer is 3,249,062 cases reported.

(iii.) A logistic model. Suppose that P satisfies: (1/P)(dP/dt) = 0.19 - 0.0002P with P(0) = 95.

Solve the differential equation and use your solution to predict the number of

cases in Hong Kong by June 12 (t = 87). Answer is 950 cases reported.

b.) What does each of the three models predict about the long term behavior of the disease? Be sure to show mathematical evidence for each prediction.

The problem i am having is showing the long term with the cases by mathematical evidence. Please help !

Explanation / Answer

(i.) A linear model. Suppose that P satisfies: dP/dt = 30.2. with P(0) = 95.

dP/dt = 30.2

dP = 30.2*dt

P = 30.2*t + c

at t= 0 ;

P = c = 95

so P = 30.2*t + 95

@ t= 87

P = 2722.4

(ii.) An exponential model. Suppose that P satisfies: (1/P)(dP/dt) = 0.12 with P(0) = 95.

1/P(dP/dt) = 0.12

dP/P= 0.12*dt

ln P = 0.12*t + c

P = k*e^(0.12*t)

at t=0

P = k*1 = k = 95

P=95*e^(0.12*t)

@t =87

P = 3249061.982

(iii.) A logistic model. Suppose that P satisfies: (1/P)(dP/dt) = 0.19 - 0.0002P with P(0) = 95.

(1/P)dP/dt = 0.19 - 0.0002P

dP/P*(0.19 - 0.0002P ) = dt

1/P*(0.19 - 0.0002P ) = A/P + B/(0.19 - 0.0002P ) = {A*(0.19 - 0.0002P ) + B*P} / {P*(0.19 - 0.0002P )}

A*0.19 = 1

A = 1/0.19

(B- 0.0002) = 0

B = 0.0002

so 1/P*(0.19 - 0.0002P ) = 1/0.19*P + 0.0002/(0.19 - 0.0002P )

dP/0.19*P + 0.0002dP/(0.19 - 0.0002P ) = dt

dP/0.19*P + dP/(950-P) = dt

(ln P )/0.19 - ln (950-P) = t + c

(ln P ) - 0.19 ln (950-P) =0.19* t + d

P/(950-P)^0.19 = k*e^(0.19t)

at t= 0

95/(950-95)^0.19 = k

k = 26.342 ;

P/(950-P)^0.19 = 26.342*e^(0.19t)

@t=87

P/(950-P)^0.19 = 3977683735.3 ;

P = 950 ;

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