- #1

zaboda42

- 32

- 0

a) Find the general solution of the differential equation.

b) Find the particular solution of the differential equation given that 5000 people have the disease when first noticed, that is, when t=0, P=5.

c) Estimate how many people will eventually have the disease. Explain your answer.

Here's what i did:

**(a)**

dP/[P(30 - P)] = 0.03 dt.

Now 1/[P(30 - P)] = 1/30[1/P + 1/(30 - P)] so the DE becomes

[1/P + 1/(30 - P)] dP = 30(0.03) dt and integrating gives

ln|P| - ln|(30 - P)| = 0.9t + c

ln(P/(30 - P)) = 0.9t + c

P/(30 - P) = Ae^(0.9t) where A = e^c

P = Ae^(0.9t)*(30 - P)

P[1 + Ae^(0.9t)] = 30Ae^(0.9t)

so the general solution is

P(t) = 30Ae^(0.9t)/[1 + Ae^(0.9t)].

**(b)**

When t = 0, P = 5 so

5 = 30A/[1 + A]

5[1 + A] = 30A

25A = 5

so A = 0.2 and the particular solution is

P(t) = 6e^(0.9t)/[1 + 0.2e^(0.9t)].

**(c)**

The number eventually having the disease will be the limit of P as t--> infinity.

Now P(t) = 6e^(0.9t)/[1 + 0.2e^(0.9t)]

= 6/[e^(-0.9t) + 0.2]

--> 6/0.2 = 30 as t--> infinity since e^(-0.9t)-->0 as t-->infinity.

So 30 000 people will eventually have the disease.I've tried to do the work, please just let me know if i have anything wrong. Thanks guys.