Question

[14 marks] A cruise ship has docked in Fremantle harbour∗. Ofthe500peopleonboard,10have been diagnosed with the COVID-19 virus. The spread of the virus to other people on the ship is governed by the initial value problem dI dt= (1/500)I(500−I) with I(0) = 10 where I(t) is the number of people on board who have the virus and t is time (in days).

(a) Solve the initial value problem for I(t).

(b) Allowing for rounding, how long before everyone on the boat has the virus (in other words, solve I(t) = 499.5)?

(c) Use Euler’s method to compute a numerical solution for t ∈ [0,10] using a step size of ∆t = 0.5.

(d) Plot both of your solutions (from (a) and (c)) on the same set of axes and comment. ∗Please note, this is a completely made up question—it is for educational purposes only!

Answer 1

at - 500 $(0) I I(500-I I(0) = 10 dI 1/500-I) = dto Integrating on both sides, we get d1 Constant. ln (I) + ln (500- I - t + sooc let 5000 = K = constant en(I) – ln (500-I = tek en( I ) =t+k I - - Két (k. - Constant - ek) I= 500k, et - I ket I (0)=10 2 10 = sook, (6) - 10 (k.)(e) 10=490 k, → [k, ata - I = sovet - I et 49

03 I (49tet) - sooet sooet 49 tet 1) Ilt-499.5 =+=? 499.5 = scoot 49+ et 244755 + 499.50 = 500et oset = 24475.5 et = 48951 t = ln (48951) = 10.799 days It = 10.8 days (c) Exter's equation f(I,t) = d = 2100 . . Innya In + hf(tn, In h = 0.5 . I, 10+ 0.5 +10,10) & t=0 Ij = 1040-5( 10440) I = 14.9 To=10

n	Tn	f(Tn,In)	In
0	0.0000	9.8000	10.0000
1	0.5000	14.4560	14.9000
2	1.0000	21.1487	22.1280
3	1.5000	30.5635	32.7023
4	2.0000	43.3791	47.9841
5	2.5000	59.9648	69.6736
6	3.0000	79.7934	99.6560
7	3.5000	100.6028	139.5527
8	4.0000	117.7649	189.8541
9	4.5000	124.9968	248.7366
10	5.0000	117.5006	311.2350
11	5.5000	96.2071	369.9853
12	6.0000	68.4923	418.0888
13	6.5000	43.1211	452.3350
14	7.0000	24.7416	473.8955
15	7.5000	13.3565	486.2663
16	8.0000	6.9559	492.9445
17	8.5000	3.5519	496.4225
18	9.0000	1.7951	498.1984
19	9.5000	0.9024	499.0960
20	10.0000	0.4524	499.5472

Analytic Data Euler Data 0.0000 2.0000 4.0000 6.0000 8.000010.00001 2.0000 tin days