AMP1514 Assignment 01 solutions PDF

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BAR CODELearn without limits. universityof south africaTutorial Letter 201/0/Mathematical ModellingAPMYear moduleDepartment of Mathematical SciencesSolutions to Assignment 01APM1514/201/0/APM1514-TL-201 SolutionsQuestion 1 [20 marks]1 Autonomous. Not First Order. 1 Autonomous. Not First Order. 1 Aut...

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APM1514/201/0/2021

Tutorial Letter 201/0/2021 Mathematical Modelling

APM1514 Year module Department of Mathematical Sciences

Solutions to Assignment 01

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university of south africa

APM1514-TL-201 Solutions Question 1 [20 marks] 1.1 Autonomous. Not First Order. 1.2 Autonomous. Not First Order. 1.3 Autonomous. Not First Order. 1.4 Autonomous. First Order. 1.5 Autonomous. First Order. 1.6 Autonomous. First Order. 1.7 Autonomous. Not First Order. 1.8 Autonomous. Not First Order. 1.9 Autonomous. Not First Order. 1.10 Autonomous. First Order. Question 2 [16 marks] 2.1 P (n + 1) = (1)n+1P (n). Given that P (0) = 3. ∴ P (1) = (1)0+1P (0) = 3 ∴ P (2) = (1)1+1P (1) = 3 ∴ P (3) = (1)2+1P (2) = (3) = 3 ∴ P (4) = (1)3+1P (3) = 3 2.2 P (n + 1) = P (n) + sin(nπ). Given that P (0) = 0. ∴ P (0) = 0 ∴ P (1) = 0 + sin(0π) = 0 ∴ P (2) = P (1) + sin(π) = 0 ∴ P (3) = P (2) + sin(2π) = 0 ∴ P (4) = P (3) + sin(3π) = 0

2.3 P (n + 1) =

p

P (n). Given that P (0) = 4.

1

∴ P (1) = | ± 2| = 2 We consider only the positive input in the recursive updating and not the imaginary solution: p p ∴ P (2) = P (1) = 2 p 1 ∴ P (3) = P (2) = 24 1

∴ P (4) = 28

2.4 P (n + 1) = P (n)[P (n)  1]. Given that P (0) = 2. ∴ P (1) = P (0)[P (0)  1] = 2 ∴ P (2) = P (1)[P (1)  1] = 2 ∴ P (3) = P (2)[P (2)  1] = 2 ∴ P (4) = P (3)[P (3)  1] = 2 2.5 P (t) = ln |10 cos(π nt )|, where, t = 0, 1, . . . , n, and n 2 N. ∴ P (0) = ln |10 cos(π n0 )| = ln 10, 8n 2 N ∴ P (1) = ln |10 cos(π n1 )| for n = 2 we have P (1) as undefined and generally approaches ln 10 as n ! 1. ∴ P (2) = ln |10 cos(π n2 )| for n = 4 we have P (2) as undefined and generally approaches ln 10 as n ! 1. ∴ P (3) = ln |10 cos(π n3 )| for n = 6 we have P (3) as undefined and generally approaches ln 10 as n ! 1. ∴ P (4) = ln |10 cos(π n4 )| for n = 8 we have P (4) as undefined and generally approaches ln 10 as n ! 1. 1

2.6 P (t) = t−2 e− t . ∴ P (0) is undefined. P (1) = e−1 . P (2) = 4√1e . P (3) = 2.7 P (n + 1) =

n+1 P (n) n!

+ 2, given that P (0) = 1. 2

1 √ . 9 3e

P (4) =

1√ . 16 4 e

∴ P (1) = 11 P (0) + 2 = 3, ∴ P (2) = 12 P (1) + 2 = 8, ∴ P (3) = 23 P (2) + 2 = 14. ∴ P (4) = 64 P (3) + 2 = 1 nn

2.8 P (n + 1) =

34 . 3

+ (n2  n)2−n , given that P (0) = 1.

∴ P (1) is undefined, ∴ P (2) =

1 1

+ (0) = 1,

∴ P (3) =

1 4

+

∴ P (4) =

1 27

2 4

= 34 .

+ (3) 41 =

85 . 108

Question 3 [6 marks] 3.1 an+1 = 3an + 1, We first evaluate a1 , a 2 , a3 , by using a0 : a1 = 3a0 + 1, a2 = 3[3a0 + 1] + 1 = 32 a0 + 3 + 1, a3 = 3[32 a0 + 3 + 1] + 1 = 3 3 a0 + 32 + 3 + 1. In general we obtain: n

an = 3 a0 + 3 n−1 + . . . + 1 = 3n a0 +

n−1 X

3n ,

n=0







 3n  1 3n  1 n an = 3 a0 + = 3 a0 + . 31 2 n

3.2 an+1 = an2  10,

3

We first evaluate a1 , a 2 , a3 , by using a0 : a1 = a02  10, a2 = [a02  10]2  10,  2 a3 = [a02  10]2  10  10.

In general we obtain a nested solution given by: i2 h  2 an = . . . [a02  10]2  10  10 . . . 10. | {z } (n−1)times

3.3 =  Pt , Z Z dP dt = , P t dP dt

ln |P | =  ln |t| + b, let A = eb , Therefore, the general solution is P =

A t

3.4 dP dt

Z

= (1 + 2P ), Z 2.dP =  1.dt, 1 + 2P

1 2 1 ln |1 + 2P | = t + a, 2 let b = 2a, ln |1 + 2P | = 2t + b, let C = eb , 1 + 2p = e−2t+b 1 + 2p = Ce−2t 4

Thus, the general solution is p = 21 + De−2t , where D =

C 2

.

Question 4 [10 marks] 4.1 P (n + 1) =

P (n) , 3

with P (0) = 2,

We first evaluate P (1), P (2), P (2), by using P (0) = 2: P (1) =

P (0) 2 = , 3 3

P (1) 2 = 2, 3 3 2 P (2) = 3. P (3) = 3 3 Therefore, the particular solution is P (2) =

P (n) =

P (0) 2 = n. n 3 3

4.2 P (n + 1) = P (n) + 2, with P (0) = 1, We first evaluate P (1), P (2), P (2), by using P (0) = 1: P (1) = P (0) + 2 = 1, P (2) = (P (0) + 2) + 2 = P (0) + 2(2) = 5, P (3) = P (0) + 2(3) = 7. Therefore, the particular solution is P (n) = P (0) + 2n = 1 + 2n. Question 5 [20 marks] 5.1 1, 2P  P = 0 P (1, 2  1) = 0 5

0, 2P = 0 P =0 Only one equilibrium pont exists at zero. 5.2 P = 0, 2P + 1 0, 8P = 1 5 P = = 1, 25 4 Only one equilibrium pont exists at 1,25. 5.3 P = P 2  2P  2 P 2  3P  2 = 0 p 3± 9+8 P = 2

P1 =

√ 3+ 9+8 , 2

or P2 =

√ 3− 9+8 2

Thus, two equilibrium ponts exist. 5.4 P4  P = 0

P (P 3  1) = 0

P (P  1)(P 2 + P + 1) = 0 P = 0, or P = 1 or √ P = −1± 2 1−4 , this P however has no solution. Therefore, only solutions P = 0 and P = 1, are regarded as equilibrium ponts that exist. 5.5

p p p p 2 P = P P p p p P ( P  2) = 0 p p p P = 0, or P = 2 P = 0, or P = 2 6

Therefore, solutions P = 0 and P = 2, are equilibrium ponts that exist. Question 6 [16 marks] 6.1 P (n + 1) = P (n), P (1) = P (0), P (2) = P (1) = (P (0)) = P (0), P (3) = P (2) = P (0), The general solution is P (n) = (1)n P (0). 6.2 P (n + 1) = 5P (n), P (1) = 5P (0), P (2) = 52 P (0), P (3) = 53 P (0), The general solution is P (n) = 5n P (0). 6.3 P (n + 1) = 0, 01P (n), P (1) = (0, 01)P (0), P (2) = (0, 01)2 P (0), P (3) = (0, 01)3 P (0), The general solution is P (n) = (0, 01)n P (0). 6.4 P (n + 1) = 0, 2P (n), P (1) = (0, 2)P (0), P (2) = (0, 2)2 P (0), P (3) = (0, 2)3 P (0), 7

The general solution is P (n) = (1)n (0, 2)n P (0). 6.5

1 P (n + 1) = P (n), 8 1 P (1) = P (0), 8 ✓ ◆ 1 P (2) = 2 P (0), 8 ✓ ◆ 1 P (3) = 3 P (0), 8

The general solution is P (n) =

✓

◆ P (0) . 8n

Question 7 [12 marks] 7.1 P (n + 1) = P (n), with P (0) = a5, Assume that a 6= 0 and let r = 1 (meaning it is not r > 0): ✓ ◆ 5 n , P (n + 1) = (1) a such that as n ! 1, for a > 0, we have 8 5 > < f or n 2 even ! , a P (n + 1) ! > : f or n 2 odd ! 5, a OR

for a < 0, we have

8

8 5 > < f or n 2 even ! , a P (n + 1) ! > : f or n 2 odd ! 5. a

Therefore, the solution diverges to the value ± a5 depending on your initial assumption of a. 7.2 P (n + 1) = 2P (n), with P (0) = 1, Let r = 2: P (n + 1) = 2n (1) As n ! 1, we have P (n+1) ! 1. Meaning that the solution grows without bound. 7.3 P (n + 1) = 41P (n), with P (0) = 2, Since r =

1 4

satisfies the inequality: 0 < r < 1. ✓ ◆n 1 (2) P (n + 1) = 4

Therefore, as n ! 1, we have P (n + 1) ! 0. Meaning that the solution converges to zero.

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