SOLUTION: MATH 275 College of San Mateo Week 1 Ordinary Differential Equation HW

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I did not middle some figures since I don't own mathlab, so I can't paint the presentation lines.

Logistic Population Models after a while Harvesting
1. Logistic Enlargement after a while Inwavering Harvesting
𝑑𝑝
𝑝
= π‘˜π‘ (1 βˆ’ ) βˆ’ π‘Ž
𝑑𝑑
𝑁
Represents a logistic design of population enlargement after a while inwavering harvesting at a reprimand a. For a = a1,
what get betide to the fish population for a multitudinous modetrounce stipulations? (Note: This equation is
autonomous, so you can use service of the peculiar techniques that are suited for autonomous
equations.)

2. Logistic Enlargement after a while Occasional Harvesting
𝑑𝑝
𝑝
= π‘˜π‘ (1 βˆ’ ) βˆ’ π‘Ž(1 + sin 𝑏𝑑)
𝑑𝑑
𝑁
Is a non-autonomous equation that considers occasional harvesting. What do the parameters a and b
represent? Let b = 1. If a = a1, what get betide to the fish population for multitudinous modetrounce stipulations?

3. Consider the identical equation as in Part 2 over, but let a=a2. What get betide to the fish
population for multitudinous modetrounce stipulations after a while this appreciate of a?
Use the aftercited choices for the parameters.
Choice

k

N

a1

a2

2

0.50

2

0.21

0.25

8

0.20

5

0.24

0.25

1. Logistic Enlargement after a while Inwavering Harvesting
𝑑𝑝
𝑝
= π‘˜π‘ (1 βˆ’ ) βˆ’ π‘Ž
𝑑𝑑
𝑁
Represents a logistic design of population enlargement after a while inwavering harvesting at a reprimand a. For a = a1,
what get betide to the fish population for a multitudinous modetrounce stipulations? (Note: This equation is
autonomous, so you can use service of the peculiar techniques that are suited for autonomous
equations.)
𝑑𝑝
𝑝
= π‘˜π‘ (1 βˆ’ ) βˆ’ π‘Ž
𝑑𝑑
𝑁
π‘€β„Žπ‘’π‘Ÿπ‘’:
𝑝 = π‘π‘œπ‘π‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘›
π‘˜ = π‘”π‘Ÿπ‘œπ‘€π‘‘β„Ž π‘Ÿπ‘Žπ‘‘π‘’ π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘
𝑁 = π‘π‘Žπ‘Ÿπ‘Ÿπ‘¦π‘–π‘›π‘” π‘π‘Žπ‘π‘Žπ‘π‘–π‘‘π‘¦ π‘œπ‘“ π‘‘β„Žπ‘’ π‘’π‘›π‘£π‘–π‘Ÿπ‘œπ‘›π‘šπ‘’π‘›π‘‘
π‘Ž = β„Žπ‘Žπ‘Ÿπ‘£π‘’π‘ π‘‘π‘–π‘›π‘” π‘Ÿπ‘Žπ‘‘π‘’
𝑑𝑝
= π‘Ÿπ‘Žπ‘‘π‘’ π‘œπ‘“ π‘β„Žπ‘Žπ‘›π‘”π‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘œπ‘π‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› π‘€π‘–π‘‘β„Ž π‘Ÿπ‘’π‘ π‘π‘’π‘π‘‘ π‘‘π‘œ π‘β„Žπ‘Žπ‘›π‘”π‘’ 𝑖𝑛 π‘‘π‘–π‘šπ‘’
𝑑𝑑
𝑡𝒐𝒕𝒆: π‘“π‘œπ‘Ÿ π‘‘β„Žπ‘–π‘  π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘›, π‘‘β„Žπ‘’ π’Œ, 𝑡 π‘Žπ‘›π‘‘ π’‚πŸ π‘Žπ‘Ÿπ‘’ π‘π‘Žπ‘Ÿπ‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿπ‘ π‘”π‘–π‘£π‘’π‘› 𝑖𝑛 π‘‘β„Žπ‘’ π‘‘π‘Žπ‘π‘™π‘’.
The equation is as-well autonomous as the equation is simply hanging on wavering P as waverings k,
N and a=a1 are all inwavering appreciates.
𝑑𝑝
𝑝
= π‘˜π‘ (1 βˆ’ ) βˆ’ π‘Ž
𝑑𝑑
𝑁
πΉπ‘œπ‘Ÿ πΆβ„Žπ‘œπ‘–π‘π‘’ π‘›π‘œ. 2:
𝑑𝑝
𝑝
= (0.50)𝑝 (1 βˆ’ ) βˆ’ 0.21
𝑑𝑑
2
𝐹𝑖𝑛𝑑𝑖𝑛𝑔 π‘‘β„Žπ‘’ π‘π‘Ÿπ‘–π‘‘π‘–π‘π‘Žπ‘™ π‘π‘œπ‘–π‘›π‘‘π‘ :
0 = 0.50𝑝 βˆ’ 0.25𝑝2 βˆ’ 0.21
[0 = 0.50𝑝 βˆ’ 0.25𝑝2 βˆ’ 0.21](βˆ’100)
0 = 25𝑝2 βˆ’ 50𝑝 + 21
0 = (5𝑝 βˆ’ 3)(5𝑝 βˆ’ 7)
5𝑝 βˆ’ 3 = 0 π‘Žπ‘›π‘‘ 5𝑝 βˆ’ 7 = 0
𝑝=

3 7
,
5 5

At p(t)=3/5 and p(t)=7/5, the reprimand of modify in population throughout all modifys of interval, get be
zero.

π’…π’†π’•π’†π’“π’Žπ’Šπ’π’Šπ’π’ˆ 𝒕𝒉𝒆 𝒗𝒆𝒓𝒕𝒆𝒙 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒖𝒓𝒗𝒆:
𝑆𝑒𝑑 π‘‘β„Žπ‘’ π‘žπ‘’π‘Žπ‘‘π‘Ÿπ‘Žπ‘‘π‘–π‘ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› π‘–π‘›π‘‘π‘œ 𝑖𝑑𝑠 π‘£π‘’π‘Ÿπ‘‘π‘’π‘₯ π‘“π‘œπ‘Ÿπ‘š π‘‘π‘œ 𝑖𝑑𝑒𝑛𝑑𝑖𝑓𝑦 π‘‘β„Žπ‘’ π‘£π‘’π‘Ÿπ‘‘π‘’π‘₯ (𝐻, 𝐾)
𝑑𝑝
= βˆ’0.25(𝑝 βˆ’ 1)2 + 0.04
𝑑𝑑
𝐻 = 1 π‘Žπ‘›π‘‘ 𝐾 = 0.04
π‘‡β„Žπ‘–π‘  𝑠𝑖𝑔𝑛𝑖𝑓𝑖𝑒𝑠 π‘‘β„Žπ‘Žπ‘‘ π‘Žπ‘‘ 𝑝(𝑑) = 1, π‘‘β„Žπ‘’

𝑑𝑝
𝑖𝑠 π‘Žπ‘‘ 𝑖𝑑𝑠 β„Žπ‘–π‘”β„Žπ‘’π‘ π‘‘ π‘Žπ‘‘ 0.04
𝑑𝑑

We can then use the equilibria to indicate what would betide to the fish population at multitudinous
modetrounce stipulations (Po).
0 π‘ƒπ‘œ =

3
5

3
5

π‘‡β„Žπ‘’ πΉπ‘–π‘ β„Žπ‘’π‘  𝑀𝑖𝑙𝑙 𝑑𝑖𝑒 π‘œπ‘’π‘‘.
π‘‡β„Žπ‘’ πΉπ‘–π‘ β„Ž π‘π‘œπ‘π‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› 𝑀𝑖𝑙𝑙 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘› π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘.

3
7
π‘‡β„Žπ‘’ πΉπ‘–π‘ β„Ž π‘ƒπ‘œπ‘π‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› 𝑀𝑖𝑙𝑙 π‘–π‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘’.
5
5
7
π‘ƒπ‘œ =
π‘‡β„Žπ‘’ πΉπ‘–π‘ β„Ž π‘ƒπ‘œπ‘π‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› 𝑀𝑖𝑙𝑙 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘› π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘.
5
7
π‘ƒπ‘œ >
π‘‡β„Žπ‘’ πΉπ‘–π‘ β„Ž π‘ƒπ‘œπ‘π‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› 𝑀𝑖𝑙𝑙 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘› π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘.
5
πΉπ‘œπ‘Ÿ πΆβ„Žπ‘œπ‘–π‘π‘’ π‘›π‘œ. 8:
𝑑𝑝
𝑝
= (0.20)𝑝 (1 βˆ’ ) βˆ’ 0.24
𝑑𝑑
5
𝐹𝑖𝑛𝑑𝑖𝑛𝑔 π‘‘β„Žπ‘’ π‘π‘Ÿπ‘–π‘‘π‘–π‘π‘Žπ‘™ π‘π‘œπ‘–π‘›π‘‘π‘ :
0 = 0.20𝑝 βˆ’ 0.04𝑝2 βˆ’ 0.24
[0 = 0.20𝑝 βˆ’ 0.04𝑝2 βˆ’ 0.24](βˆ’100)
0 = 4𝑝2 βˆ’ 20𝑝 + 24
0 = (2𝑝 βˆ’ 6)(2𝑝 βˆ’ 4)
2𝑝 βˆ’ 6 = 0 π‘Žπ‘›π‘‘ 2𝑝 βˆ’ 4 = 0
𝑝 = 3 ,2
At p(t)=3...


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