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FOR PART C, can you explain how they got the quadratic equation

FOR PART C, can you explain how they got the quadratic equation

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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FOR PART C, can you explain how they got the quadratic equation

**Problem 4:** The growth of a certain population (in millions) over time, \( P(t) \), is modeled by the differential equation: \[ P' = 6P - P^2 - h. \] The parameter \( h \) in the equation is the harvesting rate (in millions per year). (a) **Determine if the population will grow, be steady, or decline in the following cases:** i) \( h=5, \, P(0) = 1, \) ii) \( h=5, \, P(0) = 5, \) iii) \( h=10, \, P(0) = 5. \) **Solution:** I need to compute the derivative in each case. i) \( P' = 6 \times 1 - 1^2 - 5 = 0 \) so the population stays constant. ii) \( P' = 30 - 25 - 5 = 0 \) so also here the population stays constant. iii) \( P' = 30 - 25 - 10 = -5 \) so the population will decline. (b) **What should the harvest rate be if we want \( P = 4 \) to be an equilibrium population?** I need that \( 0 = 24 - 16 - h \) this means that \( h = 8. \) (c) **For what values of the harvest rate, \( h \), does a stable equilibrium population exist?** Equilibrium population is a solution of a quadratic equation \( 0 = 6P - P^2 - h. \) This equation has a solution if the discriminant \( 36 - 4h \) is non-negative. This gives a condition \( h \leq 9. \) If \( h = 9 \) then there is only one equilibrium population and the function \( 6P - P^2 - h \) does not change sign around this point. So this is an unstable equilibrium. For \( h < 9 \) there are two equilibria solutions, as we saw in the lecture the large one is stable in this case. So \( h < 9. \)
Transcribed Image Text:**Problem 4:** The growth of a certain population (in millions) over time, \( P(t) \), is modeled by the differential equation: \[ P' = 6P - P^2 - h. \] The parameter \( h \) in the equation is the harvesting rate (in millions per year). (a) **Determine if the population will grow, be steady, or decline in the following cases:** i) \( h=5, \, P(0) = 1, \) ii) \( h=5, \, P(0) = 5, \) iii) \( h=10, \, P(0) = 5. \) **Solution:** I need to compute the derivative in each case. i) \( P' = 6 \times 1 - 1^2 - 5 = 0 \) so the population stays constant. ii) \( P' = 30 - 25 - 5 = 0 \) so also here the population stays constant. iii) \( P' = 30 - 25 - 10 = -5 \) so the population will decline. (b) **What should the harvest rate be if we want \( P = 4 \) to be an equilibrium population?** I need that \( 0 = 24 - 16 - h \) this means that \( h = 8. \) (c) **For what values of the harvest rate, \( h \), does a stable equilibrium population exist?** Equilibrium population is a solution of a quadratic equation \( 0 = 6P - P^2 - h. \) This equation has a solution if the discriminant \( 36 - 4h \) is non-negative. This gives a condition \( h \leq 9. \) If \( h = 9 \) then there is only one equilibrium population and the function \( 6P - P^2 - h \) does not change sign around this point. So this is an unstable equilibrium. For \( h < 9 \) there are two equilibria solutions, as we saw in the lecture the large one is stable in this case. So \( h < 9. \)
Expert Solution
Step 1: Write the given information:

The growth of a population over time P(t) is modeled by the differential equation P apostrophe equals 6 P minus P squared minus h.

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Explain how they got quadratic equation in part c).

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