The equation N ( t ) = 1000 1 + C e − 2 t is the general solution of the differential equation d N d t = 1 500 N ( 1000 − N ) and evaluate N ¯ = lim t → ∞ N ( t ) .

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Answer 1

Question

Chapter 9.1, Problem 66E

(A)

To determine

To show: The equation N(t)=10001+Ce−2t is the general solution of the differential equation dNdt=1500N(1000−N) and evaluate N¯=limt→∞N(t).

(B)

To determine

To graph: The particular solutions satisfying the initial conditions N(0)=200 and N(0)=2000.

(C)

To determine

To discuss: The long-term behavior of the population of the animal.

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Answer 2

Question

Chapter 9.1, Problem 66E

(A)

To determine

To show: The equation N(t)=10001+Ce−2t is the general solution of the differential equation dNdt=1500N(1000−N) and evaluate N¯=limt→∞N(t).

(B)

To determine

To graph: The particular solutions satisfying the initial conditions N(0)=200 and N(0)=2000.

(C)

To determine

To discuss: The long-term behavior of the population of the animal.

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Answer 3

Question

Chapter 9.1, Problem 66E

(A)

To determine

To show: The equation N(t)=10001+Ce−2t is the general solution of the differential equation dNdt=1500N(1000−N) and evaluate N¯=limt→∞N(t).

(B)

To determine

To graph: The particular solutions satisfying the initial conditions N(0)=200 and N(0)=2000.

(C)

To determine

To discuss: The long-term behavior of the population of the animal.

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Answer 4

Question

Chapter 9.1, Problem 66E

(A)

To determine

To show: The equation N(t)=10001+Ce−2t is the general solution of the differential equation dNdt=1500N(1000−N) and evaluate N¯=limt→∞N(t).

(B)

To determine

To graph: The particular solutions satisfying the initial conditions N(0)=200 and N(0)=2000.

(C)

To determine

To discuss: The long-term behavior of the population of the animal.

Expert Solution & Answer

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Answer 5

Chapter 9.1, Problem 66E

(A)

To determine

To show: The equation N(t)=10001+Ce−2t is the general solution of the differential equation dNdt=1500N(1000−N) and evaluate N¯=limt→∞N(t).

(B)

To determine

To graph: The particular solutions satisfying the initial conditions N(0)=200 and N(0)=2000.

(C)

To determine

To discuss: The long-term behavior of the population of the animal.

Answer 6

Chapter 9.1, Problem 66E

(A)

To determine

To show: The equation N(t)=10001+Ce−2t is the general solution of the differential equation dNdt=1500N(1000−N) and evaluate N¯=limt→∞N(t).

Answer 7

Chapter 9.1, Problem 66E

(A)

To determine

To show: The equation N(t)=10001+Ce−2t is the general solution of the differential equation dNdt=1500N(1000−N) and evaluate N¯=limt→∞N(t).

Answer 8

(B)

To determine

To graph: The particular solutions satisfying the initial conditions N(0)=200 and N(0)=2000.

Answer 9

(B)

To determine

To graph: The particular solutions satisfying the initial conditions N(0)=200 and N(0)=2000.

Answer 10

(C)

To determine

To discuss: The long-term behavior of the population of the animal.

Answer 11

(C)

To determine

To discuss: The long-term behavior of the population of the animal.

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

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Answer 15

Ch. 9.1 - Show that y=Cx+1 is the general solution of the...Ch. 9.1 - Prob. 2MP Ch. 9.1 - Prob. 3MP Ch. 9.1 - Prob. 4MP Ch. 9.1 - Prob. 5MP Ch. 9.1 - Prob. 1ED Ch. 9.1 - Prob. 2ED Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E

The equation N ( t ) = 1000 1 + C e − 2 t is the general solution of the differential equation d N d t = 1 500 N ( 1000 − N ) and evaluate N ¯ = lim t → ∞ N ( t ) .

Solution Summary: The author analyzes the equation N(t)=10001+Ce-2t. The equilibrium size of the population is 1000.

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To show: The equation N(t)=10001+Ce−2t is the general solution of the differential equation dNdt=1500N(1000−N) and evaluate N¯=limt→∞N(t).

To graph: The particular solutions satisfying the initial conditions N(0)=200 and N(0)=2000.

To discuss: The long-term behavior of the population of the animal.

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