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Differential Equations 6 Copyright © Cengage Learning. All rights reserved.

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Presentation on theme: "Differential Equations 6 Copyright © Cengage Learning. All rights reserved."— Presentation transcript:

1 Differential Equations 6 Copyright © Cengage Learning. All rights reserved.

2 More on Separation of Variables AB/BC and the Logistic Equation BC only 1 Day AB or BC Copyright © Cengage Learning. All rights reserved. 6.3

3 Tomorrow MOCK AP TEST 11:45 ATTENDANCE IS MANDATORY

4 Calculus Warm-Up 2/19/13 Use integration to find the general solution to the differential equation:

5 Calculus Cool Down 11/16/15 A 150 degree object is placed in a freezer whose temperature is kept at a constant 30 degrees. In 6 minutes the object has cooled to 80 degrees. How much longer till the object is 50 degrees?

6 Separable Differential Equations A separable differential equation can be expressed as the product of a function of x and a function of y. Example: Multiply both sides by dx and divide both sides by y 2 to separate the variables. (Assume y 2 is never zero.)

7 Separable Differential Equations A separable differential equation can be expressed as the product of a function of x and a function of y. Example: Combined constants of integration

8 Example: Separable differential equation Combined constants of integration Separable Differential Equations

9 Example continued: We now have y as an implicit function of x. We can find y as an explicit function of x by taking the tangent of both sides. Notice that we can not factor out the constant C, because the distributive property does not work with tangent.

10 Example: Find the general solution of First, separate the variables. y’s on one side, x’s on the other. Second, integrate both sides. Solve for y. Write in exponential form.or

11 Example: Finding a Particular Solution Given the initial condition y(0) = 1, find the particular solution of the equation Separate the variables: Now, integrate both sides.

12 u = x 2 du = 2x dx du/2 =x dx Now find C at (0,1) or by multiplying by 2, you get

13 Example: Finding a Particular Solution Curve Find the equation of the curve that passes through the point (1,3) and has a slope of y/x 2 at any point (x,y). Because the slope is y/x 2, you have Now, separate the variables. Rewrite in exponential form. at the point (1,3), C = ? So, the equation is or

14 Example: Wildlife Population (MMM #3 pg. 222) The rate of change of the number of coyotes N(t) in a population is directly proportional to 650 - N(t), where t is the time in years. When t = 0, the population is 300, and when t = 2, the population has increased to 500. Find the population when t = 3. Because the rate of change of the population is proportional to 650 - N(t), we can write the following differential equation. Separate variables Integrate

15 Write in exponential form. Using N = 300 when t = 0, you can conclude that C = 350, which produces Then, using N = 500 when t = 2, it follows that So, the model for the coyote population is Example: Wildlife Population solution con’t.

16 When t = 3, you can approximate the population to be N = 650 – 350e –0.4236(3) ≈ 552 coyotes. The model for the population is shown in Figure 6.14. Note that N = 650 is the horizontal asymptote of the graph and is the carrying capacity of the model. Figure 6.14 cont'd

17 Logistic Differential Equation BC Only

18 Logistic Differential Equation The exponential growth model was derived from the fact that the rate of change of a variable y is proportional to the value of y. You observed that the differential equation dy/dt = ky has the general solution y = Ce kt. This describes a growth rate proportional to the amount present. Exponential growth is unlimited, yet real-life populations do not increase forever. There is some limiting factor such as food or living space. There is often a maximum population that can be reached, called the carrying capacity. A more realistic population model is the logistic growth model, where the growth rate is proportional to both the amount present and the carrying capacity.

19 A model that is often used to describe this type of growth is the logistic differential equation: where k and L are positive constants. A population that satisfies this equation does not grow without bound, but approaches the carrying capacity L as t increases. From the equation, you can see that if y is between 0 and the carrying capacity L, then dy/dt > 0, and the population increases. Logistics Differential Equation

20 Logistic Differential Equation If y is greater than L, then dy/dt < 0, and the population decreases. The graph of the function y is called the logistic curve, as shown in Figure 6.17. Figure 6.17

21 Example – Deriving the General Solution Solve the logistic differential equation Solution: Begin by separating variables.

22 Example – Solution con’t Solving this equation for y produces y =

23 Logistics Growth Model All solutions of the logistic differential equation are of the form y = Logistics Differential Equation

24 Example – Analyzing a logistic Differential Equation A state game commission releases 40 elk into a game refuge. After 5 years the elk population is 104. The commission believes that the environment can support no more than 4000 elk. The growth rate of the elk population p is: The elk population follows a logistic growth model. a)Where would horizontal asymptotes occur on the graph of this model? b)Where would a point of inflection occur?

25

26 Example – Solving a logistic Differential Equation A state game commission releases 40 elk into a game refuge. After 5 years the elk population is 104. The commission believes that the environment can support no more than 4000 elk. The growth rate of the elk population p is: a)Write a model for the elk population in terms of t. b)Use it to estimate the elk population after 15 years. c)Find the limit of the model as

27 Example – Solving a logistic Differential Equation Since solutions are of the form: The solution of the equation is of the form:

28 You try: Logistic Growth Model Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears. Assuming a logistic growth model, when will the bear population reach 50? 75? 100? When is the bear population increasing most rapidly?

29 Years Bears We can graph this equation and find the intersection point with 50, 75 and 100 to find the solutions. y=50 at 22 years y=75 at 33 years y=100 at 75 years Bear population is increasing most rapidly at 22 years

30 BC Homework 6.3 p.429 13-21 odds, 49-53 odds, 71-79 odds

31 BC Homework 6.3 Day 1: p.429 1-23 odd 6.3 Day 2: p.429 45-53 odds, 71-79 odds

32 Example – Solving a logistic Differential Equation At time t=0, a bacterial culture weighs 1 gram. 2 hours later, the culture weighs 2 grams. The maximum weight of the culture is 10 grams. a)Write a logistic equation that models the weight of the culture. b)Find the weight after 5 hours. c)When will the weight reach 8 grams? d)Write a logistic differential that models the growth rate of the culture’s weight, then repeat part b using Euler’s Method with a step size of h=1. e)At what time is the culture’s weight increasing most rapidly?

33 Solution: See 6.3 #80.

34 AB Homework Day 1: Pg. 429 1-23 odds, 45, 47, 53 Day 2: MMM pg. 212-213


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