1 What is a differential equation

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1 Math 10B - Calculus by Hughes-Hallett, et al. Chapter 11 - Differential Equations Prepared by Jason Gaddis 1 What is a differential equation Remark 1.1. We have seen basic differential equations already in this course. In this chapter, we study these more in-depth and learn how to solve more complex differential equations. Remark 1.2. Let P represent the population at time t (this is short-hand for just writing P (t). Then the rate that P changes is proportional to the population at any time. This gives the differential equation, = kp, k > 0, where k is known as the growth constant. Remark 1.3. Suppose a new species is introduced to an island. At first the population would grow according to its growth rate. However, after time, the population would begin to level off as resources became scarce. (Draw picture). Assume the rate of growth is equal to (perhaps proportional) to the difference between the max population and the current population. We could then model this situation as = M P, (1) where P = P (t) is the population at time t and M is max population the island can maintain. Example 1.4. Suppose in the previous remark, we have M = 500. Show that P (t) = Ce t is a solution to the differential equation. We have = M P Ce t = 500 (500 + Ce t ). Suppose now that the initial population is 100. What is the value of C? 100 = P (0) = C C =

2 Remark 1.5. As we see from the previous example, differential equations have a general family of solutions. Only given (possibly multiple) initial conditions can we determine the precise value of C. Definition 1.6. The order of a differential equation is determined by the highest derivative appearing in the equation. Example 1.7. Determine for which values of ω the function y = cos ωt satisfies the second-order diff eq: d 2 y + 9y = 0. 2 In alternate notation we could write this equation as We have y + 9y = 0. y = ω sin ωt y = ω 2 cos ωt This gives 0 = y + 9y = ( ω 2 cos ωt) + 9(cos ωt) = (9 ω 2 ) cos ωt Thus, either ω = ±3 or else ω = π 2, 3π 2. 2 Slope fields 3 Euler s method 2

3 4 Separation of variables Example 4.1. Solve the differential equation dy dx = x y. We utilize a method called separation of variables. Cross-multiplication gives y dy = x dx. Integrating both sides, we find y dy = x dx y 2 2 = x2 2 + k Thus, solutions are of the form x 2 + y 2 = C where C = 2k. Remark 4.2. The general strategy of separation of variables is to move all x-terms to one side and y-terms to the other side. We then integrate both sides of the equation. Example 4.3. Use separation of variables to solve the differential equation dy dx = ky. Example 4.4. Solve the intial value problem with z = 1 when y = 0. dz dy = zy, 3

4 5 Growth and Decay Remark 5.1. Evaluations! Remark 5.2. Recall our standard model for populations growth: Let P represent the population at time t. Then the rate that P changes is proportional to the population at any time. This gives the differential equation, = kp, k 0. A value of k > 0 represents growth whereas a value k < 0 represents decay. (Think: half-life of a radioactive element.) Recall that every solution to the above equation is of the form Where P 0 is the value of P at t = 0. P = P 0 e kt, Example 5.3. Review solving for P with separation of variables. Mention absolute value, that is, do the negative case and show that it does not affect the solution. Remark 5.4. We want to study examples similar to the one above. Example 5.5 (Continuously compounded interest). Let B 0 be an initial balance in an account earning r% interest compounded continuously. The rate of growth of the account is proportional to the current balance B of the account. We have db = rb. Solving this differential equation gives B = B 0 e P rt. Example 5.6 (Radioactive decay). After a certain drug is fully absorbed, the quantity of drug in the body decreases at a rate proportional to the amount left in the body. Suppose the half-life of this drug is 3 hours and the usual dose is 10 mg. Write a differential equation modelling this situation and solve it. 4

5 Let A be the amount (in mg) of drug still in the body at time t. Write The solution to this system is da = ra. A = A 0 e rt, and we are given that A 0 = 10. The half-life is 3 hours. This means that A(3) = 1 2 A 0. Thus, 5 = A(3) = 10e 3r, so e 3r = 1 2 2, which implies r = ln Thus, the equation has the form A = 10e.231t. Remark 5.7 (Newton s Law of Heating a Cooling). The temperature of a hot (resp. cold) object decreases (resp. increases) at a rate proportional to the difference between its temperature and that of its surroundings. Example 5.8. Suppose a bottle of OJ has temperature 45 F is taken out of a fridge and left in a 65 F room. Write an equation and solve it. By Newton s Law of Cooling, the rate of change of temp is equal to α (a constant) times the temperature difference. Thus, if H is the current temp of the object, then dh = α(h 65). The temp is falling so α < 0. Let s replace α with k for k > 0. Then, Solving gives, dh = k(h 65). H 65 = Be kt for some constant B. Since H(0) = 45, then B = 20. Suppose after 3 hours, the temp is 60 F. Find k. 5

6 6 Applications and modeling Remark 6.1. Temperatrue of a yam? Remark 6.2. We first consider ice forming on the surface of a lake. The rate at which ice formes is inversely propertional to the thickness of the ice sheet. Let y = y(t) represent the thickness of ice and t be time. Then dy = k y for k > 0. Solving this gives, y dy = k. Then y2 2 = kt + C so y2 = 2kt + 2C. If we assume y(0) = 0, then y = 2kt. Remark 6.3. We now consider a situation where a company is taking in revenue but simultaneously making payroll payments. We assume that the net worth of the company is increasing according to the difference of rate revenue is earned and the rate payroll payments are made. Suppose the company earns revenue at a rate of 2% of its net worth. This makes sense, since companies with a high net worth can expand and take in more revenue. Suppose payroll obligations are $100 million a year. Let W = W (t) represent the net worth of the company in millions of dollars. Then we have the differential equation: dw We can solve this by separation, = 0.02W 100. dw = 0.02W 100 = 0.02(W 5000) dw = 0.02 W 5000 ln W 5000 = 0.02t + k W = Ce 0.02t. Suppose the initial net worth is 1 billion dollars. How many years before the company is bankrupt? 6

7 We solve for C using the initial condition, and we get C = Now we have W = e 0.02t. We want to solve for t with W = 0. We have 0 = e 0.02t e 0.02t = 5/4 0.02t = ln(5/4) ln(5) ln(4) t = 11.2 years

8 7 The Logistic Model Example 7.1. Zombies! A zombie outbreak has struck the UCSD campus! The rate of transfer of the zombie virus is proportional to the number of students infected and the number not infected. We can model this situation with the differential equation = kp (N P ), where N is the number of students (approx 30,000) and k is the growth rate. Initially, 1000 students (out of 30,000) are infected. Thus, P 0 = Suppose the virus is spreading at a continuous rate of 10% and hour. How long until the entire student body is infected? Let s change units, so that the population is out of thousands. Then N = 30 and P 0 = 1. Given the DE =.1P (30 P ). Separation gives P (30 P ) =.1. Using partial fraction decomposition, we have ( ) P + 1 =.1 30 P 1 (ln P ln 30 P ) =.1t + C 30 ln 30 P P =.1t C 30 P = ±e C e.1t. P Let A = ±e C, then 30 P = Ae.1t. P We have P = 1 when t = 0, so A = 29. Thus, and so 30 P = e.1t P = e.1t. 8

9 Note that, as t, P 30. Thus, L is considered the limiting value. We could ask: How long until half the student body is infected? That is, at what time t will P = 15? Solving for t gives t = ln(29) hrs. Remark 7.2. Our initial model for population growth was the DE: = kp, k > 0. A better model for population growth is given by = r(p )P where r(p ) is the variable growth rate. Population growth rate typically decreases as the population grows. Setting r(p ) = k ap, a, k > 0 gives the Logistic Equation = (k ap )P. Letting L = k/a we obtain the conventional form ( = kp 1 P ). L We call k the initial growth rate and L the carrying capacity (limiting value). Using a method similar to the previous example, we get that the general solution to the logistic differential equation: ( = kp 1 P ) L is the logistic function P = L 1 + Ae kt with A = L P 0 P 0. 9

10 Example 7.3. A certain piece of dubious information about the cancellation of final exams began to spread one day on a college campus with a population of 80,000 students. Assume that initially one thousand students heard the rumor on the radio. Within a day 10,000 students had heard the rumor. Assume that the increase of the number x (in thousands) who had heard the rumor is proportional to the number of people who have heard the rumor and the number of people who have not heard the rumor. Determine x(t). Let t be days, then x(0) = 1 and x(1) = 10. The carrying capacity is L = 80, and we must determine the rate r. Then we have dx (1 = r x ) x. 80 Separating variables we get dx ( ) 1 x = r. 80 x Using partial fraction decomposition we have ( 1 x 1/80 ) 1 x dx = r. 80 Integration gives ln x ln 1 x = rt + C ln x 80 1 x = rt + C 80 x 1 x 80 = ke rt. The initial condition gives that 1 = x(0) = k, so k = 1. To solve for r, we use x(1) = 10 to get = e r 80 7 = er r = ln(80/7). Thus our solution is x 1 x 80 = 80 7 et. 10

11 Remark 7.4. Let s review how to solve the Logistic model DE and then consider additional applications of the Logistic equation. We have ( = kp 1 P ) = k P (L P ). L L Separating the variables, we have P (L P ) = k L. Using partial fraction decomposition, we get, ( 1 1 L P + 1 ) = L P k L. The 1 L cancel, and so we have ln P ln L P = kt + C ln L P ln P = kt C ln L P P = kt C L P = ±e C e kt P L P = 1 + Ae kt P = L 1 + Ae kt Definition 7.5. Recall the heating/cooling problems from the previous section. The general solution to these was H = T + Ae kt. As t, H T. We call T the equilibrium solution. A similar situation occurs in the Logistic model. As t, P L, so L is an equilibrium solution (also called the limiting value). 11

12 Example 7.6. Let P be the number of animals in a colony at time t. The growth of the population satisfies 1000 P = 100 P. The initial population is 200 individuals. Will there ever be more than 200 or less than 100 individuals in the colony? We rewrite the equation as = (100 P )P = 1 10 Thus, k =.1 and L = 100. Solving, we get P = To determine A, we use the formula, A = L P 0 P 0 = Ae.1t ( 1 P ) P. 100 = 1 2. Thus, 100 P = 1.5e.1t. Now we ask: does there exist a t such that P > 200? Solving for this gives e.1t > 1. At t = 0, this is an equality, but then the exponential function is decreasing so it can never be greater. What about P < 100? Solving for this gives e.1t < 0 which is impossible. Remark 7.7. A related equation is the Threshold Equation, based on the model that, if a population falls below a certain capacity (the threshold level T ) it will die off. The equation is given by dy (1 = r y ) y. T This differential equation can be solved in a manner similar to the Logistic equation. 12