Chapter 3 Modeling with First-Order Differential Equations 3.1 Linear Models 2014F/M3630 1. (Population dynamics) The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t. After 3 hours, it is observed that 400 bacteria are present. After 9 hours, 2,000 bacteria are present. What was the initial number of bacteria? (Answer) Let P (t) be the number of bacteria after t hours. Then dP = kP , P (3) = 400, P (9) = 2, 000 dt √ √ , and solution to this differential equation is P (t) = 80 5et ln(5)/6 = 80 5 · (5)t/6 . √ Answer is P (0) = 80 5 counts . √ Graph of P (t) = 80 5et ln(5)/6 2. (Radioactive decay) Initially 100 mg of radioactive substance was present. After 6 hours, the mass decreased by 3%. If the rate of decay is proportional to the amount of the substance present at the time, (a) find the half-life of the radioactive. ( half life : the amount of time it takes for a unstable chemical substance to reduce its amount present by half) (b) find the amount remaining after 24 hours. (Answers) Let x(t) be the amount of the radioactive substance remaining after t hours. Then dx = kx, x(0) = 100, x(6) = 97 dt , and solution to this differential equation is x(t) = 100et ln(0.97)/6 . (a) The half life of the substance is the solution t∗ of the equation ∗ ln(0.97)/6 100et and that is t∗ = = 50 6 ln(0.5) . ln(0.97) (b) x(24) = 100e3 ln(0.97) = 100(0.97)3 Graph of x(t) = 100et ln(0.97)/6 3. (Newton’s law of cooling and warming) A thermometer reading 70o F is placed in an oven preheated to a constant temperature. Through a glass window in the oven door, an observer records that the thermometer reads 110o F after 1/2 minute, and 145o F after 1 min. How hot is the oven? (Answer) Let T (t) be the temperature of the thermometer, Tm be the temperature in the oven which is a constant. Then dT = k(T − Tm ) dt for some constant k. It is a separable type differential equation, and answer is T (t) = Tm + Cekt for some constant C.    T (1/2) = 110    T (1) = 145      T (0) = 70 Due to given conditions, : Tm + Cek/2 = 110 From (3), C = 70 − Tm , and putting this into (1) and (2) yields   Tm + (70 − Tm )ek/2 = 110 (4) Tm + (70 − Tm )ek = 145 (5) 110 − Tm k/2 (4) gives e = , and substituting this into (5) gives 70 − Tm (1) : Tm + Cek = 145 (2) : Tm + C = 70 (3)  Tm = 390F o . 1 4. (Mixture) 1 2 lb/gal of salt is pumped into the tank at a rate of 6 gal/min. The well-mixed solution is then pumped out at a slower rate of 4 gal/min. Find the number of gallons of salt in the tank after 30 minutes. A large tank is partially filled with 100 gals of fluid in which 10 pounds of salt is dissolved. Brine containing (Answer) Let x(t) be the amount of salt(lbs) in the tank after t hours. Then dx dt = input rate − output rate = 6· 1 x(t) −4· 2 100 + 2t and x(0) = 10. This is a first order linear differential equation x0 + R 2 x = 3, and an integrating factor is µ(t) = e 1/(t+50)dt = (t+50)2 . t + 50 Using the integrating factor, we derive the solution Graph of x(t) = t + 50 − 100, 000 (t + 50)2 C , x(0) = 10 (t + 50)2 100, 000 x(t) = t + 50 − (t + 50)2 x(t) = t + 50 + The answer is x(30) = 80 − 100, 000/802 = 64.375 pounds. Section 3.2 Nonlinear Models 2014F/M3630 1. (Logistic Model) Suppose a virus-caused disease is being spread in a town of large population N (constant). Let P (t) be the population of the virus contracted people. We call dP dt P ; relative growth rate The relative growth rate decreases as P (t) increases because when P (t) is big, there are less people to be contracted, thereby slowing down the rate dP/dt. Based on this theory and assuming that the relation between and P is linear, we have the following the logistic model for population dynamics : dP = P (a − bP ) dt dP dt P (0.0.1) (e.g. 1) The number N (t) of people in a community who are exposed to a particular advertisement after t days from the release of the advertisement is governed by the logistic equation (1). Initially, N (0) = 500 and it is observed that N (1) = 1000. Also, it is predicted that the number of people who will see the advertisement will eventually approach 50,000 but it cannot exceed it. Find the function N (t). − ln(50 − N ) + ln(N ) = 50kt + C; (Answer) Let the unit for N be a thousand. Then we are given ln N = 50kt + C; 50 − N N 50kt+C 50kt =e = Ae ; N = 50Ae50kt − Ae50kt N ; 50 − N 50kt 50Ae 50e0.7t ; N (t) = N (t) = 1 + Ae50kt 99 + e0.7t dN = N (a − bN ), N (0) = .5, N (1) = 1 dt Because we have an equilibrium solution N (t) = 50, dN = kN (50 − N ), N (0) = .5, N (1) = 1 dt This is a separable type. dN 1 1 1 = kdt; + dN = kdt (0 < N < 50); N (50 − N ) 50 50 − N N Z Z 1 1 dN = 50kdt; + 50 − N N (Graph of N (t) = 2 50e0.7t . Notice N (t) → 50 as t → ∞) 99 + e0.7t