Chapter 3. Modeling with First-Order Differential Equations

Transcription

1 Chapter 3 Moeling with First-Orer Differential Equations i

2 GROWTH AND DECAY: The initial-value problem x = kx, x(t 0 ) = x 0, (1) where k is a constant of proportionality, serves as a moel for iverse phenomena involving either growth or ecay. We saw in Section 1.3 that in biological applications the rate of growth of certain populations (bacteria, small animals) over short perios of time is proportional to the population present at time t. Knowing the population at some arbitrary initial time t 0, we can then use the solution of (1) to preict the population in the future that is, at times t > t 0. The constant of proportionality k in (1) can be etermine from the solution of the initial-value problem, using a subsequent measurement of x at a time t 1 > t 0. In physics an chemistry (1) is seen in the form of a first-orer reaction that is, a reaction whose rate, or velocity, x/ is irectly proportional to the amount x of a substance that is unconverte or remaining at time t. The ecomposition, or ecay, of U-238 (uranium) by raioactivity into Th-234 (thorium) is a first-orer reaction. EXAMPLE 1 (Bacterial Growth) A culture initially has P 0 number of bacteria. At t = 1 h the number of bacteria is measure to be 3 2 P 0. If the rate of growth is proportional to the number of bacteria P(t) present at time t, etermine the time necessary for the number of bacteria to triple. SOLUTION: We first solve the ifferential equation in (1), with the symbol x replace by P. With t 0 = 0 the initial conition is P(0) = P 0. We then use the empirical observation that P(1) = 3 2 P 0 to etermine the constant of proportionality k. Notice that the ifferential equation P/ = kp is both separable an linear. When it is put in the stanar form of a linear first-orer DE, P kp = 0, we can see by inspection that the integrating factor is e kt. Multiplying both sies of the equation by this term an integrating gives, in turn, [e kt P] = 0 an e kt P = c. Therefore, P(t) = ce kt. At t = 0 it follows that P 0 = ce 0 = c, so P(t) = P 0 e kt. At t = 1 we have 3 P 2 0 = P 0 e k or e k = 3. From the last equation we get k = ln 3 = , so P(t) = P 2 2 0e t. To fin the time at which the number of bacteria has triple, we solve 3P 0 = P 0 e t for t. It follows that t = ln 3, or See Figure 1. t = ln h. Figure 1: Time in which population triples 55

3 Exercises 3.1: Page 89 (6) Initially 100 milligrams of a raioactive substance was present. After 6 hours the mass ha ecrease by 3%. If the rate of ecay is proportional to the amount of the substance present at time t, fin the amount remaining after 24 hours. Solution: Let A = A(t) be the amount present at time t. From A = ka an A(0) = 100, we obtain, by solving the ifferential equation in the last initial-value problem by separation of variables, the following A = k A ln A = kt + c 1 A = e kt+c 1 = ce kt, where c = e c 1. Using A(0) = 100, we fin that c = 100. Thus, the solution of the initial-value problem is A = 100e kt. After 6 hours the mass ha ecrease by 3%. This means that Then, A(6) = 100e 6k 100e 6k = 97 k = 1 6 ln(0.97) A(24) = 100e 24(1 6 ln(0.97)) = 100e ln(0.97)4 = 100 (0.97) 4 = mg. Notice in Example 1 that the actual number P 0 of bacteria present at time t = 0 playe no part in etermining the time require for the number in the culture to triple. The time necessary for an initial population of, say, 100 or 1,000,000 bacteria to triple is still approximately 2.71 hours. As shown in Figure 2, the exponential function e kt increases as t increases for k > 0 an ecreases as t increases for k < 0. Thus problems escribing growth (whether of populations, bacteria, or even capital) are characterize by a positive value of k, whereas problems involving ecay (as in raioactive isintegration) yiel a negative k value. Accoringly, we say that k is either a growth constant (k > 0 ) or a ecay constant (k < 0 ). Figure 2: growth (k > 0 ) an ecay (k < 0 ) 56

4 SERIES CIRCUITS: For a series circuit containing only a resistor an an inuctor, Kirchhoff s secon law states that the sum of the voltage rop across the inuctor (L(i/)) an the voltage rop across the resistor (ir) is the same as the impresse voltage (E(t)) on the circuit. See Figure 3. Thus, we obtain the linear ifferential equation for the current i(t), L i + Ri = E(t), (2) where L an R are constants known as the inuctance an the resistance, respectively. The current i(t) is also calle the response of the system. The voltage rop across a capacitor with capacitance C is given by q(t)/c, where q is the charge on the capacitor. Hence, for the series circuit shown in Figure 4, Kirchhoff s secon law gives Ri + 1 q = E(t). (3) C But current i an charge q are relate by i = q/, so (3) becomes the linear ifferential Equation R q + 1 q = E(t). (4) C Figure 3: LR series circuit Figure 4: RC series circuit EXAMPLE 2 (Series Circuit) A 12-volt battery is connecte to a series circuit in which the inuctance is 1 2 resistance is 10 ohms. Determine the current i if the initial current is zero. henry an the SOLUTION: From (2) we see that we must solve 1 i + 10i = 12, 2 subject to i(0) = 0. First, we multiply the ifferential equation by 2 an rea off the integrating factor e 20t. We then obtain [e20t i] = 24e 20t. Integrating each sie of the last equation an solving for i gives i(t) = ce 20t. Now i(0) = 0 implies that 0 = 6 + c or c = 6. Therefore, the response is

5 i(t) = e 20t. Exercises 3.1: Pages (29) A 30-volt electromotive force is applie to an LR series circuit in which the inuctance is 0.1 henry an the resistance is 50 ohms. Fin the current i(t) if i(0) = 0. Determine the current as t. Solution: Assume that L i + Ri = E(t). Comparing to the ifferential equation, we see from the question that L = 0.1, R = 50 an E(t) = 30 Hence, we see that we must solve 0.1 i + 50i = 30 or i + 500i = 300 subject to i(0) = 0. First, we multiply the ifferential equation by 10 an rea off the integrating factor e 500t. We then obtain [e500t i] = 300e 500t. Integrating each sie of the last equation an solving for i gives i(t) = ce 500t. Now i(0) = 0 implies that 0 = 3 + c or c = 3. Therefore, the response is 5 5 Therefore, i(t) = e 500t. lim i(t) = lim (3 t t e 500t ) = (31) A 100-volt electromotive force is applie to an RC series circuit in which the resistance is 200 ohms an the capacitance is 10 4 fara. Fin the charge q(t) on the capacitor if q(0) = 0. Fin the current i(t). Solution: Assume that R q + 1 q = E(t), C where R = 200, C = 10 4, an E(t) = 100, so that 200 q + 1 q = q + 50 q = 1, Linear in q (5) 2 58

6 From this form we ientify P(t) = 50 an f(t) = 1 2 an further observe that P an f are continuous on (, ). Hence, the integrating factor is I. F. = e (50) = e 50t. Now, we multiply (5) by e 50t an rewrite [e50t q] = 1 2 e50t. q 50t e + e50t (50)q = 1 2 e50t as It follows from integration e 50t q = e50t + c, or q = ce 50t Use q(0) = 0, we have c = Hence, Since q = e 50t. i = q = 1 2 e 50t. 59

7 Exercises 3.1: Pages (1) The population of a community is known to increase at a rate proportional to the number of people present at time t. If an initial population P 0 has ouble in 5 years, how long will it take to triple? To quaruple? (4) 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 observe that 400 bacteria are present. After 10 hours 2000 bacteria are present. What was the initial number of bacteria? (32) A 200-volt electromotive force is applie to an RC series circuit in which the resistance is 1000 ohms an the capacitance is fara. Fin the charge q(t) on the capacitor if i(0) = 0.4. Determine the charge an current at t = s. Determine the charge as t. 60