Even-Numbered Homework Solutions

Even-Numbered Homework Solutions Chapter 1 1.1 8. Using the decay-rate parameter you computed in 1.1.7, determine the time since death if: (a) 88% of the original C-14 is still in the material The decay-rate parameter from 1.1.7 is λ = ln2. Then use the following equation to determine the time of 5230 death: ln2 0.88 = e 5230 t ln(0.88) = ln2 t t = 964.54 years 5230 (b) 12% of the original C-14 is still in the material As above: ln2 0.12 = e 5230 t ln(0.12) = ln2 t t = 15998.01 years 5230 14. Suppose two students memorize lists according to the model dl = 2(1 L). (a) If one of the students knows one-half of the list at time t = 0 and the other knows none of the list, which student is learning more rapidly at this instant? First, determine dl for each student. Call the student who knows one-half of the list Student A and the other Student B. For Student A, dl dl = 2(1 0.5) = 2(0.5) = 1. For Student B, = 2(1 0) = 2(1) = 2. So, Student B is learning more rapidly at this instant. (b) Will the student who starts out knowing none of the list ever catch up to the student who starts out knowing one-half of the list? Student B will never catch up to Student A with the given model. 16. Using the table given in 1.1.16: (a) Let s(t) = s 0 e kt be an exponential function. Show that the graph of ln s(t) as a function of t is a line. What is its slope and vertical intercept? Take the natural log of both sides of the equation: ln(s(t)) = ln(s 0 e kt ) = ln(s 0 )+ln(e kt ) = kt+ln(s 0 ). This equation is in point-slope form, so the slope is k and the vertical intercept is y = ln(s 0 ). Since s 0 = 5669, y = 8.64 is the vertical intercept. (b) Is spending on education in the U.S. rising exponentially fast? If so, what is the growth-rate coefficient? In the year 1910, t = 10 and s 10 = 10081. Then ln(s(10)) = ln(10081) = k(10)+ln(5669). So: 10k = ln(10081) ln(5669) k = 0.0576. Yes, the spending on education is rising exponentially fast, and the growth-rate coefficient is k = 0.0575. (Note that your k may be different, depending on which s t you used to determine it.) 1

18. Suppose that the growth-rate parameter k = 0.3 and the carrying capacity N = 2500 in the logistic population model of Exercise 17. Suppose P(0) = 2500. 1.2 (a) If 100 fish are harvested each year, what does the model predict for the long-term behavior of the fish population? In other words, what does a qualitative analysis of the model yield? If 100 fish are harvested every year, then our model is dp ( = 0.3P 1 P 2500 Set dp then dp around 2104. dp = 0 to get critical values at P 396 and P 2104. If 0 < P < 396, then dp > 0. If P > 2104, then ) 100 = 0.00012P 2 +0.3P 100. < 0. If 396 < P < 2104 < 0. Since P(0) = 2500, the fish population will decrease until it levels off (b) If one-third of the fish are harvested each year, what does the model predict for the long-term behavior of the fish population? If one-third of the fish are harvested every year, then our model is dp ( = 0.3P 1 P ) 1 P. The only 2500 3 nonnegative critical value is P = 0. If P > 0, then dp < 0. So the population will decrease until the fish are extinct. 6. Find the general solution of the differential equation: This is a separable differential equation. So: = t4 y 1 y = t4 1 y = t 4 = t4 y. ln(y) = 1 5 t5 +C, where C is a constant (and all constants are lumped together on right) y = e 1 5 t5 +C y = e c e 1 5 t5 y = Ke 1 5 t5, where e C = K, a constant. 30. Solve the initial value problem: = t y t 2. y(0) = 4. y Note that this is a separable differential equation. So: = t y t 2 y = t y(1 t 2 ) t y = (1 t 2 ) y = t (1 t 2 ) 1 2 y2 = 1 2 ln( 1 t2 )+C y 2 = ln( 1 t 2 )+C y = ln( 1 t 2 )+C. The initial condition is y(0) = y 0 = 4. So, 4 = ln( 1 0 )+C = 0+C, so C = 16. Then our final solution is y = ln( 1 t 2 )+16. 2

32. Solve the given initial-value problem: = ty2 +2y 2, y(0) = 1. Simplify to = y2 (t+2). Then note that this is a separable differential equation. So: = y2 (t+2) 1 = t+2 y2 1 y = 1 2 t2 +2t+C 2 y = t2 +4t+C y = 2 t 2 +4t+C. The initial condition is y(0) = 1. So, 1 = 34. Solve the given initial-value problem: 2 2 0 2, so C = 2. Then our final solution is y = +4(0)+C 3 t 2 +4. = 1 y2, y(0) = 2. y Note that this is a separable differential equation. So: = 1 y2 y y 1 y2 = 1 2 ln(1 y2 ) = t+c ln(1 y 2 ) = 2t+C 1 y 2 = Ce 2t y 2 = 1 Ce 2t The initial condition is y(0) = 2. So, ( 2) 2 = 4 = 1 Ce 2(0) = 1 C, so C = 3. Then our final solution is y = 1+3e 2t. 40. (a) Suppose N = 200, k 1 = 0.1, k 2 = 0.1, E = 400 and C(0) = 150. What will the person s cholesterol level be after 2 days on this diet? We have dc = k 1 1(C 0 C)+k 2 E k 1 (C 0 C)+k 2 E dc = 1 k 1 ln k 1 (C 0 C)+k 2 E = t+b, where B is a constant of integration ln k 1 (C 0 C)+k 2 E = k 1 (t+b) k 1 (C 0 C)+k 2 E = Be k1t k 1 C = Be k1t k 2 E k 1 C 0 C(t) = 1 k 1 (Be k1t k 2 E k 1 C 0 ). Now we can plug in the parameters and initial condition to find B: C(0) = 150 = 1 (B 0.1(400) 0.1(200)) B = 45. 0.1 So, C(t) = 10(45e 0.1t 60), which means that C(2) = 231.57. (b) With the initial conditions as above, what will the person s cholesterol level be after 5 days on this diet? C(5) = 10(45e 0.5 60) = 327.06. (c) What will the person s cholesterol level be after a long time on this diet? Simplify C(t) to C(t) = 450 e 0.1t +600. As t, 450 e 0.1t 0, so after a long time on this diet the person s cholesterol level will be 600. 3

(d) Let E = 100. The initial cholesterol level at the starting time of the new diet is the result of part (c). What will the person s cholesterol level be after 1 day on the new diet, after 5 days on the new diet, and after a very long time on the new diet? Find our new B: C(0) = 600 = 1 (B 0.1(100) 0.1(200)) B = 30. 0.1 Then C(1) = 10( 30e 0.1t 30) = 300(e 0.1t +1) = 300(e 0.1 +1) = 571.45 and C(5) = 300(e 0.5 +1) = 481.96. As t, C(t) 300. (e) Change k 2 to k 2 = 0.075. With the cholesterol level from part (c), what will the person s cholesterol level be after 1 day, after 5 days, and after a very long time? Find our new B: C(0) = 600 = 1 (B 0.075(400) 0.1(200)) B = 10. 0.1 Then C(1) = 100e 0.1t +500 = 100e 0.1 +500 = 590.48 and C(5) = 100e 0.5 +500 = 560.65. As t, C(t) 500. 42. Let Q(t) represent the number of teaspoons of hot sauce at time t in minutes. Then Q(0) = 12. Note that dq represents the amount of hot sauce per cup at time t, so dq = Q 32, since 1 of the total amount of hot sauce 32 will be in each cup, and there are 32 cups in 2 gallons. This is a separable differential equation: dq = Q 32 1 1 dq = Q 32 ln(y) = t 32 +C y = Ke t 32. Now we utilize our initial condition: 12 = Ke 0 = K. So, our equation is y = 12e t 32. We are trying to dilute the chili to the intended number ( ) of teaspoons, which is 4. So, 1 4 = 12e t 32 ln = t t = 35.155 minutes. 3 32 Thus, it takes approximately 35.155 minutes and 36 cups (since one cup per minute is taken from the pot) to properly dilute the chili. 4

1.3 8. = 2y t 10. = (t+1)y 5

12. Suppose the constant function y(t) = 2 for all t is a solution of the differential equation = f(t,y). (a) What does this tell you about the function f(t,y)? At y = 2, f(t,y) = 0 for all t. (b) What does this tell you about the slope field? y = 2 has slope 0. (c) What does this tell you about solutions with initial conditions y(0) 2? Solutions with initial conditions y(0) 2 will not cross the line y = 2. 16. Determine the equation that corresponds to each slope field. (a) iii (b) viii (c) v (d) vi 1.4 6. dw = (3 w)(w +1), w(0) = 0, 0 t 5, t = 0.5 k t k w k m k 0 0 0 787 1 0.5 1.5 5415 2 1.0 3.38 7507 3 1.5 2.55 7560 4 2.0 3.35 6344 5 2.5 2.59 6344 6 3.0 3.32 6344 7 3.5 2.62 6344 8 4.0 3.31 6344 9 4.5 2.65 6344 10 5.0 3.29 6344 6

8. = e2/y, y(1) = 2, 1 t 3, t = 0.5 k t k w k m k 0 1 2 787 1 1.5 3.35 5415 2 2.0 4.27 7507 3 2.5 5.07 7560 4 3.0 5.81 6344 1.5 10. Consider the differential equation = 2 y. (a) Show that y(t) = 0 for all t is an equilibrium solution. = 2 y 2 y = 0 y = 0 y(t) = 0 (b) Find all solutions. Case 1: y > 0 y > 0 y = y = 2 y Case 2: y < 0 1 2 y = 1 2 (2) y = t+c, where C is a constant y = t+c y = (t+c) 2 y > 0 y = y = 2 y 1 2 y = y = t+d, where D is a constant y = t D y = ( t D) 2 = (t+d) 2 y = (t+d) 2 7

12. (a) Show that y 1 (t) = 1 t 1 and y 2(t) = 1 are solutions of = y2. ( ) d 1 We have: = 1 ( ) 2 1 t 1 (t 1) 2 = = y 2 t 1 1. The same steps show that y 2 is also a solution of the differential equation. (b) What can you say about the solutions of = y2 for which the initial condition y(0) satisfies the inequality 1 < y(0) < 1 2? Note that y 1 (0) = 1 0 1 = 1 and y 2(0) = 1 y 1 (0) < y(0) < y 2 (0), so 1 t 1 < y(t) < 1. 0 2 = 1 2. So, 1 < y(0) < 1 2 implies that 14. = 1 (y +1)(),y(0) = 0 (a) Find a formula for the solution. = 1 (y +1)() (y +1) = 1 1 2 y2 +y = ln +C, where C is a constant Since y(0) = 0, 0 = ln 0 2 +C = ln(2)+c C = ln(2). So: 1 2 y2 +y = ln ln(2) = ln 2 Now we put all terms on the left side and multiply by two to get rid of the fraction coefficient: 1 2 y2 +y ln 2 = 0 2y 2 +2y 2ln 2 = 0 Use the quadratic formula to find a formula for y: 2± 2 2 4(1)( 2)ln 2 2± 4+8ln 2 2±2 1+2ln 2 y = = = 2(1) 2 2 = 1± 1+2ln 2 (b) State the domain of definition of the solution. Since 1+2ln 2 cannot be negative, we have: 1+2ln 2 0 ln 2 1 2 e 1 2 2 8

2e 1 2 t 2e 1 2 +2 Thus, the domain is all real numbers such that t 2e 1 2 +2. 1.6 2. Sketch the phase lines for the given differential equation. Identify the equilibrium points as sinks, sources, or nodes. 14. Note that = y2 4y 12 = (y +2)(y 6). Setting this equal to zero gives y = 2 and y = 6 as equilibrium points. If y < 2, you will find that is positive. For 2 < y < 6, is negative. For y > 6, Thus, y = 6 is a source and y = 2 is a sink. Drawing the phase line is straightforward. is positive. 14.jpg 9

34. 34.jpg 1.8 12. Solve the given initial value problem: 2y = 7e2t,y(0) = 3 Following the example on page 120, we guess that the general solution of the homogeneous equation is y(t) = ke 2t. We guess y p (t) = αe 2t, with α as the undetermined coefficient. But, like the example, we find that upon substituting y p (t) into 2y, we get zero. Instead, try y p(t) = αte 2t. The derivative of y p (t) is α(1+2t)e 2t. Then we have: 2y p = α(1+2t)e 2t 2αte 2t = αe 2t α = 7. So, y(t) = ke 2t +7te 2t. Use the initial condition to find k: The solution is thus y(t) = 3e 2t +7te 2t. y(0) = 3 3 = ke 0 k = 3. 18. Consider the nonhomogeneous linear equation = y +2. (a) Compute an equilibrium solution for this equation. = 0 0 = y +2 y = 2 Thus y = 2 is an equilibrium solution. (b) Verify that y(t) = 2 e t is a solution for this equation. = d (2 e t ) = 0 ( 1)e t = e t Now note that e t = 2 (2 e t ) = 2 y. Thus, y(t) is a solution. 22. Find the general solution and the solution that satisfies the initial condition y(0) = 0. 10

+y = t3 +sin(3t) Guess that y 1 (t) = αsin(3t)+βcos(3t) and y 2 (t) = at 3 +bt 2 +ct+d. Solve for α and β: Equating coefficients gives: +y = sin(3t) 3αcos(3t) 3βsin(3t) + αsin(3t) + βcos(3t) = sin(3t) (3α+β)cos(3t)+( 3β +α 1)sin(3t) = 0 3α+β = 0 and 3β +α 1 = 0 α = 1 10 Now we need to solve for a,b,c, and d: and β = 3 10. +y = t3 (3at 2 +2bt+c)+(at 3 +bt 2 +ct+d) t 3 = 0 (a+1)t 3 +(3a+b)t 2 +(2b+c)t+(c+d) = 0 Equating coefficients gives: a = 1, 3a+b = 0, 2b+c = 0, and c+d = 0 b = 3,c = 6, and d = 6. Now, we know that y h (t) is given by y h (t) = ke t. So, y(t) = ke t + 1 10 sin(3t) 3 10 cos(3t)+t3 3t 2 +6t 6. Our initial condition y(0) = 0 yields k = 63 10. 1.9 8. Solve the given initial-value problem. Note that µ(t) = e ( 1 = 1 t+1 y +4t2 +4t,y(1) = 10 ) t+1 = 1 by natural log rules. Then: 1+t 1 t+1 y (t+1) 2 = 4t2 +4t 1+t d ( ) y = 4t t+1 ( ) y d t+1 = 4t y t+1 = 2t2 +C, where C is a constant y = (2t 2 +C)(t+1) Using the initial value y(1) = 10, we get: 10 = (2+C)(2) = 2C +4 C = 3. 16. Determine the general solution to the equation and express it with as few integrals as possible. 11

= y +4cos(t2 ) The integrating factor is given by µ(t) = e 1 = e t. Then we have: (e t ) (e t )y = e t 4cos(t 2 ) e t y d = e t 4cos(t 2 ) e t y d = 4 cos(t 2 )e t e t y = 4 cos(t 2 )e t y = 4e t cos(t 2 )e t 12