Solution for each chapter and Solution manual For Elementary Differential Equations 12e Boyce
- 2. Soluon manual For Elementary Differenal Equaons 12e Boyce Hello all , We have all what you need with best price Our email : storetestbanks@gmail.com Our website : testbanks-store.com
- 3. 1.1 CHAPTER 1 Introduction 1. For > 3∕2, the slopes are negative, therefore the solutions are decreasing. For < 3∕2, the slopes are positive, hence the solutions are increasing. The equilibrium solution appears to be () = 3∕2, to which all other solutions converge. 2. For > 3∕2, the slopes are positive, therefore the solutions increase. For < 3∕2, the slopes are negative, therefore, the solutions decrease. As a result, diverges from 3∕2 as → ∞ if (0) 3∕2. For > −1∕2, the slopes are negative, therefore the solutions decrease. For < −1∕2, the slopes are positive, therefore, the solutions increase. As a result, → −1∕2 as → ∞. 3. 1
- 4. 2 CHAPTER 1 Introduction 4. For > −1∕2, the slopes are positive, and hence the solutions increase. For < −1∕2, the slopes are negative, and hence the solutions decrease. All solutions diverge away from the equilibrium solution () = −1∕2. 5. For all solutions to approach the equilibrium solution () = 2∕3, we must have ′ < 0 for > 2∕3, and ′ > 0 for < 2∕3. The required rates are satisfied by the differential equation ′ = 2 − 3. 6. For solutions other than () = 2 to diverge from = 2, () must be an increasing func- tion for > 2, and a decreasing function for < 2. The simplest differential equation whose solutions satisfy these criteria is ′ = − 2. 7. For = 0 and = 4 we have ′ = 0 and thus = 0 and = 4 are equilibrium solutions. For > 4, ′ < 0 so if (0) > 4 the solution approaches = 4 from above. If 0 < (0) < 4, then ′ > 0 and the solutions “grow” to = 4 as →∞. For (0) < 0 we see that ′ < 0 and the solutions diverge from 0. 8. Note that ′ = 0 for = 0 and = 5. The two equilibrium solutions are () = 0 and () = 5. Based on the direction field, ′ > 0 for > 5; thus solutions with initial values greater than 5 diverge from the solution () = 5. For 0 < < 5, the slopes are negative, and hence solu- tions with initial values between 0 and 5 all decrease toward the solution () = 0. For < 0, the slopes are all positive; thus solutions with initial values less than 0 approach the solution () = 0.
- 5. 1.1 3 9. Since ′ = 2 , = 0 is the only equilibrium solution and ′ > 0 for all . Thus → 0 if the initial value is negative; diverges from 0 if the initial value is positive. 10. Observe that ′ = 0 for = 0 and = 2. The two equilibrium solutions are () = 0 and () = 2. Based on the direction field, ′ > 0 for > 2; thus solutions with initial values greater than 2 diverge from () = 2. For 0 < < 2, the slopes are also positive, and hence solutions with initial values between 0 and 2 all increase toward the solution () = 2. For < 0, the slopes are all negative; thus solutions with initial values less than diverge from the solution 0 () = 0. 11. -() ′ = 2 − . 12. From Figure 1.1.6 we can see that = 2 is an equilibrium solution and thus (c) and (j) are the only possible differential equations to consider. Since ∕ > 0 for > 2, and ∕ < 0 for < 2 we conclude that (c) is the correct answer: ′ = − 2. 13. -() ′ = −2 − . 14. -() ′ = 2 + . 15. From Figure 1.1.9 we can see that = 0 and = 3 are equilibrium solutions, so (e) and (h) are the only possible differential equations. Furthermore, we have ∕ < 0 for > 3 and for < 0, and ∕ > 0 for 0 < < 3. This tells us that (h) is the desired differential equation: ′ = (3 − ). 16. -() ′ = ( − 3). 17. (a) Let () denote the amount of chemical in the pond at time . The amount will be measured in grams and the time will be measured in hours. The rate at which the chemical is entering the pond is given by 300 gal/h ⋅ .01 g/gal = 3 g/h. The rate at which the chemical leaves the pond is given by 300 gal/h ⋅ ∕106 g/gal = (3 × 10−4 ) g/h. Thus the differential equation is given by ∕ = 3 − (3 × 10−4 ). (b) The equilibrium solution occurs when ′ = 0, or = 104 grams. Since ′ > 0 for < 104 g and ′ < 0 for > 104 g, all solutions approach the equilibrium solution independent of the amount present at = 0. (c) Let () denote the amount of chemical in the pond at time . From part (a) the function () satisfies the differential equation ∕ = 3 − (3 × 10−4 ). Thus in terms of the concentration () = ()∕106 , ∕ = (1∕106 )(∕) = (1∕106 )(3 − (3 × 10−4 )) = (3 × 10−6 ) − (10−6 )(3 × 10−4 ) = (3 × 10−6 ) − (3 × 10−4 ).
- 6. 4 CHAPTER 1 Introduction 18. The surface area of a spherical raindrop of radius is given by = 42 . The volume of a spherical raindrop is given by = 43 ∕3. Therefore, we see that the surface area = 2∕3 for some constant . If the raindrop evaporates at a rate proportional to its surface area, then ∕ = −2∕3 for some > 0. 19. The difference between the temperature of the object and the ambient temperature is − 70 ( in ◦ F). Since the object is cooling when > 70, and the rate constant is = 0.05 min−1 , the governing differential equation for the temperature of the object is ∕ = −.05 ( − 70). 20. (a) Let () be the total amount of the drug (in milligrams) in the patient’s body at any given time (hr). The drug enters the body at a constant rate of 500 mg/hr. The rate at which the drug leaves the bloodstream is given by 0.4 (). Hence the accumulation rate of the drug is described by the differential equation ∕ = 500 − 0.4 (mg/hr). (b) Based on the direction field, the amount of drug in the bloodstream approaches the equilib- rium level of 1250 mg (within a few hours). 21. (a) Following the discussion in the text, the differential equation is (∕) = − 2 , or equivalently, ∕ = − 2 ∕. (b) After a long time, ∕ ≈ 0. Hence the object attains a terminal velocity given by √ ∞ = ∕ . (c) Using the relation ∞ 2 = , the required drag coefficient is = 2∕49 kg/s. (d) 22. All solutions become asymptotic to the line = − 3 as → ∞.
- 7. 1.2 5 23. The solutions approach −∞, 0, or ∞ depending upon the value of (0). 24. The solutions approach −∞, ∞, or oscillate depending upon the value of (0). 25. For all (0), there is a number that depends on the value of (0) such that () = 0 and the solution does not exist for > . 1. (a) The differential equation can be rewritten as 5 − = . Integrating both sides of this equation results in − ln |5 − | = + 1, or equivalently, 5 − = − . Applying the initial condition (0) = 0 results in the specification of the constant as = 5 − 0. Hence the solution is () = 5 + (0 − 5)− . 1.2
- 8. 6 CHAPTER 1 Introduction All solutions appear to converge to the equilibrium solution () = 5. (b) The differential equation can be rewritten as 5 − 2 = . Integrating both sides of this equation results in −(1∕2) ln |5 − 2| = + 1, or equivalently, 5 − 2 = −2 . Applying the initial condition (0) = 0 results in the specification of the constant as = 5 − 20. Hence () = 5∕2 + (0 − 5∕2)−2 . All solutions appear to converge to the equilibrium solution () = 5∕2, at a faster rate than in part (a). (c) Rewrite the differential equation as 10 − 2 = . Integrating both sides of this equation results in −(1∕2) ln |10 − 2| = + 1, or equivalently, 5 − = −2 . Applying the initial condition (0) = 0 results in the specification of the constant as = 5 − 0. Hence () = 5 + (0 − 5)−2 . All solutions appear to converge to the equilibrium solution () = 5, at a faster rate than in part (a), and at the same rate as in part (b). 2. (a) The differential equation can be rewritten as − 5 = .
- 9. 1.2 7 Integrating both sides of this equation results in ln | − 5| = + 1, or equivalently, − 5 = . Applying the initial condition (0) = 0 results in the specification of the constant as = 0 − 5. Hence the solution is () = 5 + (0 − 5) . All solutions appear to diverge from the equilibrium solution () = 5. (b) Rewrite the differential equation as 2 − 5 = . Integrating both sides of this equation results in (1∕2) ln |2 − 5| = + 1, or equivalently, 2 − 5 = 2 . Applying the initial condition (0) = 0 results in the specification of the constant as = 20 − 5. So the solution is () = (0 − 5∕2)2 + 5∕2. All solutions appear to diverge from the equilibrium solution () = 5∕2. (c) The differential equation can be rewritten as 2 − 10 = . Integrating both sides of this equation results in (1∕2) ln |2 − 10| = + 1, or equivalently, − 5 = 2 . Applying the initial condition (0) = 0 results in the specification of the con- stant as = 0 − 5. Hence the solution is () = 5 + (0 − 5)2 . All solutions appear to diverge from the equilibrium solution () = 5.
- 10. 8 CHAPTER 1 Introduction 3. (a) Rewrite the differential equation as − = , which is valid for ∕. Integrating both sides results in −(1∕) ln | − | = + 1, or equivalently, − = − . Hence the general solution is () = ( − − )∕. Note that if = ∕, then ∕ = 0, and () = ∕ is an equilibrium solution. (b) (c) (i) As increases, the equilibrium solution gets closer to () = 0, from above. The con- vergence rate of all solutions is . As solution increases, the solutions converge to the equilibrium quicker. (ii) As increases, then the equilibrium solution () = ∕ also becomes larger. In this case, the convergence rate remains the same. (iii) If and both increase but ∕ =constant, then the equilibrium solution () = ∕ remains the same, but the convergence rate of all solutions increases. 4. (a) The equilibrium solution satisfies the differential equation ∕ = 0. Setting − = 0, we obtain () = ∕. (b) Since ∕ = ∕, it follows that ∕ = ( + ) − = . 5. (a) Rewrite Eq.(ii) as (1∕)∕ = and thus ln || = + , or 1 = . (b) If = 1() + , then ∕ = 1∕. Substituting both these into Eq.(i) we get 1∕ = (1 + ) − . Since 1∕ = 1, this leaves − = 0 and thus = ∕. Hence = 1() + ∕ is the solution to Eq(i). (c) Substitution of 1 = shows this is the same as that given in Eq.(17). 6. (a) Consider the simpler equation 1∕ = −1. As in the previous solutions, rewrite the equation as (1∕1)1 = − . Integrating both sides results in 1() = − . (b) Now set () = 1() + , and substitute into the original differential equation. We find that −1 + 0 = −(1 + ) + . That is, − + = 0, and hence = ∕. (c) The general solution of the differential equation is () = − + ∕. This is exactly the form given by Eq.(17) in the text. Invoking an initial condition (0) = 0, the solution may also be expressed as () = ∕ + (0 − ∕)− . 7. (a) The general solution is () = 900 + ∕2 , that is, () = 900 + (0 − 900)∕2 . With 0 = 850, the specific solution becomes () = 900 − 50∕2 . This solution is a decreasing expo- nential, and hence the time of extinction is equal to the number of months it takes, say , for the population to reach zero. Solving 900 − 50∕2 = 0, we find that = 2 ln(900∕50) ≈ 5.78 months. (b) The solution, () = 900 + (0 − 900)∕2 , is a decreasing exponential as long as 0 < 900. Hence 900 + (0 − 900)∕2 = 0 has only one root, given by 900 = 2 ln months. 900 − 0
- 11. 1.2 9 (c) The answer in part (b) is a general equation relating time of extinction to the value of the initial population. Setting = 12 months, the equation may be written as 900∕(900 − 0) = 6 , which has solution 0 ≈ 897.8. Since 0 is the initial population, the appropriate answer is 0 = 898 mice. 8. (a) The solution to this problem is given by Eq.(26), which has a limiting velocity of 49 m/s. Substituting = 48.02 (which is 98% of 49) into Eq.(26) yields ∕5 = 0.02. Solving for we have = −5 ln(0.02) = 19.56 s. (b) Use Eq.(29) with = 19.56. 9. (a) If the drag force is proportional to 2 then = 98 − 2 is the net force acting on the falling mass ( = 10 kg). Thus 10∕ = 98 − 2 , which has a limiting velocity of 2 = 98∕. Setting 2 = 492 gives = 98∕492 and hence ∕ = (492 − 2 )∕(10∕) = (492 − 2 )∕245. (b) From part (a) we have ∕(492 − 2 ) = (1∕245), which after integration yields (1∕49) tanh −1 (∕49) = ∕245 + 0. Setting = 0 and = 0, we have 0 = 0 + 0, or 0 = 0. Thus tanh −1 (∕49) = ∕5, or () = 49 tanh(∕5) m/s. (c) (d) In the graph of part (c), the solution for the linear drag force lies below the solution for the quadratic drag force. This latter solution approaches equilibrium faster since, as the velocity increases, there is a larger drag force. (e) Note that ∫ tanh() = ln(cosh ) + . The distance the object falls is given by = 245 ln(cosh(∕5)). (f) Solving 300 = 245 ln(cosh(∕5)) gives ≈ 9.48 s. 10. (a,b) The general solution of the differential equation is () = − . Given that (0) = 100 mg, the value of the constant is given by = 100. Hence the amount of thorium-234 present at any time is given by () = 100 − . Furthermore, based on the hypothesis, setting = 1 results in 82.04 = 100 − . Solving for the rate constant, we find that = − ln(82.04∕100) ≈ .19796/week or ≈ .02828/day. (c) Let be the time that it takes the isotope to decay to one-half of its original amount. From part (a), it follows that 50 = 100 − , in which = .19796/week. Taking the natural logarithm of both sides, we find that ≈ 3.5014 weeks or ≈ 24.51 days. 11. The general solution of the differential equation ∕ = − is () = 0− , in which 0 = (0) is the initial amount of the substance. Let be the time that it takes the sub- stance to decay to one-half of its original amount, 0. Setting = in the solution, we have 0.5 0 = 0− . Taking the natural logarithm of both sides, it follows that − = ln(0.5) or = ln 2. 12. (a) Rewrite the differential equation as ∕( − ) = − and then integrate to find ln | − | = − + . Thus − = ±− . For = 0 we have 0 − = ± and thus () = + (0 − )− . (b) Set () − = (0 − )∕2 in the solution in part (a). We obtain 1∕2 = − , or = ln 2.
- 12. 10 CHAPTER 1 Introduction 13. (a) Rewrite the differential equation as ∕( − ) = −(1∕), thus, upon integrat- ing and simplifying, we get = −∕ + . (0) = 0 implies that the solution of the differential equation is () = (1 − −∕ ). (b) As → ∞, the exponential term vanishes, and the limiting value is = . (c) In this case ∕ + ∕ = 0. (1) = . The solution of this differential equation is () = −∕ , so (1) = −1∕ = , or = 1∕ . Thus () = 1∕ −∕ = −(−1)∕ for ≥ 1. 14. (a) The accumulation rate of the chemical is (0.01)(300) grams per hour. At any given time , the concentration of the chemical in the pond is ()∕106 grams per gallon. Consequently, the chemical leaves the pond at a rate of (3 × 10−4 )() grams per hour. Hence, the rate of change of the chemical is given by = 3 − 0.0003 () g/hr. Since the pond is initially free of the chemical, (0) = 0. (b) The differential equation can be rewritten as ∕(10000 − ) = 0.0003 . Integrating both sides of the equation results in − ln |10000 − | = 0.0003 + . Taking the exponential of both sides gives 10000 − = −0.0003 . Since (0) = 0, the value of the constant is . = 10000 Hence the amount of chemical in the pond at any time is () = 10000(1 − −0.0003 ) grams. Note that 1 year = 8760 hours. Setting = 8760, year is the amount of chemical present after one (8760) ≈ 9277.77 grams, that is, 9.27777 kilograms. (c) With the accumulation rate now equal to zero, the governing equation becomes ∕ = −0.0003 () g/hr. Resetting the time variable, we now assign the new initial value as (0) = 9277.77 grams. (d) The solution of the differential equation in part (c) is () = 9277.77 −0.0003 . Hence, one year after the source is removed, the amount of chemical in the pond is (8760) ≈ 670.1 grams. (e) Letting be the amount of time after the source is removed, we obtain the equation 10 = 9277.77 −0.0003 . Taking the natural logarithm of both sides, −0.0003 = ln(10∕9277.77) or ≈ 22, 776 hours ≈ 2.6 years.