MATH 307 SOLUTIONS-MANUAL-1.pdf

MATH 307 SOLUTIONS-MANUAL-1.pdf - CHAPTER 1 Chapter One...

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Unformatted text preview: —————————————————————————— CHAPTER 1. —— Chapter One Section 1.1 1. For , the slopes are negative, and hence the solutions decrease. For , the slopes are positive, and hence the solutions increase. The equilibrium solution appears to be , to which all other solutions converge. 3. For , the slopes are tive, and hence the solutions increase. For , the slopes are negative, and hence the solutions decrease. All solutions appear to diverge away from the equilibrium solution . 5. For , the slopes are tive, and hence the solutions increase. For , the slopes are negative, and hence the solutions decrease. All solutions diverge away from ________________________________________________________________________ page 1 —————————————————————————— CHAPTER 1. —— the equilibrium solution . 6. For , the slopes are tive, and hence the solutions increase. For , the slopes are negative, and hence the solutions decrease. All solutions diverge away from the equilibrium solution . 8. For all solutions to approach the equilibrium solution , we must have for , and for . The required rates are satisfied by the differential equation . 9. For solutions other than to diverge from function for , and a decreasing function for equation whose solutions satisfy these criteria is . 10. For solutions other than for , and for equation . , must be an increasing . The simplest differential to diverge from , we must have . The required rates are satisfied by the differential 12. Note that for and . The two equilibrium solutions are and . Based on the direction field, for ; thus solutions with initial values greater than diverge from the solution . For , the slopes are negative, and hence solutions with initial values between and all decrease toward the ________________________________________________________________________ page 2 —————————————————————————— CHAPTER 1. —— solution values less than . For , the slopes are all positive; thus solutions with initial approach the solution . 14. Observe that for and . The two equilibrium solutions are and . Based on the direction field, for ; thus solutions with initial values greater than diverge from . For , the slopes are also positive, and hence solutions with initial values between and all increase toward the solution . For , the slopes are all negative; thus solutions with initial values less than diverge from the solution . 16. Let any given time be the total amount of the drug in milligrams in the patient's body at . The drug is administered into the body at a constant rate of The rate at which the drug leaves the bloodstream is given by accumulation rate of the drug is described by the differential equation Hence the Based on the direction field, the amount of drug in the bloodstream approaches the equilibrium level of 18. Following the discussion in the text, the differential equation is ________________________________________________________________________ page 3 —————————————————————————— CHAPTER 1. —— or equivalently, After a long time, Using the relation Hence the object attains a terminal velocity given by , the required drag coefficient is 19. All solutions appear to approach a linear asymptote verify that is a solution. . It is easy to 20. ________________________________________________________________________ page 4 —————————————————————————— CHAPTER 1. —— All solutions approach the equilibrium solution 23. All solutions appear to diverge from the sinusoid which is also a solution corresponding to the initial value , . 25. All solutions appear to converge to . First, the rate of change is small. The slopes eventually increase very rapidly in magnitude. 26. ________________________________________________________________________ page 5 ————————————————————————— CHAPTER 1. —— The direction field is rather complicated. Nevertheless, the collection of points at which the slope field is zero, is given by the implicit equation The graph of these points is shown below: The y-intercepts of these curves are at , . It follows that for solutions with initial values , all solutions increase without bound. For solutions with initial values in the range and , the slopes remain negative, and hence these solutions decrease without bound. Solutions with initial conditions in the range initially increase. Once the solutions reach the critical value, given by the equation , the slopes become negative and remain negative. These solutions eventually decrease without bound. ________________________________________________________________________ page 6 —————————————————————————— CHAPTER 1. —— Section 1.2 1 The differential equation can be rewritten as Integrating both sides of this equation results in . Applying the initial condition the constant as . Hence the solution is , or equivalently, results in the specification of All solutions appear to converge to the equilibrium solution 1 Rewrite the differential equation as Integrating both sides of this equation results in equivalently, . Applying the initial condition the constant as . Hence the solution is , or results in the specification of All solutions appear to converge to the equilibrium solution than in Problem 1a 2 , but at a faster rate The differential equation can be rewritten as ________________________________________________________________________ page 7 —————————————————————————— CHAPTER 1. —— Integrating both sides of this equation results in . Applying the initial condition the constant as . Hence the solution is , or equivalently, results in the specification of All solutions appear to diverge from the equilibrium solution 2 Rewrite the differential equation as Integrating both sides of this equation results in . Applying the initial condition the constant as . Hence the solution is , or equivalently, results in the specification of All solutions appear to diverge from the equilibrium solution 2 . . . The differential equation can be rewritten as Integrating both sides of this equation results in . Applying the initial condition the constant as . Hence the solution is , or equivalently, results in the specification of ________________________________________________________________________ page 8 —————————————————————————— CHAPTER 1. —— All solutions appear to diverge from the equilibrium solution 3 . . Rewrite the differential equation as , which is valid for equivalently, Note that if . Integrating both sides results in . Hence the general solution is , then , and is an equilibrium solution. , or As increases, the equilibrium solution gets closer to , from above. Furthermore, the convergence rate of all solutions, that is, , also increases. As increases, then the equilibrium solution also becomes larger. In this case, the convergence rate remains the same. If and both increase but constant , then the equilibrium solution remains the same, but the convergence rate of all solutions increases. 5 . Consider the simpler equation write the equation as . As in the previous solutions, re- Integrating both sides results in Now set , and substitute into the original differential equation. We find that ________________________________________________________________________ page 9 —————————————————————————— CHAPTER 1. —— . That is, , and hence . . The general solution of the differential equation is exactly the form given by Eq. in the text. Invoking an initial condition the solution may also be expressed as This is , 6 . The general solution is , that is, . With , the specific solution becomes . This solution is a decreasing exponential, and hence the time of extinction is equal to the number of months it takes, say , for the population to reach zero. Solving , we find that months. The solution, , is a decreasing exponential as long as . Hence has only one root, given by . The answer in part is a general equation relating time of extinction to the value of the initial population. Setting months , the equation may be written as , which has solution answer is mice . . Since is the initial population, the appropriate 7 . The general solution is . Based on the discussion in the text, time is measured in months . Assuming month days , the hypothesis can be expressed as . Solving for the rate constant, , with units of per month . . days months . The hypothesis is stated mathematically as N/30 . It follows that , and hence the rate constant is given by The units are understood to be per month . 9 . Assuming no air resistance, with the positive direction taken as downward, Newton's Second Law can be expressed as in which is the gravitational constant measured in appropriate units. The equation can be ________________________________________________________________________ page 10 —————————————————————————— CHAPTER 1. —— written as initial velocity . , with solution The object is released with an . Suppose that the object is released from a height of units above the ground. Using the fact that , in which is the downward displacement of the object, we obtain the differential equation for the displacement as With the origin placed at the point of release, direct integration results in . Based on the chosen coordinate system, the object reaches the ground when . Let be the time that it takes the object to reach the ground. Then . Using the quadratic formula to solve for , The positive answer corresponds to the time it takes for the object to fall to the ground. The negative answer represents a previous instant at which the object could have been launched upward with the same impact speed , only to ultimately fall downward with speed , from a height of units above the ground. . The impact speed is calculated by substituting . into in part That is, 10 ,b . The general solution of the differential equation is Given that mg , the value of the constant is given by . Hence the amount of thorium-234 present at any time is given by . Furthermore, based on the hypothesis, setting results in Solving for the rate constant, we find that /week or /day . . Let be the time that it takes the isotope to decay to one-half of its original amount. From part , it follows that , in which /week . Taking the natural logarithm of both sides, we find that weeks or s. 11. The general solution of the differential equation is , in which is the initial amount of the substance. Let be the time that it takes the substance to decay to one-half of its original amount , . Setting in the solution, we have . Taking the natural logarithm of both sides, it follows that or ________________________________________________________________________ page 11 —————————————————————————— CHAPTER 1. —— 12. The differential equation governing the amount of radium-226 is , with solution Using the result in Problem 11, and the fact that the half-life years , the decay rate is given by per year . The amount of radium-226, after years, is therefore Let be the time that it takes the isotope to decay to of its original amount. Then setting , and , we obtain Solving for the decay time, it follows that or years . 13. The solution of the differential equation, with As , is , the exponential term vanishes, and hence the limiting value is . 14 . The accumulation rate of the chemical is grams per hour . At any given time , the concentration of the chemical in the pond is grams per gallon . Consequently, the chemical leaves the pond at a rate of grams per hour . Hence, the rate of change of the chemical is given by gm/hr . Since the pond is initially free of the chemical, . . The differential equation can be rewritten as Integrating both sides of the equation results in . Taking the natural logarithm of both sides gives . Since , the value of the constant is . Hence the amount of chemical in the pond at any time is grams . Note that year hours . Setting , the amount of chemical present after one year is grams , that is, kilograms . . With the accumulation rate now equal to zero, the governing equation becomes gm/hr . Resetting the time variable, we now assign the new initial value as grams . . The solution of the differential equation in Part is Hence, one year after the source is removed, the amount of chemical in the pond is grams . ________________________________________________________________________ page 12 —————————————————————————— CHAPTER 1. —— . Letting be the amount of time after the source is removed, we obtain the equation Taking the natural logarithm of both sides, or hours years . 15 . It is assumed that dye is no longer entering the pool. In fact, the rate at which the dye leaves the pool is kg/min gm per hour . Hence the equation that governs the amount of dye in the pool is gm/hr . The initial amount of dye in the pool is grams . . The solution of the governing differential equation, with the specified initial value, is . The amount of dye in the pool after four hours is obtained by setting . That is, grams . Since size of the pool is gallons , the concentration of the dye is grams/gallon . . Let be the time that it takes to reduce the concentration level of the dye to grams/gallon . At that time, the amount of dye in the pool is grams . Using the answer in part , we have . Taking the natural logarithm of both sides of the equation results in the required time hours . . Note that . Consider the differential equation . Here the parameter corresponds to the flow rate, measured in gallons per minute . Using the same initial value, the solution is given by In order to determine the appropriate flow rate, set and . (Recall that gm of ________________________________________________________________________ page 13 —————————————————————————— CHAPTER 1. —— dye has a concentration of gm/gal ). We obtain the equation Taking the natural logarithm of both sides of the equation results in the required flow rate gallons per minute . ________________________________________________________________________ page 14 —————————————————————————— CHAPTER 1. —— Section 1.3 1. The differential equation is second order, since the highest derivative in the equation is of order two. The equation is linear, since the left hand side is a linear function of and its derivatives. 3. The differential equation is fourth order, since the highest derivative of the function is of order four. The equation is also linear, since the terms containing the dependent variable is linear in and its derivatives. 4. The differential equation is first order, since the only derivative is of order one. The dependent variable is squared, hence the equation is nonlinear. 5. The differential equation is second order. Furthermore, the equation is nonlinear, since the dependent variable is an argument of the sine function, which is not a linear function. 7. Also, . Hence and 9. . Substituting into the differential equation, we have . Hence the given function is a solution. . Thus 10. and Clearly, is a solution. Likewise, , , , . Substituting into the left hand side of the equation, we find that . Hence both functions are solutions of the differential equation. 11. and hand side of the equation, we have . Substituting into the left Likewise, and . Substituting into the left hand side of the differential equation, we have . Hence both functions are solutions of the differential equation. 12. and . Substituting into the left hand side of the differential equation, we have 2 . Likewise, and . Substituting into the left hand side of the equation, we have 2 2 2 ________________________________________________________________________ page 15 —————————————————————————— CHAPTER 1. —— 2 2 Hence both functions are solutions of the 2 differential equation. 13. and . Substituting into the left hand side of the differential equation, we have . Hence the function is a solution of the differential equation. 15. Let . Then results in . Since The roots of this equation are 17. equation, we have equation , and substitution into the differential equation , we obtain the algebraic equation and , that is, . Substituting into the differential . Since , we obtain the algebraic . The roots are , . 18. Let . Then , and the derivatives into the differential equation, we have , we obtain the algebraic equation that . Clearly, the roots are , . Substituting . Since By inspection, it follows and 20. and . Substituting the derivatives into the differential equation, we have . After some algebra, it follows that . For , we obtain the algebraic equation The roots of this equation are and 21. The order of the partial differential equation is two, since the highest derivative, in fact each one of the derivatives, is of second order. The equation is linear, since the left hand side is a linear function of the partial derivatives. 23. The partial differential equation is fourth order, since the highest derivative, and in fact each of the derivatives, is of order four. The equation is linear, since the left hand side is a linear function of the partial derivatives. 24. The partial differential equation is second order, since the highest derivative of the function is of order two. The equation is nonlinear, due to the product on the left hand side of the equation. 25. It is evident that derivatives are and Likewise, given , the second ________________________________________________________________________ page 16 —————————————————————————— CHAPTER 1. —— 2 2 Adding the partial derivatives, 2 2 . Hence is also a solution of the differential equation. 27. Let . Then the second derivatives are It is easy to see that . Likewise, given Clearly, is also a solution of the partial differential equation. 28. Given the function It follows that Hence , we have , the partial derivatives are . is a solution of the partial differential equation. ________________________________________________________________________ page 17 —————————————————————————— CHAPTER 1. —— 29 . The path of the particle is a circle, therefore polar coordinates are intrinsic to the problem. The variable is radial distance and the angle is measured from the vertical. Newton's Second Law states that F a In the tangential direction, the equation of motion may be expressed as , in which the tangential acceleration, that is, the linear acceleration along the path is is positive in the direction of increasing . Since the only force acting in the tangential direction is the component of weight, the equation of motion is Note that the equation of motion in the radial direction will include the tension in the rod . . Rearranging the terms results in the differential equation ________________________________________________________________________ page 18 —————————————————————————— CHAPTER 2. —— Chapter Two Section 2.1 1 Based on the direction field, all solutions seem to converge to a specific increasing function. The integrating factor is , and hence It follows that all solutions converge to the function 2 . All slopes eventually become positive, hence all solutions will increase without bound. The integrating factor is , and hence evident that all solutions increase at an exponential rate. It is 3 ________________________________________________________________________ page 18 —————————————————————————— CHAPTER 2. —— . All solutions seem to converge to the function The integrating factor is , and hence clear that all solutions converge to the specific solution 4 It is . . . Based on the direction field, the solutions eventually become oscillatory. The integrating factor is in which function 5 , and hence the general solution is is an arbitrary constant. As becomes large, all solutions converge to the . ________________________________________________________________________ page 19 —————————————————————————— CHAPTER 2. —— . All slopes eventually become positive, hence all solutions will increase without bound. The integrating factor is can be written as , that is, sides of the equation results in the general solution all solutions will increase exponentially. The differential equation Integration of both It follows that 6 All solutions seem to co...
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