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qxd 9/15/05 12:06 PM Page v PREFACE General Character and Purpose of the Instructor’s Manual This Manual contains: (I) Detailed solutions of the even-numbered problems. (II) General comments on the purpose of each section and its classroom use, with mathematical and didactic information on teaching practice and pedagogical aspects.Some of the comments refer to whole chapters (and are indicated accordingly). Changes in Problem Sets The major changes in this edition of the text are listed and explained in the Preface of the book. They include global improvements produced by updating and streamlining chapters as well as many local improvements aimed at simplification of the whole text. Speedy orientation is helped by chapter summaries at the end of each chapter, as in the last edition, and by the subdivision of sections into subsections with unnumbered headings. Resulting effects of these changes on the problem sets are as follows.The problems have been changed.

The large total number of more than 4000 problems has been retained, increasing their overall usefulness by the following: • Placing more emphasis on modeling and conceptual thinking and less emphasis on technicalities, to parallel recent and ongoing developments in calculus. • Balancing by extending problem sets that seemed too short and contracting others that were too long, adjusting the length to the relative importance of the material in a section, so that important issues are reflected sufficiently well not only in the text but also in the problems.Thus, the danger of overemphasizing minor techniques and ideas is avoided as much as possible. • Simplification by omitting a small number of very difficult problems that appeared in the previous edition, retaining the wide spectrum ranging from simple routine problems to more sophisticated engineering applications, and taking into account the “algorithmic thinking” that is developing along with computers. • Amalgamation of text, examples, and problems by including the large number of more than 600 worked-out examples in the text and by providing problems closely related to those examples. Addition of TEAM PROJECTS, CAS PROJECTS, and WRITING PROJECTS, whose role is explained in the Preface of the book. • Addition of CAS EXPERIMENTS, that is, the use of the computer in “experimental mathematics” for experimentation, discovery, and research, which often produces unexpected results for open-ended problems, deeper insights, and relations among practical problems. These changes in the problem sets will help students in solving problems as well as in gaining a better understanding of practical aspects in the text.

It will also enable instructors to explain ideas and methods in terms of examples supplementing and illustrating theoretical discussions—or even replacing some of them if so desired. imfm. qxd 9/15/05 12:06 PM Page vi vi Instructor’s Manual “Show the details of your work.

” This request repeatedly stated in the book applies to all the problem sets. Of course, it is intended to prevent the student from simply producing answers by a CAS instead of trying to understand the underlying mathematics. Orientation on ComputersComments on computer use are included in the Preface of the book.

Software systems are listed in the book at the beginning of Chap. 19 on numeric analysis and at the beginning of Chap. 24 on probability theory. ERWIN KREYSZIG im01. qxd 9/21/05 10:17 AM Page 1 Part A. ORDINARY DIFFERENTIAL EQUATIONS (ODEs) CHAPTER 1 First-Order ODEs Major Changes There is more material on modeling in the text as well as in the problem set. Some additions on population dynamics appear in Sec.

1. 5. Electric circuits are shifted to Chap. 2, where second-order ODEs will be available.This avoids repetitions that are unnecessary and practically irrelevant. Team Projects, CAS Projects, and CAS Experiments are included in most problem sets. SECTION 1.

1. Basic Concepts. Modeling, page 2 Purpose. To give the students a first impression what an ODE is and what we mean by solving it.

Background Material. For the whole chapter we need integration formulas and techniques, which the student should review. General Comments This section should be covered relatively rapidly to get quickly to the actual solution methods in the next sections.Equations (1)–(3) are just examples, not for solution, but the student will see that solutions of (1) and (2) can be found by calculus, and a solution y e x of (3) by inspection. Problem Set 1.

1 will help the student with the tasks of Solving y ? (x) by calculus Finding particular solutions from given general solutions Setting up an ODE for a given function as solution Gaining a first experience in modeling, by doing one or two problems Gaining a first impression of the importance of ODEs without wasting time on matters that can be done much faster, once systematic methods are available.Comment on “General Solution” and “Singular Solution” Usage of the term “general solution” is not uniform in the literature. Some books use the term to mean a solution that includes all solutions, that is, both the particular and the singular ones. We do not adopt this definition for two reasons. First, it is frequently quite difficult to prove that a formula includes all solutions; hence, this definition of a general solution is rather useless in practice. Second, linear differential equations (satisfying rather general conditions on the coefficients) have no singular solutions (as mentioned in the text), so that or these equations a general solution as defined does include all solutions.

For the latter reason, some books use the term “general solution” for linear equations only; but this seems very unfortunate. 1 im01. qxd 9/21/05 10:17 AM Page 2 2 Instructor’s Manual SOLUTIONS TO PROBLEM SET 1. 1, page 8 2.

6. 10. 12. 14. 16. y e 3x/3 c 4.

y (sinh 4x)/4 c Second order. 8. First order. y ce0. 5x, y(2) ce 2, c 2/e, y (2/e)e0. 5x 0. 736e0. 5x y ce x x 1, y(0) c 1 3, c 2, y 2e x x 1 1 1 y c sec x, y(0) c/cos 0 c _ , y _ sec x 2 2 2 Substitution of y cx c into the ODE gives y Similarly, y _ x 2, 1 4 2 y y _ x, 1 2 y c2 xc thus (cx _x 2 1 4 c 2) 0.

_x 2 1 4 1 x(_ x) 2 0. 18. In Prob. 17 the constants of integration were set to zero. Here, by two integrations, 1 y g, v y gt c1, y _gt 2 c1t c2, y(0) c2 y0, 2 and, furthermore, v(0) c1 v0, hence y _ gt 2 1 2 v0 t y0, as claimed.

Times of fall are 4. 5 and 6. 4 sec, from t 100/4. 9 and 200/4. 9. 20. y ky. Solution y y0 ekx, where y0 is the pressure at sea level x 0.

Now 1 y(18000) y0 ek 18000 _ y0 (given). From this, 2 1 1 1 ek 18000 _ , y(36000) y0 ek 2 18000 y0(ek 18000)2 y0(_ )2 _ y0. 2 2 4 22.For 1 year and annual, daily, and continuous compounding we obtain the values ya(1) 1060.

00, yd(1) yc(1) respectively. Similarly for 5 years, ya(5) 1000 1. 065 1338. 23, yc(5) yd(5) 1000e 0. 06 5 1000(1 0.

06/365)365 1061. 84, 1061. 83, 1000e0.

06 1000(1 0. 06/365)365 5 1349. 83, 1349. 86. We see that the difference between daily compounding and continuous compounding is very small. The ODE for continuous compounding is yc ryc. SECTION 1. 2.

Geometric Meaning of y ? (x, y). Direction Fields, page 9 Purpose. To give the student a feel for the nature of ODEs and the general behavior of fields of solutions.This amounts to a conceptual clarification before entering into formal manipulations of solution methods, the latter being restricted to relatively small—albeit important—classes of ODEs. This approach is becoming increasingly important, especially because of the graphical power of computer software. It is the analog of conceptual studies of the derivative and integral in calculus as opposed to formal techniques of differentiation and integration.

Comment on Isoclines These could be omitted because students sometimes confuse them with solutions. In the computer approach to direction fields they no longer play a role. im01. qxd 9/21/05 0:17 AM Page 3 Instructor’s Manual 3 Comment on Order of Sections This section could equally well be presented later in Chap. 1, perhaps after one or two formal methods of solution have been studied. SOLUTIONS TO PROBLEM SET 1.

2, page 11 2. Semi-ellipse x 2/4 y 2/9 13/9, y 0. To graph it, choose the y-interval large enough, at least 0 y 4.

1 4. Logistic equation (Verhulst equation; Sec. 1. 5). Constant solutions y 0 and y _ . 2 _ , decreasing for y(0) _ . 1 1 For these, y 0. Increasing solutions for 0 y(0) 2 2 6.

The solution (not of interest for doing the problem) is obtained by using dy/dx x c 1/(dx/dy) and solving 1 2/(tan _ y 2 x/dy 1/(1 sin y) by integration, 2 c)/(x c)). 1); thus y 2 arctan ((x 8. Linear ODE. The solution involves the error function. 12.

By integration, y c 1/x. 16. The solution (not needed for doing the problem) of y 1/y can be obtained by separating variables and using the initial condition; y 2/2 t c, y 2t 1. 18.

The solution of this initial value problem involving the linear ODE y y t 2 is t 2 y 4e t 2t 2. 20. CAS Project. (a) Verify by substitution that the general solution is y 1 ce x. Limit y 1 (y(x) 1 for all x), increasing for y(0) 1, decreasing for y(0) 1. (b) Verify by substitution that the general solution is x 4 y4 c.

More “squareshaped,” isoclines y kx. Without the minus on the right you get “hyperbola-like” curves y 4 x 4 const as solutions (verify! ). The direction fields should turn out in perfect shape. (c) The computer may be better if the isoclines are complicated; but the computer may give you nonsense even in simpler cases, for instance when y(x) becomes imaginary. Much will depend on the choice of x- and y-intervals, a method of trial and error. Isoclines may be preferable if the explicit form of the ODE contains roots on the right.

SECTION 1. 3. Separable ODEs. Modeling, page 12 Purpose.To familiarize the student with the first “big” method of solving ODEs, the separation of variables, and an extension of it, the reduction to separable form by a transformation of the ODE, namely, by introducing a new unknown function. The section includes standard applications that lead to separable ODEs, namely, 1. the ODE giving tan x as solution 2. the ODE of the exponential function, having various applications, such as in radiocarbon dating 3.

a mixing problem for a single tank 4. Newton’s law of cooling 5. Torricelli’s law of outflow. im01. qxd 9/21/05 10:17 AM Page 4 4 Instructor’s ManualIn reducing to separability we consider 6. the transformation u ODEs.

y/x, giving perhaps the most important reducible class of Ince’s classical book [A11] contains many further reductions as well as a systematic theory of reduction for certain classes of ODEs. Comment on Problem 5 From the implicit solution we can get two explicit solutions y c (6x)2 representing semi-ellipses in the upper half-plane, and y c (6x)2 representing semi-ellipses in the lower half-plane. [Similarly, we can get two explicit solutions x(y) representing semi-ellipses in the left and right half-planes, respectively. On the x-axis, the tangents to the ellipses are vertical, so that y (x) does not exist. Similarly for x (y) on the y-axis. This also illustrates that it is natural to consider solutions of ODEs on open rather than on closed intervals. Comment on Separability An analytic function ? (x, y) in a domain D of the xy-plane can be factored in D, ? (x, y) g(x)h(y), if and only if in D, ? xy? ?x ? y [D.

Scott, American Math. Monthly 92 (1985), 422–423]. Simple cases are easy to decide, but this may save time in cases of more complicated ODEs, some of which may perhaps be of practical interest.You may perhaps ask your students to derive such a criterion. Comments on Application Each of those examples can be modified in various ways, for example, by changing the application or by taking another form of the tank, so that each example characterizes a whole class of applications. The many ODEs in the problem set, much more than one would ordinarily be willing and have the time to consider, should serve to convince the student of the practical importance of ODEs; so these are ODEs to choose from, depending on the students’ interest and background.Comment on Footnote 3 Newton conceived his method of fluxions (calculus) in 1665–1666, at the age of 22.

Philosophiae Naturalis Principia Mathematica was his most influential work. Leibniz invented calculus independently in 1675 and introduced notations that were essential to the rapid development in this field. His first publication on differential calculus appeared in 1684. SOLUTIONS TO PROBLEM SET 1.

3, page 18 2. dy/y 2 (x 1 y 2)dx. The variables are now separated. Integration on both sides gives _x 2 1 2 2x c*. Hence y x2 2 4x c . im01. qxd 9/21/05 10:17 AM Page 5 Instructor’s Manual 5 4.

Set y 9x y . Then y v 9 v v2. 9x. By substitution into the given ODE you obtain By separation, dv v v 3 2 9 dx. Integration gives 1 v arctan 3 3 x c*, v y 9x, 3 tan (3x c) 9x. arctan 3x c and from this and substitution of y v 3 tan (3x u c), 6. Set u y/x. Then y xu, y of u on both sides gives y 4x y y x xu .

Substitution into the ODE and subtraction 4 u 4 . u u xu u, xu Separation of variables and replacement of u with y/x yields 2u du 8 dx, x u2 8 ln x c, y2 x 2 (8 ln x c). 8. u y/x, y xu, y u and divide by x 2 to get xy xu xu . Substitute u into the ODE, drop xu on both sides, x 2u _ x 2u2 1 2 xu, u _ u2.

1 2Separate variables, integrate, and solve algebraically for u: du u2 Hence y xu c _ dx, 1 2 1 u _ (x 1 2 c*), 2x x u 2 c x . . 10. By separation, y dy 4x dx.

By integration, y 2 4x 2 c. The initial condition y(0) 3, applied to the last equation, gives 9 0 c. Hence y 2 4x 2 9. 12. Set u y/x. Then y u xu .

Divide the given ODE by x 2 and substitute u and u into the resulting equation. This gives 2u(u xu ) 3u2 1. Subtract 2u2 on both sides and separate the variables. This gives 2xuu u2 1, 2u du u2 1 u2 xu dx .

x Integrate, take exponents, and then take the square root: ln (u2 Hence y y x 5x 1. x cx c 1 1. 2, c 5.This gives the answer From this and the initial condition, y(1) 1) ln x c*, 1 cx, u cx 1. im01.

qxd 9/21/05 10:17 AM Page 6 6 Instructor’s Manual 14. Set u y/x. Then y xu, y u xu . Substitute this into the ODE, subtract u on both sides, simplify algebraically, and integrate: xu Hence y 2 2x 2 cos (x 2) u 2x 2(sin (x 2) uu 2x cos (x 2), u2/2 sin (x 2) 1 (sin _ 2 c. 0, c). By the initial condition, y xu y 6 5 4 3 2 1 c), c x 2 sin (x 2). –4 –3 –2 –1 0 –1 –2 –3 –4 –5 –6 1 2 3 4 x Problem Set 1.

3. Problem 14. First five real branches of the solution 16. u y/x, y integrate: u Hence xu, y u xu u 4x 4 cos2u.Simplify, separate variables, and 4x 3 dx, c).

c) 16). 0, c 16. Answer: tan u x4 c. 4x 3 cos2u, y du/cos2u xu x arctan (x 4 From the initial condition, y(2) y 18. Order terms: dr (1 d Separate variables and integrate: dr r 1 b sin b cos 2 arctan (16 x arctan (x 4 b cos ) br sin . d , ln r ln (1 b cos ) c*.

im01. qxd 9/21/05 10:17 AM Page 7 Instructor’s Manual 7 Take exponents and use the initial condition: r c(1 b cos ), r( 2 ) c(1 b 0) , c . Hence the answer is r (1 b cos ). 20. On the left, integrate g(w) over w from y0 to y. On the right, integrate ? (t) over t from x0 to x. In Prob.

19, y x wew dw 0 2 (t 1) dt. 22. Consider any straight line y ax through the origin.

Its slope is y/x a. The slope of a solution curve at a point of intersection (x, ax) is y g(y/x) g(a) const, independent of the point (x, y) on the straight line considered. 24. Let kB and kD be the constants of proportionality for the birth rate and death rate, kB y kD y, where y(t) is the population at time t. By separating respectively. Then y variables, integrating, and taking exponents, dy/y (kB kD) dt, ln y (kB kD)t c*, y ce(kB kD)t . 26.

The model is y Ay ln y with A 0. Constant solutions are obtained from y 0 when y 0 and 1.Between 0 and 1 the right side is positive (since ln y 0), so that the solutions grow. For y 1 we have ln y 0; hence the right side is negative, so that the solutions decrease with increasing t.

It follows that y 1 is stable. The general solution is obtained by separation of variables, integration, and two subsequent exponentiations: dy/(y ln y) ln y ce A dt, At ln (ln y) y exp (ce At At c*, , ). 28.

The temperature of the water is decreasing exponentially according to Newton’s law of cooling. The decrease during the first 30 min, call it d1, is greater than that, d2, during the next 30 min. Thus d1 d2 190 110 80 as measured.Hence the temperature at the beginning of parking, if it had been 30 min earlier, before the arrest, would have been greater than 190 80 270, which is impossible. Therefore Jack has no alibi.

30. The cross-sectional area A of the hole is multiplied by 4. In the particular solution, 15.

00 0. 000332t is changed to 15. 00 4 0.

000332t because the second term contains A/B. This changes the time t 15. 00/0. 000332 when the tank is empty, to t 15. 00/(4 0. 000332), that is, to t 12. 6/4 3.

1 hr, which is 1/4 of the original time. 32. According to the physical information given, you have S 0. 15S . Now let * 0. This gives the ODE dS/d 0. 5S.

Separation of variables yields the general solution S S0 e0. 15 with the arbitrary constant denoted by S0. The angle should be so large that S equals 1000 times S0. Hence e0. 15 1000, (ln 1000)/0. 15 46 7.

3 2 , that is, eight times, which is surprisingly little. Equally remarkable is that here we see another application of the ODE y ky and a derivation of it by a general principle, namely, by working with small quantities and then taking limits to zero. im01. qxd 9/21/05 10:17 AM Page 8 8 Instructor’s Manual 36. B now depends on h, namely, by the Pythagorean theorem, B(h) r2 (R2 (R h)2 ) (2Rh h2). Hence you can use the ODE h 26.

6(A/B) h in the text, with constant A as before and the new B. The latter makes the further calculations different from those in Example 5. From the given outlet size A 5 cm2 and B(h) we obtain dh dt Now 26. 56 5/ By integration, 26.

56 5 (2Rh h2) h. 42. 27, so that separation of variables gives (2Rh1/2 _ Rh3/2 4 3 h3/2 ) dh _h5/2 2 5 42. 27 dt. 42. 27t c. From this and the initial condition h(0) _ R5/2 _R5/2 4 2 3 5 R we obtain 0.

9333R5/2 c. Hence the particular solution (in implicit form) is _ Rh3/2 _h5/2 4 2 42. 27t 0. 9333R5/2. 3 5 The tank is empty (h 0 For R 42. 27t 1m 0) for t such that hence t 0. 9333 5/2 R 42. 7 37 [min].

0. 0221R5/2. 0. 9333R5/2; 100 cm this gives t 0. 0221 1005/2 2210 [sec] The tank has water level R/2 for t in the particular solution such that 4 R3/2 R 3/2 3 2 2 R5/2 5 25/2 0. 9333R5/2 42. 27t. The left side equals 0.

4007R5/2. This gives t 0. 4007 0. 9333 5/2 R 42. 27 0. 01260R5/2. For R 100 this yields t 1260 sec 21 min.

This is slightly more than half the time needed to empty the tank. This seems physically reasonable because if the water level is R/2, this means that 11/16 of the total water volume has flown out, and 5/16 is left—take into account that the velocity decreases monotone according to Torricelli’s law.R r h R=h Problem Set 1. 3. Tank in Problem 36 im01. qxd 9/21/05 10:17 AM Page 9 Instructor’s Manual 9 SECTION 1. 4. Exact ODEs.

Integrating Factors, page 19 Purpose. This is the second “big” method in this chapter, after separation of variables, and also applies to equations that are not separable. The criterion (5) is basic. Simpler cases are solved by inspection, more involved cases by integration, as explained in the text. Comment on Condition (5) Condition (5) is equivalent to (6 ) in Sec. 10. 2, which is equivalent to (6) in the case of two variables x, y. Simple connectedness of D follows from our assumptions in Sec.

. 4. Hence the differential form is exact by Theorem 3, Sec. 10. 2, part (b) and part (a), in that order. Method of Integrating Factors This greatly increases the usefulness of solving exact equations. It is important in itself as well as in connection with linear ODEs in the next section.

Problem Set 1. 4 will help the student gain skill needed in finding integrating factors. Although the method has somewhat the flavor of tricks, Theorems 1 and 2 show that at least in some cases one can proceed systematically—and one of them is precisely the case needed in the next section for linear ODEs. SOLUTIONS TO PROBLEM SET 1. , page 25 2. (x x y) dx (y y over x: x) dy 0. Exact; the test gives u _x 2 1 2 1 on both sides.

Integrate xy k(y). _ y2 1 2 Differentiate this with respect to y and compare with N: uy x k y x. Thus k y, k c*. 1 1 1 Answer: _ x 2 xy _ y 2 _ (x y)2 c; thus y x c.

2 2 2 y x 4. Exact; the test gives e e on both sides. Integrate M with respect to x to get u xey ye x k(y). Differentiate this with respect to y and equate the result to N: uy xey ex k N xey e x. Hence k 0, k const. Answer: xey ye x c. 6.

Exact; the test gives e x sin y on both sides. Integrate M with respect to x: u e x cos y k(y). x Differentiate: uy x sin y x k. Equate this to N e sin y. Hence k 0, k const.

Answer: e cos y c. 8. Exact; 1/x 2 1/y 2 on both sides of the equation. Integrate M with respect to x: u x2 x y y x k(y). Differentiate this with respect to y and equate the result to N: uy Answer: x2 x y y x y2 c. x y2 1 x k N, k 2y, k y 2. im01. qxd 9/21/05 10:17 AM Page 10 10 Instructor’s Manual 10.

Exact; the test gives 2x sin (x 2) on both sides. Integrate N with respect to y to get u y cos (x 2) l(x). Differentiate this with respect to x and equate the result to M: ux 2xy sin (x 2) l M 2xy sin (x 2), l 0. Answer: y cos (x 2) c. 12.

Not exact.Try Theorem 1. In R you have Py so that R Qx 1, F ex y 1 ex y(x 1) xex y 1 Q e x, and the exact ODE is (ey ye x) dx (xey e x) dy 0. FP with The test gives ey respect to x gives e x on both sides of the equation. Integration of M u xey ye x k(y). FQ: e x. Differentiate this with respect to y and equate it to N uy xey e x k N xey Hence k 0.

Answer: xey ye x c. 14. Not exact; 2y y.

Try Theorem 1; namely, R (Py Qx)/Q (2y y)/( xy) y2 ) dx x3 3/x. y dy x2 Hence F 1/x 3. The exact ODE is (x 0.

FQ The test gives 2y/x 3 on both sides of the equation. Obtain u by integrating N with respect to y: u y2 2x 2 l(x).Thus y 2/2x 2 x2 x 2/2 y2 x2 c ux y2 x3 l M x y2 . x3 Hence l x, l x 2/2, condition y(2) 1: c*. Multiply by 2 and use the initial 3.

75 because inserting y(2) 1 into the last equation gives 4 0. 25 3. 75. 16. The given ODE is exact and can be written as d(cos xy) 0; hence cos xy c, or you can solve it for y by the usual procedure. y(1) gives 1 c. Answer: cos xy 1. 18.

Try Theorem 2. You have R* (Qx Py)/P [ 1 cos xy y x sin xy ( x sin xy x )] y2 P 1 . y Hence F* y.

This gives the exact ODE (y cos xy x) dx (y x cos xy) dy 0. im01. qxd 9/21/05 10:17 AM Page 11 Instructor’s Manual 11In the test, both sides of the equation are cos xy to x: 1 u sin xy _ x 2 k(y).

Hence 2 xy sin xy. Integrate M with respect uy x cos xy k (y). y; hence Equate the last equation to N 1 k y 2/2.

Answer: sin xy _ x 2 2 20. Not exact; try Theorem 2: R* (Qx Py)/P [1 y _y 2 1 2 x cos xy. This shows that k c. sin2 y 2x cos y sin y)]/P (cos2 y 2 [2 sin y 2x cos y sin y]/P x cos y)/(sin y cos y 2 tan y. ln (1/cos2 y); hence F* 0. 1/cos2 y. x cos2 y) 2(sin y)(sin y 2(sin y)/cos y Integration with respect to y gives The resulting exact equation is (tan y 2 ln (cos y) x) dx x dy cos2 y The exactness test gives 1/cos2 y on both sides.Integration of M with respect to x yields x 1 u x tan y _ x 2 k(y).

From this, uy k. 2 cos2 y 1 Equate this to N x/cos2 y to see that k 0, k const. Answer: x tan y _x 2 c. 2 22.

(a) Not exact. Theorem 2 applies and gives F* 1/y from 1 R* (Qx Py)/P (0 cos x)/(y cos x) . y Integrating M in the resulting exact ODE cos x dx with respect to x gives u sin x k(y). From this, uy k N 1 . y2 1 dy y2 0 Hence k 1/y.

Answer: sin x 1/y c. Note that the integrating factor 1/y could have been found by inspection and by the fact that an ODE of the general form ? (x) dx g(y) dy 0 is always exact, the test resulting in 0 on both sides. b) Yes. Separation of variables gives dy/y 2 cos x dx. By integration, 1/y sin x c* in agreement with the solution in (a). (d) seems better than (c). But this may depend on your CAS.

In (d) the CAS may draw vertical asymptotes that disturb the figure. From the solution in (a) or (b) the student should conclude that for each nonzero y(x0) y0 there is a unique particular solution because sin x0 1/y0 c. im01. qxd 9/21/05 10:17 AM Page 12 12 Instructor’s Manual 24. (A) ey cosh x c.

(B) R* tan y, F 1/cos y. Separation: 2 dy/cos y (C) R 2/x, F 1/x 2, x 2v dv/(1 (1 y 2/x 2x) dx, c. v dx/x, 3/4 tan y x x2 c. /x, and separation: x2 y2 cx; x 9/4 v2) cx divide by x. (D) Separation is simplest. y R* 3/y, F*(y) y 3.

. R 9/(4x), F(x) , x 3y 4 c. SECTION 1. 5. Linear ODEs. Bernoulli Equation.

Population Dynamics, page 26 Purpose. Linear ODEs are of great practical importance, as Problem Set 1. 5 illustrates (and even more so are second-order linear ODEs in Chap.

2). We show that the homogeneous ODE of the first order is easily separated and the nonhomogeneous ODE is solved, once and for all, in the form of an integral (4) by the method of integrating factors. Of course, in simpler cases one does not need (4), as our examples illustrate.

Comment on Notation We write y p(x)y r(x). p(x) seems standard. r(x) suggests “right side. ” The notation y p(x)y q(x) used in some calculus books (which are not concerned with higher order ODEs) would be shortsighted here because later, in Chap. 2, we turn to second-order ODEs y p(x)y q(x)y r(x), where we need q(x) on the left, thus in a quite different role (and on the right we would have to choose another letter different from that used in the first-order case). Comment on Content Bernoulli’s equation appears occasionally in practice, so the student should remember how to handle it.A special Bernoulli equation, the Verhulst equation, plays a central role in population dynamics of humans, animals, plants, and so on, and we give a short introduction to this interesting field, along with one reference in the text. Riccati and Clairaut equations are less important than Bernoulli’s, so we have put them in the problem set; they will not be needed in our further work.

Input and output have become common terms in various contexts, so we thought this a good place to mention them. Problems 37–42 express properties that make linearity important, notably in obtaining new solutions from given ones.The counterparts of these properties will, of course, reappear in Chap. 2.

Comment on Footnote 5 Eight members of the Bernoulli family became known as mathematicians; for more details, see p. 220 in Ref. [GR2] listed in App. 1. im01.

qxd 9/21/05 10:17 AM Page 13 Instructor’s Manual 13 SOLUTIONS TO PROBLEM SET 1. 5, page 32 4. The standard form (1) is y y e 4x [ e 4y 4x x, so that (4) gives x dx c] ce 4x x/4 1/16. 6. The standard form (1) is y c 3 y x 2 from the initial condition, y x 3 1 .

From this and (4) we obtain, with x3 [ x 3x 3 dx c] x 3[x c] x 2 2x 3. 8. From (4) with p y e 2x 2, h 2x, r 4 cos 2x we obtain c] e 2x [ e 2x 4 cos 2x dx e 2x(cos 2x sin 2x) c]. It is perhaps worthwhile mentioning that integrals of this type can more easily be evaluated by undetermined coefficients.

Also, the student should verify the result by differentiation, even if it was obtained by a CAS. From the initial condition we obtain 1 y(_ ) 4 ce /2 0 1 2; hence c e /2 . The answer can be written y 10.

In (4) we have p y e /2 2x cos 2x sin 2x. 4x 2; hence h e 4×3/3 4x 3/3, so that (4) gives 3 [ e (4x /3) x 2/ 2 (4x 2 x) dx c]. The integral can be evaluated by noting that the factor of the exponential function under the integral sign is the derivative of the exponent of that function.We thus obtain 3 3 2 3 2 y e 4x /3 [e(4x /3) x /2 c] ce 4x /3 e x /2. 12. y tan x dy y 4 2(y 2 4).

Separation of variables gives cos x dx. sin x By integration, ln y 4 2 ln sin x c*. Taking exponents on both sides gives y 1 y(_ ) 2 4 c sin2 x, y c sin2 x 4. The desired particular solution is obtained from the initial condition c 4 tan x, h cos x e cos x 0, c 4. Answer: y 4 4 sin2 x. 14. In (4) we have p y ln (cos x), eh 0.

01x 1/cos x, so that (4) gives [ 100 e 0. 01x (cos x) [ dx c] c] cos x. im01. qxd 9/21/05 10:17 AM Page 14 14 Instructor’s Manual The initial condition gives y(0) 100 c 0; hence c solution is y 100(1 e 0.

1x) cos x. 100. The particular The factor 0. 01, which we included in the exponent, has the effect that the graph of y shows a long transition period. Indeed, it takes x 460 to let the exponential function e 0. 01x decrease to 0. 01. Choose the x-interval of the graph accordingly.

16. The standard form (1) is y Hence h 3 y cos2 x 1 . cos2 x 3 tan x, and (4) gives the general solution y e 3 tan x [ e3 tan x dx cos2 x c] . To evaluate the integral, observe that the integrand is of the form _ (3 tan x) e 3 tan x; 1 3 that is, _ (e3 tan x) . 1 3 1 Hence the integral has the value _ e3 tan x.

This gives the general solution 3 3 tan x _ 3 tan x 1 y e [1 e c] _ ce 3 tan x. 3 3 The initial condition gives from this 1 1 4 y( _ ) _ ce 3 _ ; 4 3 3 _ e3 3 tan x. 1 The answer is y 3 hence c e 3. 18. Bernoulli equation. First solution method: Transformation to linear form. Set y 1/u.

Then y y u /u2 1/u 1/u2. Multiplication by u2 gives the linear ODE in standard form u u 1. General solution u ce x 1. Hence the given ODE has the general solution y 1/(ce x 1). 1 we obtain 2, Answer: y 1/(1 2e x).

From this and the initial condition y(0) y(0) 1/(c 1) 1, c Second solution method: Separation of variables and use of partial fractions. y y(y Integration gives ln y 1 ln y ln j y y 1 j x c*. 1) ( 1 y 1 1 ) dy y dx. Taking exponents on both sides, we obtain y y 1 1 1 y ce x, 1 y 1 ce x, y 1 1 ce x . We now continue as before. im01. qxd 9/21/05 10:17 AM Page 15 Instructor’s Manual 15 20. Separate variables, integrate, and take exponents: cot y dy and sin y Now use the initial condition y(0) 1 Answer: y arcsin (e arctan x). 22. First solution method: by setting z z ( 2 sin 2y)y . ce _ : 1 2 ce 0, c 1. arctan x dx/(x 2 1), ln sin y . arctan x c* cos 2y (linearization): From z we have From the ODE, _z 1 2 xz 2x. This is a linear ODE.Its standard form is z 2xz 4x. In (4) this gives p 2x, h x 2. Hence (4) gives the solution in terms of z in the form z ex [ e 2 x2 ( 4x) dx c] ex [2e 2 x2 c] 2 cex . 2 From this we obtain the solution y _ arccos z 1 2 _ arccos (2 1 2 cex ). 2 Second solution method: Separation of variables. By algebra, y sin 2y Separation of variables now gives sin 2y dy 2 cos 2y x dx. Integrate: _ ln 2 1 2 cos 2y _ x2 1 2 x( cos 2y 2). c*. Multiply by 2 and take exponents: ln 2 Solve this for y: cos 2y 24. Bernoulli ODE. Set u 3y 2 to obtain 2 cex , 2 cos 2y x2 2c*, 2 cos 2y cex . 2 y _ arccos (2 1 2 cex ). 2 3 and note that u 3y 2y 3x 2y 3 e 3y 2y . Multiply the given ODE by x3 sinh x. In terms of u this gives the linear ODE u In (4) we thus have h u and y u1/3. e x3 3x 2u e x3 sinh x. x 3. The solution is [ ex e 3 x3 sinh x dx c] e x3 [cosh x c] im01. qxd 9/21/05 10:17 AM Page 16 16 Instructor’s Manual 26. The salt content in the inflow is 50(1 cos t). Let y(t) be the salt content in the tank to be determined. Then y(t)/1000 is the salt content per gallon. Hence (50/1000)y(t) is the salt content in the outflow per minute. The rate of change y equals the balance, y Thus y 0. 05y general solution y 50(1 e e 0. 5t ln Out 50(1 cos t) 0. 05, h cos t) dt a cos t ce 0. 05y. 0. 05t, and (4) gives the c) b sin t) 0. 05t cos t). Hence p ( e0. 05t 50(1 (e0. 05t (1000 a cos t 0. 05t c) 1000 b sin t where a 2. 5/(1 0. 052) 2. 494 and b 50/(1 0. 052) 49. 88, which we obtained by evaluating the integral. From this and the initial condition y(0) 200 we have y(0) 1000 a c 200, c 200 1000 a 802. 5. Hence the solution of our problem is y(t) 1000 2. 494 cos t 49. 88 sin t 802. 5e 0. 05t . Figure 20 shows the solution y(t). The last term in y(t) is the only term that depends on the initial condition (because c does).It decreases monotone. As a consequence, y(t) increases but keeps oscillating about 1000 as the limit of the mean value. This mean value is also shown in Fig. 20. It is obtained as the solution of the ODE y 0. 05y 50. Its solution satisfying the initial condition is y 1000 800e 0. 05t . 28. k1(T Ta) follows from Newton’s law of cooling. k2(T Tw) models the effect of heating or cooling. T Tw calls for cooling; hence k2(T Tw) should be negative in this case; this is true, since k2 is assumed to be negative in this formula. Similarly for heating, when heat should be added, so that the temperature increases.The given model is of the form T kT K k1C cos ( /12)t. This can be seen by collecting terms and introducing suitable constants, k k1 k2 (because there are two terms involving T ), and K k1A k2 Tw P. The general solution is T cekt K/k L( k cos ( t/12) ( /12) sin ( t/12)), 2 where L k1C/(k 2 /144). The first term solves the homogeneous ODE T kT and decreases to zero. The second term results from the constants A (in Ta), Tw, and P. The third term is sinusoidal, of period 24 hours, and time-delayed against the outside temperature, as is physically understandable. 30. y ky(1 y) ? y), where k 0 and y is the proportion of infected persons. Equilibrium solutions are y 0 and y 1. The first, y 0, is unstable because im01. qxd 9/21/05 10:17 AM Page 17 Instructor’s Manual 17 ?(y) 0 if 0 y 1 but ? (y) 0 for negative y. The solution y 1 is stable because ? (y) 0 if 0 y 1 and ? (y) 0 if y 1. The general solution is y 1 ce kt 1 . It approaches 1 as t * . This means that eventually everybody in the population will be infected. 32. The model is y Ay By 2 Hy Ky By 2 y(K By) where K A H. Hence the general solution is given by (9) in Example 4 with A replaced by K A H.The equilibrium solutions are obtained from y 0; hence they are y1 0 and y2 K/B. The population y2 remains unchanged under harvesting, and the fraction Hy2 of it can be harvested indefinitely—hence the name. 34. For the first 3 years you have the solution y1 4/(5 3e 0. 8t ) from Prob. 32. The idea now is that, by continuity, the value y1(3) at the end of the first period is the initial value for the solution y2 during the next period. That is, y2(3) Now y2 is the solution of y this gives y2 Check the continuity at t y y1(3) 4/(5 3e 2. 4 ). y 2 (no fishing! ). Because of the initial condition 4/(4 e3 t 3e0. 6 t). by calculating y2(3) 4/(4 e0 3e 2. 4 ). Similarly, for t from 6 to 9 you obtain y3 4/(5 e4. 8 0. 8t e1. 8 0. 8t 3e 6: 3e 0. 6 0. 8t ). This is a period of fishing. Check the continuity at t y3(6) This agrees with y2(6) 36. y1 1/u1, u1(0) y1 1/y1(0) u 1/u 2 1 4/(5 4/(4 e0 e e 3 3 5. 4 ). 3e 5. 4 ). 0. 5, 0. 8y1 y12 1, 0. 8/u1 1/u12, u1 u1 1. 25 1. 25 for 0 t 3. u 2 u2 1, u2 1 c2 e c1e 0. 8u1 0. 8t 0. 8t 0. 75e 1 3 1/y1 2. 4 c2 e t. The continuity condition is u1(3) 1. 25 0. 75e . u2(3) im01. qxd 9/21/05 10:17 AM Page 18 18 Instructor’s Manual For c2 this gives c2 This gives for 3 e 3( 1 t 6 u2 Finally, for 6 t 1 0. 5e3 t 1. 25 0. 75e 2. 4 ) 0. 25e 3 t 0. 75e0. 6. 0. 75e0. 6 0. 8u3 0. 8t 1/y2. 1, whose general solution is 9 we have the ODE is u 3 u3 1. 25 c3 e 1 . 3 5. 4 c3 is determined by the continuity condition at t u3(6) This gives c3 e4. 8( 1. 25 0. 25e 4. 8 6, namely, 0. 25e 3 1. 25 c3 e 4. 8 u2(6) 1 0. 25e t 0. 75e 5. 4 . 0. 25e 1. 8 0. 75e 0. 6 ) 0. 75e . 0. 8t Substitution gives the solution for 6 u3 1. 25 ( 0. 25e 4. 8 9: 0. 75e 0. 6 0. 25e1. 8 )e 1/y3. 38. Substitution gives the identity 0 0. These problems are of importance because they show why linear ODEs are preferable over nonlinear ones in the modeling process.Thus one favors a linear ODE over a nonlinear one if the model is a faithful mathematical representation of the problem. Furthermore, these problems illustrate the difference between homogeneous and nonhomogeneous ODEs. 40. We obtain (y1 y2) p(y1 y2) y 1 y 2 py1 py2 (y 1 r 0. 42. The sum satisfies the ODE with r1 r2 on the right. This is important as the key to the method of developing the right side into a series, then finding the solutions corresponding to single terms, and finally, adding these solutions to get a solution of the given ODE.For instance, this method is used in connection with Fourier series, as we shall see in Sec. 11. 5. 44. (a) y Y v reduces the Riccati equation to a Bernoulli equation by removing the term h(x). The second transformation, v 1/u, is the usual one for transforming a Bernoulli equation with y 2 on the right into a linear ODE. Substitute y Y 1/u into the Riccati equation to get Y u /u2 p(Y pY u /u2 Multiplication by u2 gives u u the linear ODE as claimed. 1/u) gY 2 py1) r (y 2 py2) g(Y 2 2Y/u 1/u2) h. Since Y is a solution, Y h. There remains g(2Y/u g(2Yu p)u 1/u2). 1).Reshuffle terms to get g, p/u pu (2Yg im01. qxd 9/21/05 10:17 AM Page 19 Instructor’s Manual 19 (b) Substitute y Now substitute y 1 u /u2 Y x (2x 3 x to get 1 2x 4 1/u. This gives 1)(x 1/u) x x 2(x 2 x4 x4 x 1/u2) 1, which is true. x4 x 1. x 2/u2. 2x/u Most of the terms cancel on both sides. There remains u /u2 1/u Multiplication by u2 finally gives u u x 2. The general solution is u ce x x2 2x 2 and y x 1/u. Of course, instead performing this calculation we could have used the general formula in (a), in which 2Yg p 2x( x 2) 2x 3 1 1 and g x 2. (c) Substitution of Y 2Yg Also, g u 2 shows that this is a solution. In the ODE for u you find p 2x 2 ( sin x) (3 3u 2x 2 sin x) 3. sin x. Hence the ODE for u is u ce 3x sin x. Solution: and y x2 1/u. 0. 1 cos x 0. 3 sin x (e) y y xy y /y 2 by the chain rule. Hence y (x 1/y 2) 0. (A) From y 0 we obtain by integrations y cx a. Substitution into the given ODE gives cx a xc 1/c; hence a 1/c. This is a family of straight lines. (B) The other factor is zero when x 1/y 2. By integration, y 2×1/2 c*. 1/2 Substituting y and y x into the given equation y xy 1/y , we obtain 2×1/2 c* x x 1/ 2 1/x 1/ 2 ; hence c* 0.This gives the singular solution y 2 x, a curve, to which those straight lines in (A) are tangent. (f) By differentiation, 2y y y xy y 0, y (2y x) 0, (A) y 0, y cx a. By substitution, c 2 xc cx a 0, a c 2, y cx c 2, a family of straight lines. (B) y x/2, y x 2/4 c*. By substitution into the given ODE, 2 2 2 x /4 x /2 x /4 c* 0, c* 0, y x 2/4, the envelope of the family; see Fig. 6 in Sec. 1. 1. SECTION 1. 6. Orthogonal Trajectories. Optional, page 35 Purpose. To show that families of curves F(x, y, c) 0 can be described by ODEs ? (x, y) and the switch to y 1/? x, y) produces as general solution the orthogonal y trajectories. This is a nice application that may also help the student to gain more self-confidence, skill, and a deeper understanding of the nature of ODEs. We leave this section optional, for reasons of time. This will cause no gap. The reason why ODEs can be applied in this fashion results from the fact that general solutions of ODEs involve an arbitrary constant that serves as the parameter of this one-parameter family of curves determined by the given ODE, and then another general solution similarly determines the one-parameter family of the orthogonal trajectories.Curves and their orthogonal trajectories play a role in several physical applications (e. g. , in connection with electrostatic fields, fluid flows, and so on). im01. qxd 9/21/05 10:17 AM Page 20 20 Instructor’s Manual SOLUTIONS TO PROBLEM SET 1. 6, page 36 2. xy c, and by differentiation, y xy 0; hence y y/x. The ODE of the trajectories is y x/y. By separation and integration, y2/2 x 2/2 c*. Hyperbolas. (So are the given curves. ) 4. By differentiation, 2yy 4x; hence y 2x/y. Thus the ODE of the trajectories is y y/2x.By separating, integrating, and taking exponents on both sides, dy/y 6. ye 3x dx/2x, ln y _ ln x 1 2 c**, y c*/ x. c. Differentiation gives (y 3y)e 3x 0. 3y. For the trajectories we obtain _ x c**, 1 _x c*. 2 y 3 3 Hence the ODE of the given family is y _, 1 _ y2 1 y 1/(3y), yy 3 2 8. 2x 2yy 0, so that the ODE of the curves is y x/y. Hence the ODE of the trajectories is y y/x. Separating variables, integrating, and taking exponents gives hyperbolas as trajectories; namely, y /y 1/x, 1 ln y ln x c**, xy c*. 10. xy 1/2 ?, c or x 2y c. By differentiation, 2xy 1 x 2y 2y


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