Solution Manuals Of ADVANCED ENGINEERING

im01.qxd9/21/0510:17 AMPage 3Instructor’s Manual3Comment on Order of SectionsThis section could equally well be presented later in Chap. 1, perhaps after one or twoformal methods of solution have been studied.SOLUTIONS TO PROBLEM SET 1.2, page 112. Semi-ellipse x 2/4 y 2/9 13/9, y 0. To graph it, choose the y-interval largeenough, at least 0 y 4.4. Logistic equation (Verhulst equation; Sec. 1.5). Constant solutions y 0 and y 12.For these, y 0. Increasing solutions for 0 y(0) 12, decreasing for y(0) 12.6. The solution (not of interest for doing the problem) is obtained by usingdy/dx 1/(dx/dy) and solvingx c 2/(tan 12 y 1);dx/dy 1/(1 sin y) by integration,thus y 2 arctan ((x 2 c)/(x c)).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 byseparating variables and using the initial condition; y 2/2 t c, y 2t . 118. The solution of this initial value problem involving the linear ODE y y t 2 isy 4e t t 2 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 fory(0) 1.(b) Verify by substitution that the general solution is x 4 y 4 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 inperfect shape.(c) The computer may be better if the isoclines are complicated; but the computermay give you nonsense even in simpler cases, for instance when y(x) becomesimaginary. Much will depend on the choice of x- and y-intervals, a method of trialand error. Isoclines may be preferable if the explicit form of the ODE contains rootson the right.SECTION 1.3. Separable ODEs. Modeling, page 12Purpose. To familiarize the student with the first “big” method of solving ODEs, theseparation of variables, and an extension of it, the reduction to separable form by atransformation 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 solution2. the ODE of the exponential function, having various applications, such as inradiocarbon dating3. a mixing problem for a single tank4. Newton’s law of cooling5. Torricelli’s law of outflow.

im01.qxd9/21/0510:17 AMPage 44Instructor’s ManualIn reducing to separability we consider6. the transformation u y/x, giving perhaps the most important reducible class ofODEs.Ince’s classical book [A11] contains many further reductions as well as a systematictheory of reduction for certain classes of ODEs.Comment on Problem 5From the implicit solution we can get two explicit solutionsy c (6x) 2representing semi-ellipses in the upper half-plane, andy c (6x) 2representing semi-ellipses in the lower half-plane. [Similarly, we can get two explicitsolutions 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. Similarlyfor x (y) on the y-axis.This also illustrates that it is natural to consider solutions of ODEs on open rather thanon closed intervals.Comment on SeparabilityAn analytic function ƒ(x, y) in a domain D of the xy-plane can be factored in D,ƒ(x, y) g(x)h(y), if and only if in D,ƒxyƒ ƒx ƒy[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 perhapsbe of practical interest. You may perhaps ask your students to derive such a criterion.Comments on ApplicationEach of those examples can be modified in various ways, for example, by changing theapplication or by taking another form of the tank, so that each example characterizes awhole class of applications.The many ODEs in the problem set, much more than one would ordinarily be willingand have the time to consider, should serve to convince the student of the practicalimportance of ODEs; so these are ODEs to choose from, depending on the students’interest and background.Comment on Footnote 3Newton 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 wereessential to the rapid development in this field. His first publication on differential calculusappeared in 1684.SOLUTIONS TO PROBLEM SET 1.3, page 182. dy/y 2 (x 2)dx. The variables are now separated. Integration on both sides gives 1 12x 2 2x c*.yHencey 2.x 2 4x c

im01.qxd9/21/0510:17 AMPage 5Instructor’s Manual54. Set y 9x v. Then y v 9x. By substitution into the given ODE you obtainy v 9 v2.dv dx.v 9By separation,2Integration gives1varctan x c*,33arctanv 3x c3and from this and substitution of y v 9x,v 3 tan (3x c),y 3 tan (3x c) 9x.6. Set u y/x. Then y xu, y u xu . Substitution into the ODE and subtractionof u on both sides givesy 4xy4 u xu u,yxuxu 4.uSeparation of variables and replacement of u with y/x yields2u du 8dx,xu2 8 ln x c,y 2 x 2 (8 ln x c).8. u y/x, y xu, y u xu . Substitute u into the ODE, drop xu on both sides,and divide by x 2 to getxy xu x 2u 12x 2u2 xu,u 12u2.Separate variables, integrate, and solve algebraically for u:du 12 dx,u2 1 12(x c*),uHencey xu u 2.c x2x.c x10. By separation, y dy 4x dx. By integration, y 2 4x 2 c. The initial conditiony(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 andu into the resulting equation. This gives2u(u xu ) 3u2 1.Subtract 2u2 on both sides and separate the variables. This givesdx2u du .2xu 12xuu u2 1,Integrate, take exponents, and then take the square root:ln (u2 1) ln x c*,u2 1 cx,u cx . 1Hencey xu x cx . 1From this and the initial condition, y(1) c 1 2, c 5. This gives the answer .y x 5x 1

im01.qxd9/21/0510:17 AMPage 66Instructor’s Manual14. Set u y/x. Then y xu, y u xu . Substitute this into the ODE, subtract u onboth sides, simplify algebraically, and integrate:xu 2x 2cos (x 2)uuu 2x cos (x 2),u2/2 sin (x 2) c.Hence y 2 2x 2(sin (x 2) c). By the initial condition, (sin 12 c), c 0,y xu x 2 sin ( x �5–6Problem Set 1.3. Problem 14. First five real branches of the solution16. u y/x, y xu, y u xu u 4x 4 cos2u. Simplify, separate variables, andintegrate:u 4x 3 cos2u,du/cos2u 4x 3 dx,tan u x 4 c.Hencey xu x arctan (x 4 c).From the initial condition, y(2) 2 arctan (16 c) 0, c 16. Answer:y x arctan (x 4 16).18. Order terms:dr(1 b cos ) br sin .dSeparate variables and integrate:drb sin r1 b cosd ,ln r ln (1 b cos ) c*.

im01.qxd9/21/0510:17 AMPage 7Instructor’s Manual7Take exponents and use the initial condition:r c(1 b cos ),r( ) c(1 b 0) ,2c .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 ƒ(t) over t fromx0 to x. In Prob. 19, wey (t 1) dt.xw2dw 1022. Consider any straight line y ax through the origin. Its slope is y/x a. The slopeof 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,respectively. Then y kB y kD y, where y(t) is the population at time t. By separatingvariables, integrating, and taking exponents,dy/y (kB kD) dt,ln y (kB kD)t c*,y ce(kB k )tD.26. The model is y Ay ln y with A 0. Constant solutions are obtained fromy 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. Thegeneral solution is obtained by separation of variables, integration, and two subsequentexponentiations:dy/(y ln y) A dt, Atln y ce,ln (ln y) At c*,y exp (ce At).28. The temperature of the water is decreasing exponentially according to Newton’s lawof 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 thetemperature 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 Jackhas 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 termcontains A/B. This changes the time t 15.00/0.000332 when the tank is empty, tot 15.00/(4 0.000332), that is, to t 12.6/4 3.1 hr, which is 1/4 of the originaltime.32. According to the physical information given, you haveS 0.15S .Now let * 0. This gives the ODE dS/d 0.15S. Separation of variables yieldsthe general solution S S0 e0.15 with the arbitrary constant denoted by S0. Theangle 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 anda derivation of it by a general principle, namely, by working with small quantitiesand then taking limits to zero.

im01.qxd9/21/0510:17 AMPage 88Instructor’s Manual36. B now depends on h, namely, by the Pythagorean theorem,B(h) r 2 (R2 (R h)2 ) (2Rh h2).Hence you can use the ODEh 26.56(A/B) h in the text, with constant A as before and the new B. The latter makes the furthercalculations different from those in Example 5.From the given outlet size A 5 cm2 and B(h) we obtaindh5 26.56 h .dt (2Rh h2)Now 26.56 5/ 42.27, so that separation of variables gives(2Rh1/2 h3/2 ) dh 42.27 dt.By integration,4 Rh3/2 2h5/2 42.27t c.35From this and the initial condition h(0) R we obtain4 R5/2 2R5/2 0.9333R5/2 c.35Hence the particular solution (in implicit form) is4 Rh3/2 2h5/2 42.27t 0.9333R5/2.35The tank is empty (h 0) for t such that0 42.27t 0.9333R5/2;hencet 0.9333 5/2R 0.0221R5/2.42.27For R 1 m 100 cm this givest 0.0221 1005/2 2210 [sec] 37 [min].The tank has water level R/2 for t in the particular solution such that42 R5/2R3/2R 3/2 0.9333R5/2 42.27t.325 25/2The left side equals 0.4007R5/2. This givest 0.4007 0.9333 5/2R 0.01260R5/2. 42.27For R 100 this yields t 1260 sec 21 min. This is slightly more than half thetime needed to empty the tank. This seems physically reasonable because if the waterlevel is R/2, this means that 11/16 of the total water volume has flown out, and 5/16is left—take into account that the velocity decreases monotone according toTorricelli’s law.RR hrhProblem Set 1.3. Tank in Problem 36

im01.qxd9/21/0510:17 AMPage 9Instructor’s Manual9SECTION 1.4. Exact ODEs. Integrating Factors, page 19Purpose. This is the second “big” method in this chapter, after separation of variables, andalso applies to equations that are not separable. The criterion (5) is basic. Simpler casesare 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 twovariables x, y. Simple connectedness of D follows from our assumptions in Sec. 1.4. Hencethe differential form is exact by Theorem 3, Sec. 10.2, part (b) and part (a), in that order.Method of Integrating FactorsThis greatly increases the usefulness of solving exact equations. It is important in itselfas well as in connection with linear ODEs in the next section. Problem Set 1.4 will helpthe student gain skill needed in finding integrating factors. Although the method hassomewhat the flavor of tricks, Theorems 1 and 2 show that at least in some cases one canproceed systematically—and one of them is precisely the case needed in the next sectionfor linear ODEs.SOLUTIONS TO PROBLEM SET 1.4, page 252. (x y) dx (y x) dy 0. Exact; the test gives 1 on both sides. Integratex y over x:u 12x 2 xy k(y).Differentiate this with respect to y and compare with N:uy x k y x.Thusk y,k 12 y 2 c*.Answer: 12x 2 xy 12 y 2 12(x y)2 c; thus y x c.4. Exact; the test gives ey e x on both sides. Integrate M with respect to x to getu xey ye x k(y).Differentiate this with respect to y and equate the result to N:uy xey e x 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).uy e x sin y k .Differentiate:Equate this to N e x sin y. Hence k 0, k const. Answer: e x 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 xy k(y).yxDifferentiate this with respect to y and equate the result to N:uy x1 k N,2 yxk 2y,Answer:x2 xy y 2 c.yxk y 2.

im01.qxd9/21/0510:17 AMPage 1010Instructor’s Manual10. Exact; the test gives 2x sin (x 2) on both sides. Integrate N with respect to y to getu 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 havePy Qx ex y 1 ex y(x 1) xex y 1 Qso that R 1, F e x, and the exact ODE is(ey ye x) dx (xey e x) dy 0.The test gives ey e x on both sides of the equation. Integration of M FP withrespect to x givesu xey ye x k(y).Differentiate this with respect to y and equate it to N FQ:uy xey e x k N xey e x.Hence k 0. Answer: xey ye x c.14. Not exact; 2y y. Try Theorem 1; namely,R (Py Qx)/Q (2y y)/( xy) 3/x.HenceF 1/x 3.The exact ODE is(x y2y) dx 2 dy 0.x3xThe test gives 2y/x 3 on both sides of the equation. Obtain u by integrating N FQwith respect to y:u y2 l(x).2x 2Thusux y2y2 l M x .x3x3Hence l x, l x 2/2, y 2/2x 2 x 2/2 c*. Multiply by 2 and use the initialcondition y(2) 1:y2x 2 2 c 3.75xbecause 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, oryou can solve it for y by the usual procedure. y(1) gives 1 c.Answer: cos xy 1.18. Try Theorem 2. You haveR* (Qx Py)/P [ 1xcos xy x sin xy ( x sin xy 2 )]yyHence F* y. This gives the exact ODE(y cos xy x) dx (y x cos xy) dy 0.P 1.y

im01.qxd9/21/0510:17 AMPage 11Instructor’s Manual11In the test, both sides of the equation are cos xy xy sin xy. Integrate M with respectto x:u sin xy 1 x 2 k(y).Henceu x cos xy k (y).y2Equate the last equation to N y x cos xy. This shows that k y; hencek y 2/2. Answer: sin xy 12x 2 12y 2 c.20. Not exact; try Theorem 2:R* (Qx Py)/P [1 (cos2 y sin2 y 2x cos y sin y)]/P [2 sin2 y 2x cos y sin y]/P 2(sin y)(sin y x cos y)/(sin y cos y x cos2 y) 2(sin y)/cos y 2 tan y.Integration with respect to y gives 2 ln (cos y) ln (1/cos2 y); hence F* 1/cos2 y.The resulting exact equation isx(tan y x) dx dy 0.cos2 yThe exactness test gives 1/cos2 y on both sides. Integration of M with respect to xyieldsxu x tan y 12x 2 k(y).From this,uy k .cos2 yEquate this to N x/cos2 y to see that k 0, k const. Answer: x tan y 1x 2 c.222. (a) Not exact. Theorem 2 applies and gives F* 1/y fromR* (Qx Py)/P (0 cos x)/(y cos x) 1.yIntegrating M in the resulting exact ODEcos x dx 1dy 0y2with respect to x givesu sin x k(y).From this,uy k N 1.y2Hence k 1/y. Answer: sin x 1/y c.Note that the integrating factor 1/y could have been found by inspection and by thefact that an ODE of the general formƒ(x) dx g(y) dy 0is always exact, the test resulting in 0 on both sides.(b) Yes. Separation of variables givesdy/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 maydraw vertical asymptotes that disturb the figure.From the solution in (a) or (b) the student should conclude that for each nonzeroy(x0) y0 there is a unique particular solution becausesin x0 1/y0 c.

im01.qxd9/21/0510:17 AMPage 1212Instructor’s Manual24. (A) ey cosh x c.(B) R* tan y, F 1/cos y. Separation:dy/cos2 y (1 2x) dx,tan y x x 2 c.(C) R 2/x, F 1/x 2, x y 2/x c. v y/x, and separation:2v dv/(1 v2) dx/x,x 2 y 2 cx;divide by x.(D) Separation is simplest. y cx 3/4. R 9/(4x), F(x) x 9/4, x 3y 4 c.R* 3/y, F*(y) y 3.SECTION 1.5. Linear ODEs. Bernoulli Equation. Population Dynamics,page 26Purpose. 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 thehomogeneous ODE of the first order is easily separated and the nonhomogeneous ODEis solved, once and for all, in the form of an integral (4) by the method of integratingfactors. Of course, in simpler cases one does not need (4), as our examples illustrate.Comment on NotationWe writey p(x)y r(x).p(x) seems standard. r(x) suggests “right side.” The notationy p(x)y q(x)used in some calculus books (which are not concerned with higher order ODEs) wouldbe shortsighted here because later, in Chap. 2, we turn to second-order ODEsy 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 wouldhave to choose another letter different from that used in the first-order case).Comment on ContentBernoulli’s equation appears occasionally in practice, so the student should rememberhow to handle it.A special Bernoulli equation, the Verhulst equation, plays a central role in populationdynamics of humans, animals, plants, and so on, and we give a short introduction to thisinteresting field, along with one reference in the text.Riccati and Clairaut equations are less important than Bernoulli’s, so we have putthem 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 thisa good place to mention them.Problems 37–42 express properties that make linearity important, notably in obtainingnew solutions from given ones. The counterparts of these properties will, of course,reappear in Chap. 2.Comment on Footnote 5Eight members of the Bernoulli family became known as mathematicians; for more details,see p. 220 in Ref. [GR2] listed in App. 1.

im01.qxd9/21/0510:17 AMPage 13Instructor’s Manual13SOLUTIONS TO PROBLEM SET 1.5, page 324. The standard form (1) is y 4y x, so that (4) gives y e 4x [ e 4xx dx c] ce 4x x/4 1/16.13y 3 . From this and (4) we obtain, withxxc 2 from the initial condition,6. The standard form (1) is y y x 3 [ x 3x 3 dx c] x 3[x c] x 2 2x 3.8. From (4) with p 2, h 2x, r 4 cos 2x we obtain y e 2x [ e 2x 4 cos 2x dx c] e 2x[e 2x(cos 2x sin 2x) c].It is perhaps worthwhile mentioning that integrals of this type can more easily beevaluated by undetermined coefficients. Also, the student should verify the result bydifferentiation, even if it was obtained by a CAS. From the initial condition we obtainy( 14 ) ce /2 0 1 2;hencec e /2.The answer can be writteny e /2 2x cos 2x sin 2x.10. In (4) we have p 4x 2; hence h 4x 3/3, so that (4) gives y e 4x /3 [ e (4x33/3) x 2/ 2(4x 2 x) dx c].The int

ENGINEERING MATHEMATICS imfm.qxd 9/15/05 12:06 PM Page i. imfm.qxd 9/15/05 12:06 PM Page ii. INSTRUCTOR’S MANUAL FOR ADVANCED ENGINEERING MATHEMATICS NINTH EDITION ERWIN KREYSZIG Professor of Mathematics Ohio St