(Solution Manual) Advanced Engineering Mathematics

Kreyszig, 10

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Solution Manual

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Solution Manual for Advanced Engineering Mathematics, The downloadable file consists of 417 pages, and the total size of the file is 4.05 MB. Purchase now only for highly discounted price.

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SOLUTIONS MANUAL FOR ADVANCED ENGINEERING MATHEMATICS TENTH EDITION ERWIN KREYSZIG HERBERT KREYSZI G EDWARD J. NORMINTO NPart 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.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 of what an ODE is and what we mean by solving it. The role of initial conditions should be emphasized since, in most cases, solving an engineering problem of a physical nature usually means finding the solution of an initialvalue problem (IVP). Further points to stress and illustrate by examples are: The fact that a general solution represents a family of curves. The distinction between an arbitrary constant, which in this chapter will always be denoted by c, and a fixed constant (usually of a physical or geometric nature and given in most cases). The examples of the text illustrate the following.Example 1: the verification of a solution Examples 2 and 3: ODEs that can actually be solved by calculus with Example 2 giving an impression of exponential growth (Malthus!) and decay (radioactivity and furtherapplications in later sections) Example 4: the straightforward solution of an IVPExample 5: a very detailed solution in all steps of a physical IVP involving a physical constant k Background Material. For the whole chapter we need integration formulas and techniques from calculus, which the student should review. General Comments on Text 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. Instead of (3), one could perhaps take a third-orderlinear ODE with constant coefficients or an Euler–Cauchy equation, both not of great interest.The present (3) is included to have a nonlinear ODE (a concept that will be mentioned laterwhen we actually need it); it is not too difficult to verify that a solution is with arbitrary constants a,b,c,d.y/H11005ax/H11001b cx/H11001d 1im01_demo.qxd 8/2/10 7:57 PM Page 1Problem Set 1.1 will help the student with the tasks of Solving by calculus Finding particular solutions from given general solutionsSetting up an ODE for a given function as solution, e.g., Gaining a first experience in modeling, by doing one or two problemsGaining 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 allsolutions, that is, both the particular and the singular ones. We do not adopt this definition for two reasons. First, it is frequentlyquite difficult to prove that a formula includes allsolutions; 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 (asmentioned in the text), so that for these equations a general solution as defined does includeall solutions. For the latter reason, some books use the term “general solution” for linearequations only; but this seems very unfortunate. SOLUTIONS TO PROBLEM SET 1.1, page 8 2. 4.6. 8. 10. 12.14.16.Substitution of into the ODE gives . Similarly, . 18. . 20.kfollows from . Answer: . Since the decay is exponential, would give . (y 0>2)>2/H110050.25y036,000 /H110052#18,000 e35,000 k/H110050.26y0e18,000 k/H110051 2, k/H11005(ln 1 2)>18,000 /H11005/H11002 0.000039e/H115463.6k/H110051 2˛, k/H110050.19254, (a) e/H11546k/H110050.825, (b) 3.012#10/H1154631y/H110051 4 x2, yr/H110051 2 x, thus 1 4 x2/H11002x (1 2 x)/H110011 4 x2/H110050yr2/H11002xyr/H11001y/H11005c2/H11002xc/H11001(cx/H11002c2) /H110050y/H11005cx/H11002c2y/H110054/H110024 sin2 xy2/H110024x2/H1100512y/H11005pe/H115462.5x2y/H11005/H110021 0.23 e/H115460.2x/H11001c1x2/H11001c2x/H11001c3y/H11005a cos x/H11001b sin xy/H11005ce/H115461.5xy/H11005e/H11546x2>2/H11001cy/H11005exyr/H11005f (x)2 Instructor’s Manualim01_demo.qxd 8/2/10 7:57 PM Page 2SECTION 1.2. Geometric Meaning of . Direction Fields, Euler’s Method, 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 formalmanipulations of solution methods, the latter being restricted to relatively small—albeitimportant—classes of ODEs. This approach is becoming increasingly important, especiallybecause 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 differentiationand integration. 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. Euler’s method has been included for essentially two reasons, namely, as an early eye opener to the possibility of numerically obtaining approximate values of solutions by step-by-step computations and, secondly, to enhance the student’s conceptual geometricunderstanding of the nature of an ODE. Furthermore, the inaccuracy of the method will motivate the development of much more accurate methods by practically the same basic principle (in Sec. 21.1). Problem Set 1.2 will help the student with the tasks of: Drawing direction fields and approximate solution curves Handling your CAS in selecting appropriate windows for specific tasksA first look at the important Verhulst equation (Prob. 4)Bell-shaped curves as solutions of a simple ODEOutflow from a vessel (analytically discussed in the next section)Discussing a few types of motion for given velocity (Parachutist, etc.)Comparing approximate solutions for different step size SOLUTIONS TO PROBLEM SET 1.2, page 11 2.Ellipses . If your CAS does not give you what you expected, change the given point. 4.Verhulst equation, to be discussed as a population model in Sec. 1.5. The given points correspond to constant solutions , an increasing solution through , and a decreasing solution through . 6.Solution , not needed for doing the problem. 8.ODE of the bell-shaped curves . 10.ODE of the outflow from a vessel, to be discussed in Sec. 1.3. 12. , not needed to do the problem. 14. , not needed to do the problem. 16. (a) Your PC may give you fields of varying quality, depending on the choice of the region graphed, and good choices are often obtained only after some trial and error.Enlarging generally gives more details. Subregions where is large are usuallycritical and often tend to give nonsense.ƒ yrƒy(x) /H11005sin (x/H110011 4 p)y/H1100521t/H110011y/H11005ce/H11546x2y(x)/H11005/H11002 arctan [1>(x/H11001c)](0, 3) (0, 1)[(0, 0) and (0, 2)]x2/H110011 4 y2/H11005cyr/H11549f (x, y)Instructor’s Manual 3im01_demo.qxd 8/2/10 7:57 PM Page 3(b) . Your CAS will produce the direction field well, even at points of the x-axis where the tangents of solution curves are vertical. (c) (not needed for doing the problem). (d) by remembering calculus. 18. . The computed value for shows that its error has decreased by about a factor 10. This is typical for this “first-order method” (Euler’s method), as will beseen in Sec. 21.1.x/H110050.1 y/H11005e xy/H11005ce/H11546x>2y2/H11001x2/H11005c2x/H1100118yyr/H1100504 Instructor’s Manual Error Error in Prob. 17 0.01 1.010000 1.010050 0.000050 0.02 1.020100 1.020201 0.0001010.03 1.030301 1.030455 0.0001540.04 1.040604 1.040811 0.0002070.05 1.051010 1.051271 0.0002610.06 1.061520 1.061837 0.0003170.07 1.072135 1.072508 0.0003730.08 1.082857 1.083287 0.0004300.09 1.093685 1.094174 0.0004890.10 1.104622 1.105171 0.000549 0.005171y(x n) yn xn Error 0 1.0000 0 0.2 1.0000 0.00000.4 0.9984 0.00830.6 0.9729 0.04530.8 0.8502 0.09721.0 0.5541 0.05411.2 0.2471 0.03961.4 0.1205 0.03631.6 0.0647 0.02231.8 0.0373 0.01302.0 0.0227 0.0076/H11002/H11002/H11002/H11002y n xn20.The error is first negative, then positive, and finally decreases as the solution (which is decreasing for all positive x) approaches the limit 0. The computed values are: 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 atransformation of the ODE, namely, by introducing a new unknown function. The section includes standard applications that lead to separable ODEs, namely, 1–3. Three simple separable ODEs with solutions involving , an exponential function, (bell-shaped curves) 4.The ODE of the exponential function, having various applications, such as in radiocarbon datinge /H11546x 2tan xim01_demo.qxd 8/2/10 7:57 PM Page 45.A mixing problem for a single tank 6.Newton’s law of cooling 7.Torricelli’s law of outflow In reducing to separability we consider 8.The transformation , giving perhaps the most important reducible class of ODEs 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 representing semi-ellipses in the upper half-plane, andrepresenting semi-ellipses in the lower half-plane. [Similarly, we can get two explicit solutions representing semi-ellipses in the left and right half-planes, respectively.] Onthe x-axis, the tangents to the ellipses are vertical, so that does not exist. Similarly for 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 in a domain Dof the xy-plane can be factored in D, , if and only if in D, [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 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 awhole 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 practicalimportance 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 calculusappeared in 1684.f xy f/H11005fx fyf (x, y)/H11005g(x)h(y)f (x, y)xr(y)yr(x)x(y)y/H11005/H110022c/H11002(6x)2y/H11005/H110012c/H11002(6x)2u/H11005y>xInstructor’s Manual 5im01_demo.qxd 8/2/10 7:57 PM Page 5SOLUTIONS TO PROBLEM SET 1.3, page 18 2. , so that multiplication by 4 gives the answer . These are curves that lie between a circle and a square, outside the circle and inside the square that touch the circle at the points of intersection with theaxes. The figure shows a quarter of such a curve for . c/H110051y 4/H11001x4/H11005cy3 dy/H11005/H11002 x3 dx, 1 4 y4/H11005/H110021 4 x4/H11001c~6 Instructor’s Manual y 11 0 t Sec. 1.3. Prob. 2. Quarter of the solution curve 4.Separation, integration, and taking exponents gives , and . 6.Separation of variables, integration, and taking the reciprocal gives . 8.From the ODE and the suggested transformation we obtain . Separation of variables and integration gives . This implies and gives the answer 10.From the transformation and the ODE we have . Separation of variables, integration, and again using the transformation gives . 12.Separation of variables and integration gives .dy 1/H110014y2/H11005dx and 1 2 arctan 2 y/H11005x/H11001c~du/H11005dx>x, u/H11005ln x/H11001c, y/H11005ux/H11005x (ln x/H11001c)yr/H11005urx/H11001u/H110051/H11001y x/H110051/H11001u, hence urx/H110051y/H11005v/H110024x/H110052 tan (2 x/H11001c)/H110024x.v/H110052 tan (2 x/H11001c)dv v2/H110014/H11005dx and 1 2 arctan v 2/H11005x/H11001c~yr/H11005vr/H110024/H11005v2, hence vr/H11005v2/H110014dy y2/H11005e2x/H115461 dx, /H110021 y/H110051 2 e2x/H115461/H11001c~ y/H110052 c/H11002e2x/H115461y/H11005c1sin 2pxln ƒyƒ/H110051 2 ln ƒsin 2pxƒ/H11001c dy>y/H11005p cot 2px dx,im01_demo.qxd 8/2/10 7:57 PM Page 6Hence arctan . Solving for ygives the general solution and from the initial condition. 14. . The general solution is and from the initial condition. 16.From the transformation and the ODE we have . Hence . Separation of variables and integration gives hence . From this and the transformation we obtain . From the initial condition we get , so that the answer is . 18.On the left, integrate gfrom to y. On the right, integrate over xfrom to x. In Prob. 12, . 20.Let and be the constants of proportionality for the birth rate and death rate, respectively. Then , where is the population at time t. By separating variables, integrating, and taking exponents, , ln . 22.The acceleration is , and the distance traveled is 5.5 meters. This is obtained as follows. Since (i.e., we count time from the instant theparticle enters the accelerator), we have for a motion of constant acceleration (A)and the velocity is . From the given data we thus obtain and so that . Finally, with this a and that b, from (A) we get . s(10 /H115463)/H110059#106#10/H115466 2/H11001103#10/H115463/H110055.5 [m]a/H11005103(104/H11002103)/H11005107/H11002106/H110059/H11554106v(10/H115463)/H1100510/H115463a/H11001103/H11005104v(0)/H11005b/H11005103v(t)/H11005sr(t)/H11005at/H11001bs(t)/H11005a t2 2/H11001bts(0)/H110050a/H110059/H11554106 meters>sec2y/H11005(kB/H11002kD)t/H11001c*, y/H11005ce(kB/H11546kD)tdy>y/H11005(kB/H11002kD) dty(t) yr/H11005kBy/H11002kDykD kB/H20910y 3w dw/H11005/H20910x 2 (/H110024t) dtx0 f (x) y0y/H11005tan x/H11002x/H110012y(0)/H110052/H11001tan c/H110050 and c/H110050y/H11005v/H11002x/H110012/H110052/H11002x/H11001tan (x/H11001c)v/H11005tan (x/H11001c)dv v2/H110011/H11005dx and arctan v/H11005x/H11001cvr/H11005v2/H110011y/H11005v/H11002x/H110012 and yr/H11005vr/H110021/H11005v2c/H11005r0 r/H11005ce/H11546t2 dr r/H11005/H11002 2t dt, ln r/H11005/H11002 t2/H11001c~c/H11005/H11002 2y/H110051 2 tan (2 x/H11001c)2y/H110052x/H11001cInstructor’s Manual 7im01_demo.qxd 8/2/10 7:57 PM Page 724.Let be the amount of salt in the tank at time t. Then each gallon contains lb of salt. gal of water run in during a short time , and is the loss of salt during . Thus Answer: . 26.The model is ln y with . Constant solutions are obtained from when . Between 0 and 1 the right side is positive (since ln ), sothat the solutions grow. For we have ln ; hence the right side is negative,so that the solutions decrease with increasing t. It follows that is stable. The general solution is obtained by separation of variables, integration, and two subsequentexponentiations: , 28.This follows from the inquality . 30.Acceleration . Hence (initial speed of further flight end speed upon return from peak), (heightreached after the 10 sec). At the peak, , , say; thus for the further flight(measured from the peak), (see before).This gives the further flight time to the peak and the furtherheight , approximately. Answer: 32. in Fig. 15 is the weight (the force of attraction acting on the body). Its component parallel to the surface in , and . Hence thefriction is , and it acts against the direction of motion. From this andNewton’s second law, noting that the acceleration is ( vthe velocity), we obtain The mass mdrops out, and two integrations give . Since the slide is 10 meters long, the last equation with gives the time . From this we obtain the answer . 34. TEAM PROJECT. (a) Note that at the origin, , so that is undefined at the origin. (b) . (c) . Here the student should learn that cmust not appear in the ODE. . y>x/H11005cy r>x/H11002y>x2/H110050, yr/H11005y>xy/H11005cx(xy)r/H11005y/H11001xyr/H110050, yr/H11005/H11002 y>xyr x>y/H110050>0v/H110053.203#2.50 /H110058.01 [meters>sec]t/H1100512#10>3.203 /H110052.50s/H1100510v/H110053.203 t and s/H110053.203 t2 2 /H110053.203 m. /H11005m#9.80(0.500 /H110020.2#0.866) m dv dt/H11005mg sin a/H110020.2mg cos adv>dt0.2mg cos aN/H11005mg cos a mg sin aW/H11005mg1167 /H110016245 /H110057412 [m]. s(t1)/H110054.9t12/H110056245t/H11005t1/H11005350>9.8/H1100535.7s(t)/H11005(g>2)t2/H110054.9t2, v(t)/H110059.8t/H11005350s/H110050 v/H110050y(10) /H110057000>6/H110051167 /H11005yr/H110057t2>2, y/H110057t3>6, yr(10) /H11005350 ys/H110057t1>26/H110050.016 /H110220.010 /H110221>27/H110050.0078 ln y/H11005ce/H11546At, y/H11005exp ( ce/H11546At). dy>(y ln y)/H11005/H11002 A dt, ln (ln y)/H11005/H11002 At/H11001c*y/H110051y/H110220 y/H110221y/H110210 y/H110050 and 1yr/H110050 A/H110220 yr/H11005/H11002 Ayy(60) /H11005100e/H115460.3/H1100574 [lb]100e/H115460.005 t. ¢y>¢t/H11005/H11002 y>200, yr/H11005/H11002 0.005 y, y(t)/H11005 ¢t/H11002¢y/H110052¢t (y>400) /H11005¢t y>200 ¢t 2¢ty>400 y(t)8 Instructor’s Manualim01_demo.qxd 8/2/10 7:57 PM Page 8(d)The right sides and are the slopes of the curves. Orthogonality is important and will be discussed further in Sec. 1.6. (e)No. 36. Team Project. Bnow depends on h, namely, by the Pythagorean theorem, . Hence you can use the ODE in the text, with constant Aas before and the new B. The latter makes further calculations different from those in Example 5. From the given outlet size and we obtain . Now , so that separation of variables gives By integration,From this and the initial conditions we obtain . Hence the particular solution (in implict form) is . The tank is empty for tsuch that ; hence . For this gives . The tank has water level for tin the particular solution such that . The left side equals . This gives . For this yields . This is slightly more than half the time needed toempty the tank. This seems physically reasonablebecause if the water level is R2, this means that of the total water volume has flown out, and isleft—take into account that the velocity decreasesmonotone according to Torricelli’s law.5>1611>16 >t/H110051260 sec /H1100521 min R/H11005100t/H110050.4007 /H110020.9333 /H1100242.27 R5>2/H110050.01260 R5>20.4007 R5>24 3 R R3>2 23>2/H110022 5 R5>2 25>2/H110050.9333 R5>2/H1100242.27 tR>2t/H110050.0221#1005>2/H110052210 [sec] /H1100537 [min]R/H110051 m/H11005100 cmt/H110050.9333 42.27 R5>2/H110050.0221 R5>20/H11005/H11002 42.27 t/H110010.9333 R5>2(h/H110050)4 3 Rh3>2/H110022 5 h5>2/H11005/H11002 42.27 t/H110010.9333 R5>24 3 R5>2/H110022 5 R5>2/H110050.9333 R5>2/H11005ch(0)/H11005R4 3 Rh3>2/H110022 5 h5>2/H11005/H11002 42.27 t/H11001c.(2Rh1>2/H11002h3>2) dh/H11005/H11002 42.27 dt.26.56#5>p/H1100542.27dh dt/H11005/H11002 26.56#5 p(2Rh/H11002h2) 1hB(h) A/H110055 cm2hr/H11005/H11002 26.56( A>B)1hB(h)/H11005pr2/H11005p(R2/H11002(R/H11002h)2)/H11005p(2Rh/H11002h2)y>x /H11002x>yInstructor’s Manual 9 RR = h r h Problem Set 1.3. Tank in Team Project 36im01_demo.qxd 8/2/10 7:57 PM Page 9SECTION 1.4. Exact ODEs. Integrating Factors, page 20 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. Simplercases are solved by inspection, more involved cases by integration, as explained inthe 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 Dfollows from our assumptions in Sec. 1.4. 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 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. In Example 2, exactness is seen from . In Example 3, separation of variables gives . SOLUTIONS TO PROBLEM SET 1.4, page 26 2.Exact, . Note that an ODE is always exact. 4.Exact. The test gives . By integration, . Hence 6.The new ODE is . It is exact, . M y/H11005Nx/H110056(y/H110011)x/H1154643(y/H110011)2x/H115464 dx/H110022(y/H110011)x/H115463 dy/H110050uu/H110053re3u/H11001cr/H110053re3u, cr/H110050, c/H11005constu/H11005/H20910e3u dr/H11005re3u/H11001c(u)3e3u/H110053e3uf (x) dx/H11001g(y) dy/H110050x4/H11001y4/H11005cdy y/H11005dx x, y/H11005cx0 0x (/H11002sin y cosh x)/H11005/H11002 sin y sinh x0 0y cos y sinh x/H110011/H11005/H11002 sin y sinh x.s10 Instructor’s Manualim01_demo.qxd 8/2/10 7:57 PM Page 10The general solution is . 8.Exact; the test gives on both sides. Integrate Mwith respect to x: Differentiate: . Equate this to . Hence const . Answer: . 10. is exact because . By inspection or systematically, . 12. shows exactness. By integration, gives . Answer: . 14.The integrating factor gives the exact ODE . The general solution is and from the initial condition. 16. Team Project. (a) (b) (c) , and separation: divide by x. (d)Separation is simplest. 18. CAS Project. (a) Theorem 1 does not apply. Theorem 2 gives The exact ODE is as one could have seen by inspection—any equation of the form f (x) dx/H11001g(y) dy/H110050y/H115462 dy/H11002sin x dx/H110050,1 F dF dy/H11005/H110021 y2 sin x (0/H110012y sin x)/H11005/H110022 y, F/H11005exp/H20910/H110022 y dy/H110051 y2.R*/H110053>y, F*(y)/H11005y3.x3y4/H11005c. F(x)/H11005x/H115469>4, R/H11005/H11002 9>(4x), y/H11005cx/H115463>4.2v dv>(1/H11002v2)/H11005dx>x, x2/H11002y2/H11005cx;R/H11005/H11002 2>x, F/H110051>x2, x/H11002y2>x/H11005c, v/H11005y>xdy>cos2 y/H11005/H11002 (1/H110012x) dx, tan y/H11005/H11002 x/H11002x2/H11001c.R*/H11005tan y, F/H110051>cos y. Separation:ey cosh x/H11005cc/H110051xa/H115451yb/H115451/H11005c(a/H110011) xayb/H115451 dx/H11001(b/H110011) xa/H115451yb dy/H11005d(xa/H115451yb/H115451)/H110050y/H110052e/H11546x2c/H110052 y(0)/H110052yex2/H11005c.(2xyex2)y/H110052xex2/H11005(ex2)xy sin ( x/H11001y)/H11005c /H11005[y cos ( x/H11001y)/H11001sin (x/H11001y)]x [y cos ( x/H11001y)]y/H11005cos ( x/H11001y)/H11002y sin ( x/H11001y)y cos ( x/H11001y) dx/H11001[y cos ( x/H11001y)/H11001sin (x/H11001y)] dy/H110050ex cos y/H11005c kr/H110050, k/H11005 N/H11005/H11002 ex sin yuy/H11005/H11002 ex sin y/H11001kr u/H11005ex cos y/H11001k(y)./H11002ex sin y(y/H110011) 2x/H115463/H11005cInstructor’s Manual 11im01_demo.qxd 8/2/10 7:57 PM Page 11is exact! We now obtain (b)Yes, . (c)The vertical asymptotes that some CAS programs draw disturb the graph. From the solution in (b) the student should conclude that for each initial condition with there is a unique particular solution because from (b), . (d) . SECTION 1.5. Linear ODEs. Bernoulli Equation. Population Dynamics, page 27 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 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 Notation We write seems standard, suggests “ right side. ” The notation 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 where we need 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 thisinteresting field, along with one reference in the text.q(x)y s/H11001p(x)yr/H11001q(x)y/H11005r(x),yr/H11001p(x)y/H11005q(x)r(x) p(x)yr/H11001p(x)y/H11005r(x).y/H110130c/H33522/H110051/H11002y0 cos x0 y0y0/HS110050y(x0)/H11005y0yr/H11005y2 sin x, dy y2/H11005sin x dx, /H110021 y/H11005/H11002 cos x/H11001c, y/H110051 cos x/H11001c/H33522˛ u/H11005cos x/H110021 y/H11005c. uy/H11005kr(y)/H110051 y2, k/H11005/H110021 y, u/H11005/H20910/H11002sin x dx/H11005cos x/H11001k(y)12 Instructor’s Manualim01_demo.qxd 8/2/10 7:57 PM Page 12Riccati 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 15–20 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 7 Eight members of the Bernoulli family became known as mathematicians; for more details, see p. 220 in Ref. [GenRef 2] listed in App. 1. Examples in the Text. The examples in the text concern the following. Example 1 illustrates the use of the integral formula (4) for the linear ODE (1).Example 2 deals with the RL-circuit for which the underlying physics is rather simple and straightforward and the solution exhibits exponential approach to a constant value . Several particular solutions are shown in Fig. 19. Example 3 on hormone level is an input–output problem, eventually giving a periodic steady-state solution, after an exponential term has decreased to zero, theoretically as , practically after a very short time, as shown in Fig. 20. Example 4 concerns the logistic or Verhulst ODE, perhaps the practically most important case of a Bernoulli ODE. The Bernoulli ODE is reduced to a linear ODE by setting , giving (10). Example 5 concerns population dynamics, based on Malthus’s and Verhulst’s ODEs, both of which are autonomous. This concept is defined in connection with (13) and willbe of central interest in the theory and application of systems of ODEs in Chap. 4, inparticular, in Sec. 4.5 when we shall discuss the Lotka–Volterra population model. Problem Set 1.5 stikes a balance between formal problems (3–13) for linear ODEs, experimentation (Prob. 14), some basic theory (15–21), formal problems (22–28) fornonlinear ODEs, a project (29) on transformation, two ODEs of lesser importance (Clairautand Riccati ODEs in Team Project 30, showing singular solutions), and, finally, a varietyof modeling problems (31–40) taken from various fields. SOLUTIONS TO PROBLEM SET 1.5, page 34 4.The standard form (1) is , so that (4) gives 6.From (4) with we obtain . 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 bydifferentiation, even if it was obtained by a CAS. From the initial condition we obtain hence The answer can be written y/H110052e p>2/H115462x/H11001cos 2 x/H11001sin 2 x.c/H110052ep>2. y (1 4 p)/H11005ce/H11002p>2/H110010/H110011/H110053;y/H11005e/H115462xc/H20910e2x 4 cos 2 x dx/H11001cd/H11005e/H115462x[e2x(cos 2 x/H11001sin 2 x/H11001c]p/H110052, h/H110052x, r/H110054 cos 2 x/H11005ce2x/H110012x/H110011. y/H11005e2x c/H20910e/H115462x (/H110024x) dx/H11001cdyr/H110022y/H11005/H11002 4xu/H11005y1/H11546a (a/HS110051)t:/H11009(48>11 A)Instructor’s Manual 13im01_demo.qxd 8/2/10 7:57 PM Page 138.In (4) we have , so that (4) gives The initial condition gives ; hence c 100. The particular solution is The factor 0.01, which we include in the exponent, has the effect that the graph of y shows a long transition period. Indeed, it takes to let the exponential function decrease to 0.01. Choose the x-interval of the graph accordingly. 10.The standard form (1) is Hence , and (4) gives the general solution To evaluate the integral, observe that the integrand is of the form that is,Hence the integral has the value . This gives the general solution . The initial condition gives from this hence The answer is . 12. is the general solution. The initial condition gives . 14. CAS Experiment (a) if is undefined at , the point at which the “waves” of accumulate; the factor xmakes them smaller and smaller. Experiment with various x-intervals. (b) need not be an integer. Try . Try and see how the “waves” near 0 become larger and larger. 16.Substitution gives the identity . 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 ODEover a nonlinear one if the model is a faithful mathematical representation of theproblem. Furthermore, these problems illustrate the difference between homogeneousand nonhomogeneous ODEs.0/H110050n/H11005/H11002 1n/H11005 1 2 y/H11005xn3sin (1>x)/H11001c4. y(2>p)/H11005(2>p)n. nsin (1>x) x/H110050y(2>p)/H110052>p. y y/H11005x sin (1>x)/H11001cx. c/H110050c/H110051 y/H11005cx/H115464 /H11001 x4y/H110051 3/H11001e3/H115463 tan xc/H11005e3. y(1 4p)/H110051 3/H11001ce/H110023/H110054 3 ˛;y/H11005e/H115463 tan x [1 3 e3 tan x/H11001c]/H110051 3/H11001ce/H115463 tan x1 3 e3 tan x1 3 (e3 tan x)r.1 3 (3 tan x)r e3 tan x;y/H11005e/H115463 tan xc/H20910e3 tan x cos2 x dx/H11001cd.h/H110053 tan xyr/H110013 cos2 x y/H110051 cos2 x .e/H115460.01xx/H11005460y/H11005100 (1/H11002e/H110020.01x ) cos x./H11005 y(0)/H11005/H11002 100/H11001c/H110050y/H11005(cos x)c/H20910cos x cos x e/H110020.01x dx/H11001cd/H11005[/H11002100 e/H110020.01x/H11001c] cos x.p/H11005tan x, h/H11005/H11002 ln (cos x), eh/H110051>cos x14 Instructor’s Manualim01_demo.qxd 8/2/10 7:57 PM Page 1418.We obtain 20.The sum satisfies the ODE with on the right. This is important as the key to the method of developing the right side into a series, then finding the solutionscorresponding to single terms, and finally, adding these solutions to get a solution ofthe given ODE. For instance, this method is used in connection with Fourier series,as we shall see in Sec. 11.5. 22. Bernoulli equation. First solution method: Transformation to linear form. Set Then . Multiplication by gives the linear ODE in standard form . General solution . Hence the given ODE has the general solution . From this and the initial condition we obtain . Second solution method: Separation of variables and use of partial fractions. . Integration gives ln . Taking exponents on both sides, we obtain . We now continue as before. 24. ; hence 26.This ODE can simply be solved by separating variables, hence with from the initial condition. As an alternative, we can regard it as an ODE for the unknown function and solve it by (4) with xand yinterchanged.x/H11005x(y)c/H11005/H11002 1y/H11005arcsin [ cˆ1x/H1100212] or x/H110051/H11001c sin ycot y dy/H11005dx>1x/H1100212, ln ƒsin yƒ/H11005ln ƒx/H110021ƒ/H11001c/H33522u/H11005e/H115462xc/H11002/H20910 e2x 2x dx/H11001cd/H110051 2/H11002x/H11001ce/H115462x, y/H110051uu/H11005y2, yyr/H11001y2/H11005/H11002 x, 1 2 ur/H11001u/H11005/H11002 x, ur/H110012u/H11005/H11002 2xy/H110021 y/H110051/H110021 y/H11005c/H33522ex, 1 y/H110051/H11002c/H33522ex, y/H110051 1/H11001cexƒy/H110021ƒ/H11002ln ƒyƒ/H11005ln `y/H110021 y`/H11005x/H11001c*dy y(y/H110021)/H11005a1 y/H110021/H110021 yb dy/H11005dxy(0)/H110051>(c/H110011)/H11005/H110021 3, c/H11005/H11002 4, Answer: y/H110051>(1/H110024ex)y(0)/H11005/H110021 3,y/H110051>(cex/H110011)u/H11005cex/H110011 ur/H11002u/H11005/H11002 1/H11002u2yr/H11001y/H11005/H11002 ur>u2/H110011>u/H110051>u2y/H110051>u.r1/H11001r2 /H110050. /H11005r/H11002r /H11005(yr1/H11001py1)/H11002(yr2/H11001py2) ( y1/H11002y2)r/H11001p(y1/H11002y2)/H11005yr1/H11002yr2/H11001py1/H11002py2Instructor’s Manual 15im01_demo.qxd 8/2/10 8:20 PM Page 1528.Using the given transformation , we obtain the linear ODE which we can solve by (4) with zinstead of y, . From this we obtain . 30. Team Project. (a) reduces the Riccati equation to a Bernoulli equation by removing the term . The second transformation, , is the usual one fortransforming a Bernoulli equation with on the right into a linear ODE. Substitute into the Riccati equation to get . Since Yis a solution, . There remains . Multiplication by gives . Reshuffle terms to get , the linear ODE as claimed. (b)Substitute to get , which is true. Now substitute . This gives . Most of the terms cancel on both sides. There remains . Multiplication by finally gives . The general solution is and . Of course, instead performing this calculation we could have used the general formula in (a), in which and . (c)By differentiation, . By substitution, , a family of straight lines. (B) . By substitution into the given ODE, , the envelope of the family; see Fig. 6 in Problem Set 1.1. 32. follows from Newton’s law of cooling. models the effect of heating or cooling. calls for cooling; hence should be negativein this case; this is true, since is assumed to be negative in this formula. Similarlyfor heating, when heat should be added, so that the temperature increases. The given model is of the form . T r/H11005kT/H11001K/H11001k1C cos (p>12)tk2k2(T/H11002Tw) T/H11022Twk2(T/H11002Tw) k1(T/H11002Ta)x2>4/H11002x2>2/H11001x2>4/H11001c*/H110050, c*/H110050, y/H11005x2>4yr/H11005x>2, y/H11005x2>4/H11001c*c2/H11002xc/H11001cx/H11001a/H110050, a/H11005/H11002 c2, y/H11005cx/H11002c2y/H11005cx/H11001a2yrys/H11002yr/H11002xys/H11001yr/H110050, ys(2yr/H11002x)/H110050. (A) ys/H110050,/H11002g/H11005/H11001 x22Yg/H11002p/H110052x(/H11002x2)/H110012x3/H110011/H110051y/H11005x/H110011>uu/H11005ce/H11546x/H11001x2/H110022x/H110012ur/H11001u/H11005x2/H11002u2/H11002ur>u2/H110021>u/H11005/H11002 x2>u21/H11002ur>u2/H11002(2x3/H110011)(x/H110011>u)/H11005/H11002 x2(x2/H110012x>u/H110011>u2)/H11002x4/H11002x/H110011y/H11005x/H110011>u1/H110022x4/H11002x/H11005/H11002 x4/H11002x4/H11002x/H110011 y/H11005Y/H11005xur/H11001(2Yg/H11002p)u/H11005/H11002 gur/H11002pu/H11005/H11002 g(2Yu/H110011) /H11002u2/H11002ur>u2/H11001p>u/H11005g ( ˛2Y>u/H110011>u2)Y r/H11001pY/H11005gY2/H11001hYr/H11002ur>u2/H11001p1Y/H110011>u2/H11005g ˛(Y2/H110012Y>u/H110011>u22/H11001hy/H11005Y/H110011>uy2v/H110051>u h(x)y/H11005Y/H11001vy/H110051zz/H11005xe/H11546x a/H209101 x exxex dx/H11001cb/H11005xe/H11546x11 2 e2x/H11001c2/H11005cxe/H11546x/H110011 2 xexzr/H11001a1/H110021 xb z/H11005xex,y2/H11005z16 Instructor’s Manualim01_demo.qxd 8/2/10 7:57 PM Page 16This can be seen by collecting terms and introducing suitable constants, (because there are two terms involving T), and . The general solution is , where . The first term solves the homogeneous ODE and decreases to zero. The second term results from the constants A(in ), , and P. The third term is sinusoidal, of period 24 hours, and time-delayed against the outsidetemperature, as is physically understandable. 34. , where and yis the proportion of infected persons. Equilibrium solutions are and . The first, , is unstable because for negative y. The solution is stable because . The general solution is It approaches 1 as . This means that eventually everybody in the population will be infected. 36.The model is where . Hence the general solution is given by (12) in Example 4 with A replaced by . The equilibrium solutions are obtained from ; hencethey are and . The population remains unchanged underharvesting, and the fraction of it can be harvested indefinitely—hence the name. 38.For the first 3 years you have the solution from Prob. 36. The idea now is that, by continuity, the value at the end of the first period is the initial value for the solution during the next period. That is, . Now is the solution of (no fishing!). Because of the initial condition this gives . Check the continuity at by calculating . Similarly, for tfrom 6 to 9 you obtain . This is a period of fishing. Check the continuity at : . This agrees with . y 2(6)/H110054>(4/H11001e/H115463/H110023e/H115465.4)y3(6)/H110054>(5/H11002e0/H11001e/H115463/H110023e/H115465.4)t/H110056y3/H110054>(5/H11002e4.8/H115460.8t/H11001e1.8/H115460.8t/H110023e/H115460.6/H115460.8t)y2(3)/H110054>(4/H11001e0/H110023e/H115462.4)t/H110053y2/H110054>(4/H11001e3/H11546t/H110023e0.6/H11546t)yr/H11005y/H11002y2y2y2(3)/H11005y1(3)/H110054>(5/H110023e/H115462.4)y2y1(3)y1/H110054>(5/H110023e/H115460.8t)Hy2y2 y2/H11005K>B y1/H110050yr/H110050 K/H11005A/H11002HK/H11005A/H11002Hyr/H11005Ay/H11002By2/H11002Hy/H11005Ky/H11002By2/H11005y(K/H11002By)t:/H11009y/H110051 1/H11001ce/H11546kt.f (y)/H110220 if 0 /H11021y/H110211 and f (y)/H110210 if y/H110221y/H110051 f (y)/H110220 if 0 /H11021y/H110211 but f (y)/H110210y/H110050 y/H110051 y/H110050k/H110220 yr/H11005ky(1/H11002y)/H11005f (y)Tw TaTr/H11005kT L/H11005k1C>(k2/H11001p2>144)T/H11005cekt/H11002K>k/H11001L(/H11002k cos (pt>12)/H11001(p>12) sin (pt>12))K/H11005/H11002 k1A/H11002k2Tw/H11001Pk/H11005k1/H11001k2Instructor’s Manual 17im01_demo.qxd 8/2/10 7:57 PM Page 1740.Let ydenote the amount of fresh air measured in cubic feet. Then the model is obtained from the balance equation “Inflow minus Outflow equals the rate of change ”; that is, . The general solution of this linear ODE is . The initial condition is (initially no fresh air) and gives ; hence . The particular solution of our problem is . This equals 90% if tis such that thus if . SECTION 1.6. Orthogonal Trajectories. Optional , page 36 Purpose. To show that families of curves can be described by ODEs and the switch to produces as general solution the orthogonal 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 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 generalsolution 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). Problem Set 1.6 should help the student to obtain skill in representing families of curves (Probs. 1–3), finding trajectories (4–10), and understanding some basic physical andgeometric applications of trajectories (11–16). This will also involve the Cauchy–Riemannequations, which are basic in complex analysis. SOLUTIONS TO PROBLEM SET 1.6, page 38 2. gives a circle of radius rwith center on the cubic parabola. Since this center has distance , we have . 4. . Note that these curves and their OTs are congruent. This is typical of ODEs with fnot depending on y. y r/H11005f (x)yr/H110052x, y~r/H11005/H11002 1>(2x), y~/H11005/H110021 2 ln x/H11001c~r2/H11005c2/H11001c6r/H110052c2/H11001c6(x0, y0)/H11005(c, c3) (x/H11002c)2/H11001(y/H11002c3)2/H11002r2/H110050y~r/H11005/H11002 1>f (x, y~) yr/H11005f (x, y)F (x, y, c)/H110050t/H11005(ln 0.1)>(/H110020.03) /H1100577 [min]e/H115460.03t/H110050.1y/H1100520,000(1 /H11002e/H115460.03t)c/H11005/H11002 20,000 y(0)/H11005c/H1100120,000 /H110050y(0)/H110050y/H11005ce/H115460.03t/H1100120,000yr/H11005600/H11002 600 20,000 y/H11005600/H110020.03y18 Instructor’s Manualim01_demo.qxd 8/2/10 7:57 PM Page 186.Differentiating the given formula, we obtain . Thus . This is the differential equation of the given hyperbolas. Hence the differential equation of the orthogonal trajectories is . Separation of variables and integration gives . Answer: The hyperbolas are the orthogonal trajectories of the given hyperbolas. 8.Squaring the given formula, differentiating, and solving algebraically for , we obtain . This is the differential equation of the given curves. Hence the differential equation of the orthogonal trajectories is . By separation of variables and integration we obtain ln . Taking exponents gives the answer . 10. . Solve algebraically for 2 c: . Differentiation gives . By algebra, . Solve for : /H20882 . This is the ODE of the given family. Hence the ODE of the trajectories is . y~r/H11005y~2/H11002x2 2xy~/H110051 2 ay~ x/H11002x y~bay2/H11002x2 y2b/H11005/H110022xy y2/H11002x2yr/H11005/H110022x yyryra/H11002x2 y2/H110011b/H11005/H110022x y˛2x y/H11002x2yr y2/H11001yr/H110050x2/H11001y2 y/H11005x2 y/H11001y/H110052cx2/H11001y2/H110022cy/H110050y~/H11005c*e/H115462x ƒy~ƒ/H11005/H11002 2x/H11001c~y~r/H11005/H11002 2y~y2/H11002x/H11005c, 2yyr/H110051, yr/H110051 2yyrx2/H11002y~2/H11005c*y~ dy/H11005x dx, 1 2 y~2/H110051 2 x2/H11001c~y~r/H11005x y~yr/H11005/H11002y xxyr/H11001y/H110050Instructor’s Manual 19im01_demo.qxd 8/2/10 7:57 PM Page 19To solve this equation, set . Then . Subtract uon both sides to get . Now separate variables, integrate, and take exponents, obtaining . Write and multiply by on both sides of the last equation. This gives . The answer is . Note that the given circles all have their centers on the y-axis and pass through the origin. The result shows that their orthogonal trajectories are circles, too, with centerson the x-axis and passing through the origin. 12.Setting gives from the equation ; hence and , which verifies that those circles all pass through and 1,each of them simultaneously through both points. Subtracting on both sides of thegiven equation, we obtain . Emphasize to your class that the ODE for the given curves must always be free of c. Having accomplished this, we can now differentiate. This gives . This is the ODE of the given curves. Replacing with and ywith , we obtain the ODE of the trajectories: /H20882 . Multiplying this by , we get . Multiplying this by , we obtain .2y~y~r x/H110011/H110021 x2/H11002y~2 x2/H11005d dx ay~2 xb/H110011/H110021 x2/H110050y~2>x22xy~r y~/H11001x2/H110021 y~2/H110021/H110050y~r(/H11002y~r)/H1100502x y~/H11002ax2/H110021 y~2/H110021by~/H110021>y~r yr2x y/H11002ax2/H110021 y2/H110021b yr/H110050x2/H11001y2/H110022cy/H110051, x2/H11001y2/H110021/H110052cy, x2/H110021 y/H11001y/H110052cc2/H110021 x/H110051 x/H11005/H11002 1x2/H11001c2/H110051/H11001c2x2/H11001(y/H11002c) 2/H110051/H11001c2y/H110050(x/H11002c3)2/H11001y~2/H11005c32y~2/H11001x2/H11005c2xx2u/H11005y~>x2u du u2/H110011/H11005/H11002dx x, ln (u2/H110011)/H11005/H11002 ln ƒxƒ/H11001c1, u2/H110011/H11005c2 xxur/H11005/H11002u2/H110011 2uy~r/H11005xur/H11001u/H110051 2 au/H110021 ubu/H11005y~>x20 Instructor’s Manualim01_demo.qxd 8/2/10 7:57 PM Page 20By integration, . Thus, . We see that these are the circles dashed in Fig. 25, as claimed. 14.By differentiation, . Hence the ODE of the orthogonal trajectories is . By separation, . Integration and taking exponents gives ln . This shows that the ratio has substantial influence on the form of the trajectories. For the given curves are circles, and we obtain straight lines as trajectories. gives quadratic parabolas. For higher integer values of we obtain parabolas of higher order. Intuitively, the “flatter” the ellipses are, the more rapidlythe trajectories must increase to have orthogonality. Note that our discussion also covers families of parabolas; simply interchange the roles of the curves and their trajectories. Note further that, in the light of the present answer, our example in the text turns out to be typical. 16. . Since cis just an additive constant, the statement about the curves follows; these curves are obtained from any one of them by translation in the y-direction. Similarly for the OTs, whose ODE is with the function on the right independent of . SECTION 1.7. Existence and Uniqueness of Solutions for Initial Value Problems, page 38 Purpose. To give the student at least some impression of the theory that would occupy a central position in a more theoretical course on a higher level. Short Courses. This section can be omitted. Comment on Iteration Methods Iteration methods were used rather early in history, but it was Picard who made them popular. Proofs of the theorems in this section (given in books of higher level, e.g., [A11])are based on the Picard iteration (see CAS Project 6). Iterations are well suited for the computer because of their modest storage demand and usually short programs in which the same loop or loops are used many times, with differentdata. Because integration is generally not difficult for a CAS, Picard’s method has gainedsome popularity during the past few decades.y~y~ r/H11005/H11002 1>f (x)y/H11005/H20848f (x) dx/H11001ca2>b2a2>b2/H110052a2/H11005b2a2>b2ƒy~ƒ/H11005a2 b2 ln ƒxƒ/H11001c**, y~/H11005c*xa2>b2dy~ y~/H11005a2 b2 dx xy~r/H11005a2y~ b2x2x a2/H110012yyr b2/H110050, yr/H11005/H110022x>a2 2y>b2/H11005/H11002b2x a2yy~2/H11001(x/H11002c*)2/H11005c*2/H110021y~2/H11001x2/H110011/H110052c*xy~2 x/H11001x/H110011 x/H110052c*Instructor’s Manual 21im01_demo.qxd 8/2/10 7:57 PM Page 21Example 1 is simple, involving only , and is typical inasmuch as it illustrates that the actual interval of existence is much larger than the interval guaranteed by ExistenceTheorem 1. Example 2 shows that IVPs violating uniqueness can be constructed relatively easily.Lipschitz and Hölder conditions play a basic role in the theory of PDEs on a level substantially higher than that of our Chap.12. SOLUTIONS TO PROBLEM SET 1.7, page 42 2.The initial condition is given at the point . The coefficient of is 0 at that point, so from the ODE we already see that something is likely to go wrong. Separatingvariables, integrating, and taking exponents gives . This last expression is the general solution. It shows that for any c. Hence the initial condition cannot be satisfied. This does not contradict the theoremsbecause we first have to write the ODE in standard form: . This shows that fis not defined when (to which the initial condition refers). 4.For we still get no solution, violating the existence as in Prob. 2. For we obtain infinitely many solutions, because cremains unspecified. Thus in this case the uniqueness is violated. Neither of the two theorems is violated in either case. 6. CAS Project. (b) (c) (d) . It approximates . General solution . (e) would be a good candidate to begin with. Perhaps you write the initial choice as ; then corresponds to the choice in the text, and you see howthe expressions in aare involved in the approximations. The conjecture is true for any choice of a constant (or even of a continuous function of x). It was mentioned in footnote 10 that Picard used his iteration for proving his existence and uniqueness theorems. Since the integrations involved in the methodcan be handled on the computer quite efficiently, the method has gained in importancein numerics. 8.The student should get an understanding of the “intermediate” position of a Lipschitz condition between continuity and (partial) differentiability. The student should also realize that the linear equation is basically simpler than the nonlinear one. The calculation is straightforward because we have f (x, y)/H11005r(x)/H11002p(x)ya/H110050 y0/H11001ayr/H11005yy/H11005(x/H11001c)2y/H110050 y/H11005(x/H110021)2, y/H110050y(x)/H110051 1/H110022x/H110051/H110012x/H110014x2/H110018x3/H11001. . .y0/H110051, y1/H110051/H110012x, y2/H110051/H110012x/H110014x2/H110018x3 3, . . .yn/H11005x2 2!/H11001x3 3!/H11001. . ./H11001xn/H115451 (n/H110011)!, y/H11005ex/H11002x/H110021k/H110050 k/HS110050x/H110052yr/H11005f (x, y)/H110052y x/H110022y(1)/H110051y(2)/H110050dy y/H110052 dx x/H110022, ln ƒyƒ/H110052 ln ƒx/H110022ƒ/H11001c*, y/H11005c(x/H110022)2yr x/H110052y/H11005tan x22 Instructor’s Manualim01_demo.qxd 8/2/10 7:57 PM Page 22and this implies that (A) .This becomes a Lipschitz condition if we note that the continuity of for implies that is bounded, say for all these x. Taking absolute values on both sides of (A) now gives . 10.By separation and integration, . Taking exponents gives the general solution . From this we can see the answers: No solution if . A unique solution if equals any .Infinitely many solutions if . This does not contradict the theorems because is not defined when . SOLUTIONS TO CHAP. 1 REVIEW QUESTIONS AND PROBLEMS, page 43 12. . Note that the solution curves are congruent. 14. . The figure also shows the solution curves through . y(/H110021)/H110051], (1, 0.1), (1, 1), and (1, 2)(/H110021, 1) [thus, y/H11005x 2/H11001cxy/H11005tanh ( x/H11001c)x/H110050 or 1f (x, y)/H110052x/H110021 x2/H11002xy(0)/H110050 or y(1)/H110050y0 and x0/HS110050 or x0/HS110051 y(x0)y(0)/H11005k/HS110050 or y(1)/H11005k/HS110050y/H11005c(x2/H11002x)dy y/H110052x/H110021 x2/H11002x dx, ln ƒyƒ/H11005ln ƒx2/H11002xƒ/H11001c*ƒf (x, y2)/H11002f (x, y1)ƒ/H11017Mƒy2/H11002y1ƒƒp(x)ƒ/H11017M p(x) ƒx/H11002x0ƒ/H11017ap(x)f (x, y2)/H11002f (x, y1)/H11005/H11002 p(x)(y2/H11002y1)Instructor’s Manual 23 1 –1 –22 1 –1 0 –2 2y xy(x) Problem 14. Direction field of xyr/H11005y/H11001x ˛2im01_demo.qxd 8/2/10 7:57 PM Page 2316.Solution . Computations: y/H110051>(1/H110014e/H11546x)24 Instructor’s Manual Error 0 0.2000 0 0.1 0.2160 0.00050.2 0.2329 0.00100.3 0.2509 0.00150.4 0.2696 0.00210.5 0.2893 0.00260.6 0.3098 0.00320.7 0.3312 0.00370.8 0.3534 0.00410.9 0.3762 0.00461.0 0.3997 0.0048yn xn 18. . 20.This Bernoulli equation (a Verhulst equation if ) can be reduced to linear form, as shown in Example 4 of Sec. 1.5 (except for the notation). The general solutionis (see (12) in Sec. 1.5) 22.The general solution of this linear differential equation is obtained as explained in Sec. 1.6, From this and the initial condition we have . Answer : 24.To solve this Bernoulli equation we set . Then , Substitution into the given ODE gives . We now multiply by , obtaining . General solution: . u/H11005ce x/H110012 ur/H11002u/H11005/H11002 2/H110022u3>2/H110021 2 u/H115463>2 ur/H110011 2 u/H115461>2/H11005u/H115463>2yr/H11005/H110021 2u/H115463>2 ur. y/H11005u/H115461>2u/H11005y/H115462y/H11005(x/H110024.3)e/H115462x2.c/H11005/H11002 4.3 y(0)/H11005/H11002 4.3y/H11005e/H115462x2 a/H20910e2x2e/H115462x2 dx/H11001cb/H110051x/H11001c2e/H115462x2y/H110051 ce/H11546ax/H11002b>a.b/H110210y/H11005ce0.4x/H1100225 cos x/H1100210 sin xProblem 16. Solution curve and computed values0.4 0.35 0.3 0.25 0.2 0.2 0 0.4 0.6 0.8 1 xyim01_demo.qxd 8/2/10 7:57 PM Page 24