Lista 3 Exercícios Livro com Resposta-2021 2

2.3 LINEAR EQUATIONS e 61 (ii) Occasionally, a first-order differential equation is not linear in one variable but in linear in the other variable. For example, the differential equation oe dx x+y is not linear in the variable y. But its reciprocal dx py dx 2 T=xty or [TT x=y dy dy is recognized as linear in the variable x. You should verify that the integrating factor e/~4 = e-Y and integration by parts yield the explicit solution x = —y? — 2y — 2 + ce’ for the second equation. This expression is then an implicit solution of the first equation (iii) Mathematicians have adopted as their own certain words from engineer- ing, which they found appropriately descriptive. The word transient, used ear- lier, is one of these terms. In future discussions the words input and output will occasionally pop up. The function fin (2) is called the input or driving func- tion; a solution y(x) of the differential equation for a given input is called the output or response. (iv) The term special functions mentioned in conjunction with the error func- tion also applies to the sine integral function and the Fresnel sine integral introduced in Problems 55 and 56 in Exercises 2.3. “Special Functions” is actually a well-defined branch of mathematics. More special functions are studied in Section 6.4. EX E RC | S E S 2 . 3 Answers to selected odd-numbered problems begin on page ANS-2. In Problems 1—24 find the general solution of the given dif- 18 a + ayy =] ferential equation. Give the largest interval J over which the » COS SINT (cos'x)y = general solution is defined. Determine whether there are any dy transient terms in the general solution. 19. «+ 1) ax + (x + 2)y = 2xe™ dy dy dy 1. = =5y 2. = +2y=0 20. (x + 2h — =5 — 8y — 4xy dx , dx , ¢ ) dx y y 3, Ma yeex 4.32 4 y= m1. + rsec 9 = cos 0 * dx “dx ' 2 2 ' 3 dP 5. y' + 3xry =x 6. y' + 2xy =x 22,7 + UP =P + 4t— 2 7. xy! +xy=1 8. yy =2y 49° +5 dy 4 23. x= + (3x + lhy =e dy ae dy dx 9x - y=x’*sinx 10. x— + 2y =3 dx dx > dy > 24. (x? — 1)— + 2y =(« + 1) dy 3 dy 5 dx VW. x + 4y=x°-x 12. (1 +x)>-xy=x+x dx dx 13. x2y’ +x(¢ + Dy =e? In Problems 25—36 solve the given initial-value problem. Give the largest interval J over which the solution is defined 14. xy’ + (1 + x)y = e * sin 2x d y = 15. ydx — 4(x + y®) dy =0 25. a =xt5y, y(0) =3 16. y dx = (ye” — 2x) dy dy 1 26. — = 2x — 3y, y(0) = 3 dy . dx 17. cos x — + (sinx)y = 1 , dx 27. xy’ +y=e, yO) =2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 62 e CHAPTER 2. FIRST-ORDER DIFFERENTIAL EQUATIONS dx Discussion Problems 28. y— —x=2y, yl) =5 dy 44. Reread the discussion following Example 2. Construct a 29, ia + Ri=E, i(0) =i L,R,E, iyconstants linear first-order differential equation for which all dt nonconstant solutions approach the horizontal asymp- aT tote y = 4asx—>™, 30. a KP Tn)s 10) = To. ky Tn» To constants 45. Reread Example 3 and then discuss, with reference dy to Theorem 1.2.1, the existence and uniqueness of a 3h x2 + y=4x+1, yd) =8 solution of the initial-value problem consisting of dx xy’ — 4y = x®%e* and the given initial condition. ! y= YP wo) = — 32. y+ ary = et, WO) = 1 (a) v0) =0 — (b) y(0) = yo, yo > 0 dy (x0) = y ) 33. (x +S ty =Inx, y(1) = 10 (©) yo) = Yo. x0 > 0 Yo > 0 46. Reread Example 4 and then find the general solution of dy : . . : ~ 34. x(x + p2 +xy=1, ye=1 the differential equation on the interval (—3, 3). dx 47. Reread the discussion following Example 5. Construct a 35. y' — (sinx)y =2sinx, y( /2)=1 linear first-order differential equation for which all solu- 5 tions are asymptotic to the line y = 3x — 5 as x >, 36. y’ + (t y= (0) = -1 . we : y (tan x)y = cos’x, y(0) 48. Reread Example 6 and then discuss why it is technically . incorrect to say that the function in (9) is a “solution” of In Problems 37—40 proceed as in Example 6 to solve the the IVP on the interval [0, ~). given initial-value problem. Use a graphing utility to graph . . . . the continuous function y(x). 49. (a) Construct a linear first-order differential equation of the form xy’ + ao(x)y = g(x) for which y, = c/x 37 dy + 2y = f(x), y(0) = 0, where and y= x°. Give an interval on which ta) IS > y = x3 + c/x? is the general solution of the DE. jl, OSxS3 (b) Give an initial condition y(xo) = yo for the DE SQ) = 0, x>3 found in part (a) so that the solution of the IVP dy is y=x>—1/x3. Repeat if the solution is 38. 2 + y = f(x), yO) = 1, where y =x° + 2/x3. Give an interval J of definition of dx each of these solutions. Graph the solution curves. Is fo) = { 0SxS1 there an initial-value problem whose solution is , —1, x>1 defined on —, 2%)? dy _ _ (c) Is each IVP found in part (b) unique? That is, can 39. = + 2xy = fla), y(0) = 2, where there be more than one IVP for which, say, x O<x 1 y =x° — 1/x3, x in some interval J, is the solution? SO) =40 0, x21 50. In determining the integrating factor (3), we did not use dy a constant of integration in the evaluation of fP(x) dx. 40. (1 + x°) ax + Ixy = f(x), y(0) = 0, where Explain why using [P(x) dx + c) has no effect on the solution of (2). x, Osx 1 : . . . f@ = ~ 51. Suppose P(x) is continuous on some interval J and a is a ~%, x21 number in J. What can be said about the solution of the 41. Proceed in a manner analogous to Example 6 to solve the initial-value problem y’ + P(x)y = 0, y(a) = 0? initial-value problem y’ + P(x)y = 4x, y(0) = 3, where Mathematical Models P(x) = 32 0=x=1, Lo . . (x) = ~2x, x>1 52. Radioactive Decay Series The following system > . of differential equations is encountered in the study of the Use a graphing utility to graph the continuous function decay of a special type of radioactive series of elements: yQ). dx 42. Consider the initial-value problem y’ + e*y = f(x), dt Tax y(0) = 1. Express the solution of the IVP for x > 0 asa dy nonelementary integral when f(x) = 1. What is the so- a Aw — Apy, lution when f(x) = 0? When f(x) = e*? where A, and A are constants. Discuss how to solve this 43. Express the solution of the initial-value problem system subject to x(0) = xo, y(0) = yo. Carry out your y' — 2xy = 1, y(1) = 1, in terms of erf(x). ideas. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 2.4. EXACTEQUATIONS e 63 53. Heart Pacemaker A heart pacemaker consists of a defined to be 1 at = 0. Express the solution y(x) of switch, a battery of constant voltage Eo, a capacitor with the initial-value problem x*y'’ + 2x?y = 10 sinx, constant capacitance C, and the heart as a resistor with y(1) = 0 in terms of Si(x). constant resistance R. When the switch is closed, the (b) Use a CAS to graph the solution curve for the IVP capacitor charges; when the switch is open, the capacitor forx >0 discharges, sending an electrical stimulus to the heart. . find the value of th : During the time the heart is being stimulated, the voltage (c) Use a CAS to ind t e value of the absolute maxi- E across the heart satisfies the linear differential equation mum of the solution y(x) for x > 0. dE 1 56. (a) The Fresnel sineintegral is defined by a RCE’ S(x) = Jp sin(@P/2) dt. Express the solution y(x) . of the initial-value problem y’— (sin x?)y = 0, Solve the DE, subject to E(4) = Eo. y(0) = 5, in terms of S(x). b) Use a CAS to graph the solution curve for the IVP Computer Lab Assignments ») on (—2, ©), erep 54. (a) Express the solution of the initial-value problem (c) It is known that S(x) —} as x > © and S(x) > -5 y’ — 2xy = -1,9(0) = V_ /2, in terms of erfe(x). as x —> —% , What does the solution y(x) approach (b) Use tables or a CAS to find the value of y(2). Use a as x > ©? As x—> —%0? CAS to graph the solution curve for the IVP on (d) Use a CAS to find the values of the absolute (-%, ©). maximum and the absolute minimum of the 55. (a) The sine integralfunction is defined by solution y(x). Si(x) = Jj (sin t/t) dt, where the integrand is = Differential of a Function of Two Variables _ If z = f(x, y) is a function of two variables with continuous first partial derivatives in a region R of the xy-plane, then its differential is —, —, dz = ays Lay, (1) ox oy In the special case when f(x, y) = c, where c is a constant, then (1) implies ) —, Harty =o. (2) Ox oy Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 2.4 EXACT EQUATIONS e 69 (ii) In some texts on differential equations the study of exact equations precedes that of linear DEs. Then the method for finding integrating factors just discussed can be used to derive an integrating factor for y’ + P(x)y = f(x). By rewriting the last equation in the differential form (P@)y — f(x)) dx + dy = 0, we see that M,—N, | e Wo PO: From (13) we arrive at the already familiar integrating factor e/?* used in Section 2.3. EX E RC | S E S 2 . 4 Answers to selected odd-numbered problems begin on page ANS-2. In Problems 1-20 determine whether the given differential 19. (4Py — 150? — y)dt + (t4 + 3y? — dy =0 equation is exact. If it is exact, solve it. 1 1 y t . 2x - + By + Tdy= 20. (-+ 5-3 dt+|ye+>=—s dy=0 1. (Qx — 1)dx + By + 7) dy=0 (: 2 P4+y (se P4y ly 2. (2x + y) dx — (x + 6y) dy =0 . ae In Problems 21-26 solve the given initial-value problem. 3. (5x + 4y) dx + (4x — 8y3) dy = 0 (Sx + 4y) dx + (Ax ~ 8y") dy 2. (x t+ ye dx + Qxry +2 -l)dy=0, y(I=1 . (siny — y si + + »>—y)dy= 4. (sin y — ysin x) dx + (cosx + xcosy — y)dy=0 22. (e+ y)dx +Q4+x+ye)dy=0, y(0)=1 5. (2xy? — 3) dx + (2x*y + 4)dy =0 Oxy" — 3)dx + Ox'y + 4) dy 23. (4y + 2t — 5) dt + (6y + 4t— l)dy=0, y(-l)=2 1 dy y . 2 6. (29-1 + cos3e B+ 4X — aye + aysindx=0 By -P dy t | _ vO S dx x y 24. ( y day? yd) =1 2_ 2 2 _ Ixy) dy = 7. (x — y) dx + (x° — 2xy) dy = 0 25. (y? cos x — 3x2y — 2x) dx y + Qysinx —2x°+Iny)dy=0, y(0)=e 8. {1+ Inx += dx =(1 — Inx)dy 1 dy * 26. (43 + cosx — 2xy oe y(y + sin x), y(0) = 1 9. (x — y3 + y* sin x) dx = (3xy? + 2y cos x) dy It+y dx 10. (x3 + y3) dx + 3xy* dy =0 In Problems 27 and 28 find the value of k so that the given differential equation is exact. 1 11. (yiIny — ee”) dx + (2 +xIny dy=0 27. (y3 + kxy* — 2x) dx + xy? + 20x?y7) dy =0 12. (3x?y + e”) dx + (x3 + xe” — 2y) dy = 0 28. (6xy? + cos y) dx + (2kx*y? — x sin y) dy = 0 dy In Problems 29 and 30 verify that the given differential 13. x ada 2xe* — y + 6x7 equation is not exact. Multiply the given differential equa- tion by the indicated integrating factor p(x, y) and verify that 14 (: _3 4y dy ys 3 1 the new equation is exact. Solve. y dx * 29. (—xy sin x + 2y cos x) dx + 2x cos x dy = 0; 1 d. B(x, y) = xy 15. (es Troe a+ vy =0 1+ 9x" dy 30. (x? + 2xy — y?) dx + (9? + 2xy — x?) dy = 0; By) = + yy? 16. (Sy — 2x)y’ — 2y=0 : : _ In Problems 31-36 solve the given differential equation by 17. (tan x ~ sin x sin y) dx + cos x cos y dy = 0 finding, as in Example 4, an appropriate integrating factor. 18. (2y sinx cos x — y + 2y*e*”) dx 31. (2y? + 3x) dx + 2xy dy =0 = (x — sin’x — 4xye*”) dy 32. yat yt l)dx+(x+ 2y)dy=0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 70 e CHAPTER 2. FIRST-ORDER DIFFERENTIAL EQUATIONS 33. 6xy dx + (4y + 9x2) dy = 0 Mathematical Model 45. Falling Chain A portion of a uniform chain of length 2 Pp ig 34. cos x dx + (: +~ sinxdy =0 8 ft is loosely coiled around a peg at the edge of a high y horizontal platform, and the remaining portion of the 35. (10 — 6y + e**) dx — 2dy =0 chain hangs at rest over the edge of the platform. See 36. (y? + xy?) dx + (Sy? — xy + y3 sin y) dy = 0 Figure 2.4.2. Suppose that the length of the overhang- ing chain is 3 ft, that the chain weighs 2 lb/ft, and that In Problems 37 and 38 solve the given initial-value problem the aan dircefion oS Gownward Siarting at r= 0 by finding as in Example 4, an appropriate integrating factor. SECONES, ME WEISDE OF the ovemangms Porton causes the chain on the table to uncoil smoothly and to fall to 37. x dx + (x’y + 4y)dy =0, y(4)=0 the floo . If x(#) denotes the length of the chain over- 3 3 _ _ hanging the table at time t > 0, then v = dx/dt is its 38. G+ y — S)dx = (y +2y) dy, yO) =1 velocity. When all resistive forces are ignored, it can 39. (a) Show that a one-parameter family of solutions of be shown that a mathematical model relating v to x is the equation given by (Axy + 3x?) dx + (2y + 22%) dy =0 wi 4 p= 328, 3 2 2 dx isx? + 2x*y+ yo =c. (b) Show that the initial conditions y(0) = —2 and (a) Rewrite this model in differential form. Proceed as y(1) = 1 determine the same implicit solution in Problems 31—36 and solve the DE for v in terms . of x by finding an appropriate integrating factor. (c) Find explicit solutions yj(x) and y2(x) of the dif- Find an explicit solution v(x). fe tial ti i rt h that y\(0) = —2 : : : : : erentia equa ton m pa (a) Suen ine ¥10) (b) Determine the velocity with which the chain leaves and y2(1) = 1. Use a graphing utility to graph y\(x) y and yo(x) the platform. 19(x). peg Discussion Problems 40. Consider the concept of an integrating factor used in platform edge Problems 29—38. Are the two equations M dx + Ndy = 0 and wM dx + wN dy = 0 necessarily equivalent in the x(t) sense that a solution of one is also a solution of the other? | Discuss. 41. Reread Example 3 and then discuss why we can con- FIGURE 2.4.2 Uncoiling chain in Problem 45 clude that the interval of definition of the explicit solution of the IVP (the blue curve in Figure 2.4.1) is Computer Lab Assianments (-1,) p 9 . . 46. Streamlines 42. Discuss how the functions M(x, y) and M(x, y) can be . . . . found so that each differential equation is exact. Carry (a) The solution of the differential equation out your ideas. 1 2xy ae [rs Bo |. 0 a dy | qy = (a) M(x, y) dx + (se + 2xy + Y dy =0 (7? +)? (e+ y’)? » “yan x is a family of curves that can be interpreted as (b) (yi + ety dx + N(x, y)dy = 0 streamlines of a fluid flow around a circular object . . . . whose boundary is described by the equation 43. Differential equations are sometimes solved by x? + y? = 1. Solve this DE and note the solution having a clever idea. Here is a little exercise in f(x,y) = fore = 0. I : Although the — differential ti — ; Ve y) ‘i. "y dy ‘ 0 «hot exact chow hon (b) Use a CAS to plot the streamlines for 2 = + + + + i the rearrangement (x dx + y dy) /Vx? + y? = dx and c 0, +02, +04, *0.6, and +0.8 in three the observation !d(x2 + y2) = xdx + y dy can lead to different ways. First, use the contourplot of a CAS. a solution 2 Second, solve for x in terms of the variable y. Plot , the resulting two functions of y for the given values 44. True or False: Every separable first-order equation of c, and then combine the graphs. Third, use the dy/dx = g(x)h(y) is exact. CAS to solve a cubic equation for y in terms of x. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 74 e CHAPTER 2. FIRST-ORDER DIFFERENTIAL EQUATIONS The last equation is separable. Using partial fractions du 1 1 1 ——— = dx or =|— - — | du = dx (u — 3)(u + 3) 6}u-3 ut+3 and then integrating yields y 1 in| =xto¢ or ua 3_ e&*6e, = ce, replace e%' by ¢ 6 lu+3 ut 3 a Ve Solving the last equation for u and then resubstituting gives the solution /\\ 3(1 + ce) 3(1 + ce%) Ai ‘ 1 — ce® “ y= ax t 1 — ce" (6) Finally, applying the initial condition y(0) = 0 to the last equation in (6) gives c = —1. Figure 2.5.1, obtained with the aid of a graphing utility, shows the graph of 31 — e& the particular solution y = 2x + 30 <9) in dark blue, along with the graphs of FIGURE 2.5.1 — Solutions of DE in 1+ e* Example 3 some other members of the family of solutions (6). => EX E RC | S E S 2 . 5 Answers to selected odd-numbered problems begin on page ANS-2. Each DE in Problems 1-14 is homogeneous. Each DE in Problems 15—22 is a Bernoulli equation. In Problems 1-10 solve the given differential equation by In Problems 15—20 solve the given differential equation by using an appropriate substitution. using an appropriate substitution. 1. («— y)dx+xdy=0 2. («+ y)dx+xdy=0 dy 1 dy (x — y) dx + xdy (x + y)dx + xdy I. x + y= 5 16, © y= ey" 3.xdxt+(y—2x)dy=0 4 ydx=2(x+y) dy * » * dy dy 5. (9? + yx) dx — x? dy =0 17,2 = yayi- 1) 18. x2 - (1 +x)y = xy? dx dx 6. (y? + yx) dx + x? dy =0 dy dy 9.224 yr 20.30 +2) 2 = 20 - 1) 7 dy _y7x dt dt dx ytx In Problems 21 and 22 solve the given initial-value problem. dy _x+3y 3, 2 = dy 4 1 dx 3xty 21. x8 2xy = 3y4, yl) =5 9. -—y dx + |x + Vxy)dy =0 dy y ( y) y 22. ye + ys? = 1, y(0) =4 x dy 10. x2 =y+ Vey, x>0 dx Each DE in Problems 23-30 is of the form given in (5). In Problems 11—14 solve the given initial-value problem. In Problems 23~—28 solve the given differential equation by using an appropriate substitution. dy UW. xy’ =y— x, yl) =2 dy dy 1l-x-y xy dx y x y(1) 23, 2 =(+y +41) yg, @ = y dx dx x+y 12. (2 + 297) = ay (-1)=1 dy dy . 9 dy yy 25, 2 = tan?(x + y) 2. = sin(x + y) dx dx 13. (x + ye") dx — xe" dy =0, y(1) =0 dy dy 27. = =24+Vy-2x4+3 28—=1+4+e 7° 14. ydx + x(Inx—Iny— 1)dy=0, y(l)=e dx dx Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 2.6 ANUMERICALMETHOD e 75 In Problems 29 and 30 solve the given initial-value problem. particular solution y, of the equation. Show that the d substitution y = y; + u reduces Riccati’s equation 29, & = cos(x + y), y(0) = /4 to a Bernoulli equation (4) with n=2. The dx Bernoulli equation can then be reduced to a linear 30 dy __3x + 2y yl) = -1 equation by the substitution w = u~!. “dx 3x4 2y +2’ (b) Find a one-parameter family of solutions for the differential equation Discussion Problems dy = 4 — 1 + y 31. Explain why it is always possible to express any homoge- dx rox y thot differential equation M(x, y) dx + M(x, y) dy = Oin where y; = 2/x is a known solution of the equation. e form dy y 36. Determine an appropriate substitution to solve ==F(-. ' dx (: xy! = y In(xy). You might start by proving that Math tical Model M(x, y) =x"M(1,y/x) and N(x, y) = x*N(1, 9/2). aoe oes 32. Put the homogeneous differential equation 37. Falling Chain In Problem 45 in Exercises 2.4 we saw that a mathematical model for the velocity v of a chain (5x7 — 2y*) dx — xy dy = 0 slipping off the edge of a high horizontal platform is into the form given in Problem 31. dv . : . . xv— + v? = 32x. 33. (a) Determine two singular solutions of the DE in dx Problem 10. In that problem you were asked to solve the DE by con- (b) If the initial condition y(5) = 0 is as prescribed in verting it into an exact equation using an integrating Problem 10, then what is the largest interval J over factor. This time solve the DE using the fact that it is a which the solution is defined? Use a graphing util- Bernoulli equation. ity to graph the solution curve for the IVP. 38. Population Growth In the study of population dy- 34. In Example 3 the solution y(x) becomes unbounded as namics one of the most famous models for a growing x — +, Nevertheless, y(x) is asymptotic to a curve as but bounded population is the logistic equation x— — and to a different curve as x — %. What are the . dP equations of these curves? a = P(a — bP), : : : _ 2 35. ie Gite equation dy / dx = P(x) + Ody + Rady where a and b are positive constants. Although we is known as Riccati’s equation. will come back to this equation and solve it by an (a) A Riccati equation can be solved by a succession alternative method in Section 3.2, solve the DE this of two substitutions provided that we know a first time using the fact that it is a Bernoulli equation. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANS-2 e ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS EXERCISES 2.2 (PAGE 51) 31. y= 2x + 1 + 5/x; (0, %) ,-_l _— 1 o-3x 33. (x + Dy = xInx — x + 21; (0, %) 1. y 3cos5x + ¢ 3. y= ze +e 35, y = —2 + 3e-e% (—v6, 00) 5. y =cx4 7. —3e°27 =2e* +e 1 yO 3 tq — e® <x< 9. 18inx —i8 =!y+42y4Inlyl +e 37. y=4? ey 3 px = ay + 2y + Inly| Ve = Der, x>3 11. 4cosy = 2x + sin2x +c 13. (e° +1) 74+ 2(e? + Ib =c a+3e", 0=<x<1 ce! 39. y= (: 3) Poy] 15. S = cek 17. P= ——_ 2e + 3Je™, 1 + ce _ -2x 19. (y+ 3Pe*=c(xv+4)prey 21. y = sin(}x + c) 4. y= 2x— 1+ 4e™%, O=xs1 Gein) 4x? Inx + (1 + 4e7)x?, x>1 eT tx 23. x = tan(41 — } 25. y = ——— : : ~ * an( i7) y x 43, y =e"! + 5 Vire® (erf(x) — erf(1)) ii} Xo ES 27. y= tx + 3 V1 = x 29. y= lace 53. E(t) = Eye ARE <x Xr 3l.y=—-Vx4+ x - 1; (-%, — 1%), U EXERCISES 2.4 (PAGE 69) e 33. y= —In(Z2 — e*); (—%, In2) a 3 — gael Le-xt+3y+Iysc 334+ 4xy—2yt=c 2 35. (a) y= 2,9 = —2y = 23 + etl 5. x?y? -—3x+4y=c 7. not exact a 37. y = —1 andy = 1 are singular solutions of Problem 21; 9. xy + y’cosx — $x? =c oO y = 0 of Problem 22 11. not exact x 39. y= 1 _ x x 3 a 13. xy — 2xe* + 2e*-—2x° =c Qa 41..y=1+ 4 tan (5.x) 15. «°y? — tan”! 3x =c a 45. y =tanx —secx tc 17. —In|cos x| + cosxsiny =c S Var 19. fy -—S5h-tyt+y=c 47, y=[-l+cQ1+ > y= ( a) ie +xrytay-y=4 49. y =2V Vie“ _ ee + 4 a> Vie" _ ¢ 23. dry +P — 51+ 3y2-y=8 O 57, y(x) = (4h/L?)x? + a 25. y?sinx —y—x° +yIny—y=0 Oo 27. k = 10 29. x?y? cosx =c a 31. Py tec 33. 3xryi t yt =e Oo EXERCISES 2.3 (PAGE 61) 35. -2ye+ Wert x se uw = Sx > a hy=ce(-%,) 37. e& (2 + 4) = 20 n »—1,3x -x (_ sap kG : ee 3. y = 7e* + ce™, (—%, ©); ce™ is transient 39. (c) yx) = -e —- Ve a a O 5Sy=H= 5 + ce’, (—%, ©); ce is transient Pe , . yi(x) = x? + VF -— + 4 B 7. y =x !Inx + cx !, (0, ©); solution is transient ui 9. y=cx — x cos x, (0, ©) x 9 = 13 4 “4: . 45. (a) v(x) = 8,/5 -S (b) 12.7 ft/s ms dl. y=5x° — 5x + cx", (0, ©); cx “is transient 30 x < 13. y=hx%e* + cx %e™, (0, ©); cx 7e™ is transient 15. x = 2y® + cy4, (0, %) EXERCISES 2.5 (PAGE 74) 17. y = sinx + ccos x, (—7/2, 7/2) 1. y + x In|x| = cx 19. (x + De*y = x* +c, (—1, ©); solution is transient (ee eves Pech) 3. (x ynlx — y| = y + ey) 21. (sec 6 + tan 0)r = 6 — cos 6 + c, (—7/2, 7/2) 23. y =e > + cx !e73*, (0, ); solution is transient 5. x + yIn|x| = cy 25. y= —ty _ A + 18 65x, (—2, 2) 7. In(x? + y*) + 2 tan” '(y/x) =c 9. 4x = y(In|y| — c)? 11. y? + 3x3 In|x| = 8x3 27. y= x le + (2 — ex !; (0,%) 13. In|x| = e/* — 1 15. =1+ex3 3 = 1 3x vy = 00 iE, ; 82) ety (oe 5 17. y x+3+ ce 19. e ct . R 0 R 5 > 21. y3 _ 2x7! + Bye 23. y= —x — 1+ tan(x + c) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ° ANS-3 25. 2y—2x+sin2(x+ y)=c 23. A(t) = 1000 — 1000e~1/! 27. 4(y — 2x + 3)=(x+ 0 25. A(t) = 1000 — 10¢ — 5(100 — 1)?; 100 min 29. —cot(x + y) + ese(x ty) =x+ V2-1 27. 64.38 lb 2 29, i(t) = 2 — 26-5 j > hasta 35. (b) y=— + (-}x4+ cx 3)! (b) y x ( ax rT cx ) 31. q(t) = iw _ mes i(t) = se" 60 — 60e71° 0=rx<20 33. i(t) = { tno EXERCISES 2.6 (PAGE 79) 60(e° — le", t> 20 1. y2 = 2.9800, y4 = 3.1151 _ mg — M8) him 3. yio = 2.5937, yao = 2.6533; y = e* 35. fa) = + (», k ) 5. ys = 0.4198, yio = 0.4124 meg 7. ys = 0.5639, yo = 0.5565 (b) v> 7s I> ~ 9. ys = 1.2194, yio = 1.2696 ~~ 13. Euler: yjo9 = 3.8191, yao = 5.9363 (c) s() = mam (x _ 78m 2 RK4: yio = 42.9931, yoo = 84.0132 k k k = ym, me x CHAPTER 2 IN REVIEW (PAGE 80) K\o YU e 1. —A/k, a repeller for k > 0, an attractor for k < 0 3 ~m . k 3. true 39. (a) v() = a(t + r) — ft 2 By 4k 4k k 4p ww 5. — = xsiny 9 co dx 9 7. true , (c) 33; seconds a 9. y = c\e* a dy 41. (a) P(t) = Pye a i. mz? (sinx)y = x 43. (a) Ast > %, x(t) > r/k el dy 5 5 (b) x(t) = r/k — (r/kye™5 (In 2)/k = 13. —=(y- 1? (y - 3 >) dx (y y¥ (y — 3) 47. (c) 1.988 ft > 15. semi-stable for n even and unstable for n odd; a semi-stable for n even and asymptotically stable Qa for n odd. EXERCISES 3.2 (PAGE 100) Oo 19, 2x + sin 2x =2 In(y? +1)t+e 1. (a) N = 2000 Qa 21. (6x + ly? = -3x3 +¢ 2000 ¢ 5 23. O=cr!+HXe(-1 4 5Int) (b) NO = F999 q go NUO) = 1834 4 25. y=t4+ e+ 4)4 3. 1,000,000; 5.29 mo ~ y= 4(Py — 1) — (Py — 4e** 27. y = csc x, (7, 277) 5. (b) P(t) = (Po ) = Po je ° 29. (b) y = (x + 2 Vyy ~ 40), Gy — 2 Vo. 9) Po“ DN Po He 2 (c) For 0 < Po < 1, time of extinction is ee 1 4(o= 1) = t= —3m a n EXERCISES 3.1 (PAGE 90) ° S 5 3 V3 2Py) — 5 1. 7.9 yr; 10 yr 7. P= s+ SB san -S + tan! (2=4)| 3. 760; approximately 11 persons/yr 2 2 2 v3 5. llh time of extinction is 7. 136.5h _ 2 5 _,(2Po—5 9. 1(15) = 0.00098/p or approximately 0.1% of Io = V3 tan V3 + tan V3 11. 15,600 years 13. T(1) = 36.67° F; approximately 3.06 min 9. 29.3 g;X —> 60 as 1—> ~; 0 g of A and 30 g of B 15. approximately 82.1 s; approximately 145.7 s 4A, \? 17. 390° 11. (a) h() = (va - “a ; lis 0 <t = VHA, /4A, 19. about 1.6 hours prior to the discovery of the body ue 21. A(t) = 200 — 17027150 (b) 576 v10 s or 30.36 min ; 13. (a) approximately 858.65 s or 14.31 min (b) 243 s or 4.05 min Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.