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Complete Solutions Manual to Accompany

A First Course in Differential Equations with Modeling Applications ELEVENTH EDITION, METRIC VERSION

And

Differential Equations with Boundary-Value Problems NINTH EDITION, METRIC VERSION

Dennis G. Zill Loyola Marymount University, Los Angeles, CA

Prepared by

Roberto Martinez Loyola Marymount University, Los Angeles, CA

Metric Version Prepared by

Aly El-Iraki

©.

Professor Emeritus, Alexandria University, Egypt

Australia • Brazil • Mexico • Singapore • United Kingdom • United States

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Chapter 1 Introduction to Diﬀerential Equations

1.1

Deﬁnitions and Terminology

1. Second order; linear 2. Third order; nonlinear because of (dy/dx)4 3. Fourth order; linear 4. Second order; nonlinear because of cos(r + u) 5. Second order; nonlinear because of (dy/dx)2 or

1 + (dy/dx)2

6. Second order; nonlinear because of R2 7. Third order; linear 8. Second order; nonlinear because of ẋ2 9. Writing the diﬀerential equation in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear in y because of y 2 . However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is linear in x. 10. Writing the diﬀerential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is linear in v. However, writing it in the form (v + uv − ueu )(du/dv) + u = 0, we see that it is nonlinear in u. 11. From y = e−x/2 we obtain y = − 12 e−x/2 . Then 2y + y = −e−x/2 + e−x/2 = 0. 12. From y = 65 − 65 e−20t we obtain dy/dt = 24e−20t , so that dy + 20y = 24e−20t + 20 dt

6 6 −20t − e 5 5

= 24.

13. From y = e3x cos 2x we obtain y = 3e3x cos 2x−2e3x sin 2x and y = 5e3x cos 2x−12e3x sin 2x, so that y − 6y + 13y = 0. 1

CHAPTER 1

INTRODUCTION TO DIFFERENTIAL EQUATIONS

14. From y = − cos x ln(sec x + tan x) we obtain y = −1 + sin x ln(sec x + tan x) and y = tan x + cos x ln(sec x + tan x). Then y + y = tan x. 15. The domain of the function, found by solving x+2 ≥ 0, is [−2, ∞). From y = 1+2(x+2)−1/2 we have (y − x)y = (y − x)[1 + (2(x + 2)−1/2 ] = y − x + 2(y − x)(x + 2)−1/2 = y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2 = y − x + 8(x + 2)1/2 (x + 2)−1/2 = y − x + 8. An interval of definition for the solution of the differential equation is (−2, ∞) because y is not defined at x = −2. 16. Since tan x is not defined for x = π/2 + nπ, n an integer, the domain of y = 5 tan 5x is {x 5x = π/2 + nπ} or {x x = π/10 + nπ/5}. From y = 25 sec2 5x we have y = 25(1 + tan2 5x) = 25 + 25 tan2 5x = 25 + y 2 . An interval of definition for the solution of the differential equation is (−π/10, π/10). Another interval is (π/10, 3π/10), and so on. 17. The domain of the function is {x 4 − x2 = 0} or {x x = −2 and x = 2}. From y = 2x/(4 − x2 )2 we have 2 1 = 2xy 2 . y = 2x 4 − x2 An interval of definition for the solution of the differential equation is (−2, 2). Other intervals are (−∞, −2) and (2, ∞). √ 18. The function is y = 1/ 1 − sin x , whose domain is obtained from 1 − sin x = 0 or sin x = 1. Thus, the domain is {x x = π/2 + 2nπ}. From y = − 12 (1 − sin x)−3/2 (− cos x) we have 2y = (1 − sin x)−3/2 cos x = [(1 − sin x)−1/2 ]3 cos x = y 3 cos x. An interval of definition for the solution of the differential equation is (π/2, 5π/2). Another one is (5π/2, 9π/2), and so on.

1.1 Definitions and Terminology

19. Writing ln(2X − 1) − ln(X − 1) = t and diﬀerentiating 4

implicitly we obtain dX 1 dX 2 − =1 2X − 1 dt X − 1 dt 2 1 dX − =1 2X − 1 X − 1 dt

–4

–2

2X − 2 − 2X + 1 dX =1 (2X − 1) (X − 1) dt

–4

dX = −(2X − 1)(X − 1) = (X − 1)(1 − 2X). dt Exponentiating both sides of the implicit solution we obtain 2X − 1 = et X −1 2X − 1 = Xet − et (et − 1) = (et − 2)X X=

et − 1 . et − 2

Solving et − 2 = 0 we get t = ln 2. Thus, the solution is defined on (−∞, ln 2) or on (ln 2, ∞). The graph of the solution defined on (−∞, ln 2) is dashed, and the graph of the solution defined on (ln 2, ∞) is solid. y

20. Implicitly diﬀerentiating the solution, we obtain 4

dy − 4xy + 2y =0 −2x dx dx 2 dy

−x dy − 2xy dx + y dy = 0 2

2xy dx + (x2 − y)dy = 0.

–4

–2

2 –2

y 2 − 2x2 y − 1

= 0 Using the quadratic formula to solve √ 2 √ 2 4 4 for y, we get y = 2x ± 4x + 4 /2 = x ± x + 1 . √ Thus, two explicit solutions are y1 = x2 + x4 + 1 and √ y2 = x2 − x4 + 1 . Both solutions are deﬁned on (−∞, ∞). The graph of y1 (x) is solid and the graph of y2 is dashed.

–4

CHAPTER 1

INTRODUCTION TO DIFFERENTIAL EQUATIONS

21. Diﬀerentiating P = c1 et / 1 + c1 et we obtain 1 + c1 et c1 et − c1 et · c1 et 1 + c1 et − c1 et c1 et dP = = dt 1 + c1 et 1 + c1 et (1 + c1 et )2 =

c1 et 1 + c1 et

1−

c1 et = P (1 − P ). 1 + c1 et

22. Diﬀerentiating y = 2x2 − 1 + c1 e−2x we obtain 2

dy 2 = 4x − 4xc1 e−2x , so that dx

dy 2 2 + 4xy = 4x − 4xc1 e−2x + 8x3 − 4x + 4c1 xe−x = 8x3 dx 23. From y = c1 e2x + c2 xe2x we obtain 4c2 xe2x , so that

dy d2 y = (4c1 + 4c2 )e2x + = (2c1 + c2 )e2x + 2c2 xe2x and dx dx2

dy d2 y + 4y = (4c1 + 4c2 − 8c1 − 4c2 + 4c1 )e2x + (4c2 − 8c2 + 4c2 )xe2x = 0. −4 2 dx dx 24. From y = c1 x−1 + c2 x + c3 x ln x + 4x2 we obtain dy = −c1 x−2 + c2 + c3 + c3 ln x + 8x, dx d2 y = 2c1 x−3 + c3 x−1 + 8, dx2 and

d3 y = −6c1 x−4 − c3 x−2 , dx3

so that x3

2 dy d3 y 2 d y + 2x −x + y = (−6c1 + 4c1 + c1 + c1 )x−1 + (−c3 + 2c3 − c2 − c3 + c2 )x dx3 dx2 dx

+ (−c3 + c3 )x ln x + (16 − 8 + 4)x2 = 12x2 In Problems 25–28, we use the Product Rule and the derivative of an integral ((12) of this section): ˆ x d g(t) dt = g(x). dx a ˆ x −3t ˆ x −3t dy e−3x 3x e e 3x 3x dt we obtain = 3e dt + · e or 25. Diﬀerentiating y = e t dx t x 1 1 ˆ x −3t 1 e dy = 3e3x dt + , so that dx t x 1 ˆ x −3t ˆ x −3t 1 e e dy 3x 3x − 3xy = x 3e dt + − 3x e dt x dx t x t 1 1 ˆ x −3t ˆ x −3t e e dt + 1 − 3xe3x dt = 1 = 3xe3x t t 1 1

1.1 Definitions and Terminology

√

ˆ x

cos t cos x √ √ dt + √ · x or 26. Differentiating y = x x t 4 4 ˆ x 1 cos t dy √ dt + cos x, so that = √ dx 2 x 4 t ˆ x ˆ x √ 1 cos t cos t dy √ dt + cos x − x √ dt √ − y = 2x 2x dx 2 x 4 t t 4 ˆ x ˆ x √ √ cos t cos t √ dt + 2x cos x − x √ dt = 2x cos x = x t t 4 4 ˆ ˆ x dy 5 sin x 10 10 x sin t sin t 5 10 dt we obtain =− 2 − 2 dt + · or 27. Differentiating y = + x 1 t dx x x 1 t x x ˆ xx 5 10 10 sin x sin t dy =− 2 − 2 dt + , so that dx x x 1 t x2 ˆ ˆ 5 5 10 x sin t 10 x sin t 10 sin x 2 dy 2 +x + xy = x − 2 − 2 dt + + dt x dx x x 1 t x2 x x 1 t ˆ x ˆ x sin t sin t dt + 10 sin x + 5 + 10 dt = 10 sin x = −5 − 10 t t 1 1 ˆ x ˆ x dy 2 2 2 −x2 −x2 t2 −x2 −x2 = −2xe −2xe e dt we obtain et dt+ex ·e−x 28. Differentiating y = e +e dx 0 ˆ x0 dy −x2 −x2 t2 = −2xe or − 2xe e dt + 1, so that dx 0 ˆ x ˆ x dy −x2 −x2 t2 −x2 −x2 t2 + 2xy = −2xe − 2xe e dt + 1 + 2x e +e e dt dx 0 0 ˆ x ˆ x 2 −x2 −x2 t2 −x2 −x2 − 2xe e dt + 1 + 2xe + 2xe et dt = 1 = −2xe 1 cos t dy √ dt we obtain = √ dx 2 x t

29. From

−x2 , x < 0 x≥0 x2 ,

y =

−2x, 2x,

we obtain

x<0 x≥0

so that xy − 2y = 0. 30. The function y(x) is not continuous at x = 0 since lim y(x) = 5 and lim y(x) = −5. Thus, y (x) does not exist at x = 0.

x→0−

x→0+

31. Substitute the function y = emx into the equation y + 2y = 0 to get (emx ) + 2(emx ) = 0 memx + 2emx = 0 emx (m + 2) = 0 Now since emx > 0 for all values of x, we must have m = −2 and so y = e−2x is a solution.

CHAPTER 1

INTRODUCTION TO DIFFERENTIAL EQUATIONS

32. Substitute the function y = emx into the equation 5y − 2y = 0 to get 5(emx ) − 2(emx ) = 0 5memx − 2emx = 0 emx (5m − 2) = 0 Now since emx > 0 for all values of x, we must have m = 2/5 and so y = e2x/5 is a solution. 33. Substitute the function y = emx into the equation y − 5y + 6y = 0 to get (emx ) − 5(emx ) + 6(emx ) = 0 m2 emx − 5memx + 6emx = 0 emx (m2 − 5m + 6) = 0 emx (m − 2)(m − 3) = 0 Now since emx > 0 for all values of x, we must have m = 2 or m = 3 therefore y = e2x and y = e3x are solutions. 34. Substitute the function y = emx into the equation 2y + 7y − 4y = 0 to get 2(emx ) + 7(emx ) − 4(emx ) = 0 2m2 emx + 7memx − 4emx = 0 emx (2m2 + 7m − 4) = 0 emx (m + 4)(2m − 1) = 0 Now since emx > 0 for all values of x , we must have m = −4 or m = 1/2 therefore y = e−4x and y = ex/2 are solutions. 35. Substitute the function y = xm into the equation xy + 2y = 0 to get x · (xm ) + 2(xm ) = 0 x · m(m − 1)xm−2 + 2mxm−1 = 0 (m2 − m)xm−1 + 2mxm−1 = 0 xm−1 [m2 + m] = 0 xm−1 [m(m + 1)] = 0 The last line implies that m = 0 or m = −1 therefore y = x0 = 1 and y = x−1 are solutions.

1.1 Definitions and Terminology

36. Substitute the function y = xm into the equation x2 y − 7xy + 15y = 0 to get x2 · (xm ) − 7x · (xm ) + 15(xm ) = 0 x2 · m(m − 1)xm−2 − 7x · mxm−1 + 15xm = 0 (m2 − m)xm − 7mxm + 15xm = 0 xm [m2 − 8m + 15] = 0 xm [(m − 3)(m − 5)] = 0 The last line implies that m = 3 or m = 5 therefore y = x3 and y = x5 are solutions. In Problems 37–40, we substitute y = c into the diﬀerential equations and use y = 0 and y = 0 37. Solving 5c = 10 we see that y = 2 is a constant solution. 38. Solving c2 + 2c − 3 = (c + 3)(c − 1) = 0 we see that y = −3 and y = 1 are constant solutions. 39. Since 1/(c − 1) = 0 has no solutions, the diﬀerential equation has no constant solutions. 40. Solving 6c = 10 we see that y = 5/3 is a constant solution. 41. From x = e−2t + 3e6t and y = −e−2t + 5e6t we obtain dx = −2e−2t + 18e6t dt

and

dy = 2e−2t + 30e6t . dt

Then x + 3y = (e−2t + 3e6t ) + 3(−e−2t + 5e6t ) = −2e−2t + 18e6t =

dx dt

5x + 3y = 5(e−2t + 3e6t ) + 3(−e−2t + 5e6t ) = 2e−2t + 30e6t =

dy . dt

and

42. From x = cos 2t + sin 2t + 15 et and y = − cos 2t − sin 2t − 15 et we obtain

and

1 dx = −2 sin 2t + 2 cos 2t + et dt 5

and

1 dy = 2 sin 2t − 2 cos 2t − et dt 5

d2 x 1 = −4 cos 2t − 4 sin 2t + et dt2 5

and

d2 y 1 = 4 cos 2t + 4 sin 2t − et . dt2 5

Then 1 1 d2 x 4y + et = 4(− cos 2t − sin 2t − et ) + et = −4 cos 2t − 4 sin 2t + et = 2 5 5 dt and

1 1 d2 y 4x − et = 4(cos 2t + sin 2t + et ) − et = 4 cos 2t + 4 sin 2t − et = 2 . 5 5 dt

CHAPTER 1

INTRODUCTION TO DIFFERENTIAL EQUATIONS

43. (y )2 + 1 = 0 has no real solutions because (y )2 + 1 is positive for all differentiable functions y = φ(x). 44. The only solution of (y )2 + y 2 = 0 is y = 0, since if y = 0, y 2 > 0 and (y )2 + y 2 ≥ y 2 > 0. 45. The first derivative of f (x) = ex is ex . The first derivative of f (x) = ekx is kekx . The differential equations are y = y and y = ky, respectively. 46. Any function of the form y = cex or y = ce−x is its own second derivative. The corresponding differential equation is y − y = 0. Functions of the form y = c sin x or y = c cos x have second derivatives that are the negatives of themselves. The differential equation is y + y = 0. √ 47. We first note that 1 − y 2 = 1 − sin2 x = cos2 x = | cos x|. This prompts us to consider values of x for which cos x < 0, such as x = π. In this case dy dx

x=π

d (sin x) = dx

= cos x x=π = cos π = −1, x=π

but

√ 1 − y 2 |x=π = 1 − sin2 π = 1 = 1. Thus, y = sin x will only be a solution of y = 1 − y 2 when cos x > 0. An interval of definition is then (−π/2, π/2). Other intervals are (3π/2, 5π/2), (7π/2, 9π/2), and so on. 48. Since the first and second derivatives of sin t and cos t involve sin t and cos t, it is plausible that a linear combination of these functions, A sin t + B cos t, could be a solution of the differential equation. Using y = A cos t − B sin t and y = −A sin t − B cos t and substituting into the differential equation we get y + 2y + 4y = −A sin t − B cos t + 2A cos t − 2B sin t + 4A sin t + 4B cos t = (3A − 2B) sin t + (2A + 3B) cos t = 5 sin t Thus 3A − 2B = 5 and 2A + 3B = 0. Solving these simultaneous equations we find A = 15 13 15 10 and B = − 10 . A particular solution is y = sin t − cos t. 13 13 13 49. One solution is given by the upper portion of the graph with domain approximately (0, 2.6). The other solution is given by the lower portion of the graph, also with domain approximately (0, 2.6). 50. One solution, with domain approximately (−∞, 1.6) is the portion of the graph in the second quadrant together with the lower part of the graph in the first quadrant. A second solution, with domain approximately (0, 1.6) is the upper part of the graph in the first quadrant. The third solution, with domain (0, ∞), is the part of the graph in the fourth quadrant.

1.1 Definitions and Terminology

51. Diﬀerentiating (x3 + y 3 )/xy = 3c we obtain xy(3x2 + 3y 2 y ) − (x3 + y 3 )(xy + y) =0 x2 y 2 3x3 y + 3xy 3 y − x4 y − x3 y − xy 3 y − y 4 = 0 (3xy 3 − x4 − xy 3 )y = −3x3 y + x3 y + y 4 y =

y 4 − 2x3 y y(y 3 − 2x3 ) . = 2xy 3 − x4 x(2y 3 − x3 )

52. A tangent line will be vertical where y is undeﬁned, or in this case, where x(2y 3 − x3 ) = 0. This gives x = 0 or 2y 3 = x3 . Substituting y 3 = x3 /2 into x3 + y 3 = 3xy we get 1 1 x x3 + x3 = 3x 2 21/3 3 3 3 x = 1/3 x2 2 2 x3 = 22/3 x2 x2 (x − 22/3 ) = 0. Thus, there are vertical tangent lines at x = 0 and x = 22/3 , or at (0, 0) and (22/3 , 21/3 ). Since 22/3 ≈ 1.59, the estimates of the domains in Problem 50 were close. √ √ 53. The derivatives of the functions are φ 1 (x) = −x/ 25 − x2 and φ 2 (x) = x/ 25 − x2 , neither of which is deﬁned at x = ±5. 54. To determine if a solution curve passes through (0, 3) we let t = 0 and P = 3 in the equation P = c1 et /(1 + c1 et ). This gives 3 = c1 /(1 + c1 ) or c1 = − 32 . Thus, the solution curve P =

(−3/2)et −3et = 1 − (3/2)et 2 − 3et

passes through the point (0, 3). Similarly, letting t = 0 and P = 1 in the equation for the one-parameter family of solutions gives 1 = c1 /(1 + c1 ) or c1 = 1 + c1 . Since this equation has no solution, no solution curve passes through (0, 1). 55. For the first-order differential equation integrate f (x). For the second-order differential equa´ ´ tion integrate twice. In the latter case we get y = ( f (x)dx) dx + c1 x + c2 . 56. Solving for y using the quadratic formula we obtain the two differential equations y =

1 2 + 2 1 + 3x6 x

and y =

1 2 − 2 1 + 3x6 , x

so the diﬀerential equation cannot be put in the form dy/dx = f (x, y).

CHAPTER 1

INTRODUCTION TO DIFFERENTIAL EQUATIONS

57. The differential equation yy − xy = 0 has normal form dy/dx = x. These are not equivalent because y = 0 is a solution of the first differential equation but not a solution of the second. 58. Differentiating we get y = c1 + 2c2 x and y = 2c2 . Then c2 = y /2 and c1 = y − xy , so y = y − xy x +

y 1 x2 = xy − x2 y 2 2

and the diﬀerential equation is x2 y − 2xy + 2y = 0. 59. (a) Since e−x is positive for all values of x, dy/dx > 0 for all x, and a solution, y(x), of the diﬀerential equation must be increasing on any interval. 2

(b)

lim

x→−∞ dx

= lim e−x = 0 and lim 2

x→∞ dx

x→−∞

= lim e−x = 0. Since dy/dx approaches 0 as 2

x→∞

x approaches −∞ and ∞, the solution curve has horizontal asymptotes to the left and to the right.

dy dx

d 2 2 e−x = −2xe−x . dx

Since the second derivative is positive for x < 0 and negative for x > 0, the solution curve is concave up on (−∞, 0) and concave down on (0, ∞).

(d) 60. (a) The derivative of a constant solution y = c is 0, so solving 5 − c = 0 we see that c = 5 and so y = 5 is a constant solution. (b) A solution is increasing where dy/dx = 5 − y > 0 or y < 5. A solution is decreasing where dy/dx = 5 − y < 0 or y > 5. 61. (a) The derivative of a constant solution is 0, so solving y(a − by) = 0 we see that y = 0 and y = a/b are constant solutions. (b) A solution is increasing where dy/dx = y(a − by) = by(a/b − y) > 0 or 0 < y < a/b. A solution is decreasing where dy/dx = by(a/b − y) < 0 or y < 0 or y > a/b.

1.1

Definitions and Terminology

(c) Using implicit diﬀerentiation we compute d2 y = y(−by ) + y (a − by) = y (a − 2by). dx2 Thus d2 y/dx2 = 0 when y = a/2b. Since d2 y/dx2 > 0 for 0 < y < a/2b and d2 y/dx2 < 0 for a/2b < y < a/b, the graph of y = φ(x) has a point of inﬂection at y = a/2b. (d)

y y = a/b

y=0

62. (a) If y = c is a constant solution then y = 0, but c2 + 4 is never 0 for any real value of c. (b) Since y = y 2 + 4 > 0 for all x where a solution y = φ(x) is defined, any solution must be increasing on any interval on which it is defined. Thus it cannot have any relative extrema. (c) Using implicit differentiation we compute d2 y/dx2 = 2yy = 2y(y 2 + 4). Setting d2 y/dx2 = 0 we see that y = 0 corresponds to the only possible point of inflection. Since d2 y/dx2 < 0 for y < 0 and d2 y/dx2 > 0 for y > 0, there is a point of inflection where y = 0. (d)

CHAPTER 1

INTRODUCTION TO DIFFERENTIAL EQUATIONS

63. In Mathematica use Clear[y] y[x ]:= x Exp[5x] Cos[2x] y[x] y''''[x] − 20y'''[x] + 158y''[x] − 580y'[x] + 841y[x]//Simplify The output will show y(x) = e5x x cos 2x, which verifies that the correct function was entered, and 0, which verifies that this function is a solution of the differential equation. 64. In Mathematica use Clear[y] y[x ]:= 20Cos[5Log[x]]/x − 3Sin[5Log[x]]/x y[x] xˆ3 y'''[x] + 2xˆ2 y''[x] + 20x y'[x] − 78y[x]//Simplify The output will show y(x) = 20 cos(5 ln x)/x − 3 sin(5 ln x)/x, which verifies that the correct function was entered, and 0, which verifies that this function is a solution of the differential equation.

1.2

Initial-Value Problems

1. Solving −1/3 = 1/(1 + c1 ) we get c1 = −4. The solution is y = 1/(1 − 4e−x ). 2. Solving 2 = 1/(1 + c1 e) we get c1 = −(1/2)e−1 . The solution is y = 2/(2 − e−(x+1) ) . 3. Letting x = 2 and solving 1/3 = 1/(4 + c) we get c = −1. The solution is y = 1/(x2 − 1). This solution is defined on the interval (1, ∞). 4. Letting x = −2 and solving 1/2 = 1/(4 + c) we get c = −2. The solution is y = 1/(x2 − 2). √ This solution is defined on the interval (−∞, − 2 ). 5. Letting x = 0 and solving 1 = 1/c we get c = 1. The solution is y = 1/(x2 + 1). This solution is defined on the interval (−∞, ∞). 6. Letting x = 1/2 and solving −4 = 1/(1/4 + c) we get c = −1/2. The solution is y = √ √ 1/(x2 − 1/2) = 2/(2x2 − 1). This solution is defined on the interval (−1/ 2 , 1/ 2 ). In Problems 7–10, we use x = c1 cos t + c2 sin t and x = −c1 sin t + c2 cos t to obtain a system of two equations in the two unknowns c1 and c2 .

1.2 Initial-Value Problems

7. From the initial conditions we obtain the system c1 = −1c2 = 8 The solution of the initial-value problem is x = − cos t + 8 sin t. 8. From the initial conditions we obtain the system c 2 = 0 − c1 = 1 The solution of the initial-value problem is x = − cos t. 9. From the initial conditions we obtain √ √ 1 1 1 3 3 c1 + c2 = − c2 + =0 2 2 2 2 2 √ Solving, we ﬁnd c1 = 3/4 and c2 = 1/4. The solution of the initial-value problem is √ x = ( 3/4) cos t + (1/4) sin t. 10. From the initial conditions we obtain

√

√ √ 2 2 c1 + c2 = 2 √2 √2 √ 2 2 c1 + c2 = 2 2. [6pt] − 2 2

Solving, we find c1 = −1 and c2 = 3. The solution of the initial-value problem is x = − cos t + 3 sin t. In Problems 11–14, we use y = c1 ex + c2 e−x and y = c1 ex − c2 e−x to obtain a system of two equations in the two unknowns c1 and c2 . 11. From the initial conditions we obtain c1 + c 2 = 1 c1 − c2 = 2. Solving, we find c1 = 32 and c2 = − 12 . The solution of the initial-value problem is y = 3 x 1 −x 2e − 2e . 12. From the initial conditions we obtain ec1 + e−1 c2 = 0 ec1 − e−1 c2 = e. Solving, we find c1 = 12 and c2 = − 12 e2 . The solution of the initial-value problem is 1 1 1 1 y = ex − e2 e−x = ex − e2−x . 2 2 2 2

CHAPTER 1

INTRODUCTION TO DIFFERENTIAL EQUATIONS

13. From the initial conditions we obtain e−1 c1 + ec2 = 5 e−1 c1 − ec2 = −5. Solving, we find c1 = 0 and c2 = 5e−1 . The solution of the initial-value problem is y = 5e−1 e−x = 5e−1−x . 14. From the initial conditions we obtain c1 + c2 = 0 c1 − c2 = 0. Solving, we find c1 = c2 = 0. The solution of the initial-value problem is y = 0. 15. Two solutions are y = 0 and y = x3 . 16. Two solutions are y = 0 and y = x2 . (Also, any constant multiple of x2 is a solution.) 2 ∂f = y −1/3 . Thus, the differential equation will have a unique ∂y 3 solution in any rectangular region of the plane where y = 0.

17. For f (x, y) = y 2/3 we have

18. For f (x, y) =

√

xy we have ∂f /∂y = 12

x/y . Thus, the diﬀerential equation will have a

unique solution in any region where x > 0 and y > 0 or where x < 0 and y < 0. ∂f 1 y we have = . Thus, the diﬀerential equation will have a unique solution x ∂y x in any region where x = 0.

19. For f (x, y) =

20. For f (x, y) = x + y we have

∂f = 1. Thus, the diﬀerential equation will have a unique ∂y

solution in the entire plane. 21. For f (x, y) = x2 /(4 − y 2 ) we have ∂f /∂y = 2x2 y/(4 − y 2 )2 . Thus the diﬀerential equation will have a unique solution in any region where y < −2, −2 < y < 2, or y > 2. 22. For f (x, y) =

∂f −3x2 y 2 x2 we have . Thus, the diﬀerential equation will have a = 3 1+y ∂y (1 + y 3 )2

unique solution in any region where y = −1. 23. For f (x, y) =

y2 x2 + y 2

we have

∂f 2x2 y = . Thus, the diﬀerential equation will have a ∂y (x2 + y 2 )2

unique solution in any region not containing (0, 0).

1.2 Initial-Value Problems

24. For f (x, y) = (y + x)/(y − x) we have ∂f /∂y = −2x/(y − x)2 . Thus the differential equation will have a unique solution in any region where y < x or where y > x. In Problems 25–28, we identify f (x, y) = y 2 − 9 and ∂f /∂y = y/ y 2 − 9. We see that f and ∂f /∂y are both continuous in the regions of the plane determined by y < −3 and y > 3 with no restrictions on x. 25. Since 4 > 3, (1, 4) is in the region defined by y > 3 and the differential equation has a unique solution through (1, 4). 26. Since (5, 3) is not in either of the regions defined by y < −3 or y > 3, there is no guarantee of a unique solution through (5, 3). 27. Since (2, −3) is not in either of the regions defined by y < −3 or y > 3, there is no guarantee of a unique solution through (2, −3). 28. Since (−1, 1) is not in either of the regions defined by y < −3 or y > 3, there is no guarantee of a unique solution through (−1, 1). 29. (a) A one-parameter family of solutions is y = cx. Since y = c, xy = xc = y and y(0) = c · 0 = 0. (b) Writing the equation in the form y = y/x, we see that R cannot contain any point on the y-axis. Thus, any rectangular region disjoint from the y-axis and containing (x0 , y0 ) will determine an interval around x0 and a unique solution through (x0 , y0 ). Since x0 = 0 in part (a), we are not guaranteed a unique solution through (0, 0). (c) The piecewise-defined function which satisfies y(0) = 0 is not a solution since it is not differentiable at x = 0. d tan (x + c) = sec2 (x + c) = 1+tan2 (x + c), we see that y = tan (x + c) satisfies 30. (a) Since dx the differential equation. (b) Solving y(0) = tan c = 0 we obtain c = 0 and y = tan x. Since tan x is discontinuous at x = ±π/2, the solution is not defined on (−2, 2) because it contains ±π/2. (c) The largest interval on which the solution can exist is (−π/2, π/2). 1 1 1 d is a solution of the differ= y 2 , we see that y = − − = 31. (a) Since dx x+c (x + c)2 x+c ential equation. (b) Solving y(0) = −1/c = 1 we obtain c = −1 and y = 1/(1−x). Solving y(0) = −1/c = −1 we obtain c = 1 and y = −1/(1+x). Being sure to include x = 0, we see that the interval of existence of y = 1/(1 − x) is (−∞, 1), while the interval of existence of y = −1/(1 + x) is (−1, ∞).