Skip to content
Textbook Guides

Solution Manual for Differential Equations and Linear Algebra, 4th Edition

Solution Manual for Differential Equations and Linear Algebra, 4th Edition makes studying more efficient with well-organized textbook notes.

Preview 16 of 994 pages

See the first 16 pages exactly as they appear in the document.

  1. Page 1
    Read page 1 as text

    S OLUTIONS M ANUAL D IFFERENTIAL E QUATIONS AND L INEAR A LGEBRA F OURTH E DITION Stephen W. Goode and Scott A. Annin California State University, Fullerton

  2. Page 2
  3. Page 3
    Read page 3 as text

    Contents Chapter 1 1 Chapter 2 123 Chapter 3 207 Chapter 4 256 Chapter 5 348 Chapter 6 390 Chapter 7 442 Chapter 8 534 Chapter 9 631 Chapter 10 762 Chapter 11 861 Appendix A 979 Appendix B 982 Appendix C 987

  4. Page 4
    Read page 4 as text

    1 Chapter 1 Solutions Solutions to Section 1.1 True-False Review: (a): FALSE. A derivative must involve some derivative of the function y = f ( x ), not necessarily the first derivative. (b): FALSE. The order of a differential equation is the order of the highest , not the lowest, derivative appearing in the differential equation. (c): FALSE. This differential equation has order two, since the highest order derivative that appears in the equation is the second order expression y ′′ . (d): FALSE. The carrying capacity refers to the maximum population size that the environment can support in the long run; it is not related to the initial population in any way. (e): TRUE. The value y (0) is called an initial condition to the differential equation for y ( t ). (f ): TRUE. According to Newton’s Law of Cooling, the rate of cooling is proportional to the difference between the object’s temperature and the medium’s temperature. Since that difference is greater for the object at 100 ◦ F than the object at 90 ◦ F , the object whose temperature is 100 ◦ F has a greater rate of cooling. (g): FALSE. The temperature of the object is given by T ( t ) = T m + ce − kt , where T m is the temperature of the medium, and c and k are constants. Since e − kt  = 0, we see that T ( t )  = T m for all times t . The temperature of the object approaches the temperature of the surrounding medium, but never equals it. (h): TRUE. Since the temperature of the coffee is falling, the temperature difference between the coffee and the room is higher initially, during the first hour, than it is later, when the temperature of the coffee has already decreased. (i): FALSE. The slopes of the two curves are negative reciprocals of each other. (j): TRUE. If the original family of parallel lines have slopes k for k  = 0, then the family of orthogonal tra- jectories are parallel lines with slope − 1 k . If the original family of parallel lines are vertical (resp. horizontal), then the family of orthogonal trajectories are horizontal (resp. vertical) parallel lines. (k): FALSE. The family of orthogonal trajectories for a family of circles centered at the origin is the family of lines passing through the origin. (l): TRUE. If v ( t ) denotes the velocity of the object at time t and a ( t ) denotes the velocity of the object at time t , then we have a ( t ) = v ′ ( t ), which is a differential equation for the unknown function v ( t ). (m): FALSE. The restoring force is directed in the direction opposite to the displacement from the equi- librium position. (n): TRUE. The allometric relationship B = B 0 m 3 / 4 , where B 0 is a constant, relates the metabolic rate and total body mass for any species. Problems: 1. The order is 2. 2. The order is 1.

  5. Page 5
    Read page 5 as text

    2 3. The order is 3. 4. The order is 2. 5. We compute the first three derivatives of y ( t ) = ln t : dy dt = 1 t , d 2 y dt 2 = − 1 t 2 , d 3 y dt 3 = 2 t 3 . Therefore, 2 ( dy dt ) 3 = 2 t 3 = d 3 y dt 3 , as required. 6. We compute the first two derivatives of y ( x ) = x/ ( x + 1): dy dx = 1 ( x + 1) 2 and d 2 y dx 2 = − 2 ( x + 1) 3 . Then y + d 2 y dx 2 = x x + 1 − 2 ( x + 1) 3 = x 3 + 2 x 2 + x − 2 ( x + 1) 3 = ( x + 1) + ( x 3 + 2 x 2 − 3) ( x + 1) 3 = 1 ( x + 1) 2 + x 3 + 2 x 2 − 3 (1 + x ) 3 , as required. 7. We compute the first two derivatives of y ( x ) = e x sin x : dy dx = e x (sin x + cos x ) and d 2 y dx 2 = 2 e x cos x. Then 2 y cot x − d 2 y dx 2 = 2( e x sin x ) cot x − 2 e x cos x = 0 , as required. 8. ( T − T m ) − 1 dT dt = − k = ⇒ d dt (ln | T − T m | ) = − k . The preceding equation can be integrated directly to yield ln | T − T m | = − kt + c 1 . Exponentiating both sides of this equation gives | T − T m | = e − kt + c 1 , which can be written as T − T m = ce − kt , where c = ± e c 1 . Rearranging yields T ( t ) = T m + ce − kt . 9. After 4 p.m. In the first two hours after noon, the water temperature increased from 50 ◦ F to 55 ◦ F, an increase of five degrees. Because the temperature of the water has grown closer to the ambient air temperature, the temperature difference | T − T m | is smaller, and thus, the rate of change of the temperature of the water grows smaller, according to Newton’s Law of Cooling. Thus, it will take longer for the water temperature to increase another five degrees. Therefore, the water temperature will reach 60 ◦ F more than two hours later than 2 p.m., or after 4 p.m. 10. The object temperature cools a total of 40 ◦ F during the 40 minutes, but according to Newton’s Law of Cooling, it cools faster in the beginning (since | T − T m | is greater at first). Thus, the object cooled half-way

  6. Page 6
    Read page 6 as text

    3 from 70 ◦ F to 30 ◦ F in less than half the total cooling time. Therefore, it took less than 20 minutes for the object to reach 50 ◦ F. 11. The given family of curves satisfies: x 2 + 9 y 2 = c = ⇒ 2 x + 18 y dy dx = 0 = ⇒ dy dx = − x 9 y . Orthogonal trajectories satisfy: dy dx = 9 y x = ⇒ 1 y dy dx = 9 x = ⇒ d dx (ln | y | ) = 9 x = ⇒ ln | y | = 9 ln | x | + c 1 = ⇒ y = kx 9 , where k = ± e c 1 . 1.5 -0.5 -0.4 1.0 x y(x) -0.8 0.4 0.5 0.8 -1.0 -1.5 Figure 0.0.1: Figure for Problem 11 12. Given family of curves satisfies: y = cx 2 = ⇒ c = y x 2 . Hence, dy dx = 2 cx = c ( y x 2 ) x = 2 y x . Orthogonal trajectories satisfy: dy dx = − x 2 y = ⇒ 2 y dy dx = − x = ⇒ d dx ( y 2 ) = − x = ⇒ y 2 = − 1 2 x 2 + c 1 = ⇒ 2 y 2 + x 2 = c 2 , where c 2 = 2 c 1 . 13. Given a family of curves satisfies: y = c x = ⇒ x dy dx + y = 0 = ⇒ dy dx = − y x . Orthogonal trajectories satisfy: dy dx = x y = ⇒ y dy dx = x = ⇒ d dx ( 1 2 y 2 ) = x = ⇒ 1 2 y 2 = 1 2 x 2 + c 1 = ⇒ y 2 − x 2 = c 2 , where c 2 = 2 c 1 . 14. The given family of curves satisfies: y = cx 5 = ⇒ c = y x 5 . Hence, dy dx = 5 cx 4 = 5 ( y x 5 ) x 4 = 5 y x .

  7. Page 7
    Read page 7 as text

    4 x 1 y(x) 1.6 2 2.0 1.2 0.4 0.8 -1.2 -2 -0.4 -2.0 -1.6 -1 -0.8 Figure 0.0.2: Figure for Problem 12 -2 2 -4 y(x) 2 4 -4 x -2 4 Figure 0.0.3: Figure for Problem 13 Orthogonal trajectories satisfy: dy dx = − x 5 y = ⇒ 5 y dy dx = − x = ⇒ d dx ( 5 2 y 2 ) = − x = ⇒ 5 2 y 2 = − 1 2 x 2 + c 1 = ⇒ 5 y 2 + x 2 = c 2 , where c 2 = 2 c 1 .

  8. Page 8
    Read page 8 as text

    5 1.5 -0.4 1.0 y(x) x -0.8 0.4 0.8 -1.0 -1.5 Figure 0.0.4: Figure for Problem 14 15. Given family of curves satisfies: y = ce x = ⇒ dy dx = ce x = y . Orthogonal trajectories satisfy: dy dx = − 1 y = ⇒ y dy dx = − 1 = ⇒ d dx ( 1 2 y 2 ) = − 1 = ⇒ 1 2 y 2 = − x + c 1 = ⇒ y 2 = − 2 x + c 2 . 2 1 y(x) 1 x -1 -2 -1 Figure 0.0.5: Figure for Problem 15 16. Given family of curves satisfies: y 2 = 2 x + c = ⇒ dy dx = 1 y .Orthogonal trajectories satisfy: dy dx = − y = ⇒ y − 1 dy dx = − 1 = ⇒ d dx (ln | y | ) = − 1 = ⇒ ln | y | = − x + c 1 = ⇒ y = c 2 e − x .

  9. Page 9
    Read page 9 as text

    6 1 1 3 -3 -2 2 -4 -1 2 -1 4 3 y(x) x 4 Figure 0.0.6: Figure for Problem 16 17. y = cx m = ⇒ dy dx = cmx m − 1 , but c = y x m so dy dx = my x . Orthogonal trajectories satisfy: dy dx = − x my = ⇒ y dy dx = − x m = ⇒ d dx ( 1 2 y 2 ) = − x m = ⇒ 1 2 y 2 = − 1 2 m x 2 + c 1 = ⇒ y 2 = − 1 m x 2 + c 2 . 18. y = mx + c = ⇒ dy dx = m . Orthogonal trajectories satisfy: dy dx = − 1 m = ⇒ y = − 1 m x + c 1 . 19. y 2 = mx + c = ⇒ 2 y dy dx = m = ⇒ dy dx = m 2 y . Orthogonal trajectories satisfy: dy dx = − 2 y m = ⇒ y − 1 dy dx = − 2 m = ⇒ d dx (ln | y | ) = − 2 m = ⇒ ln | y | = − 2 m x + c 1 = ⇒ y = c 2 e − 2 x m . 20. y 2 + mx 2 = c = ⇒ 2 y dy dx + 2 mx = 0 = ⇒ dy dx = − mx y . Orthogonal trajectories satisfy: dy dx = y mx = ⇒ y − 1 dy dx = 1 mx = ⇒ d dx (ln | y | ) = 1 mx = ⇒ m ln | y | = ln | x | + c 1 = ⇒ y m = c 2 x. 21. The given family of curves satisfies: x 2 + y 2 = 2 cx = ⇒ c = x 2 + y 2 2 x . Hence, 2 x + 2 y dy dx = 2 c = x 2 + y 2 x .

  10. Page 10
    Read page 10 as text

    7 Therefore, 2 y dy dx = x 2 + y 2 x − 2 x = y 2 − x 2 x , so that dy dx = y 2 − x 2 2 xy . Orthogonal trajectories satisfy: dy dx = − 2 xy y 2 − x 2 = 2 xy x 2 − y 2 . 22. u = x 2 + 2 y 2 = ⇒ 0 = 2 x + 4 y dy dx = ⇒ dy dx = − x 2 y . Orthogonal trajectories satisfy: dy dx = 2 y x = ⇒ y − 1 dy dx = 2 x = ⇒ d dx (ln | y | ) = 2 x = ⇒ ln | y | = 2 ln | x | + c 1 = ⇒ y = c 2 x 2 . x 1 y(x) 1.6 2 2.0 1.2 0.4 0.8 -1.2 -2 -0.4 -2.0 -1.6 -1 -0.8 Figure 0.0.7: Figure for Problem 22 23. m 1 = tan ( a 1 ) = tan ( a 2 − a ) = tan ( a 2 ) − tan ( a ) 1 + tan ( a 2 ) tan ( a ) = m 2 − tan ( a ) 1 + m 2 tan ( a ) . 24. d 2 y dt 2 = g = ⇒ dy dt = gt + c 1 = ⇒ y ( t ) = gt 2 2 + c 1 t + c 2 . Now impose the initial conditions. y (0) = 0 = ⇒ c 2 = 0 . dy dt (0) = ⇒ c 1 = 0 . Hence, the solution to the initial-value problem is: y ( t ) = gt 2 2 . The object hits the ground at time, t 0 , when y ( t 0 ) = 100. Hence 100 = gt 2 0 2 , so that t 0 = √ 200 g ≈ 4 . 52 s, where we have taken g = 9 . 8 ms − 2 .

  11. Page 11
    Read page 11 as text

    8 25. From d 2 y dt 2 = g , we integrate twice to obtain the general equations for the velocity and the position of the ball, respectively: dy dt = gt + c and y ( t ) = 1 2 gt 2 + ct + d , where c, d are constants of integration. Setting y = 0 to be at the top of the boy’s head (and positive direction downward), we know that y (0) = 0. Since the object hits the ground 8 seconds later, we have that y (8) = 5 (since the ground lies at the position y = 5). From the values of y (0) and y (8), we find that d = 0 and 5 = 32 g + 8 c . Therefore, c = 5 − 32 g 8 . (a). The ball reaches its maximum height at the moment when y ′ ( t ) = 0. That is, gt + c = 0. Therefore, t = − c g = 32 g − 5 8 g ≈ 3 . 98 s. (b). To find the maximum height of the tennis ball, we compute y (3 . 98) ≈ − 253 . 51 feet. So the ball is 253.51 feet above the top of the boy’s head, which is 258 . 51 feet above the ground. 26. From d 2 y dt 2 = g , we integrate twice to obtain the general equations for the velocity and the position of the rocket, respectively: dy dt = gt + c and y ( t ) = 1 2 gt 2 + ct + d , where c, d are constants of integration. Setting y = 0 to be at ground level, we know that y (0) = 0. Thus, d = 0. (a). The rocket reaches maximum height at the moment when y ′ ( t ) = 0. That is, gt + c = 0. Therefore, the time that the rocket achieves its maximum height is t = − c g . At this time, y ( t ) = − 90 (the negative sign accounts for the fact that the positive direction is chosen to be downward). Hence, − 90 = y ( − c g ) = 1 2 g ( − c g ) 2 + c ( − c g ) = c 2 2 g − c 2 g = − c 2 2 g . Solving this for c , we find that c = ± √ 180 g . However, since c represents the initial velocity of the rocket, and the initial velocity is negative (relative to the fact that the positive direction is downward), we choose c = − √ 180 g ≈ − 42 . 02 ms − 1 , and thus the initial speed at which the rocket must be launched for optimal viewing is approximately 42 . 02 ms − 1 . (b). The time that the rocket reaches its maximum height is t = − c g ≈ − − 42 . 02 9 . 81 = 4 . 28 s. 27. From d 2 y dt 2 = g , we integrate twice to obtain the general equations for the velocity and the position of the rocket, respectively: dy dt = gt + c and y ( t ) = 1 2 gt 2 + ct + d , where c, d are constants of integration. Setting y = 0 to be at the level of the platform (with positive direction downward), we know that y (0) = 0. Thus, d = 0. (a). The rocket reaches maximum height at the moment when y ′ ( t ) = 0. That is, gt + c = 0. Therefore, the time that the rocket achieves its maximum height is t = − c g . At this time, y ( t ) = − 85 (this is 85 m above the platform, or 90 m above the ground). Hence, − 85 = y ( − c g ) = 1 2 g ( − c g ) 2 + c ( − c g ) = c 2 2 g − c 2 g = − c 2 2 g .

  12. Page 12
    Read page 12 as text

    9 Solving this for c , we find that c = ± √ 170 g . However, since c represents the initial velocity of the rocket, and the initial velocity is negative (relative to the fact that the positive direction is downward), we choose c = − √ 170 g ≈ − 40 . 84 ms − 1 , and thus the initial speed at which the rocket must be launched for optimal viewing is approximately 40 . 84 ms − 1 . (b). The time that the rocket reaches its maximum height is t = − c g ≈ − − 40 . 84 9 . 81 = 4 . 16 s. 28. If y ( t ) denotes the displacement of the object from its initial position at time t , the motion of the object can be described by the initial-value problem d 2 y dt 2 = g, y (0) = 0 , dy dt (0) = − 2 . We first integrate this differential equation: d 2 y dt 2 = g = ⇒ dy dt = gt + c 1 = ⇒ y ( t ) = gt 2 2 + c 1 t + c 2 . Now impose the initial conditions. y (0) = 0 = ⇒ c 2 = 0. dy dt (0) = − 2 = ⇒ c 1 = − 2. Hence the solution to the initial-value problem is y ( t ) = gt 2 2 − 2 t . We are given that y (10) = h . Consequently, h = g (10) 2 2 − 2 · 10 = ⇒ h = 10(5 g − 2) ≈ 470 m where we have taken g = 9 . 8 ms − 2 . 29. If y ( t ) denotes the displacement of the object from its initial position at time t , the motion of the object can be described by the initial-value problem d 2 y dt 2 = g, y (0) = 0 , dy dt (0) = v 0 . We first integrate the differential equation: d 2 y dt 2 = g = ⇒ dy dt = gt + c 1 = ⇒ y ( t ) = gt 2 2 + c 1 t + c 2 . Now impose the initial conditions. y (0) = 0 = ⇒ c 2 = 0. dy dt (0) = v 0 = ⇒ c 1 = v 0 . Hence the solution to the initial-value problem is y ( t ) = gt 2 2 + v 0 t . We are given that y ( t 0 ) = h . Consequently, h = gt 2 0 + v 0 t 0 . Solving for v 0 yields v 0 = 2 h − gt 2 0 2 t 0 . 30. From y ( t ) = A cos ( ωt − φ ), we obtain dy dt = − Aω sin ( ωt − φ ) and d 2 y dt 2 = − Aω 2 cos ( ωt − φ ) . Hence, d 2 y dt 2 + ω 2 y = − Aω 2 cos ( ωt − φ ) + Aω 2 cos ( ωt − φ ) = 0 . Substituting y (0) = a , we obtain a = A cos( − φ ) = A cos( φ ). Also, from dy dt (0) = 0, we obtain 0 = − Aω sin( − φ ) = Aω sin( φ ). Since A  = 0 and ω  = 0 and | φ | < π , we have φ = 0. It follows that a = A . 31. y ( t ) = c 1 cos ( ωt ) + c 2 sin ( ωt ) = ⇒ dy dt = − c 1 ω sin ( ωt ) + c 2 ω cos ( ωt ) = ⇒ d 2 y dt 2 = − c 1 ω 2 cos ( ωt ) − c 2 ω 2 sin ( ωt ) = − ω 2 [ c 1 cos ( ωt ) + c 2 cos ( ωt )] = − ω 2 y. Consequently, d 2 y dt 2 + ω 2 y = 0 . To determine the

  13. Page 13
    Read page 13 as text

    10 amplitude of the motion we write the solution to the differential equation in the equivalent form: y ( t ) = √ c 2 1 + c 2 2 [ c 1 √ c 2 1 + c 2 2 cos ( ωt ) + c 2 √ c 2 1 + c 2 2 sin ( ωt ) ] . We can now define an angle φ by cos φ = c 1 √ c 2 1 + c 2 2 and sin φ = c 2 √ c 2 1 + c 2 2 . Then the expression for the solution to the differential equation is y ( t ) = √ c 2 1 + c 2 2 [cos ( ωt ) cos φ + sin ( ωt ) sin φ ] = √ c 2 1 + c 2 2 cos ( ωt + φ ) . Consequently the motion corresponds to an oscillation with amplitude A = √ c 2 1 + c 2 2 . 32. In this problem we have m 0 = 3g, M = 2700g, a = 1 . 5. Substituting these values into Equation (1.1.26) yields m ( t ) = 2700 { 1 − [ 1 − ( 1 900 ) 1 / 4 ] e − 1 . 5 t/ (4(2700) 1 / 4 ) } 4 . Therefore the mass of the heron after 30 days is m (30) = 2700 { 1 − [ 1 − ( 1 900 ) 1 / 4 ] e − 45 / (4(2700) 1 / 4 ) } 4 ≈ 1271 . 18 g . 33. In this problem we have m 0 = 8g, M = 280g, a = 0 . 25. Substituting these values into Equation (1.1.26) yields m ( t ) = 280 { 1 − [ 1 − ( 1 35 ) 1 / 4 ] e − t/ (16(280) 1 / 4 ) } 4 . We need to find the time, t when the mass of the rat reaches 75% of its fully grown size. Therefore we need to find t such that 75 100 · 280 = 280 { 1 − [ 1 − ( 1 35 ) 1 / 4 ] e − t/ (16(280) 1 / 4 ) } 4 . Solving algebraically for t yields t = 16 · (280) 1 / 4 · ln [ 1 − (1 / 35) 1 / 4 1 − (75 / 100)) 1 / 4 ] ≈ 140 days . Solutions to Section 1.2 True-False Review: (a): TRUE. This is condition 1 in Definition 1.2.8.

  14. Page 14
    Read page 14 as text

    11 (b): TRUE. This is the content of Theorem 1.2.12. (c): FALSE. There are solutions to y ′′ + y = 0 that do not have the form c 1 cos x + 5 c 2 cos x , such as y ( x ) = sin x . Therefore, c 1 cos x + 5 c 2 cos x does not meet the second requirement set forth in Definition 1.2.8 for the general solution. (d): FALSE. There are solutions to y ′′ + y = 0 that do not have the form c 1 cos x + 5 c 1 sin x , such as y ( x ) = cos x + sin x . Therefore, c 1 cos x + 5 c 1 sin x does not meet the second requirement set form in Definition 1.2.8 for the general solution. (e): TRUE. Since the right-hand side of the differential equation is a function of x only, we can integrate both sides n times to obtain the formula for the solution y ( x ). Problems: 1. Linear. 2. Non-linear, because of the y 2 expression on the right side of the equation. 3. Non-linear, because of the term yy ′′ on tthe left side of the equation. 4. Non-linear, because of the expression tan y appearing on the left side of the equation. 5. Linear. 6. Non-linear, because of the expression 1 y ′ on the left side of the equation. 7. y ( x ) = c 1 e − 5 x + c 2 e 5 x = ⇒ y ′ = − 5 c 1 e − 5 x + 5 c 2 e 5 x = ⇒ y ′′ = 25 c 1 e − 5 x + 25 c 2 e 5 x = ⇒ y ′′ − 25 y = (25 c 1 e − 5 x + 25 c 2 e 5 x ) − 25( c 1 e − 5 x + c 2 e 5 x ) = 0. Thus y ( x ) = c 1 e − 5 x + c 2 e 5 x is a solution of the given differential equation for all x ∈ R . 8. y ( x ) = c 1 cos 2 x + c 2 sin 2 x = ⇒ y ′ = − 2 c 1 sin 2 x + 2 c 2 cos 2 x = ⇒ y ′′ = − 4 c 1 cos 2 x − 4 c 2 sin 2 x = ⇒ y ′′ + 4 y = ( − 4 c 1 cos 2 x − 4 c 2 sin 2 x ) + 4( c 1 cos 2 x + c 2 sin 2 x ) = 0. Thus y ( x ) = c 1 cos 2 x + c 2 sin 2 x is a solution of the given differential equation for all x ∈ R . 9. y ( x ) = c 1 e x + c 2 e − 2 x = ⇒ y ′ = c 1 e x − 2 c 2 e − 2 x = ⇒ y ′′ = c 1 e x + 4 c 2 e − 2 x = ⇒ y ′′ + y ′ − 2 y = ( c 1 e x + 4 c 2 e − 2 x ) + ( c 1 e x − 2 c 2 e − 2 x ) − 2( c 1 e x + c 2 e − 2 x ) = 0. Thus y ( x ) = c 1 e x + c 2 e − 2 x is a solution of the given differential equation for all x ∈ R . 10. y ( x ) = 1 x + 4 = ⇒ y ′ = − 1 ( x + 4) 2 = − y 2 . Thus y ( x ) = 1 x + 4 is a solution of the given differential equation for x ∈ ( −∞ , − 4) or x ∈ ( − 4 , ∞ ). 11. y ( x ) = c 1 √ x = ⇒ y ′ = c 1 2 √ x = y 2 x . Thus y ( x ) = c 1 √ x is a solution of the given differential equation for all x ∈ { x : x > 0 } . 12. y ( x ) = c 1 e − x sin (2 x ) = ⇒ y ′ = 2 c 1 e − x cos (2 x ) − c 1 e − x sin (2 x ) = ⇒ y ′′ = − 3 c 1 e − x sin (2 x ) − 4 c 1 e − x cos (2 x ) = ⇒ y ′′ + 2 y ′ + 5 y = − 3 c 1 e − x sin (2 x ) − 4 c 1 e − x cos (2 x ) + 2[2 c 1 e − x cos (2 x ) − c 1 e − x sin (2 x )] + 5[ c 1 e − x sin (2 x )] = 0. Thus y ( x ) = c 1 e − x sin (2 x ) is a solution to the given differential equation for all x ∈ R . 13. y ( x ) = c 1 cosh (3 x ) + c 2 sinh (3 x ) = ⇒ y ′ = 3 c 1 sinh (3 x ) + 3 c 2 cosh (3 x ) = ⇒ y ′′ = 9 c 1 cosh (3 x ) +

  15. Page 15
    Read page 15 as text

    12 9 c 2 sinh (3 x ) = ⇒ y ′′ − 9 y = [9 c 1 cosh (3 x ) + 9 c 2 sinh (3 x )] − 9[ c 1 cosh (3 x ) + c 2 sinh (3 x )] = 0. Thus y ( x ) = c 1 cosh (3 x ) + c 2 sinh (3 x ) is a solution to the given differential equation for all x ∈ R . 14. y ( x ) = c 1 x 3 + c 2 x = ⇒ y ′ = − 3 c 1 x 4 − c 2 x 2 = ⇒ y ′′ = 12 c 1 x 5 + 2 c 2 x 3 = ⇒ x 2 y ′′ + 5 xy ′ + 3 y = x 2 ( 12 c 1 x 5 + 2 c 2 x 3 ) + 5 x ( − 3 c 1 x 4 − c 2 x 2 ) + 3 ( c 1 x 3 + c 2 x ) = 0. Thus y ( x ) = c 1 x 3 + c 2 x is a solution to the given differential equation for all x ∈ ( −∞ , 0) or x ∈ (0 , ∞ ). 15. y ( x ) = c 1 x 2 ln x = ⇒ y ′ = c 1 (2 x ln x + x ) = ⇒ y ′′ = c 1 (2 ln x + 3) = ⇒ x 2 y ′′ − 3 xy ′ + 4 y = x 2 · c 1 (2 ln x + 3) − 3 x · c 1 (2 x ln x + x ) + 4 c 1 x 2 ln x = c 1 x 2 [(2 ln x + 3) − 3(2 x ln x + 1) + 4 ln x ] = 0. Thus y ( x ) = c 1 x 2 ln x is a solution of the given differential equation for all x > 0. 16. y ( x ) = c 1 x 2 cos(3 ln x ) = ⇒ y ′ = c 1 [2 x cos(3 ln x ) − 3 x sin(3 ln x )] = ⇒ y ′′ = c 1 [ − 7 cos(3 ln x ) − 6 sin(3 ln x )] = ⇒ x 2 y ′′ − 3 xy ′ +13 y = x 2 · c 1 [ − 7 cos(3 ln x ) − 9 sin(3 ln x )] − 3 x · c 1 [2 x cos(3 ln x ) − 3 x sin(3 ln x )]+13 c 1 x 2 cos(3 ln x ) = c 1 x 2 { [ − 7 cos(3 ln x ) − 9 sin(3 ln x )] − 3[2 cos(3 ln x ) − 3 sin(3 ln x )] + 13 cos(3 ln x ) } = 0. Thus y ( x ) = c 1 x 2 cos(3 ln x ) is a solution of the given differential equation for all x > 0. 17. y ( x ) = c 1 √ x + 3 x 2 = ⇒ y ′ = c 1 2 √ x + 6 x = ⇒ y ′′ = − c 1 4 √ x 3 + 6 = ⇒ 2 x 2 y ′′ − xy ′ + y = 2 x 2 ( − c 1 4 √ x 3 + 6 ) − x ( c 1 2 √ x + 6 x ) +( c 1 √ x +3 x 2 ) = 9 x 2 . Thus y ( x ) = c 1 √ x +3 x 2 is a solution to the given differential equation for all x ∈ { x : x > 0 } . 18. y ( x ) = c 1 x 2 + c 2 x 3 − x 2 sin x = ⇒ y ′ = 2 c 1 x + 3 c 2 x 2 − x 2 cos x − 2 x sin x = ⇒ y ′′ = 2 c 1 + 6 c 2 x + x 2 sin x − 2 x cos x − 2 x cos − 2 sin x . Substituting these results into the given differential equation yields x 2 y ′′ − 4 xy ′ + 6 y = x 2 (2 c 1 + 6 c 2 x + x 2 sin x − 4 x cos x − 2 sin x ) − 4 x (2 c 1 x + 3 c 2 x 2 − x 2 cos x − 2 x sin x ) + 6( c 1 x 2 + c 2 x 3 − x 2 sin x ) = 2 c 1 x 2 + 6 c 2 x 3 + x 4 sin x − 4 x 3 cos x − 2 x 2 sin x − 8 c 1 x 2 − 12 c 2 x 3 + 4 x 3 cos x + 8 x 2 sin x + 6 c 1 x 2 + 6 c 2 x 3 − 6 x 2 sin x = x 4 sin x. Hence, y ( x ) = c 1 x 2 + c 2 x 3 − x 2 sin x is a solution to the differential equation for all x ∈ R . 19. y ( x ) = c 1 e ax + c 2 e bx = ⇒ y ′ = ac 1 e ax + bc 2 e bx = ⇒ y ′′ = a 2 c 1 e ax + b 2 c 2 e bx . Substituting these results into the differential equation yields y ′′ − ( a + b ) y ′ + aby = a 2 c 1 e ax + b 2 c 2 e bx − ( a + b )( ac 1 e ax + bc 2 e bx ) + ab ( c 1 e ax + c 2 e bx ) = ( a 2 c 1 − a 2 c 1 − abc 1 + abc 1 ) e ax + ( b 2 c 2 − abc 2 − b 2 c 2 + abc 2 ) e bx = 0 . Hence, y ( x ) = c 1 e ax + c 2 e bx is a solution to the given differential equation for all x ∈ R . 20. y ( x ) = e ax ( c 1 + c 2 x ) = ⇒ y ′ = e ax ( c 2 ) + ae ax ( c 1 + c 2 x ) = e ax ( c 2 + ac 1 + ac 2 x ) = ⇒ y ′′ = ea ax ( ac 2 ) + ae ax ( c 2 + ac 1 + ac 2 x ) = ae ax (2 c 2 + ac 1 + ac 2 x ). Substituting these into the differential equation yields y ′′ − 2 ay ′ + a 2 y = ae ax (2 c 2 + ac 1 + ac 2 x ) − 2 ae ax ( c 2 + ac 1 + ac 2 x ) + a 2 e ax ( c 1 + c 2 x ) = ae ax (2 c 2 + ac 1 + ac 2 x − 2 c 2 − 2 ac 1 − 2 ac 2 x + ac 1 + ac 2 x ) = 0 . Thus, y ( x ) = e ax ( c 1 + c 2 x ) is a solution to the given differential eqaution for all x ∈ R .

  16. Page 16
    Read page 16 as text

    13 21. y ( x ) = e ax ( c 1 cos bx + c 2 sin bx ) so, y ′ = e ax ( − bc 1 sin bx + bc 2 cos bx ) + ae ax ( c 1 cos bx + c 2 sin bx ) = e ax [( bc 2 + ac 1 ) cos bx + ( ac 2 − bc 1 ) sin bx ] so, y ′′ = e ax [ − b ( bc 2 + ac 1 ) sin bx + b ( ac 2 + bc 1 ) cos bx ] + ae ax [( bc 2 + ac 1 ) cos bx + ( ac 2 + bc 1 ) sin bx ] = e ax [( a 2 c 1 − b 2 c 1 + 2 abc 2 ) cos bx + ( a 2 c 2 − b 2 c 2 − abc 1 ) sin bx ] . Substituting these results into the differential equation yields y ′′ − 2 ay ′ + ( a 2 + b 2 ) y = ( e ax [( a 2 c 1 − b 2 c 1 + 2 abc 2 ) cos bx + ( a 2 c 2 − b 2 c 2 − abc 1 ) sin bx ]) − 2 a ( e ax [( bc 2 + ac 1 ) cos bx + ( ac 2 − bc 1 ) sin bx ]) + ( a 2 + b 2 )( e ax ( c 1 cos bx + c 2 sin bx )) = e ax [( a 2 c 1 − b 2 c 1 + 2 abc 2 − 2 abc 2 − 2 a 2 c 1 + a 2 c 1 + b 2 c 1 ) cos bx + ( a 2 c 2 − b 2 c 2 − 2 abc 1 + 2 abc 1 − 2 a 2 c 2 + a 2 c 2 + b 2 c 2 ) sin bx ] = 0 . Thus, y ( x ) = e ax ( c 1 cos bx + c 2 sin bx ) is a solution to the given differential equation for all x ∈ R . 22. y ( x ) = e rx = ⇒ y ′ = re rx = ⇒ y ′′ = r 2 e rx . Substituting these results into the given differential equation yields e rx ( r 2 − r − 6) = 0, so that r must satisfy r 2 − r − 6 = 0, or ( r − 3)( r + 2) = 0. Consequently r = 3 and r = − 2 are the only values of r for which y ( x ) = e rx is a solution to the given differential equation. The corresponding solutions are y ( x ) = e 3 x and y ( x ) = e − 2 x . 23. y ( x ) = e rx = ⇒ y ′ = re rx = ⇒ y ′′ = r 2 e rx . Substituting these results into the given differential equation yields e rx ( r 2 + 6 r + 9) = 0, so that r must satisfy r 2 + 6 r + 9 = 0, or ( r + 3) 2 = 0. Consequently r = − 3 is the only value of r for which y ( x ) = e rx is a solution to the given differential equation. The corresponding solution are y ( x ) = e − 3 x . 24. y ( x ) = x r = ⇒ y ′ = rx r − 1 = ⇒ y ′′ = r ( r − 1) x r − 2 . Substitution into the given differential equation yields x r [ r ( r − 1) + r − 1] = 0, so that r must satisfy r 2 − 1 = 0. Consequently r = − 1 and r = 1 are the only values of r for which y ( x ) = x r is a solution to the given differential equation. The corresponding solutions are y ( x ) = x − 1 and y ( x ) = x . 25. y ( x ) = x r = ⇒ y ′ = rx r − 1 = ⇒ y ′′ = r ( r − 1) x r − 2 . Substitution into the given differential equation yields x r [ r ( r − 1) + 5 r + 4] = 0, so that r must satisfy r 2 + 4 r + 4 = 0, or equivalently ( r + 2) 2 = 0. Consequently r = − 2 is the only value of r for which y ( x ) = x r is a solution to the given differential equation. The corresponding solution is y ( x ) = x − 2 . 26. y ( x ) = 1 2 x (5 x 2 − 3) = 1 2 (5 x 3 − 3 x ) = ⇒ y ′ = 1 2 (15 x 2 − 3) = ⇒ y ′′ = 15 x . Substitution into the Legendre equation with N = 3 yields (1 − x 2 ) y ′′ − 2 xy ′ + 12 y = (1 − x 2 )(15 x ) + x (15 x 2 − 3) + 6 x (5 x 2 − 3) = 0. Consequently the given function is a solution to the Legendre equation with N = 3. 27. y ( x ) = a 0 + a 1 x + a 2 x 2 = ⇒ y ′ = a 1 +2 a 2 x = ⇒ y ′′ = 4 a 2 . Substitution into the given differential equation yields (1 − x 2 )(2 a 2 ) − x ( a 1 +2 a 2 x )+4( a 0 + a 1 x + a 2 x 2 ) = 0 = ⇒ 3 a 1 x +2 a 2 +4 a 0 = 0. For this equation to hold for all x we require 3 a 1 = 0, and 2 a 2 + 4 a 0 = 0. Consequently a 1 = 0, and a 2 = − 2 a 0 . The corresponding solution to the differential equation is y ( x ) = a 0 (1 − 2 x 2 ). Imposing the normalization condition y (1) = 1 requires that a 0 = − 1. Hence, the required solution to the differential equation is y ( x ) = 2 x 2 − 1. 28. x sin y − e x = c = ⇒ x cos y dy dx + sin y − e x = 0 = ⇒ dy dx = e x − sin y x cos y .

978 more pages in the full document

Unlock it once and it stays in your library, ready to chat with or turn into flashcards.

Unlock the full document

More Mathematics study materials

See all

Study this document with CramX

Once it is in your library, every tool can work from it.