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Solution Manual For Differential Equations And Linear Algebra, 3rd Edition

Solution Manual For Differential Equations And Linear Algebra, 3rd Edition gives you all the tools you need to solve your textbook problems effectively.

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    1 Solutions to Section 1.1 True-False Review: 1. FALSE. A derivative must involve some derivative of the function y = f ( x ), not necessarily the first derivative. 2. TRUE. The initial conditions accompanying a differential equation consist of the values of y, y ′ , . . . at t = 0. 3. TRUE. If we define positive velocity to be oriented downward, then dv dt = g, where g is the acceleration due to gravity. 4. TRUE. We can justify this mathematically by starting from a ( t ) = g , and integrating twice to get v ( t ) = gt + c , and then s ( t ) = 1 2 gt 2 + ct + d , which is a quadratic equation. 5. FALSE. The restoring force is directed in the direction opposite to the displacement from the equilibrium position. 6. 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. 7. 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 6 = 0, we see that T ( t ) 6 = T m for all times t . The temperature of the object approaches the temperature of the surrounding medium, but never equals it. 8. 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. 9. FALSE. The slopes of the two curves are negative reciprocals of each other. 10. TRUE. If the original family of parallel lines have slopes k for k 6 = 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. 11. FALSE. The family of orthogonal trajectories for a family of circles centered at the origin is the family of lines passing through the origin. Problems: 1. Starting from the differential equation d 2 y dt 2 = g , where g is the acceleration of gravity and y is the unknown position function, we integrate twice to obtain the general equations for the velocity and the position of the object: dy dt = gt + c 1 and y ( t ) = gt 2 2 + c 1 t + c 2 , where c 1 , c 2 are constants of integration. Now we impose the initial conditions: y (0) = 0 implies that c 2 = 0,

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    2 and dy dt (0) = 0 implies that c 1 = 0 . Hence, the solution to the initial-value problem is y ( t ) = gt 2 2 . The object hits the ground at the time t 0 for which 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 . 2. Starting from the differential equation d 2 y dt 2 = g , where g is the acceleration of gravity and y is the unknown position function, 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. 3. Starting from the differential equation d 2 y dt 2 = g , where g is the acceleration of gravity and y is the unknown position function, 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 .

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    3 (b) The time that the rocket reaches its maximum height is t = − c g ≈ − − 42 . 02 9 . 81 = 4 . 28 s. 4. Starting from the differential equation d 2 y dt 2 = g , where g is the acceleration of gravity and y is the unknown position function, 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 . 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. 5. 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 , where g is the acceleration of gravity and y is the unknown position function. We integrate this differential equation twice to obtain the general equations for the velocity and the position of the object: dy dt = gt + c 1 and y ( t ) = gt 2 2 + c 1 t + c 2 . Now we impose the initial conditions: since y (0) = 0, we have c 2 = 0. Moreover, since dy dt (0) = − 2, we have 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 . 6. 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 .

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    4 We integrate the differential equation twice to obtain the velocity and position functions, respectively: dy dt = gt + c 1 and y ( t ) = gt 2 2 + c 1 t + c 2 . Now we impose the initial conditions. Since y (0) = 0, we have c 2 = 0. Moreover, since dy dt (0) = v 0 , we have 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 . 7. 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 6 = 0 and ω 6 = 0 and | φ | < π , we have φ = 0. It follows that a = A . 8. Taking derivatives of y ( t ) = c 1 cos ( ωt ) + c 2 sin ( ωt ), we obtain dy dt = − c 1 ω sin ( ωt ) + c 2 ω cos ( ωt ) and 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 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 .

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    5 9. 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. 10. 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. 11. 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. 12. Starting from ( T − T m ) − 1 dT dt = − k , we obtain 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 this, we conclude that T ( t ) = T m + ce − kt . 13. 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. 14. 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 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.

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    6 15. Applying implicit differentiation to the given family of curves x 2 + 4 y 2 = c with respect to x gives 2 x + 8 y dy dx = 0 . Therefore, dy dx = − x 4 y . Therefore, the collection of orthogonal trajectories satisfies: dy dx = 4 y x = ⇒ 1 y dy dx = 4 x = ⇒ d dx (ln | y | ) = 4 x = ⇒ ln | y | = 4 ln | x | + c 1 = ⇒ y = kx 4 , 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 1: Figure for Problem 15 16. Differentiation of the given family of curves y = c x with respect to x gives dy dx = − c x 2 = − 1 x · c x = − y x . Therefore, the collection of orthogonal trajectories satisfies: 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 . -2 2 -4 y(x) 2 4 -4 x -2 4 Figure 2: Figure for Problem 16

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    7 17. Solving the equation y = cx 2 for c gives c = y x 2 . Hence, differentiation leads to dy dx = 2 cx = 2 ( y x 2 ) x = 2 y x . Therefore, the collection of orthogonal trajectories satisfies: 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 . 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 3: Figure for Problem 17 18. Solving the equation y = cx 4 for c gives c = y x 4 . Hence, dy dx = 4 cx 3 = 4 ( y x 4 ) x 3 = 4 y x . Therefore, the collection of orthogonal trajectories satisfies: dy dx = − x 4 y = ⇒ 4 y dy dx = − x = ⇒ d dx (2 y 2 ) = − x = ⇒ 2 y 2 = − 1 2 x 2 + c 1 = ⇒ 4 y 2 + x 2 = c 2 , where c 2 = 2 c 1 . 1.5 -0.4 1.0 y(x) x -0.8 0.4 0.8 -1.0 -1.5 Figure 4: Figure for Problem 18 19. Implicit differentiation of the given family of curves y 2 = 2 x + c with respect to x gives 2 y dy dx = 2. That is, dy dx = 1 y . Therefore, the collection of orthogonal trajectories satisfies: dy dx = − y = ⇒ y − 1 dy dx = − 1 = ⇒ d dx (ln | y | ) = − 1 = ⇒ ln | y | = − x + c 1 = ⇒ y = ke − x ,

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    8 1 1 3 -3 -2 2 -4 -1 2 -1 4 3 y(x) x 4 Figure 5: Figure for Problem 19 where k = ± e c 1 . 20. Differentiating the given family of curves y = ce x with respect to x gives dy dx = ce x = y . Therefore, the collection of orthogonal trajectories satisfies: 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 , where c 2 = 2 c 1 . 2 1 y(x) 1 x -1 -2 -1 Figure 6: Figure for Problem 20 21. Differentiating the given family of curves y = mx + c with respect to x gives dy dx = m . Therefore, the collection of orthogonal trajectories satisfies: dy dx = − 1 m = ⇒ y = − 1 m x + c 1 . 22. Differentiating the given family of curves y = cx m with respect to x gives dy dx = cmx m − 1 . Since c = y x m , we have dy dx = my x . Therefore, the collection of orthogonal trajectories satisfies: 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 ,

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    9 where c 2 = 2 c 1 . 23. Implicit differentiation of the given family of curves y 2 + mx 2 = c with respect to x gives 2 y dy dx +2 mx = 0. That is, dy dx = − mx y . Therefore, the collection of orthogonal trajectories satisfies: 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, where c 2 = ± e c 1 . 24. Implicit differentiation of the given family of curves y 2 = mx + c with respect to x gives 2 y dy dx = m . That is, dy dx = m 2 y . Therefore, the collection of orthogonal trajectories satisfies: 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 , where c 2 = ± e c 1 . 25. Consider the coordinate curve u = x 2 + 2 y 2 (i.e. u is constant). Differentiating implicitly with respect to x gives 0 = 2 x + 4 y dy dx . Therefore, dy dx = − x 2 y . Therefore, the collection of orthogonal trajectories satisfies: 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 , where c 2 = ± e c 1 . 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 7: Figure for Problem 25 26. We have 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 ) .

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    10 Solutions to Section 1.2 True-False Review: 1. FALSE. The order of a differential equation is the order of the highest derivative appearing in the differential equation. 2. TRUE. This is condition 1 in Definition 1.2.11. 3. TRUE. This is the content of Theorem 1.2.15. 4. 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 1.2.11 for the general solution. 5. 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 1.2.11 for the general solution. 6. 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. 2, nonlinear. 2. 3, linear. 3. 2, nonlinear. 4. 2, nonlinear. 5. 4, linear. 6. 3, nonlinear. 7. We can quickly compute the first two derivatives of y ( x ): y ′ ( x ) = ( c 1 +2 c 2 ) e x cos 2 x +( − 2 c 1 + c 2 ) e x sin 2 x and y ′′ ( x ) = ( − 3 c 1 +4 c 2 ) e x cos 2 x +( − 4 c 1 − 3 c 2 ) e x sin x. Then we have y ′′ − 2 y ′ + 5 y = [( − 3 c 1 + 4 c 2 ) e x cos 2 x + ( − 4 c 1 − 3 c 2 ) e x sin x ] − 2 [( c 1 + 2 c 2 ) e x cos 2 x + ( − 2 c 1 + c 2 ) e x sin 2 x ]+5( c 1 e x cos 2 x + c 2 e x sin 2 x ) , which cancels to 0, as required. This solution is valid for all x ∈ R . 8. We can quickly compute the first two derivatives of y ( x ): y ′ ( x ) = c 1 e x − 2 c 2 e − 2 x and y ′′ ( x ) = c 1 e x + 4 c 2 e − 2 x . Then we have 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 .

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    11 9. The derivative of y ( x ) = 1 x + 4 is y ′ ( x ) = − 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 , ∞ ). 10. The derivative of y ( x ) = c 1 √ x is y ′ ( x ) = c 1 2 √ x = y 2 x . Thus y ( x ) = c 1 √ x is a solution of the given differential equation for all x > 0. 11. We can quickly compute the first two derivatives of y ( x ) = c 1 e − x sin (2 x ): y ′ ( x ) = 2 c 1 e − x cos (2 x ) − c 1 e − x sin (2 x ) and y ′′ ( x ) = − 3 c 1 e − x sin (2 x ) − 4 c 1 e − x cos (2 x ) . Therefore, we have 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 , which shows that y ( x ) = c 1 e − x sin (2 x ) is a solution to the given differential equation for all x ∈ R . 12. We can quickly compute the first two derivatives of y ( x ) = c 1 cosh (3 x ) + c 2 sinh (3 x ): y ′ ( x ) = 3 c 1 sinh (3 x ) + 3 c 2 cosh (3 x ) and y ′′ ( x ) = 9 c 1 cosh (3 x ) + 9 c 2 sinh (3 x ) . Therefore, we have 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 , which shows that y ( x ) = c 1 cosh (3 x ) + c 2 sinh (3 x ) is a solution to the given differential equation for all x ∈ R . 13. We can quickly compute the first two derivatives of y ( x ) = c 1 x 3 + c 2 x : y ′ ( x ) = − 3 c 1 x 4 − c 2 x 2 and y ′′ ( x ) = 12 c 1 x 5 + 2 c 2 x 3 . Therefore, we have 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 , which shows that y ( x ) = c 1 x 3 + c 2 x is a solution to the given differential equation for all x ∈ ( −∞ , 0) or x ∈ (0 , ∞ ). 14. We can quickly compute the first two derivatives of y ( x ) = c 1 √ x + 3 x 2 : y ′ ( x ) = c 1 2 √ x + 6 x and y ′′ ( x ) = − c 1 4 √ x 3 + 6 . Therefore, we have 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 , which shows that y ( x ) = c 1 √ x + 3 x 2 is a solution to the given differential equation for all x > 0.

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    12 15. We can quickly compute the first two derivatives of y ( x ) = c 1 x 2 + c 2 x 3 − x 2 sin x : y ′ ( x ) = 2 c 1 x + 3 c 2 x 2 − x 2 cos x − 2 x sin x and y ′′ ( x ) = 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 . 16. We can quickly compute the first two derivatives of y ( x ) = c 1 e ax + c 2 e bx : y ′ ( x ) = ac 1 e ax + bc 2 e bx and y ′′ ( x ) = 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 . 17. We can quickly compute the first two derivatives of y ( x ) = e ax ( c 1 + c 2 x ): y ′ ( x ) = e ax ( c 2 ) + ae ax ( c 1 + c 2 x ) = e ax ( c 2 + ac 1 + ac 2 x ) and y ′′ ( x ) = 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 equation for all x ∈ R . 18. We can quickly compute the first two derivatives of y ( x ) = e ax ( c 1 cos bx + c 2 sin bx ): y ′ ( x ) = 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 ] , y ′′ ( x ) = 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 ] .

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    13 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 . 19. From y ( x ) = e rx , we obtain y ′ ( x ) = re rx and y ′′ ( x ) = r 2 e rx . Substituting these results into the given differential equation yields e rx ( r 2 + 2 r − 3) = 0, so that r must satisfy r 2 + 2 r − 3 = 0, or ( r + 3)( r − 1) = 0. Consequently r = − 3 and r = 1 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 x . 20. From y ( x ) = e rx , we obtain y ′ ( x ) = re rx and y ′′ ( x ) = r 2 e rx . Substituting these results into the given differential equation yields e rx ( r 2 − 8 r + 16) = 0, so that r must satisfy r 2 − 8 r + 16 = 0, or ( r − 4) 2 = 0. Consequently the only value of r for which y ( x ) = e rx is a solution to the differential equation is r = 4. The corresponding solution is y ( x ) = e 4 x . 21. From y ( x ) = x r , we obtain y ′ ( x ) = rx r − 1 and y ′′ ( x ) = r ( r − 1) x r − 2 . Substituting these results 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 . 22. From y ( x ) = x r , we obtain y ′ ( x ) = rx r − 1 and y ′′ ( x ) = r ( r − 1) x r − 2 . Substituting these results 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 . 23. From y ( x ) = 1 2 x (5 x 2 − 3) = 1 2 (5 x 3 − 3 x ), we obtain y ′ ( x ) = 1 2 (15 x 2 − 3) and y ′′ ( x ) = 15 x . Substituting these results 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. 24. From y ( x ) = a 0 + a 1 x + a 2 x 2 , we obtain y ′ ( x ) = a 1 + 2 a 2 x and y ′′ ( x ) = 4 a 2 . Substituting these results 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. That is, 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. 25. Differentiating x sin y − e x = c implicitly with respect to x yields x cos y dy dx + sin y − e x = 0. Thus, dy dx = e x − sin y x cos y .

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    14 26. Differentiating xy 2 + 2 y − x = c implicitly with respect to x yields 2 xy dy dx + y 2 + 2 dy dx − 1 = 0. Thus, dy dx = 1 − y 2 2( xy + 1) . 27. Differentiating e xy − x = c implicitly with respect to x yields e xy [ x dy dx + y ] − 1 = 0. Therefore, we have xe xy dy dx + ye xy = 1. Hence, dy dx = 1 − ye xy xe xy . Given that y (1) = 0, we have e 0(1) − 1 = c , so that c = 0. Therefore, e xy − x = 0. Rearranging this equation and taking logarithms, we conclude that y = ln x x . 28. Differentiating e y/x + xy 2 − x = c implicitly with respect to x yields e y/x    x dy dx − y x 2    + 2 xy dy dx + y 2 − 1 = 0 . A short algebraic simplification yields dy dx = x 2 (1 − y 2 ) + ye y/x x ( e y/x + 2 x 2 y ) . 29. Differentiating x 2 y 2 − sin x = c implicitly with respect to x yields 2 x 2 y dy dx +2 xy 2 − cos x = 0. Rearranging, we obtain dy dx = cos x − 2 xy 2 2 x 2 y . Since y ( π ) = 1 π , we have π 2 ( 1 π ) 2 − sin π = c . Therefore, c = 1. Hence, x 2 y 2 − sin x = 1 so that y 2 = 1 + sin x x 2 . Since y ( π ) = 1 π , take the branch of y where x > 0, so that y ( x ) = √ 1 + sin x x . 30. By integrating dy dx = sin x with respect to x , we obtain y ( x ) = − cos x + c for all x ∈ R . 31. By integrating dy dx = x − 1 / 2 with respect to x , we obtain y ( x ) = 2 x 1 / 2 + c for all x > 0. 32. By integrating d 2 y dx 2 = xe x twice with respect to x , we obtain dy dx = xe x − e x + c 1 and y ( x ) = xe x − 2 e x + c 1 x + c 2 for all x ∈ R . 33. We consider three cases: Case 1: n = − 1 : In this case, two integrations with respect to x yields dy dx = ln | x | + c 1 and y ( x ) = x ln | x | + c 1 x + c 2 for all x ∈ ( −∞ , 0) or x ∈ (0 , ∞ ). Case 2: n = − 2 : In this case, two integrations with respect to x yields dy dx = − x − 1 + c 1 and y ( x ) = c 1 x + c 2 − ln | x | for all x ∈ ( −∞ , 0) or x ∈ (0 , ∞ ).

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    15 Case 3: n 6 = − 1 , − 2 : In this case, two integrations with respect to x yields dy dx = x n +1 n + 1 + c 1 and y ( x ) = x n +2 ( n + 1)( n + 2) + c 1 x + c 2 for all x ∈ R . 34. Integrating dy dx = ln x with respect to x yields y ( x ) = x ln x − x + c . Since y (1) = 2, we have 2 = 1(0) − 1+ c , so that c = 3. Thus, y ( x ) = x ln x − x + 3. 35. Integrating d 2 y dx 2 = cos x twice with respect to x yields dy dx = sin x + c 1 and y ( x ) = − cos x + c 1 x + c 2 . From y ′ (0) = 1, we obtain c 1 = 1, and from y (0) = 2, we obtain c 2 = 3. Thus, y ( x ) = 3 + x − cos x . 36. Integrating d 3 y dx 3 = 6 x three times with respect to x , we obtain d 2 y dx 2 = 3 x 2 + c 1 , dy dx = x 3 + c 1 x + c 2 , y ( x ) = 1 4 x 4 + 1 2 c 1 x 2 + c 2 x + c 3 . From y ′′ (0) = 4, we obtain c 1 = 4, from y ′ (0) = − 1, we obtain c 2 = − 1, and from y (0) = 1, we obtain c 3 = 1. Thus, y ( x ) = 1 4 x 4 + 2 x 2 − x + 1. 37. Integrating y ′′ ( x ) = xe x twice with respect to x , we obtain y ′ ( x ) = xe x − e x + c 1 and y ( x ) = xe x − 2 e x + c 1 x + c 2 . From y ′ (0) = 4, we obtain c 1 = 5, and from y (0) = 3, we obtain c 2 = 5. Thus, y ( x ) = xe x − 2 e x + 5 x + 5. 38. Starting with y ( x ) = c 1 e x + c 2 e − x , we find that y ′ ( x ) = c 1 e x − c 2 e − x and y ′′ ( x ) = c 1 e x + c 2 e − x . Thus, y ′′ − y = 0, so y ( x ) = c 1 e x + c 2 e − x is a solution to the differential equation on ( −∞ , ∞ ). Next we establish that every solution to the differential equation has the form c 1 e x + c 2 e − x . Suppose that y = f ( x ) is any solution to the differential equation. Then according to Theorem 1.2.15, y = f ( x ) is the unique solution to the initial-value problem y ′′ − y = 0 , y (0) = f (0) , y ′ (0) = f ′ (0) . However, consider the function y ( x ) = f (0) + f ′ (0) 2 e x + f (0) − f ′ (0) 2 e − x . This is of the form y ( x ) = c 1 e x + c 2 e − x , where c 1 = f (0)+ f ′ (0) 2 and c 2 = f (0) − f ′ (0) 2 , and therefore solves the differential equation y ′′ − y = 0. Furthermore, evaluation this function at x = 0 yields y (0) = f (0) and y ′ (0) = f ′ (0) . Consequently, this function solves the initial-value problem above. However, by assumption, y ( x ) = f ( x ) solves the same initial-value problem. Owing to the uniqueness of the solution to this initial-value problem, it follows that these two solutions are the same: f ( x ) = c 1 e x + c 2 e − x .

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