Solutions for Goode - DFQ and LA

Solutions for Goode - DFQ and LA - 1 Solutions to Section...

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Unformatted text preview: 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 = g, dt 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 1 v (t) = gt + c, and then s(t) = gt2 + ct + d, which is a quadratic equation. 2 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) = Tm + ce−kt , where Tm is the temperature of the medium, and c and k are constants. Since e−kt = 0, we see that T (t) = Tm 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 = 0, then the family of orthogonal tra1 jectories are parallel lines with slope − 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: d2 y 1. Starting from the differential equation 2 = g , where g is the acceleration of gravity and y is the unknown dt position function, we integrate twice to obtain the general equations for the velocity and the position of the object: dy gt2 = gt + c1 and y (t) = + c1 t + c2 , dt 2 where c1 , c2 are constants of integration. Now we impose the initial conditions: y (0) = 0 implies that c2 = 0, 2 and dy dt (0) = 0 implies that c1 = 0. Hence, the solution to the initial-value problem is y (t) = gt2 . 2 The object hits the ground at the time t0 for which y (t0 ) = 100. Hence 100 = s, where we have taken g = 9.8 ms −2 gt2 0 2, so that t0 = 200 g ≈ 4.52 . d2 y 2. Starting from the differential equation 2 = g , where g is the acceleration of gravity and y is the unknown dt position function, we integrate twice to obtain the general equations for the velocity and the position of the ball, respectively: dy 1 = gt + c and y (t) = gt2 + ct + d, dt 2 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 5 − 32g and 5 = 32g + 8c. Therefore, c = . 8 (a) The ball reaches its maximum height at the moment when y (t) = 0. That is, gt + c = 0. Therefore, t=− 32g − 5 c = ≈ 3.98 s. g 8g (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. d2 y 3. Starting from the differential equation 2 = g , where g is the acceleration of gravity and y is the unknown dt position function, we integrate twice to obtain the general equations for the velocity and the position of the rocket, respectively: dy 1 = gt + c and y (t) = gt2 + ct + d, dt 2 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 c time that the rocket achieves its maximum height is t = − . At this time, y (t) = −90 (the negative sign g accounts for the fact that the positive direction is chosen to be downward). Hence, −90 = y − c g = 1 c g− 2 g 2 +c − c g = c2 c2 c2 − =− . 2g g 2g √ Solving this for c, we find that c = ± 180g . 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 = − 180g ≈ −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 . 3 (b) The time that the rocket reaches its maximum height is t = − c −42.02 ≈− = 4.28 s. g 9.81 d2 y 4. Starting from the differential equation 2 = g , where g is the acceleration of gravity and y is the unknown dt position function, we integrate twice to obtain the general equations for the velocity and the position of the rocket, respectively: dy 1 = gt + c and y (t) = gt2 + ct + d, dt 2 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 c time that the rocket achieves its maximum height is t = − . At this time, y (t) = −85 (this is 85 m above g the platform, or 90 m above the ground). Hence, −85 = y − c g = 1 c g− 2 g 2 +c − c g = c2 c2 c2 − =− . 2g g 2g √ Solving this for c, we find that c = ± 170g . 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 = − 170g ≈ −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 . −40.84 c = 4.16 s. (b) The time that the rocket reaches its maximum height is t = − ≈ − g 9.81 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 d2 y dy = g, y (0) = 0, (0) = −2, dt2 dt 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 = gt + c1 dt and y (t) = gt2 + c1 t + c2 . 2 Now we impose the initial conditions: since y (0) = 0, we have c2 = 0. Moreover, since c1 = −2. Hence the solution to the initial-value problem is y (t) = Consequently, h = dy (0) = −2, we have dt gt2 − 2t. We are given that y (10) = h. 2 g (10)2 − 2 · 10 =⇒ h = 10(5g − 2) ≈ 470 m where we have taken g = 9.8 ms−2 . 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 d2 y = g, dt2 y (0) = 0, dy (0) = v0 . dt 4 We integrate the differential equation twice to obtain the velocity and position functions, respectively: dy = gt + c1 dt and y (t) = gt2 + c1 t + c2 . 2 Now we impose the initial conditions. Since y (0) = 0, we have c2 = 0. Moreover, since c1 = v0 . Hence the solution to the initial-value problem is y (t) = Consequently, h = gt2 + v0 t0 . Solving for v0 yields 0 dy (0) = v0 , we have dt gt2 + v0 t. We are given that y (t0 ) = h. 2 2h − gt2 0 . 2t0 v0 = 7. From y (t) = A cos (ωt − φ), we obtain dy = −Aω sin (ωt − φ) dt and d2 y = −Aω 2 cos (ωt − φ). dt2 Hence, d2 y + ω 2 y = −Aω 2 cos (ωt − φ) + Aω 2 cos (ωt − φ) = 0. dt2 Substituting y (0) = a, we obtain a = A cos(−φ) = A cos(φ). Also, from dy (0) = 0, we obtain 0 = dt −Aω sin(−φ) = Aω sin(φ). Since A = 0 and ω = 0 and |φ| < π , we have φ = 0. It follows that a = A. 8. Taking derivatives of y (t) = c1 cos (ωt) + c2 sin (ωt), we obtain dy = −c1 ω sin (ωt) + c2 ω cos (ωt) dt and d2 y = −c1 ω 2 cos (ωt) − c2 ω 2 sin (ωt) = −ω 2 [c1 cos (ωt) + c2 cos (ωt)] = −ω 2 y. dt2 d2 y Consequently, + ω 2 y = 0. To determine the amplitude of the motion we write the solution to the dt2 differential equation in the equivalent form: y (t) = c1 c2 + c2 1 2 c2 1 + c2 2 cos (ωt) + c2 c2 1 + c2 2 sin (ωt) . We can now define an angle φ by cos φ = c1 c2 + c2 1 2 and sin φ = c2 c2 + c2 1 2 . Then the expression for the solution to the differential equation is y (t) = c2 + c2 [cos (ωt) cos φ + sin (ωt) sin φ] = 1 2 c2 + c2 cos (ωt + φ). 1 2 Consequently the motion corresponds to an oscillation with amplitude A = c2 + c2 . 1 2 5 9. We compute the first three derivatives of y (t) = ln t: d3 y 2 = 3. 3 dt t 1 d2 y = − 2, 2 dt t dy 1 =, dt t Therefore, 2 dy dt 3 = 2 d3 y = 3, t3 dt as required. 10. We compute the first two derivatives of y (x) = x/(x + 1): dy 1 = dx (x + 1)2 d2 y 2 =− . dx2 (x + 1)3 and Then y+ x 2 x3 + 2x2 + x − 2 (x + 1) + (x3 + 2x2 − 3) 1 x3 + 2x2 − 3 d2 y = − = = = + , dx2 x + 1 (x + 1)3 (x + 1)3 (x + 1)3 (x + 1)2 (1 + x)3 as required. 11. We compute the first two derivatives of y (x) = ex sin x: dy = ex (sin x + cos x) dx Then 2y cot x − and d2 y = 2ex cos x. dx2 d2 y = 2(ex sin x) cot x − 2ex cos x = 0, dx2 as required. dT d = −k , we obtain (ln |T − Tm |) = −k . The preceding equation can dt dt be integrated directly to yield ln |T − Tm | = −kt + c1 . Exponentiating both sides of this equation gives |T − Tm | = e−kt+c1 , which can be written as 12. Starting from (T − Tm )−1 T − Tm = ce−kt , where c = ±ec1 . Rearranging this, we conclude that T (t) = Tm + 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 − Tm | 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 − Tm | 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. 6 15. Applying implicit differentiation to the given family of curves x2 + 4y 2 = c with respect to x gives 2x + 8y dy = 0. dx Therefore, x dy =− . dx 4y Therefore, the collection of orthogonal trajectories satisfies: dy 4y = dx x 1 dy 4 = y dx x =⇒ =⇒ d 4 (ln |y |) = dx x =⇒ ln |y | = 4 ln |x| + c1 =⇒ y = kx4 , where k = ±ec1 . y(x) 0.8 0.4 x -1.5 -0.5 -1.0 1.0 0.5 1.5 -0.4 -0.8 Figure 1: Figure for Problem 15 16. Differentiation of the given family of curves y = c with respect to x gives x c 1c y dy =− 2 =− · =− . dx x xx x Therefore, the collection of orthogonal trajectories satisfies: dy x = dx y =⇒ y dy =x dx =⇒ 12 y 2 d dx =x 12 1 y = x2 + c1 2 2 =⇒ where c2 = 2c1 . y(x) 4 2 x -4 -2 2 4 -2 -4 Figure 2: Figure for Problem 16 =⇒ y 2 − x2 = c2 , 7 17. Solving the equation y = cx2 for c gives c = y . Hence, differentiation leads to x2 dy 2y y = 2cx = 2 2 x = . dx x x Therefore, the collection of orthogonal trajectories satisfies: dy x =− dx 2y =⇒ 2y dy = −x dx d2 (y ) = −x dx =⇒ 1 y 2 = − x2 + c1 2 =⇒ =⇒ 2y 2 + x2 = c2 , where c2 = 2c1 . y(x) 2.0 1.6 1.2 0.8 0.4 x -1 -2 1 2 -0.4 -0.8 -1.2 -1.6 -2.0 Figure 3: Figure for Problem 17 18. Solving the equation y = cx4 for c gives c = y . Hence, x4 dy y 4y = 4cx3 = 4 4 x3 = . dx x x Therefore, the collection of orthogonal trajectories satisfies: x dy =− dx 4y =⇒ 4y dy = −x dx =⇒ d (2y 2 ) = −x dx =⇒ 1 2y 2 = − x2 + c1 2 =⇒ 4y 2 + x2 = c2 , where c2 = 2c1 . y(x) 0.8 0.4 x -1.5 1.0 -1.0 1.5 -0.4 -0.8 Figure 4: Figure for Problem 18 dy 19. Implicit differentiation of the given family of curves y 2 = 2x + c with respect to x gives 2y dx = 2. That dy 1 is, = . Therefore, the collection of orthogonal trajectories satisfies: dx y dy = −y dx =⇒ y −1 dy = −1 dx =⇒ d (ln |y |) = −1 dx =⇒ ln |y | = −x + c1 =⇒ y = ke−x , 8 y(x) 4 3 2 1 x -1 1 2 4 3 -1 -2 -3 -4 Figure 5: Figure for Problem 19 where k = ±ec1 . dy = cex = y . Therefore, the 20. Differentiating the given family of curves y = cex with respect to x gives dx collection of orthogonal trajectories satisfies: dy 1 =− dx y =⇒ y dy = −1 dx d dx =⇒ 12 y 2 = −1 =⇒ 12 y = −x + c1 2 =⇒ y 2 = −2x + c2 , where c2 = 2c1 . y(x) 2 1 x 1 -1 -1 -2 Figure 6: Figure for Problem 20 21. Differentiating the given family of curves y = mx + c with respect to x gives collection of orthogonal trajectories satisfies: dy 1 =− dx m =⇒ y=− dy dx = m. Therefore, the 1 x + c1 . m dy y = cmxm−1 . Since c = m , 22. Differentiating the given family of curves y = cxm with respect to x gives dx x dy my we have = . Therefore, the collection of orthogonal trajectories satisfies: dx x dy x =− dx my =⇒ y dy x =− dx m =⇒ d dx 12 y 2 =− x m =⇒ 12 12 y =− x +c1 2 2m =⇒ y2 = − 12 x +c2 , m 9 where c2 = 2c1 . dy 23. Implicit differentiation of the given family of curves y 2 + mx2 = c with respect to x gives 2y +2mx = 0. dx dy mx That is, =− . Therefore, the collection of orthogonal trajectories satisfies: dx y dy y = dx mx =⇒ y −1 dy 1 = dx mx d 1 (ln |y |) = dx mx =⇒ =⇒ m ln |y | = ln |x| + c1 =⇒ y m = c2 x, where c2 = ±ec1 . dy 24. Implicit differentiation of the given family of curves y 2 = mx + c with respect to x gives 2y = m. dx m dy = . Therefore, the collection of orthogonal trajectories satisfies: That is, dx 2y 2y dy =− dx m =⇒ y −1 2 dy =− dx m =⇒ d 2 (ln |y |) = − dx m =⇒ ln |y | = − 2 x+c1 m =⇒ 2x y = c2 e− m , where c2 = ±ec1 . 25. Consider the coordinate curve u = x2 + 2y 2 (i.e. u is constant). Differentiating implicitly with respect to dy dy x x gives 0 = 2x + 4y . Therefore, = − . Therefore, the collection of orthogonal trajectories satisfies: dx dx 2y dy 2y = dx x =⇒ y −1 dy 2 = dx x d 2 (ln |y |) = dx x =⇒ =⇒ ln |y | = 2 ln |x| + c1 where c2 = ±ec1 . y(x) 2.0 1.6 1.2 0.8 0.4 x -2 1 -1 2 -0.4 -0.8 -1.2 -1.6 -2.0 Figure 7: Figure for Problem 25 26. We have m1 = tan (a1 ) = tan (a2 − a) = tan (a2 ) − tan (a) m2 − tan (a) = . 1 + tan (a2 ) tan (a) 1 + m2 tan (a) =⇒ y = c2 x2 , 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 c1 cos x + 5c2 cos x, such as y (x) = sin x. Therefore, c1 cos x + 5c2 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 c1 cos x + 5c1 sin x, such as y (x) = cos x + sin x. Therefore, c1 cos x + 5c1 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) = (c1 +2c2 )ex cos 2x +(−2c1 + c2 )ex sin 2x and y (x) = (−3c1 +4c2 )ex cos 2x +(−4c1 − 3c2 )ex sin x. Then we have y − 2y + 5y x x = [(−3c1 + 4c2 )e cos 2x + (−4c1 − 3c2 )e sin x]−2 [(c1 + 2c2 )ex cos 2x + (−2c1 + c2 )ex sin 2x]+5(c1 ex cos 2x+c2 ex sin 2x), 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) = c1 ex − 2c2 e−2x and y (x) = c1 ex + 4c2 e−2x . Then we have y + y − 2y = (c1 ex + 4c2 e−2x ) + (c1 ex − 2c2 e−2x ) − 2(c1 ex + c2 e−2x ) = 0. Thus y (x) = c1 ex + c2 e−2x is a solution of the given differential equation for all x ∈ R. 11 1 1 1 = −y 2 . Thus y (x) = is y (x) = − is a solution of the given x+4 (x + 4)2 x+4 differential equation for x ∈ (−∞, −4) or x ∈ (−4, ∞). 9. The derivative of y (x) = √ 10. The derivative of y (x) = c1 x is y (x) = √ c1 y √= . Thus y (x) = c1 x is a solution of the given 2x 2x differential equation for all x > 0. 11. We can quickly compute the first two derivatives of y (x) = c1 e−x sin (2x): y (x) = 2c1 e−x cos (2x) − c1 e−x sin (2x) and y (x) = −3c1 e−x sin (2x) − 4c1 e−x cos (2x). Therefore, we have y + 2y + 5y = −3c1 e−x sin (2x) − 4c1 e−x cos (2x) + 2[2c1 e−x cos (2x) − c1 e−x sin (2x)] + 5[c1 e−x sin (2x)] = 0, which shows that y (x) = c1 e−x sin (2x) is a solution to the given differential equation for all x ∈ R. 12. We can quickly compute the first two derivatives of y (x) = c1 cosh (3x) + c2 sinh (3x): y (x) = 3c1 sinh (3x) + 3c2 cosh (3x) and y (x) = 9c1 cosh (3x) + 9c2 sinh (3x). Therefore, we have y − 9y = [9c1 cosh (3x) + 9c2 sinh (3x)] − 9[c1 cosh (3x) + c2 sinh (3x)] = 0, which shows that y (x) = c1 cosh (3x) + c2 sinh (3x) is a solution to the given differential equation for all x ∈ R. 13. We can quickly compute the first two derivatives of y (x) = y (x) = − 3c1 c2 −2 x4 x and c1 c2 +: x3 x y (x) = 12c1 2c2 + 3. x5 x Therefore, we have x2 y + 5xy + 3y = x2 which shows that y (x) = x ∈ (0, ∞). 12c1 2c2 +3 5 x x + 5x − 3c1 c2 −2 4 x x +3 c1 c2 + 3 x x = 0, c2 c1 + is a solution to the given differential equation for all x ∈ (−∞, 0) or 3 x x √ 14. We can quickly compute the first two derivatives of y (x) = c1 x + 3x2 : c1 y (x) = √ + 6x 2x and c1 y (x) = − √ + 6. 4 x3 Therefore, we have √ c1 c1 √ + 6x + (c1 x + 3x2 ) = 9x2 , 2x2 y − xy + y = 2x2 − √ + 6 − x 3 2x 4x √ which shows that y (x) = c1 x + 3x2 is a solution to the given differential equation for all x > 0. 12 15. We can quickly compute the first two derivatives of y (x) = c1 x2 + c2 x3 − x2 sin x: y (x) = 2c1 x + 3c2 x2 − x2 cos x − 2x sin x and y (x) = 2c1 + 6c2 x + x2 sin x − 2x cos x − 2x cos −2 sin x. Substituting these results into the given differential equation yields x2 y − 4xy + 6y = x2 (2c1 + 6c2 x + x2 sin x − 4x cos x − 2 sin x) − 4x(2c1 x + 3c2 x2 − x2 cos x − 2x sin x) + 6(c1 x2 + c2 x3 − x2 sin x) = 2c1 x2 + 6c2 x3 + x4 sin x − 4x3 cos x − 2x2 sin x − 8c1 x2 − 12c2 x3 + 4x3 cos x + 8x2 sin x + 6c1 x2 + 6c2 x3 − 6x2 sin x = x4 sin x. Hence, y (x) = c1 x2 + c2 x3 − x2 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) = c1 eax + c2 ebx : y (x) = ac1 eax + bc2 ebx and y (x) = a2 c1 eax + b2 c2 ebx . Substituting these results into the differential equation yields y − (a + b)y + aby = a2 c1 eax + b2 c2 ebx − (a + b)(ac1 eax + bc2 ebx ) + ab(c1 eax + c2 ebx ) = (a2 c1 − a2 c1 − abc1 + abc1 )eax + (b2 c2 − abc2 − b2 c2 + abc2 )ebx = 0. Hence, y (x) = c1 eax + c2 ebx is a solution to the given differential equation for all x ∈ R. 17. We can quickly compute the first two derivatives of y (x) = eax (c1 + c2 x): y (x) = eax (c2 ) + aeax (c1 + c2 x) = eax (c2 + ac1 + ac2 x) and y (x) = eaax (ac2 ) + aeax (c2 + ac1 + ac2 x) = aeax (2c2 + ac1 + ac2 x). Substituting these into the differential equation yields y − 2ay + a2 y = aeax (2c2 + ac1 + ac2 x) − 2aeax (c2 + ac1 + ac2 x) + a2 eax (c1 + c2 x) = aeax (2c2 + ac1 + ac2 x − 2c2 − 2ac1 − 2ac2 x + ac1 + ac2 x) = 0. Thus, y (x) = eax (c1 + c2 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) = eax (c1 cos bx + c2 sin bx): y (x) = eax (−bc1 sin bx + bc2 cos bx) + aeax (c1 cos bx + c2 sin bx) = eax [(bc2 + ac1 ) cos bx + (ac2 − bc1 ) sin bx], y (x) = eax [−b(bc2 + ac1 ...
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