AEM_3e_Chapter_03

AEM_3e_Chapter_03 - 3 Higher-Order Differential Equations...

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Unformatted text preview: 3 Higher-Order Differential Equations EXERCISES 3.1 Preliminary Theory: Linear Equations 3. From y = c1 x + c2 x ln x we find y = c1 + c2 (1 + ln x). Then y (1) = c1 = 3, y (1) = c1 + c2 = −1 so that c1 = 3 and c2 = −4. The solution is y = 3x − 4x ln x. 6. In this case we have y (0) = c1 = 0, y (0) = 2c2 · 0 = 0 so c1 = 0 and c2 is arbitrary. Two solutions are y = x2 and y = 2x2 . 9. Since a2 (x) = x − 2 and x0 = 0 the problem has a unique solution for −∞ < x < 2. 12. In this case we have y (0) = c1 = 1, y (1) = 2c2 = 6 so that c1 = 1 and c2 = 3. The solution is y = 1 + 3x2 . 15. Since (−4)x + (3)x2 + (1)(4x − 3x2 ) = 0 the set of functions is linearly dependent. 18. Since (1) cos 2x + (1)1 + (−2) cos2 x = 0 the set of functions is linearly dependent. 21. Suppose c1 (1 + x) + c2 x + c3 x2 = 0. Then c1 + (c1 + c2 )x + c3 x2 = 0 and so c1 = 0, c1 + c2 = 0, and c3 = 0. Since c1 = 0 we also have c2 = 0. Thus, the set of functions is linearly independent. 24. The functions satisfy the differential equation and are linearly independent since W (cosh 2x, sinh 2x) = 2 for −∞ < x < ∞. The general solution is y = c1 cosh 2x + c2 sinh 2x. 27. The functions satisfy the differential equation and are linearly independent since W x3 , x4 = x6 = 0 for 0 < x < ∞. The general solution on this interval is y = c1 x3 + c2 x4 . 30. The functions satisfy the differential equation and are linearly independent since W (1, x, cos x, sin x) = 1 for −∞ < x < ∞. The general solution on this interval is y = c1 + c2 x + c3 cos x + c4 sin x. 33. The functions y1 = e2x and y2 = xe2x form a fundamental set of solutions of the associated homogeneous equation, and yp = x2 e2x + x − 2 is a particular solution of the nonhomogeneous equation. 29 3.1 Preliminary Theory: Linear Equations 36. (a) yp1 = 5 (b) yp2 = −2x (c) yp = yp1 + yp2 = 5 − 2x (d) yp = 1 yp1 − 2yp2 = 2 5 2 + 4x EXERCISES 3.2 Reduction of Order In Problems 3-6 we use reduction of order to find a second solution. 3. Define y = u(x) cos 4x so y = −4u sin 4x + u cos 4x, and y + 16y = (cos 4x)u − 8(sin 4x)u = 0 e−8 tan 4x dx y = u cos 4x − 8u sin 4x − 16u cos 4x or u − 8(tan 4x)u = 0. If w = u we obtain the linear first-order equation w − 8(tan 4x)w = 0 which has the integrating factor = cos2 4x. Now d [(cos2 4x)w] = 0 dx 6. Define y = u(x)e5x so y = 5e5x u + e5x u , and y − 25y = e5x (u + 10u ) = 0 e10 dx gives (cos2 4x)w = c. Therefore w = u = c sec2 4x and u = c1 tan 4x. A second solution is y2 = tan 4x cos 4x = sin 4x. y = e5x u + 10e5x u + 25e5x u or u + 10u = 0. If w = u we obtain the linear first-order equation w + 10w = 0 which has the integrating factor = e10x . Now d 10x [e w] = 0 gives e10x w = c. dx and u = c1 e−10x . A second solution is y2 = e−10x e5x = e−5x . Therefore w = u = ce−10x In Problems 9-15 we use formula (5) from the text. 9. Identifying P (x) = −7/x we have y2 = x4 A second solution is y2 = x4 ln |x|. 12. Identifying P (x) = 0 we have y2 = x1/2 ln x 1 e− 0 dx dx = x1/2 ln x − x(ln x)2 ln x = −x1/2 . e− (−7/x) dx x8 dx = x4 1 dx = x4 ln |x|. x 30 3.3 A second solution is y2 = x1/2 . Homogeneous Linear Equations with Constant Coefficients 15. Identifying P (x) = 2(1 + x)/ 1 − 2x − x2 we have y2 = (x + 1) = (x + 1) e− 2(1+x)dx/(1−2x−x2 ) (x + 1)2 dx = (x + 1) 2 eln(1−2x−x ) dx (x + 1)2 1 − 2x − x2 dx = (x + 1) (x + 1)2 2 − x = −2 − x2 − x. x+1 2 − 1 dx (x + 1)2 = (x + 1) − A second solution is y2 = x2 + x + 2. 18. Define y = u(x) · 1 so y =u, y =u and y + y = u + u = 1. dx If w = u we obtain the linear first-order equation w + w = 1 which has the integrating factor e dx [e w] = ex dx gives ex w = ex + c. = ex . Now Therefore w = u = 1 + ce−x and u = x + c1 e−x + c2 . The general solution is y = u = x + c1 e−x + c2 . EXERCISES 3.3 Homogeneous Linear Equations with Constant Coefficients 3. From m2 − m − 6 = 0 we obtain m = 3 and m = −2 so that y = c1 e3x + c2 e−2x . 6. From m2 − 10m + 25 = 0 we obtain m = 5 and m = 5 so that y = c1 e5x + c2 xe5x . 9. From m2 + 9 = 0 we obtain m = 3i and m = −3i so that y = c1 cos 3x + c2 sin 3x. 12. From 2m2 + 2m + 1 = 0 we obtain m = −1/2 ± i/2 so that y = e−x/2 [c1 cos(x/2) + c2 sin(x/2)]. 15. From m3 − 4m2 − 5m = 0 we obtain m = 0, m = 5, and m = −1 so that y = c1 + c2 e5x + c3 e−x . 18. From m3 + 3m2 − 4m − 12 = 0 we obtain m = −2, m = 2, and m = −3 so that y = c1 e−2x + c2 e2x + c3 e−3x . 21. From m3 + 3m2 + 3m + 1 = 0 we obtain m = −1, m = −1, and m = −1 so that y = c1 e−x + c2 xe−x + c3 x2 e−x . 24. From m4 − 2m2 + 1 = 0 we obtain m = 1, m = 1, m = −1, and m = −1 so that y = c1 ex + c2 xex + c3 e−x + c4 xe−x . 31 3.3 Homogeneous Linear Equations with Constant Coefficients 27. From m5 + 5m4 − 2m3 − 10m2 + m + 5 = 0 we obtain m = −1, m = −1, m = 1, and m = 1, and m = −5 so that u = c1 e−r + c2 re−r + c3 er + c4 rer + c5 e−5r . 30. From m2 + 1 = 0 we obtain m = ±i so that y = c1 cos θ + c2 sin θ. If y (π/3) = 0 and y (π/3) = 2 then √ 3 1 c1 + c2 = 0 2 2 √ 3 1 − c1 + c2 = 2, 2 2 √ √ so c1 = − 3, c2 = 1, and y = − 3 cos θ + sin θ. √ √ √ 33. From m2 + m + 2 = 0 we obtain m = −1/2 ± 7 i/2 so that y = e−x/2 [c1 cos( 7 x/2) + c2 sin( 7 x/2)]. If y (0) = 0 and y (0) = 0 then c1 = 0 and c2 = 0 so that y = 0. 36. From m3 + 2m2 − 5m − 6 = 0 we obtain m = −1, m = 2, and m = −3 so that y = c1 e−x + c2 e2x + c3 e−3x . If y (0) = 0, y (0) = 0, and y (0) = 1 then c1 + c2 + c3 = 0, so c1 = −1/6, c2 = 1/15, c3 = 1/10, and 1 1 1 y = − e−x + e2x + e−3x . 6 15 10 39. From m2 + 1 = 0 we obtain m = ±i so that y = c1 cos x + c2 sin x and y = −c1 sin x + c2 cos x. From y (0) = c1 (0) + c2 (1) = c2 = 0 and y (π/2) = −c1 (1) = 0 we find c1 = c2 = 0. A solution of the boundary-value problem is y = 0. 42. The auxiliary equation is m2 − 1 = 0 which has roots −1 and 1. By (10) the general solution is y = c1 ex + c2 e−x . By (11) the general solution is y = c1 cosh x + c2 sinh x. For y = c1 ex + c2 e−x the boundary conditions imply c1 + c2 = 1, c1 e − c2 e−1 = 0. Solving for c1 and c2 we find c1 = 1/(1 + e2 ) and c2 = e2 /(1 + e2 ) so y = ex /(1 + e2 ) + e2 e−x /(1 + e2 ). For y = c1 cosh x + c2 sinh x the boundary conditions imply c1 = 1, c2 = − tanh 1, so y = cosh x − (tanh 1) sinh x. 45. The auxiliary equation should have a pair of complex roots α ± βi where α < 0, so that the solution has the form eαx (c1 cos βx + c2 sin βx). Thus, the differential equation is (e). 48. The differential equation should have the form y + k 2 y = 0 where k = 2 so that the period of the solution is π . Thus, the differential equation is (b). −c1 + 2c2 − 3c3 = 0, c1 + 4c2 + 9c3 = 1, 32 3.4 Undetermined Coefficients EXERCISES 3.4 Undetermined Coefficients 3. From m2 − 10m + 25 = 0 we find m1 = m2 = 5. Then yc = c1 e5x + c2 xe5x and we assume yp = Ax + B . Substituting into the differential equation we obtain 25A = 30 and −10A + 25B = 3. Then A = 6 , B = 6 , 5 5 yp = 6 x + 6 , and 5 5 6 6 y = c1 e5x + c2 xe5x + x + . 5 5 6. From m2 − 8m + 20 = 0 we find m1 = 4 + 2i and m2 = 4 − 2i. Then yc = e4x (c1 cos 2x + c2 sin 2x) and we assume yp = Ax2 + Bx + C + (Dx + E )ex . Substituting into the differential equation we obtain 2A − 8B + 20C = 0 −6D + 13E = 0 −16A + 20B = 0 13D = −26 20A = 100. Then A = 5, B = 4, C = 11 10 , D = −2, E = − 12 , yp = 5x2 + 4x + 13 11 10 + −2x − 12 13 ex and ex . y = e4x (c1 cos 2x + c2 sin 2x) + 5x2 + 4x + 12 11 + −2x − 10 13 9. From m2 − m = 0 we find m1 = 1 and m2 = 0. Then yc = c1 ex + c2 and we assume yp = Ax. Substituting into the differential equation we obtain −A = −3. Then A = 3, yp = 3x and y = c1 ex + c2 + 3x. 12. From m2 − 16 = 0 we find m1 = 4 and m2 = −4. Then yc = c1 e4x + c2 e−4x and we assume yp = Axe4x . Substituting into the differential equation we obtain 8A = 2. Then A = 1 , yp = 1 xe4x and 4 4 1 y = c1 e4x + c2 e−4x + xe4x . 4 15. From m2 + 1 = 0 we find m1 = i and m2 = −i. Then yc = c1 cos x + c2 sin x and we assume yp = (Ax2 + Bx) cos x + (Cx2 + Dx) sin x. Substituting into the differential equation we obtain 4C = 0, 2A + 2D = 0, −4A = 2, and −2B + 2C = 0. Then A = − 1 , B = 0, C = 0, D = 1 , yp = − 1 x2 cos x + 1 x sin x, 2 2 2 2 and 1 1 y = c1 cos x + c2 sin x − x2 cos x + x sin x. 2 2 18. From m2 − 2m + 2 = 0 we find m1 = 1 + i and m2 = 1 − i. Then yc = ex (c1 cos x + c2 sin x) and we assume yp = Ae2x cos x + Be2x sin x. Substituting into the differential equation we obtain A +2B = 1 and −2A + B = −3. Then A = 7 , B = − 1 , yp = 7 e2x cos x − 1 e2x sin x and 5 5 5 5 7 1 y = ex (c1 cos x + c2 sin x) + e2x cos x − e2x sin x. 5 5 21. From m3 − 6m2 = 0 we find m1 = m2 = 0 and m3 = 6. Then yc = c1 + c2 x + c3 e6x and we assume yp = Ax2 + B cos x + C sin x. Substituting into the differential equation we obtain −12A = 3, 6B − C = −1, 33 3.4 Undetermined Coefficients 6 and B + 6C = 0. Then A = − 1 , B = − 37 , C = 4 1 37 , yp = − 1 x2 − 4 6 37 cos x + 1 37 sin x, and 1 6 1 y = c1 + c2 x + c3 e6x − x2 − cos x + sin x. 4 37 37 24. From m3 − m2 − 4m + 4 = 0 we find m1 = 1, m2 = 2, and m3 = −2. Then yc = c1 ex + c2 e2x + c3 e−2x and we assume yp = A + Bxex + Cxe2x . Substituting into the differential equation we obtain 4A = 5, −3B = −1, and 4C = 1. Then A = 5 , B = 1 , C = 1 , yp = 5 + 1 xex + 1 xe2x , and 4 3 4 4 3 4 y = c1 ex + c2 e2x + c3 e−2x + 5 1 x 1 2x + xe + xe . 43 4 27. We have yc = c1 cos 2x + c2 sin 2x and we assume yp = A. Substituting into the differential equation we find √ A = − 1 . Thus y = c1 cos 2x + c2 sin 2x − 1 . From the initial conditions we obtain c1 = 0 and c2 = 2 , so 2 2 √ 1 y = 2 sin 2x − . 2 30. We have yc = c1 e−2x + c2 xe−2x and we assume yp = (Ax3 + Bx2 )e−2x . Substituting into the differential equation we find A = 1 and B = 3 . Thus y = c1 e−2x + c2 xe−2x + 1 x3 + 3 x2 e−2x . From the initial conditions we 6 2 6 2 obtain c1 = 2 and c2 = 9, so y = 2e−2x + 9xe−2x + 1 3 3 2 −2x x+ x e . 6 2 33. We have xc = c1 cos ωt + c2 sin ωt and we assume xp = At cos ωt + Bt sin ωt. Substituting into the differential equation we find A = −F0 /2ω and B = 0. Thus x = c1 cos ωt + c2 sin ωt − (F0 /2ω )t cos ωt. From the initial conditions we obtain c1 = 0 and c2 = F0 /2ω 2 , so x = (F0 /2ω 2 ) sin ωt − (F0 /2ω )t cos ωt. √ √ 36. We have yc = c1 e−2x + ex (c2 cos 3 x + c3 sin 3 x) and we assume yp = Ax + B + Cxe−2x . Substituting into the differential equation we find A = 1 , B = − 5 , and C = 2 . Thus 4 8 3 √ √ 1 52 y = c1 e−2x + ex (c2 cos 3 x + c3 sin 3 x) + x − + xe−2x . 4 83 √ 23 59 17 From the initial conditions we obtain c1 = − 12 , c2 = − 24 , and c3 = 72 3 , so √ √ 59 17 √ 1 23 −2x 52 + ex − cos 3 x + 3 sin 3 x + x − + xe−2x . e 12 24 72 4 83 √ √ 39. The general solution of the differential equation y + 3y = 6x is y = c1 cos 3x + c2 sin 3x + 2x. The √ condition y (0) = 0 implies c1 = 0 and so y = c2 sin 3x + 2x. The condition y (1) + y (1) = 0 implies √ √ √ √ √ √ c2 sin 3 + 2 + c2 3 cos 3 + 2 = 0 so c2 = −4/(sin 3 + 3 cos 3 ). The solution is √ −4 sin 3x √ √ √ + 2x. y= sin 3 + 3 cos 3 y=− 42. We have yc = ex (c1 cos 3x + c2 sin 3x) and we assume yp = A on [0, π ]. Substituting into the differential equation we find A = 2. Thus, y = ex (c1 cos 3x + c2 sin 3x) + 2 on [0, π ]. On (π, ∞) we have y = ex (c3 cos 3x + c4 sin 3x). From y (0) = 0 and y (0) = 0 we obtain c1 = −2, Solving this system, we find c1 = −2 and c2 = continuity of y at x = π implies eπ (−2 cos 3π + 2 3 c1 + 3c2 = 0. . Thus y = ex (−2 cos 3x + 2 3 sin 3x) + 2 on [0, π ]. Now, 2 sin 3π ) + 2 = eπ (c3 cos 3π + c4 sin 3π ) 3 34 3.5 or 2 + 2eπ = −c3 eπ or c3 = −2e−π (1 + eπ ). Continuity of y at π implies 20 π e sin 3π = eπ [(c3 + 3c4 ) cos 3π + (−3c3 + c4 ) sin 3π ] 3 Variation of Parameters or −c3 eπ − 3c4 eπ = 0. Since c3 = −2e−π (1 + e ) we have c4 = 2 e−π (1 + eπ ). The solution of the initial-value 3 problem is ex (−2 cos 3x + 2 sin 3x) + 2, 0≤x≤π 3 y (x) = 2 π x−π (1 + e )e (−2 cos 3x + 3 sin 3x), x > π . EXERCISES 3.5 Variation of Parameters The particular solution, yp = u1 y1 + u2 y2 , in the following problems can take on a variety of forms, especially where trigonometric functions are involved. The validity of a particular form can best be checked by substituting it back into the differential equation. 3. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and W= Identifying f (x) = sin x we obtain u1 = − sin2 x u2 = cos x sin x. Then u1 = 1 1 1 1 sin 2x − x = sin x cos x − x 4 2 2 2 cos x sin x = 1. − sin x cos x 1 u2 = − cos2 x. 2 and y = c1 cos x + c2 sin x + 1 1 1 sin x cos2 x − x cos x − cos2 x sin x 2 2 2 1 = c1 cos x + c2 sin x − x cos x. 2 6. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and W= Identifying f (x) = sec2 x we obtain u1 = − sin x cos2 x cos x sin x = 1. − sin x cos x u2 = sec x. 35 3.5 Variation of Parameters Then u1 = − 1 = − sec x cos x u2 = ln | sec x + tan x| and y = c1 cos x + c2 sin x − cos x sec x + sin x ln | sec x + tan x| = c1 cos x + c2 sin x − 1 + sin x ln | sec x + tan x|. 9. The auxiliary equation is m2 − 4 = 0, so yc = c1 e2x + c2 e−2x and W= e2x 2e2x e−2x −2e−2x = −4. Identifying f (x) = e2x /x we obtain u1 = 1/4x and u2 = −e4x /4x. Then 1 ln |x|, 4 1 x e4t u2 = − dt 4 x0 t u1 = and y = c1 e2x + c2 e−2x + 1 4 e2x ln |x| − e−2x x x0 e4t dt , t x0 > 0. 12. The auxiliary equation is m2 − 2m + 1 = (m − 1)2 = 0, so yc = c1 ex + c2 xex and W= Identifying f (x) = ex / 1 + x2 we obtain u1 = − u2 = Then u1 = − 1 ln 1 + x2 , u2 = tan−1 x, and 2 1 y = c1 ex + c2 xex − ex ln 1 + x2 + xex tan−1 x. 2 15. The auxiliary equation is m2 + 2m + 1 = (m + 1)2 = 0, so yc = c1 e−t + c2 te−t and W= Identifying f (t) = e−t ln t we obtain u1 = − u2 = Then te−t e−t ln t = −t ln t e−2t e−t −e −t ex ex xex xex + ex = e2x . x xex ex =− 2x (1 + x2 ) e 1 + x2 1 ex ex = . (1 + x2 ) 1 + x2 e2x te−t = e−2t . −te + e−t −t e−t e−t ln t = ln t. e−2t 1 1 u1 = − t2 ln t + t2 2 4 u2 = t ln t − t 36 3.5 and Variation of Parameters 1 y = c1 e−t + c2 te−t − t2 e−t ln t + 2 1 2 −t = c1 e−t + c2 te−t + t e ln t − 2 ex/2 1 x/2 2e 1 2 −t t e + t2 e−t ln t − t2 e−t 4 3 2 −t te . 4 18. The auxiliary equation is 4m2 − 4m + 1 = (2m − 1)2 = 0, so yc = c1 ex/2 + c2 xex/2 and W= √ Identifying f (x) = 1 ex/2 1 − x2 we obtain 4 xex/2 1 x/2 2 xe + ex/2 = ex . √ xex/2 ex/2 1 − x2 1 u1 = − = − x 1 − x2 x 4e 4 √ ex/2 ex/2 1 − x2 1 u2 = = 1 − x2 . 4ex 4 To find u1 and u2 we use the substitution v = 1 − x2 and the trig substitution x = sin θ, respectively: 1 3 /2 u1 = 1 − x2 12 u2 = Thus y = c1 ex/2 + c2 xex/2 + x 8 1 − x2 + 3 /2 1 sin−1 x. 8 1 1 − x2 + xex/2 sin−1 x. 8 1 x/2 1 − x2 e 12 e2x 2e 2x 1 + x2 ex/2 8 21. The auxiliary equation is m2 + 2m − 8 = (m − 2)(m + 4) = 0, so yc = c1 e2x + c2 e−4x and W= Identifying f (x) = 2e−2x − e−x we obtain u1 = 1 −4x 1 −3x −e e 3 6 1 1 u2 = e3x − e2x . 6 3 e−4x −4e−4x = −6e−2x . Then 1 −4x 1 + e−3x e 12 18 1 3x 1 2x u2 = e−e. 18 6 u1 = − y = c1 e2x + c2 e−4x − 1 −2x 1 1 1 e + e−x + e−x − e−2x 12 18 18 6 1 1 = c1 e2x + c2 e−4x − e−2x + e−x 4 9 Thus and 1 1 y = 2c1 e2x − 4c2 e−4x + e−2x − e−x . 2 9 The initial conditions imply 5 c1 + c2 − =1 36 7 2c1 − 4c2 + = 0. 18 37 3.5 Variation of Parameters Thus c1 = 25/36 and c2 = 4/9, and y= 24. Write the equation in the form 1 1 sec(ln x) y + 2y = x x x2 2 and identify f (x) = sec(ln x)/x . From y1 = cos(ln x) and y2 = sin(ln x) we compute y+ cos(ln x) W= − Now u1 = − and u2 = Thus, a particular solution is yp = cos(ln x) ln | cos(ln x)| + (ln x) sin(ln x), and the general solution is y = c1 cos(ln x) + c2 sin(ln x) + cos(ln x) ln | cos(ln x)| + (ln x) sin(ln x). tan(ln x) x 1 x so u1 = ln | cos(ln x)|, u2 = ln x. sin(ln x) x sin(ln x) cos(ln x) x = 1 . x 25 2x 4 −4x 1 −2x 1 −x −e +e. e+e 36 9 4 9 so EXERCISES 3.6 Cauchy-Euler Equation 3. The auxiliary equation is m2 = 0 so that y = c1 + c2 ln x. 6. The auxiliary equation is m2 + 4m + 3 = (m + 1)(m + 3) = 0 so that y = c1 x−1 + c2 x−3 . 9. The auxiliary equation is 25m2 + 1 = 0 so that y = c1 cos 1 5 1 5 ln x + c2 sin ln x . 12. The auxiliary equation is m2 + 7m + 6 = (m + 1)(m + 6) = 0 so that y = c1 x−1 + c2 x−6 . 15. Assuming that y = xm and substituting into the differential equation we obtain m(m − 1)(m − 2) − 6 = m3 − 3m2 + 2m − 6 = (m − 3)(m2 + 2) = 0. Thus y = c1 x3 + c2 cos √ √ 2 ln x + c3 sin 2 ln x . 18. Assuming that y = xm and substituting into the differential equation we obtain m(m − 1)(m − 2)(m − 3) + 6m(m − 1)(m − 2) + 9m(m − 1) + 3m + 1 = m4 + 2m2 + 1 = (m2 + 1)2 = 0. Thus y = c1 cos(ln x) + c2 sin(ln x) + c3 (ln x) cos(ln x) + c4 (ln x) sin(ln x). 38 3.6 21. The auxiliary equation is m2 − 2m + 1 = (m − 1)2 = 0 so that yc = c1 x + c2 x ln x and W (x, x ln x) = x x ln x = x. 1 1 + ln x Cauchy-Euler Equation Identifying f (x) = 2/x we obtain u1 = −2 ln x/x and u2 = 2/x. Then u1 = −(ln x)2 , u2 = 2 ln x, and y = c1 x + c2 x ln x − x(ln x)2 + 2x(ln x)2 = c1 x + c2 x ln x + x(ln x)2 , x > 0. 24. The auxiliary equation m(m − 1) + m − 1 = m2 − 1 = 0 has roots m1 = −1, m2 = 1, so yc = c1 x−1 + c2 x. With y1 = x−1 , y2 = x, and the identification f (x) = 1/x2 (x + 1), we get W = 2x−1 , Then u1 = W1 /W = −1/2(x + 1), gives W1 = −1/x(x + 1), and W2 = 1/x3 (x + 1). u2 = W2 /W = 1/2x2 (x + 1), and integration (by partial fractions for u2 ) 1 u1 = − ln(x + 1) 2 1 −1 1 1 u2 = − x − ln x + ln(x + 1), 2 2 2 so 1 1 1 1 yp = u1 y1 + u2 y2 = − ln(x + 1) x−1 + − x−1 − ln x + ln(x + 1) x 2 2 2 2 11 1 ln(x + 1) 11 1 = − − x ln x + x ln(x + 1) − = − + x ln 1 + 22 2 2x 22 x and y = yc + yp = c1 x−1 + c2 x − 27. The auxiliary equation is m2 + 1 = 0, so that y = c1 cos(ln x) + c2 sin(ln x) 1 1 sin(ln x) + c2 cos(ln x). x x The initial conditions imply c1 = 1 and c2 = 2. Thus y = cos(ln x) + 2 sin(ln x). The graph is given to the right. y = −c1 30. The auxiliary equation is m2 − 6m + 8 = (m − 2)(m − 4) = 0, so that yc = c1 x2 + c2 x4 and W= x2 x4 2x 4x3 = 2x5 . -1 1 x 3 − ln(x + 1) 2x 1 11 + x ln 1 + 22 x − ln(x + 1) , 2x y x > 0. and 50 -3 100 x y 0.05 Identifying f (x) = 8x4 we obtain u1 = −4x3 and u2 = 4x. Then u1 = −x4 , u2 = 2x2 , and y = c1 x2 + c2 x4 + x6 . The initial conditions imply 1 1 1 c1 + c2 = − 4 16 64 1 3 c1 + c2 = − . 2 16 1 1 Thus c1 = 16 , c2 = − 1 , and y = 16 x2 − 1 x4 + x6 . The graph is given above. 2 2 39 3.6 Cauchy-Euler Equation 33. Substituting x = et into the differential equation we obtain d2 y dy +9 + 8y = e2t . dt2 dt The auxiliary equation is m2 + 9m + 8 = (m + 1)(m + 8) = 0 so that yc = c1 e−t + c2 e−8t . Using undetermined coefficients we try yp = Ae2t . This leads to 30Ae2t = e2t , so that A = 1/30 and y = c1 e−t + c2 e−8t + 36. From 1 2t 1 e = c1 x−1 + c2 x−8 + x2 . 30 30 d2 y dy − dt2 dt − 2 x3 dy dt 1 x . d2 y dy − dt2 dt − − 2 dy 2 d2 y +3 x3 dt2 x dt d2 y 1 =2 2 dx x d3 y 1d =2 3 dx x dx = 1d x2 dx 1 d3 y x2 dt3 1 x3 d2 y dy − dt2 dt d2 y dt2 1 x − − it follows that 1d x2 dx = 1 d2 y x2 dt2 2 dy 2 d2 y +3 x3 dt2 x dt = d3 y d2 y dy −3 2 +2 3 dt dt dt Substituting into the differential equation we obtain d3 y d2 y dy −3 2 +2 −3 3 dt dt dt or d2 y dy − dt2 dt +6 dy − 6y = 3 + 3t dt d3 y d2 y dy − 6 2 + 11 − 6y = 3 + 3t. dt3 dt dt The auxiliary equation is m3 − 6m2 +11m − 6 = (m − 1)(m − 2)(m − 3) = 0 so that yc = c1 et + c2 e2t + c3 e3t . Using undetermined coefficients we try yp = A + Bt. This leads to (11B − 6A) − 6Bt = 3 + 3t, so that A = −17/12, B = −1/2, and y = c1 et + c2 e2t + c3 e3t − 17 1 17 1 − t = c1 x + c2 x2 + c3 x3 − − ln x. 12 2 12 2 40 3.7 Nonlinear Equations EXERCISES 3.7 Nonlinear Equations 3. Let u = y so that u = y . The equation becomes u = −u − 1 which is separable. Thus du = −dx =⇒ tan−1 u = −x + c1 =⇒ y = tan(c1 − x) =⇒ y = ln | cos(c1 − x)| + c2 . u2 + 1 6. Let u = y so that y = u du/dy . The equation becomes (y + 1)u du/dy = u2 . Separating variables we obtain du dy = =⇒ ln |u| = ln |y + 1| + ln c1 =⇒ u = c1 (y + 1) u y+1 dy dy =⇒ = c1 (y + 1) =⇒ = c1 dx dx y+1 =⇒ ln |y + 1| = c1 x + c2 =⇒ y + 1 = c3 ec1 x . 9. (a) y 10 −π/2 3π/2 x −10 (b) Let u = y so that y = u du/dy . The equation becomes u du/dy + yu = 0. Separating variables we obtain 1 1 du = −y dy =⇒ u = − y 2 + c1 =⇒ y = − y 2 + c1 . 2 2 When x = 0, y = 1 and y = −1 so −1 = −1/2 + c1 and c1 = −1/2. Then dy 1 1 1 dy 1 = − y2 − =⇒ 2 = − dx =⇒ tan−1 y = − x + c2 dx 2 2 y +1 2 2 1 =⇒ y = tan − x + c2 . 2 When x = 0, y = 1 so 1 = tan c2 and c2 = π/4. The solution of the initial-value problem is y = tan The graph is shown in part (a). (c) The interval of definition is −π/2 < π/4 − x/2 < π/2 or −π/2 < x < 3π/2. π1 −x. 4 2 41 3.7 Nonlinear Equations 12. Let u = y so that u = y . The equation becomes u − (1/x)u = u2 , which is a Bernoulli differential equation. Using the substitution w = u−1 we obtain dw/dx + (1/x)w = −1. An integrating factor is x, so d 1 1 1 c1 − x2 2x [xw] = −x =⇒ w = − x + c =⇒ = =⇒ u = =⇒ y = − ln c1 − x2 + c2 . dx 2 x u 2x c1 − x2 15. In this problem the thinner curve is obtained using a numerical solver, while the thicker curve is the graph of the Taylor polynomial. We look for a solution of the form 1 1 1 1 y (x) = y (0) + y (0)x + y (0)x2 + y (0)x3 + y (4) (0)x4 + y (5) (0)x5 . 2! 3! 4! 5! From y (x) = x + y − 2y we compute 2 2 y 40 30 y (x) = 2x + 2yy − 2y y (4) (x) = 2 + 2(y )2 + 2yy − 2y y (5) (x) = 6y y + 2yy Using y (0) = 1 and y (0) = 1 we find y (0) = −1, y (0) = 4, y (4) (0) = −6, y (5) (0) = 14. − 2y (4) . 20 10 An approximate solution is 1 2 1 7 y (x) = 1 + x − x2 + x3 − x4 + x5 . 2 3 4 60 0.5 1 1.5 2 2.5 3 3.5x 21. Let u = dx/dt so that d2 x/dt2 = u du/dx. The equation becomes u du/dx = −k 2 /x2 . Separating variables we obtain k2 1 k2 k2 1 u du = − 2 dx =⇒ u2 = + c =⇒ v 2 = + c. x 2 x 2 x When t = 0, x = x0 and v = 0 so 0 = (k 2 /x0 ) + c and c = −k 2 /x0 . Then 12 v = k2 2 Separating variables we have − √ 1 xx0 dx = k 2 dt =⇒ t = − x0 − x k x0 2 x dx. x0 − x 1 1 − x x0 and √ dx = −k 2 dt x0 − x . xx0 Using Mathematica to integrate we obtain t=− = 1 k 1 k x0 (x0 − 2x) x0 − x(x0 − x) − tan−1 2 2 2x x0 2 x(x0 − x) + x0 x0 − 2x tan−1 2 2 x(x0 − x) x x0 − x . 42 3.8 Linear Models: Initial-Value Problems EXERCISES 3.8 Linear Models: Initial-Value Problems √ 3. From 3 x + 72x = 0, x(0) = −1/4, and x (0) = 0 we obtain x = − 1 cos 4 6 t. 4 4 6. From 50x + 200x = 0, x(0) = 0, and x (0) = −10 we obtain x = −5 sin 2t and x = −10 cos 2t. 9. From 1 x + x = 0, x(0) = 1/2, and x (0) = 3/2 we obtain 4 √ 13 1 3 x = cos 2t + sin 2t = sin(2t + 0.588). 2 4 4 √ 12. From x + 9x = 0, x(0) = −1, and x (0) = − 3 we obtain √ 3 4π 2 x = − cos 3t − sin 3t = √ sin 3t + 3 3 3 √ and x = 2 3 cos(3t + 4π/3). If x = 3 then t = −7π/18 + 2nπ/3 and t = −π/2 + 2nπ/3 for n = 1, 2, 3, . . . . 15. For large values of t the differential equation is approximated by x = 0. The solution of this equation is the linear function x = c1 t + c2 . Thus, for large time, the restoring force will have decayed to the point where the spring is incapable of returning the mass, and the spring will simply keep on stretching. 18. (a) below 21. From 1 8x (b) from rest + x + 2x = 0, x(0) = −1, and x (0) = 8 we obtain x = 4te−4t − e−4t and x = 8e−4t − 16te−4t . If x = 0 then t = 1/4 second. If x = 0 then t = 1/2 second and the extreme displacement is x = e−2 feet. 24. (a) x = 1 e−8t 4e6t − 1 is not zero for t ≥ 0; the extreme displacement is x(0) = 1 meter. 3 (b) x = 1 e−8t 5 − 2e6t = 0 when t = 1 ln 5 ≈ 0.153 second; if x = 4 e−8t e6t − 10 = 0 then t = 3 6 2 3 0.384 second and the extreme displacement is x = −0.232 meter. 27. From 5 16 x 1 6 ln 10 ≈ + βx + 5x = 0 we find that the roots of the auxiliary equation are m = − 8 β ± 5 4 5 4β 2 − 25 . (a) If 4β 2 − 25 > 0 then β > 5/2. (b) If 4β 2 − 25 = 0 then β = 5/2. (c) If 4β 2 − 25 < 0 then 0 < β < 5/2. 30. (a) If x + 2x + 5x = 12 cos 2t + 3 sin 2t, x(0) = 1, and x (0) = 5 then xc = e−t (c1 cos 2t + c2 sin 2t) and xp = 3 sin 2t so that the equation of motion is x = e−t cos 2t + 3 sin 2t. (b) x 3 steady-state (c) x 3 x=xc+xp 2 -3 4 6 t 2 -3 4 6 t transient 43 3.8 Linear Models: Initial-Value Problems 33. From 2x + 32x = 68e−2t cos 4t, x(0) = 0, and x (0) = 0 we obtain xc = c1 cos 4t + c2 sin 4t and xp = 1 −2t cos 4t − 2e−2t sin 4t so that 2e 1 9 1 x = − cos 4t + sin 4t + e−2t cos 4t − 2e−2t sin 4t. 2 4 2 36. (a) From 100x + 1600x = 1600 sin 8t, x(0) = 0, and x (0) = 0 we obtain xc = c1 cos 4t + c2 sin 4t and xp = − 1 sin 8t so that by a trig identity 3 x= (b) If x = 1 3 2 1 2 2 sin 4t − sin 8t = sin 4t − sin 4t cos 4t. 3 3 3 3 sin 4t(2 − 2 cos 4t) = 0 then t = nπ/4 for n = 0, 1, 2, . . . . (c) If x = 8 cos 4t − 8 cos 8t = 8 (1 − cos 4t)(1 + 2 cos 4t) = 0 then t = π/3 + nπ/2 and t = π/6 + nπ/2 for 3 3 3 n = 0, 1, 2, . . . at the extreme values. Note: There are many other values of t for which x = 0. √ √ (d) x(π/6 + nπ/2) = 3/2 cm and x(π/3 + nπ/2) = − 3/2 cm. (e) x 1 t 1 -1 2 3 39. (a) From x + ω 2 x = F0 cos γt, x(0) = 0, and x (0) = 0 we obtain xc = c1 cos ωt + c2 sin ωt and xp = (F0 cos γt)/ ω 2 − γ 2 so that x=− (b) lim 45. Solving F0 F0 cos ωt + 2 cos γt. ω2 − γ 2 ω − γ2 γ →ω F0 −F0 t sin γt F0 (cos γt − cos ωt) = lim = t sin ωt. γ →ω ω2 − γ 2 −2γ 2ω + 2q + 100q = 0 we obtain q (t) = e−20t (c1 cos 40t + c2 sin 40t). The initial conditions q (0) = 5 and q (0) = 0 imply c1 = 5 and c2 = 5/2. Thus 1 20 q q (t) = e−20t 5 cos 40t + and q (0.01) ≈ 4.5676 coulombs. 0.0509 second. 5 sin 40t 2 = 25 + 25/4 e−20t sin(40t + 1.1071) The charge is zero for the first time when 40t + 1.1071 = π or t ≈ 48. Solving q + 100q + 2500q = 30 we obtain q (t) = c1 e−50t + c2 te−50t + 0.012. The initial conditions q (0) = 0 and q (0) = 2 imply c1 = −0.012 and c2 = 1.4. Thus, using i(t) = q (t) we get q (t) = −0.012e−50t + 1.4te−50t + 0.012 and i(t) = 2e−50t − 70te−50t . Solving i(t) = 0 we see that the maximum charge occurs when t = 1/35 second and q (1/35) ≈ 0.01871 coulomb. 51. The differential equation is 1 q + 20q + 1000q = 100 sin 60t. To use Example 10 in the text we identify E0 = 100 2 and γ = 60. Then X = Lγ − Z= and 1 1 1 = (60) − ≈ 13.3333, cγ 2 0.001(60) X 2 + 400 ≈ 24.0370, X 2 + R2 = 44 3.9 E0 100 = ≈ 4.1603. Z Z From Problem 50, then Linear Models: Boundary-Value Problems ip (t) ≈ 4.1603 sin(60t + φ) where sin φ = −X/Z and cos φ = R/Z . Thus tan φ = −X/R ≈ −0.6667 and φ is a fourth quadrant angle. Now φ ≈ −0.5880 and ip (t) = 4.1603 sin(60t − 0.5880). √ 54. In Problem 50 it is shown that the amplitude of the steady-state current is E0 /Z , where Z = X 2 + R2 and X = Lγ − 1/Cγ . Since E0 is constant the amplitude will be a maximum when Z is a minimum. Since R √ is constant, Z will be a minimum when X = 0. Solving Lγ − 1/Cγ = 0 for γ we obtain γ = 1/ LC . The maximum amplitude will be E0 /R. 57. In an LC-series circuit there is no resistor, so the differential equation is L d2 q 1 + q = E (t). 2 dt C √ √ Then q (t) = c1 cos t/ LC + c2 sin t/ LC + qp (t) where qp (t) = A sin γt + B cos γt. Substituting qp (t) into the differential equation we find 1 − Lγ 2 A sin γt + C 1 − Lγ 2 B cos γt = E0 cos γt. C Equating coefficients we obtain A = 0 and B = E0 C/(1 − LCγ 2 ). Thus, the charge is q (t) = c1 cos √ E0 C 1 1 t + c2 sin √ t+ cos γt. 1 − LCγ 2 LC LC √ The initial conditions q (0) = q0 and q (0) = i0 imply c1 = q0 − E0 C/(1 − LCγ 2 ) and c2 = i0 LC . The current is i(t) = q (t) or c1 1 c2 1 E0 Cγ i(t) = − √ sin γt sin √ t+ √ cos √ t− 1 − LCγ 2 LC LC LC LC = i0 cos √ 1 1 t− √ LC LC q0 − E0 C 1 − LCγ 2 sin √ 1 E0 Cγ sin γt. t− 1 − LCγ 2 LC EXERCISES 3.9 Linear Models: Boundary-Value Problems 3. (a) The general solution is w0 4 x. 24EI The boundary conditions are y (0) = 0, y (0) = 0, y (L) = 0, y (L) = 0. The first two conditions give c1 = 0 and c2 = 0. The conditions at x = L give the system y (x) = c1 + c2 x + c3 x2 + c4 x3 + 45 3.9 Linear Models: Boundary-Value Problems w0 4 L =0 24EI w0 2 2c3 + 6c4 L + L = 0. 2EI Solving, we obtain c3 = w0 L2 /16EI and c4 = −5w0 L/48EI . The deflection is w0 y (x) = (3L2 x2 − 5Lx3 + 2x4 ). 48EI c3 L2 + c4 L3 + (b) 0.2 0.4 0.6 0.8 1 x 1 y 6. (a) ymax = y (L) = w0 L4 /8EI (b) Replacing both L and x by L/2 in y (x) we obtain w0 L4 /128EI , which is 1/16 of the maximum deflection when the length of the beam is L. (c) ymax = y (L/2) = 5w0 L4 /384EI (d) The maximum deflection in Example 1 is y (L/2) = (w0 /24EI )L4 /16 = w0 L4 /384EI , which is 1/5 of the maximum displacement of the beam in part c. 9. This is Example 2 in the text with L = π . The eigenvalues are λn = n2 π 2 /π 2 = n2 , n = 1, 2, 3, . . . and the corresponding eigenfunctions are yn = sin(nπx/π ) = sin nx, n = 1, 2, 3, . . . . 12. For λ ≤ 0 the only solution of the boundary-value problem is y = 0. For λ = α2 > 0 we have y = c1 cos αx + c2 sin αx. Since y (0) = 0 implies c1 = 0, y = c2 sin x dx. Now y gives π 2 = c2 α cos α π =0 2 π (2n − 1)π = or λ = α2 = (2n − 1)2 , n = 1, 2, 3, . . . . 2 2 The eigenvalues λn = (2n − 1)2 correspond to the eigenfunctions yn = sin(2n − 1)x. α 1 −2 ± 2 15. The auxiliary equation has solutions m= For λ = −α2 < 0 we have y = e−x (c1 cosh αx + c2 sinh αx) . The boundary conditions imply y (0) = c1 = 0 y (5) = c2 e−5 sinh 5α = 0 4 − 4(λ + 1) = −1 ± α. 46 3.9 Linear Models: Boundary-Value Problems so c1 = c2 = 0 and the only solution of the boundary-value problem is y = 0. For λ = 0 we have y = c1 e−x + c2 xe−x and the only solution of the boundary-value problem is y = 0. For λ = α2 > 0 we have y = e−x (c1 cos αx + c2 sin αx) . Now y (0) = 0 implies c1 = 0, so y (5) = c2 e−5 sin 5α = 0 gives n2 π 2 , n = 1, 2, 3, . . . . 25 n2 π 2 nπ The eigenvalues λn = correspond to the eigenfunctions yn = e−x sin x for n = 1, 2, 3, . . . . 25 5 18. For λ = 0 the general solution is y = c1 + c2 ln x. Now y = c2 /x, so y (e−1 ) = c2 e = 0 implies c2 = 0. Then y = c1 and y (1) = 0 gives c1 = 0. Thus y (x) = 0. 5α = nπ or λ = α2 = For λ = −α2 < 0, y = c1 x−α + c2 xα . The boundary conditions give c2 = c1 e2α and c1 = 0, so that c2 = 0 and y (x) = 0. For λ = α2 > 0, y = c1 cos(α ln x) + c2 sin(α ln x). From y (1) = 0 we obtain c1 = 0 and y = c2 sin(α ln x). Now y = c2 (α/x) cos(α ln x), so y (e−1 ) = c2 eα cos α = 0 implies cos α = 0 or α = (2n − 1)π/2 and λ = α2 = (2n − 1)2 π 2 /4 for n = 1, 2, 3, . . . . The corresponding eigenfunctions are yn = sin 2n − 1 π ln x . 2 y 21. If restraints are put on the column at x = L/4, x = L/2, and x = 3L/4, then the critical load will be P4 . L x 24. (a) The boundary-value problem is d4 y d2 y + λ 2 = 0, dx4 dx y (0) = 0, y (0) = 0, y (L) = 0, y (L) = 0, where λ = α2 = P/EI . The solution of the differential equation is y = c1 cos αx + c2 sin αx + c3 x + c4 and the conditions y (0) = 0, y (0) = 0 yield c1 = 0 and c4 = 0. Next, by applying y (L) = 0, y (L) = 0 to y = c2 sin αx + c3 x we get the system of equations c2 sin αL + c3 L = 0 αc2 cos αL + c3 = 0. To obtain nontrivial solutions c2 , c3 , we must have the determinant of the coefficients equal to zero: sin αL L =0 α cos αL 1 or tan β = β, 47 3.9 Linear Models: Boundary-Value Problems where β = αL. If βn denotes the positive roots of the last equation, then the eigenvalues are found from √ 2 βn = αn L = λn L or λn = (βn /L)2 . From λ = P/EI we see that the critical loads are Pn = βn EI/L2 . With the aid of a CAS we find that the first positive root of tan β = β is (approximately) β1 = 4.4934, and so the Euler load is (approximately) P1 = 20.1907EI/L2 . Finally, if we use c3 = −c2 α cos αL, then the deflection curves are yn (x) = c2 sin αn x + c3 x = c2 sin βn x− L βn cos βn x . L (b) With L = 1 and c2 appropriately chosen, the general shape of the first buckling mode, y1 (x) = c2 sin is shown below. y1 4.4934 x− L 4.4934 cos(4.4934) x , L 0.2 0.4 0.6 0.8 1 x 27. The auxiliary equation is m2 + m = m(m +1) = 0 so that u(r) = c1 r−1 + c2 . The boundary conditions u(a) = u0 and u(b) = u1 yield the system c1 a−1 + c2 = u0 , c1 b−1 + c2 = u1 . Solving gives c1 = Thus u(r) = u0 − u1 b−a ab and c2 = u1 b − u0 a . b−a u0 − u1 b−a ab u1 b − u0 a + . r b−a EXERCISES 3.10 Nonlinear Models 3. The period corresponding to x(0) = 1, x (0) = 1 is approximately 5.8. The second initial-value problem does not have a periodic solution. x 10 8 6 4 2 2 -2 4 6 8 10 t 48 3.10 Nonlinear Models 6. From the graphs we see that the interval is approximately (−0.8, 1.1). x 3 2 1 5 −1 10 t 9. This is a damped hard spring, so x will approach 0 as t approaches ∞. x 2 t 2 -2 4 6 8 12. (a) x 40 20 x 40 20 20 -20 -40 40 60 80 100 t -20 20 40 60 80 100 t k 0.000465 -40 k 0.000466 The system appears to be oscillatory for −0.000465 ≤ k1 < 0 and nonoscillatory for k1 ≤ −0.000466. (b) x 3 2 1 20 -1 -2 -3 40 60 80 100 120 140 x 3 2 1 t -1 -2 -3 20 40 60 80 100 120 140 t k 0.3493 k 0.3494 49 3.10 Nonlinear Models The system appears to be oscillatory for −0.3493 ≤ k1 < 0 and nonoscillatory for k1 ≤ −0.3494. 15. (a) Intuitively, one might expect that only half of a 10-pound chain could be lifted by a 5-pound vertical force. √ (b) Since x = 0 when t = 0, and v = dx/dt = 160 − 64x/3 , we have v (0) = 160 ≈ 12.65 ft/s. (c) Since x should always be positive, we solve x(t) = 0, getting t = 0 and t = 3 4 graph of x(t) is a parabola, the maximum value occurs at tm = solving x (t) = 0.) At this time the height of the chain is x(tm ) ≈ 7.5 ft. This is higher than predicted because of the momentum generated by the force. When the chain is 5 feet high it still has a positive velocity of about 7.3 ft/s, which keeps it going higher for a while. 18. (a) There are two forces acting on the chain as it falls from the platform. One is the force due to gravity on the portion of the chain hanging over the edge of the platform. This is F1 = 2x. The second is due to the motion of the portion of the chain stretched out on the platform. By Newton’s second law this is F2 = = From d d (8 − x)2 d 8−x [mv ] = v= v dt dt 32 dt 16 8 − x dv 1 dx 1 dv −v = (8 − x) − v2 . 16 dt 16 dt 16 dt 5/2 ≈ 2.3717. Since the 5/2 . (This can also be obtained by 3 2 d [mv ] = F1 − F2 we have dt d 2x 1 dv v = 2x − (8 − x) − v2 dt 32 16 dt x dv 1 dx 1 dv + v = 2x − (8 − x) − v2 16 dt 16 dt 16 dt x dv dv + v 2 = 32x − (8 − x) + v2 dt dt x 8 dv dv dv = 32x − 8 +x dt dt dt dv = 32x. dt By the Chain Rule, dv/dt = (dv/dx)(dx/dt) = v dv/dx, so 8 dv dv dv = 8v = 32x and v = 4x. dt dx dx (b) Integrating v dv = 4x dx we get 1 v 2 = 2x2 + c. Since v = 0 when x = 3, we have c = −18. Then 2 √ v 2 = 4x2 − 36 and v = 4x2 − 36 . Using v = dx/dt, separating variables, and integrating we obtain √ dx = 2 dt x2 − 9 and cosh−1 x = 2t + c1 . 3 Solving for x we get x(t) = 3 cosh(2t + c1 ). Since x = 3 when t = 0, we have cosh c1 = 1 and c1 = 0. Thus, x(t) = 3 cosh 2t. Differentiating, we find v (t) = dx/dt = 6 sinh 2t. (c) To find the time when the back end of the chain leaves the platform we solve x(t) = 3 cosh 2t = 8. This gives t1 = 1 2 cosh−1 8 3 ≈ 0.8184 seconds. The velocity at this instant is v (t1 ) = 6 sinh cosh−1 8 3 √ = 2 55 ≈ 14.83 ft/s. 50 3.11 Solving Systems of Linear Equations (d) Replacing 8 with L and 32 with g in part (a) we have L dv/dt = gx. Then L Integrating we get we find v (x) = g2 g2 x − x0 . L L g2 (L − x2 ) . 0 L 12 2v dv dv dv g = Lv = gx and v = x. dt dx dx L = (g/2L)x2 + c. Setting x = x0 and v = 0, we find c = −(g/2L)x2 . Solving for v 0 Then the velocity at which the end of the chain leaves the edge of the platform is v (L) = EXERCISES 3.11 Solving Systems of Linear Equations 3. From Dx = −y + t and Dy = x − t we obtain y = t − Dx, Dy = 1 − D2 x, and (D2 + 1)x = 1 + t. The solution is x = c1 cos t + c2 sin t + 1 + t y = c1 sin t − c2 cos t + t − 1. 6. From (D + 1)x + (D − 1)y = 2 and 3x + (D + 2)y = −1 we obtain x = − 1 − 1 (D + 2)y , Dx = − 1 (D2 + 2D)y , 3 3 3 and (D2 + 5)y = −7. The solution is √ √ 7 y = c1 cos 5 t + c2 sin 5 t − 5 √ √ √ √ 2 5 5 2 3 x = − c1 − c2 cos 5 t + c1 − c2 sin 5 t + . 3 3 3 3 5 9. From Dx + D2 y = e3t and (D + 1)x + (D − 1)y = 4e3t we obtain D(D2 + 1)x = 34e3t and D(D2 + 1)y = −8e3t . The solution is 4 y = c1 + c2 sin t + c3 cos t − e3t 15 17 x = c4 + c5 sin t + c6 cos t + e3t . 15 3t Substituting into (D + 1)x + (D − 1)y = 4e gives (c4 − c1 ) + (c5 − c6 − c3 − c2 ) sin t + (c6 + c5 + c2 − c3 ) cos t = 0 so that c4 = c1 , c5 = c3 , c6 = −c2 , and x = c1 − c2 cos t + c3 sin t + 17 3t e. 15 12. From (2D2 − D − 1)x − (2D + 1)y = 1 and (D − 1)x + Dy = −1 we obtain (2D + 1)(D − 1)(D + 1)x = −1 and (2D + 1)(D + 1)y = −2. The solution is x = c1 e−t/2 + c2 e−t + c3 et + 1 y = c4 e−t/2 + c5 e−t − 2. 51 3.11 Solving Systems of Linear Equations Substituting into (D − 1)x + Dy = −1 gives 3 1 − c1 − c4 e−t/2 + (−2c2 − c5 )e−t = 0 2 2 so that c4 = −3c1 , c5 = −2c2 , and y = −3c1 e−t/2 − 2c2 e−t − 2. 15. Multiplying the first equation by D + 1 and the second equation by D2 + 1 and subtracting we obtain (D4 − D2 )x = 1. Then 1 x = c1 + c2 t + c3 et + c4 e−t − t2 . 2 Multiplying the first equation by D + 1 and subtracting we obtain D2 (D + 1)y = 1. Then 1 y = c5 + c6 t + c7 e−t − t2 . 2 Substituting into (D − 1)x + (D2 + 1)y = 1 gives (−c1 + c2 + c5 − 1) + (−2c4 + 2c7 )e−t + (−1 − c2 + c6 )t = 1 so that c5 = c1 − c2 + 2, c6 = c2 + 1, and c7 = c4 . The solution of the system is 1 x = c1 + c2 t + c3 et + c4 e−t − t2 2 1 y = (c1 − c2 + 2) + (c2 + 1)t + c4 e−t − t2 . 2 18. From Dx + z = et , (D − 1)x + Dy + Dz = 0, and x + 2y + Dz = et we obtain z = −Dx + et , Dz = −D2 x + et , and the system (−D2 + D − 1)x + Dy = −et and (−D2 + 1)x + 2y = 0. Then y = 1 (D2 − 1)x, Dy = 1 D(D2 − 1)x, 2 2 and (D − 2)(D2 + 1)x = −2et so that the solution is x = c1 e2t + c2 cos t + c3 sin t + et y= 3 2t c1 e − c2 cos t − c3 sin t 2 z = −2c1 e2t − c3 cos t + c2 sin t. 21. From (D + 5)x + y = 0 and 4x − (D + 1)y = 0 we obtain y = −(D + 5)x so that Dy = −(D2 + 5D)x. Then 4x + (D2 + 5D)x + (D + 5)x = 0 and (D + 3)2 x = 0. Thus x = c1 e−3t + c2 te−3t y = −(2c1 + c2 )e−3t − 2c2 te−3t . Using x(1) = 0 and y (1) = 1 we obtain c1 e−3 + c2 e−3 = 0 −(2c1 + c2 )e−3 − 2c2 e−3 = 1 or c1 + c2 = 0 2c1 + 3c2 = −e3 . Thus c1 = e3 and c2 = −e3 . The solution of the initial value problem is x = e−3t+3 − te−3t+3 y = −e−3t+3 + 2te−3t+3 . 52 CHAPTER 3 REVIEW EXERCISES 24. From Newton’s second law in the x-direction we have m In the y -direction we have m d2 y 1 dy dy = −mg − k sin θ = −mg − k = −mg − |c| . 2 dt v dt dt |c|dt/m d2 x 1 dx dx = −k cos θ = −k = −|c| . dt2 v dt dt From mD2 x + |c|Dx = 0 we have D(mD + |c|)x = 0 so that (mD + |c|)x = c1 or (D + |c|/m)x = c2 . This is a linear first-order differential equation. An integrating factor is e = e|c|t/m so that d |c|t/m x] = c2 e|c|t/m [e dt and e|c|t/m x = (c2 m/|c|)e|c|t/m + c3 . The general solution of this equation is x(t) = c4 + c3 e−|c|t/m . From (mD2 + |c|D)y = −mg we have D(mD + |c|)y = −mg so that (mD + |c|)y = −mgt + c1 or (D + |c|/m)y = −gt + c2 . This is a linear first-order differential equation with integrating factor e |c|dt/m = e|c|t/m . Thus d |c|t/m y ] = (−gt + c2 )e|c|t/m [e dt mg |c|t/m m2 g |c|t/m e|c|t/m y = − + 2e + c3 e|c|t/m + c4 te |c| c and y (t) = − mg m2 g t + 2 + c3 + c4 e−|c|t/m . |c| c CHAPTER 3 REVIEW EXERCISES 3. It is not true unless the differential equation is homogeneous. For example, y1 = x is a solution of y + y = x, but y2 = 5x is not. 6. 2π/5, since 1 x + 6.25x = 0 4 9. The set is linearly independent over (−∞, ∞) and linearly dependent over (0, ∞). 12. (a) The auxiliary equation is am(m − 1) + bm + c = am2 + (b − a)m + c = 0. If the roots are 3 and −1, then we want (m − 3)(m + 1) = m2 − 2m − 3 = 0. Thus, let a = 1, b = −1, and c = −3, so that the differential equation is x2 y − xy − 3y = 0. (b) In this case we want the auxiliary equation to be m2 + 1 = 0, so let a = 1, b = 1, and c = 1. Then the differential equation is x2 y + xy + y = 0. 15. From m3 + 10m2 + 25m = 0 we obtain m = 0, m = −5, and m = −5 so that y = c1 + c2 e−5x + c3 xe−5x . √ 18. From 2m4 + 3m3 + 2m2 + 6m − 4 = 0 we obtain m = 1/2, m = −2, and m = ± 2 i so that √ √ y = c1 ex/2 + c2 e−2x + c3 cos 2 x + c4 sin 2 x. 53 CHAPTER 3 REVIEW EXERCISES 21. Applying D(D2 + 1) to the differential equation we obtain D(D2 + 1)(D3 − 5D2 + 6D) = D2 (D2 + 1)(D − 2)(D − 3) = 0. Then y = c1 + c2 e2x + c3 e3x + c4 x + c5 cos x + c6 sin x yc and yp = Ax + B cos x + C sin x. Substituting yp into the differential equation yields 6A + (5B + 5C ) cos x + (−5B + 5C ) sin x = 8 + 2 sin x. Equating coefficients gives A = 4/3, B = −1/5, and C = 1/5. The general solution is 1 1 4 y = c1 + c2 e2x + c3 e3x + x − cos x + sin x. 3 5 5 24. The auxiliary equation is m2 − 1 = 0, so yc = c1 ex + c2 e−x and W= Identifying f (x) = 2ex /(ex + e−x ) we obtain u1 = 1 ex = ex + e−x 1 + e2x e2x e3x ex =− = −ex + . ex + e−x 1 + e2x 1 + e2x ex ex e−x −e−x = −2. u2 = − Then u1 = tan−1 ex , u2 = −ex + tan−1 ex , and y = c1 ex + c2 e−x + ex tan−1 ex − 1 + e−x tan−1 ex . 27. The auxiliary equation is m2 − 5m + 6 = (m − 2)(m − 3) = 0 and a particular solution is yp = x4 − x2 ln x so that y = c1 x2 + c2 x3 + x4 − x2 ln x. 30. (a) If y = sin x is a solution then so is y = cos x and m2 + 1 is a factor of the auxiliary equation m4 + 2m3 + 11m2 + 2m + 10 = 0. Dividing by m2 + 1 we get m2 + 2m + 10, which has roots −1 ± 3i. The general solution of the differential equation is y = c1 cos x + c2 sin x + e−x (c3 cos 3x + c4 sin 3x). (b) The auxiliary equation is m(m + 1) = m2 + m = 0, so the associated homogeneous differential equation is y + y = 0. Letting y = c1 + c2 e−x + 1 x2 − x and computing y + y we get x. Thus, the differential 2 equation is y + y = x. 33. The auxiliary equation is m2 − 2m + 2 = 0 so that m = 1 ± i and y = ex (c1 cos x + c2 sin x). Setting y (π/2) = 0 and y (π ) = −1 we obtain c1 = e−π and c2 = 0. Thus, y = ex−π cos x. 36. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and W= Identifying f (x) = sec3 x we obtain u1 = − sin x sec3 x = − sin x cos3 x cos x sin x = 1. − sin x cos x u2 = cos x sec3 x = sec2 x. 54 CHAPTER 3 REVIEW EXERCISES Then u1 = − 1 11 = − sec2 x 2x 2 cos 2 u2 = tan x. Thus y = c1 cos x + c2 sin x − = c1 cos x + c2 sin x − = c3 cos x + c2 sin x + and 1 cos x sec2 x + sin x tan x 2 1 − cos2 x 1 sec x + 2 cos x 1 sec x. 2 1 sec x tan x. 2 y = −c3 sin x + c2 cos x + The initial conditions imply c3 + 1 =1 2 1 c2 = . 2 Thus c3 = c2 = 1/2 and y= 1 1 1 cos x + sin x + sec x. 2 2 2 1 2. 39. (a) The auxiliary equation is 12m4 + 64m3 + 59m2 − 23m − 12 = 0 and has roots −4, − 3 , − 1 , and 2 3 general solution is y = c1 e−4x + c2 e−3x/2 + c3 e−x/3 + c4 ex/2 . (b) The system of equations is c1 + c2 + c3 + c4 3 1 1 −4c1 − c2 − c3 + c4 2 3 2 9 1 1 16c1 + c2 + c3 + c4 4 9 4 27 1 1 −64c1 − c2 − c3 + c4 8 27 8 73 Using a CAS we find c1 = 495 , c2 = 109 35 The = −1 =2 =5 = 0. 257 45 , c3 = − 3726 , and c4 = 385 . The solution of the initial-value problem is y=− 73 −4x 109 −3x/2 3726 −x/3 257 x/2 e e e e. + − + 495 35 385 45 42. From (D − 2)x − y = t − 2 and −3x + (D − 4)y = −4t we obtain (D − 1)(D − 5)x = 9 − 8t. Then 8 3 x = c1 et + c2 e5t − t − 5 25 y = (D − 2)x − t + 2 = −c1 et + 3c2 e5t + 16 11 + t. 25 25 and 45. The period of a spring/mass system is given by T = 2π/ω where ω 2 = k/m = kg/W , where k is the spring constant, W is the weight of the mass attached to the spring, and g is the acceleration due to gravity. Thus, √ √ √ √ the period of oscillation is T = (2π/ kg ) W . If the weight of the original mass is W , then (2π/ kg ) W = 3 55 CHAPTER 3 REVIEW EXERCISES √ √ √ √ and (2π/ kg ) W − 8 = 2. Dividing, we get W / W − 8 = 3/2 or W = 9 (W − 8). Solving for W we find 4 that the weight of the original mass was 14.4 pounds. 48. From x + βx + 64x = 0 we see that oscillatory motion results if β 2 − 256 < 0 or 0 ≤ β < 16. 51. For λ = α2 > 0 the general solution is y = c1 cos αx + c2 sin αx. Now y (0) = c1 so the condition y (0) = y (2π ) implies √ c1 = c1 cos 2πα + c2 sin 2πα λ = n or λ = n2 for n = 1, 2, 3, . . . . Since y = −αc1 sin αx + αc2 cos αx = −nc1 sin nx + nc2 cos nx, we see that y (0) = nc2 = y (2π ) for n = 1, 2, 3, . . . . Thus, the eigenvalues are n2 for n = 1, 2, 3, . . . , with corresponding eigenfunctions cos nx and sin nx. When λ = 0, the general solution is y = c1 x + c2 and the corresponding eigenfunction is y = 1. For λ = −α2 < 0 the general solution is y = c1 cosh αx + c2 sinh αx. In this case y (0) = c1 and y (2π ) = c1 cosh 2πα + c2 sinh 2πα, so y (0) = y (2π ) can only be valid for α = 0. Thus, there are no eigenvalues corresponding to λ < 0. 54. The force of kinetic friction opposing the motion of the mass in µN , where µ is the coefficient of sliding friction and N is the normal component of the weight. Since friction is a force opposite to the direction of motion and since N is pointed directly downward (it is simply the weight of the mass), Newton’s second law gives, for motion to the right (x > 0) , d2 x m 2 = −kx − µmg, dt and for motion to the left (x < 0), d2 x m 2 = −kx + µmg. dt Traditionally, these two equations are written as one expression m where fk = µmg and sgn(x ) = 1, x > 0 −1, x < 0. d2 x + fx sgn(x ) + kx = 0, dt2 and y (2π ) = c1 cos 2πα + c2 sin 2πα, which is true when α = 56 ...
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