Solutions a) The characteristic equation is r = 0 with roots ±2i, so the complementary solution is. y c = c 1 cos(2t)+c 2 sin(2t).

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1 Solutions 3.9. a) The characteristic equation is r + 4 = 0 with roots ±i, so the complementar solution is c = c cos(t)+c sin(t). b) We look for a particular solution in the form and obtain the equations p = u (t) cos(t)+u (t) sin(t), u cos(t)+u sin(t) =0, This leads to Integration ields u sin(t)+u cos(t) =4. u = sin(t), u = cos(t). u = cos(t)+c, u = sin(t)+c, p =+c cos(t)+c sin(t). c) With undetermined coefficients, we have the particular solution leading to p = A, p = p =0, p +4 p =4A =4. Hence A =.. a) The characteristic equation is r + = 0 with roots ±i, so the complementar solution is c = c cos t + c sin t. b) We look for a particular solution in the form p = u (t) cos t + u (t) sin t,

2 and obtain the equations This leads to Integration ields Hence u cos t + u sin t =0, u sin t + u cos t = cos t. u = tan t, u =. u = ln cos t, u = t. p = cos t ln cos t + t sin t. 3. We look for a particular solution of the form p = u (t)e t + u (t)t e t. This ields Consequentl, Integration ields u e t + u t e t =0, u e t + u (t +t)e t = te t. u = t, u =. u = t3 6, u = t, and p = t3 e t We look for a particular solution of the form p = u (t) sin t + u (t)t sin t. This ields Consequentl, u sin t + u t sin t =0, u cos t + u (sin t + t cos t) =t sin t. u = t, u = t.

3 Integration ields u = t3 3, u = t, and p = t3 6 sin t. 6. To follow the solution given here, ou need to know that d t t f(t, λ) dλ = f(t, t)+ dt 0 0 f (t, λ) dλ. t This is a special case of the chain rule for several variables. Define h(x, t) = x 0 f(t, λ) dλ, and observe that d h(t, t) = h(t, t)+ h(t, t) dt x t = ( x f f(t, x)+ 0 t (t, λ) dλ) x=t. We now appl this with In this fashion, we find f(t, λ) = sin((t λ))g(λ). = t t 0 cos((t λ))g(λ) dλ, = 4 sin((t λ))g(λ) dλ + g(t). 0 Hence (0) = (0) = 0 and +4 = g(t). 3

4 Solutions 3.0. a) The characteristic equation is r + ω 0 = 0 with roots r = ±iω 0,so c = c cos(ω 0 t)+c sin(ω 0 t). b) (i) We have p = A cos(ωt)+b sin(ωt), p = Aω cos(ωt) Bω sin(ωt), p + ω 0 p = A(ω 0 ω ) cos(ωt)+b(ω 0 ω ) sin(ωt) =F cos(ωt), hence A = F/(ω 0 ω ), B = 0. (ii) We have p = At cos(ω 0 t)+btsin(ω 0 t), p = ω 0 At sin(ω 0 t)+ω 0 Btcos(ω 0 t)+a cos(ω 0 t)+b sin(ω 0 t), p = ω 0At cos(ω 0 t) ω 0Btsin(ω 0 t) Aω 0 sin(ω 0 t)+b cos(ω 0 t), p + ω 0 = Aω 0 sin(ω 0 t)+b cos(ω 0 t)=f cos(ω 0 t), hence A = 0, B = F/(ω 0 ). 4. The spring constant is = 000 kg sec. Hence the differential equation is = 0 cos(8t),(0) = 0, (0) = 0. The solution is (t) = (cos(8t) cos(0t)), 8 (see formula (7)). We can rewrite the answer in the form (t) = sin(9t) sin t, 9 this makes it evident that the maximum excursion from equilibrium is /9. 7. We have the differential equation = 0 cos(8t), (0) = 0, (0) = 0.

5 We divide the differential equation b : = 0 cos(8t). The characteristic equation is r +4r + 40 = 0, which has roots r = ± 6i. Hence the complementar solution is c = c e t cos(6t)+c e t sin(6t). A particular solution can be found in the form which leads to p = A cos(8t)+b sin(8t), p = 8A sin(8t)+8b cos(8t), p = 64A cos(8t) 64B sin(8t), p +4 p +40 p =( 4A +3B) cos(8t)+( 3A 4B) sin(8t) = 0 cos(8t). Hence, 4A +3B = 0, 3A 4B =0, i.e. A = 3/0, B =/5. The general solution is now = 3 0 cos(8t)+ 5 sin(8t)+c e t cos(6t)+c e t sin(6t). We find (0) = c =0, (0) = 8 5 c +6c =0, i.e. c =3/0, c = 3/60. For t, the complementar solution tends to zero, and the particular solution remains, which oscillates periodicall. 3. a) To get this, decompose the force of gravit into a component in the direction of the spring and a component perpendicular to it, which is absorbed b the wall and does not contribute to the force balance. At the time the spring is released, we have x(0) = 0, x (0) = 0.

6 b) The differential equation while Zucchini is accelerated b the spring is 50 3 x + 50x = 50, x(0) = 0, x (0) = 0. The complementar solution is x c = c cos( 3t)+c sin( 3t), and a particular solution is x p =. Thus we have x = + c cos(4 t)+c sin(4 t). The initial conditions ield c = 0 +, c =0. Zucchini leaves the spring at the time T when x = 0; at this point in time and cos(4 T )= 0, sin(4 T )= x = 4 c sin(4 T ) = (40 4) = 5.4 ft/sec. c) After Zucchini leaves the cannon, his horizontal speed is constant, v h = 5.4 = His vertical speed is initiall and reaches zero in time 37.06/3 =.58. An equal amount of time is needed to come down again, so he travels for.36 sec and a distance of = ft. 3

7 Solutions With we have Hence, = c + c t + c 3 cos(t)+c 4 sin(t), = c c 3 sin(t)+c 4 cos(t), = 4c 3 cos(t) 4c 4 sin(t), =8c 3 sin(t) 8c 4 cos(t). (0) = c + c 3 =0, (0) = c +c 4 =, (0) = 4c 3 = 4, (0) = 8c 4 =8. The solution is c =, c =, c 3 =, c 4 =. 7. We have W () = ( ) ( 4) = 0, so the two solutions do not form a fundamental set. 8. The general solution is which leads to = c e t cos t + c e t sin t, (0) = c, (0) = c + c. For the given initial conditions, we find ȳ (t) =e t cos t + e t sin t, 0. We have ȳ (t) =e t sin t. ȳ = + 3, ȳ = + 3, ȳ 3 = + 3.

8 Hence and 0 A =, det A = 3.

9 Solutions 3.. a) λ 3 4λ = 0. b) λ 3 4λ = λ(λ 4), so the roots are 0 and ±. c) = c + c e t + c 3 e t.. a) λ 3 + λ λ =0. b) λ 3 + λ λ =(λ + )(λ ) = (λ +) (λ ), so the roots are (double) and. c) = c e t + c e t + c 3 te t. 5. a) 6λ 4 +8λ +=0. b) 6λ 4 +8λ +=(4λ +), so the roots are ±i/, each double. c) = c cos(t/) + c sin(t/) + c 3 t cos(t/) + c 4 t sin(t/). 8. a) λ 4 =0. b) λ 4 =(λ )(λ + ), so the roots are ±, ±i. c) = c e t + c e t + c 3 cos(t)+c 4 sin(t). 7. a) λ 3 +3λ +3λ +=0. b) λ 3 +3λ +3λ +=(λ +) 3,so is a triple root. c) = c e t + c te t + c 3 t e t. d) (0) = c =0, (0) = c + c =, (0) = c c +c 3 =0, so c = 0, c =, c 3 =.

Solutions to the Homework Replaces Section 3.7, 3.8

Solutions to the Homework Replaces Section 3.7, 3.8. Show that the period of motion of an undamped vibration of a mass hanging from a vertical spring is 2π L/g SOLUTION: With no damping, mu + ku = 0 has