Sample Solutions of Assignment 5 for MAT3270B: Section 3.4. In each of problems find the general solution of the given differential equation

Transcription

1 Sample Solutions of Assignment 5 for MAT370B: Section 3.4 In each of problems find the general solution of the given differential equation 7. y y + y = y + 9y = y + 9y 4y = y + y + 1.5y = 0 Answer: 7. The characteristic equation is r + = 0 Thus the possible values of r are r 1 = 1 + i and r = 1 i, and the general solution of the equation is y(t) = e t (c 1 cos t + c sin t). 1. The characteristic equation is 4r + 9 = 0 Thus the possible values of r are r 1 = 3i and r = 3i, and the general solution of the equation is y(t) = c 1 cos 3t + c sin 3t. 14. The characteristic equation is 9r + 9r 4 = 0 1

2 Thus the possible values of r are r 1 = 1 and r 3 = 4, and the general 3 solution of the equation is y(t) = c 1 e t 3 + c e 4t The characteristic equation is r + r = 0 Thus the possible values of r are r 1 = 1 + iand r = 1 i, and the general solution of the equation is y(t) = e t (c1 cos t + c sin t). 3. Consider the initial value problem 3u u + u = 0, u(0) =, u (0) = 0. a. Find the solution u(t) of this problem. b. Find the first time at which u(t) = 10. Answer: (a.) The characteristic equation is 3r r + = 0 Thus the possible values of r are r 1 = 1+ 3i and r 6 = 1 3i, and the 6 general solution of the equation is u(t) = e t 3t 3t 6 (c1 cos + c sin 6 6 ). (b.) From u(0) = 0, we get c 1 = From u (0) = 0, we get c = 3

3 3 Therefore, u(t) = e t 6 ( cos 3t 6 3t sin 3 6 ). The first time is = such that u(t) = Show that W (e λt cos µt, e λt sin µt) = µe λt Answer: Let y 1 = e λt cos µt y = e λt sin µt W (e λt cos µt, e λt sin µt) = y y 1 y 1y = e λt cos µt(λ sin µt + µ sin µt) e λt sin µt(λ cos µt µ sin µt) = µe λt 33. If the functions y 1 and y are linearly independent solutions of y + p(t)y + q(t)y = 0, show that between consecutive zeros of y 1 there is one and only one zero of y. Note that this result is illustrated by the solutions y 1 = cos t and y = sin t of the equation y + y = 0. Answer: Assume the two consecutive zeros of y 1 are t 1 and t, and t 1 < t. Then y 1 (t) < 0 or y 1 (t) > 0 for all t 1 < t < t. Case A: We consider the case y 1 (t) > 0 for all t 1 < t < t. In this case, obviously, y 1(t 1 ) > 0 and y 1(t ) < 0.

4 4 W (y 1, y ) = y y 1 y 1y 0 for all t, because y 1 and y are linearly independent solutions of y + p(t)y + q(t)y = 0, then W (y 1, y )(t) > 0 or W (y 1, y )(t) < 0 for all t. If W (y 1, y )(t) > 0 for all t. We get W (y 1, y )(t 1 ) = y 1(t 1 )y (t 1 ) > 0 W (y 1, y )(t ) = y 1(t )y (t ) > 0 Hence y (t 1 ) > 0 and y (t ) < 0, so there is one zero of y. If W (y 1, y )(t) < 0 for all t. We get W (y 1, y )(t 1 ) = y 1(t 1 )y (t 1 ) < 0 W (y 1, y )(t ) = y 1(t )y (t ) < 0 Hence y (t 1 ) < 0 and y (t ) > 0, so there is one zero of y. Case B: We consider the case y 1 (t) < 0 for all t 1 < t < t. In this case, obviously, y 1(t 1 ) < 0 and y 1(t ) > 0. W (y 1, y ) = y y 1 y 1y 0 for all t, because y 1 and y are linearly independent solutions of y + p(t)y + q(t)y = 0, then W (y 1, y )(t) > 0 or W (y 1, y )(t) < 0 for all t. If W (y 1, y )(t) > 0 for all t. We get W (y 1, y )(t 1 ) = y 1(t 1 )y (t 1 ) > 0 W (y 1, y )(t ) = y 1(t )y (t ) > 0

5 5 Hence y (t 1 ) < 0 and y (t ) > 0, so there is one zero of y. If W (y 1, y )(t) < 0 for all t. We get W (y 1, y )(t 1 ) = y 1(t 1 )y (t 1 ) < 0 W (y 1, y )(t ) = y 1(t )y (t ) < 0 Hence y (t 1 ) > 0 and y (t ) < 0, so there is one zero of y. We need to show there is only one zero of y between t 1 and t. If otherwise, then there is at least two zero of y between t 1 and t, we can select two consecutive zeros of y, say t 3 and t 4, such that t 1 < t 3 < t 1 < t 4 < t. Then from above proof we can show there is another zero of y 1, say t 5 such that t 1 < t 3 < t 1 < t 5 < t 4 < t. This contradict that t 1 and t are two consecutive zeros of y Euler Equation. An equation of the form t y + αty + βy = 0, t > 0, where α and β are real constants, is called an Euler equation. Show that the substitution x = ln t transform an Euler equation into an equation with constant coefficients. Answer: We get dx dt = 1 t d y dt = 1 d y t dx 1 dy t dx t y + αty + βy = d y + (α + 1)dy + βy = 0. dx dx

6 6 Section 3.5 Find the general solutions of the given differential equations. 1. y y + y = y + 4y + 9y = 0 Answer: 1. The characteristic equation is r r + 1 = (r 1)(r 1) = 0 Thus the possible value of r is r = 1, and the general solution of the equation is 8. The characteristic equation is y(t) = (c 1 + c t)e t. 16r + 4r + 90 Thus the possible value of r is r 1 = 3, and the general solution of the 4 equation is y(t) = (c 1 + c t)e 3 4 t. In each of the problems use the method of reduction of order to find second solution of the given equation. 3. t y 4ty + 6y = 0; t > 0; y 1 (t) = t 5. t y + 3ty + y = 0; t > 0; y 1 (t) = 1 t 6. t y t(t + )y + (t + )y = 0; t > 0; y 1 (t) = t

7 7 Answer: 3. Let y(t) = t v(t), then y = v t + tv y = v t + 4v t + v So we get. Then t 4 v = 0 v(t) = c 1 t + c and so y(t) = c 1 t 3 + c t. From y 1 (t) = t, we find the second solution is y (t) = t Let y(t) = t 1 v(t), then y = v t 1 t v y = t 1 v t v t + t 3 v So we get. Then tv + v = 0 v(t) = c 1 ln t + c and so y(t) = c 1 t 1 ln t + c t 1. From y 1 (t) = t 1, we find the second solution is y (t) = t 1 ln t. 6. Let y(t) = tv(t), then y = v t + v y = v t + v So we get. Then v v = 0 v(t) = c 1 e t + c

8 8 and so y(t) = c 1 te t + c t. From y 1 (t) = t, we find the second solution is y (t) = te t. 3. The differential equation y + δ(xy + y) = 0 arises in the study of the turbulent flow of a uniform stream past a circular cylinder. Verify that y 1 = exp( δx ) is one solution and then find the general solution in the form of an integral. Answer: Then it is easy to see y y 1 = e δx y 1 = δxe δx 1 = (δ x δ)e δx y 1 + δ(xy 1 + y 1 ) = 0. Let y = ve δx, then y = v e δx δxve δx y = v e δx δxv e δx + δ x ve δx δve δx. We get v δxv = 0, and then v(x) = x c 0 1e δx + c. Hence, x y(x) = (c 1 e δx + c )e δx. From y 1 = exp( δx ), we find the second solution is 0 y(x) = e δx x 0 e δx.

9 9 In the following problem use the method of of problem 33 to find second independent solution of the given equation. 34. t y + 3ty + y = 0; t > 0; y 1 (t) = 1 t Answer: The original equation can be written as From the Abel s theorem y + 3 t y + 1 t y = 0. We get and W (y 1, y ) = c 1 e 3dt t = c 1 t = y 1y 3 y 1y = y t + y t y + y t = c 1 t, y (t) = 1 t [c 1 ln t + c ]. 38. If a, b, c are positive constants, show that all solutions of ay + by + cy = 0 approach zero as t. Answer: The characteristic equation is Case A: b 4ac < 0 ar br + c = 0 Thus the possible values of r are r 1 = b+i 4ac b and r a = b i 4ac b, a and the general solution of the equation is y(t) = e bt 4ac b 4ac b a (c1 cos t + c sin t). a a

10 10 Hence, y(t) 0, as t, since a, b are positive constants. Case B: b 4ac = 0 Thus the possible values of r are r 1 = b, and the general solution of a the equation is y(t) = e bt a (c1 + c t). Hence, y(t) 0, as t, since a, b are positive constants. Case C: b 4ac > 0 Thus the possible values of r are r 1 = b+ b 4ac a and the general solution of the equation is y(t) = c 1 e b+ b 4ac a + c e b b 4ac a. and r = b b 4ac a, Hence, y(t) 0, as t, since a is positive constant and b + b 4ac < (a) If a > 0 and c > 0, but b = 0, show that the result of Problem 38 is no longer true, but that all solutions are bounded as t. (b) If a > 0 and b > 0, but c = 0, show that the result of Problem 38 is no longer true, but that all solutions approach a constant that depends on the initial conditions as t. Determine this constants for the initial conditions y(0) = y 0, y (0) = y ( 0). Answer: (a). a > 0 and b > 0, but c = 0 The characteristic equation is ar + cr = 0

11 Thus the possible values of r are r 1 = c i and r a = c i, and the a general solution of the equation is c c y(t) = c 1 cos a t + c sin a t. Hence, all solutions are bounded as t. 11 (b): a > 0 and b > 0, but c = 0 The characteristic equation is ar br = 0 Thus the possible values of r are r 1 = 0 and r = b, and the general a solution of the equation is y(t) = c 1 + c e bt a. Hence, y(t) c 1, as t, since a, b are positive constants. Obviously, c 1 depends on the initial conditions. From y (0) = y (0), we can get c = ay 0 b. From y(0) = y 0, we get c 1 = ay 0 b + y 0. Section 3.6 Answer: In each of the following problems, find the general solutions of the given differential equations. 1. y y + 3y = 3e t 4. y + y = sin t 7. y + 3y + y = t + 3 sin t 10. u + ω0u = cos ω 0 t 11. y + y + 4y = sinh t

12 1 Answer: 1. The characteristic equation is r r 3 = (r 3)(r + 1) = 0 Thus the possible values of r are r 1 = 3 and r = 1, and the general solution of the homogeneous equation is y(t) = c 1 e 3t + c e t. Let Y (t) = Ae t where A is a constant to be determined. On substituting to the original equation, we get 15Ae t = 3e t. So A = 1 5 and y(t) = c1e 3t + ce t 1 5 et is the general solution of the original equation. 4. The characteristic equation is r + r = r(r + ) = 0 Thus the possible values of r are r 1 = 0 and r =, and the general solution of the homogeneous equation is y(t) = c 1 + c e t. Let Y (t) = At + B sin t + C cos t where A, Band, C are constants to be determined. On substituting to the original equation, we get 4(B + C) sin t + 4(B C) cos t + A = sin t. So A = 3, B = 1 and C = 1 and y(t) = c 1 + c e t + 3t 1 (sin t + cos t) is the general solution of the original equation. 7. The characteristic equation is r + 3r + 1 = (r + 1)(r + 1) = 0

13 Thus the possible values of r are r 1 = 1 and r = 1, and the general solution of the homogeneous equation is y(t) = c 1 e t + c e t. Let Y (t) = A + Bt + Ct + D sin t + E cos t where A, B, C, D, ande are constants to be determined. On substituting to the original equation, we get 4C(A+3B)+(B+6C)t+Ct +( D 3E+D) sin t+( E+3D+E) cos t = t +3 sin t. 13 So A = 14, B = 6, C = 1, D = 3 10 and E = 9 13 and y(t) = c 1 e t + c e t + 14t 16Bt + t 3 10 sin t 9 13 cos t is the general solution of the original equation. 10. The characteristic equation is r + ω 0 = 0 Thus the possible values of r are r 1 = iω 0 and r = iω 0, and the general solution of the homogeneous equation is u(t) = c 1 cos ω 0 t + c sin ω 0 t. Let Y (t) = At cos ω 0 t + Bt sin ω 0 t where A and B are constants to be determined. On substituting to the original equation, we get So A = 0, B = 1 ω 0 Aω 0 sin ω 0 t + Bω 0 cos ω 0 t = cos ω 0 t. and u(t) = c 1 cos ω 0 t + c sin ω 0 t + is the general solution of the original equation. 11. The characteristic equation is r + r + 4 = 0 t ω 0 sin ω 0 t

14 14 Thus the possible values of r are r 1 = i and r = 1 15i, and the general solution of the homogeneous equation is u(t) = e t 15t 15t (c1 cos + c sin ). Let Y (t) = Ae t + Be t where A and B are constants to be determined. On substituting to the original equation, we get 6Ae t + 4Be t = e t + e t. So A = 1, B = 1 and 6 4 u(t) = e t 15t (c1 cos + c sin is the general solution of the original equation. 15t ) et 1 4 e t In each of the following problems, find the solutions of the given initial problem. 13. y + y y = t; y(0) = 0; y (0) = y y 3y = 3te t ; y(0) = 1; y (0) = 0 Answer: 13. The characteristic equation is r + r = 0 Thus the possible values of r are r 1 = and r = 1, and the general solution of the homogeneous equation is u(t) = c 1 e t + c e t. Let Y (t) = A + Bt where A and B are constants to be determined. On substituting to the original equation, we get B (A + Bt) = t.

15 15 So A = 1, B = 1 and y(t) = c 1 e t + c e t t 1 is the general solutions of the original equations. From y(0) = 0, we get c 1 + c = 1. From y (0) = 1, we get c 1 c = 1. Hence, c 1 = 1 and c = 1 and the solution of the initial problem is y(t) = e t 1 e t t The characteristic equation is r r 3 = 0 Thus the possible values of r are r 1 = 3 and r = 1, and the general solution of the homogeneous equation is u(t) = c 1 e 3t + c e t. Let Y (t) = (A+Bt)e t where A and B are constants to be determined. On substituting to the original equation, we get So A =, B = 1 and 3 B 3A 3Bt = 3t. y(t) = c 1 e t3 + c e t ( 3 t)et is the general solution of the original equation. From y(0) = 0, we get c 1 + c = 5 3. From y (0) = 1, we get 3c 1 c = 7 3. Hence, c 1 = 1 and c = 3 and the solution of the initial problem is y(t) = e 3t 3 e t (t + 3 )et.

16 16 Section 3.7 In each of Problems use the method of variation of parameters to find a particular solution of the differential equation. Then check your answer by using the method of undetermined coefficients.. y y y = e t 4. 4y 4y + y = 16e t Answer:. The characteristic equation is r r = 0 Thus the possible values of r are r 1 = and r = 1, and the general solutions of the homogeneous equation are u(t) = c 1 y 1 + c y = c 1 e t + c e t. is Method of variation of Parameters: W (y 1, y ) = 3e t then a particular solution of the original equation Hence, e Y (t) = e t t (e t ) dt + e t 3e t = 1 9 e t 3 te t e t (e t ) dt 3e t y(t) = c 1 e t + c e t 3 te t are the general solutions of the original equation. Method of undetermined coefficients: Let Y (t) = Ate t where A is constant to be determined. On substituting to the original equation, we get 3e t = e t.

17 17 So A = 3 and y(t) = c1e t + ce t 1 9 e t 3 te t is the general solution of the original equation. 4. The original equation can written as: The characteristic equation is y y y = 4e t r r = 0 Thus the possible values of r are r = 1, and the general solutions of the homogeneous equation are u(t) = c 1 y 1 + c y = c 1 e t + c te t. Method of variation of Parameters: W (y 1, y ) = e t then a particular solution of the original equation is t Y (t) = e t te (4e t t ) dt + te t e (4e t ) e t dt e t Hence, = t e t y(t) = c 1 e t + c te t + t e t are the general solutions of the original equation. Method of undetermined coefficients: Let Y (t) = At e t where A is constant to be determined. On substituting to the original equation, we get A = and y(t) = c 1 e t + c te t + t e t are the general solutions of the original equation.

18 18 In each of Problems find the general solution of the given differential equation. 5. y + y = tan t, 0 < t < π 8. y + 4y = 3 csc t, 0 < t < π Answer: 5.The characteristic equation is r + 1 = 0 Thus the possible values of r are r 1 = i and r = i, and the general solution of the equation is y(t) = c 1 cos t + c sin t. W (y 1, y ) = 1 then a particular solution of the original equation is Y (t) = cos t sin t tan tdt + sin t cos t tan tdt Hence, = cos t ln tan t + sec t, 0 < t < π y(t) = c 1 cos t + c sin t cos t ln tan t + sec t, 0 < t < π are the general solutions of the original equation. 8. The characteristic equation is r + 4 = 0 Thus the possible values of r are r 1 = i and r = i, and the general solution of the equation is y(t) = c 1 cos t + c sin t. W (y 1, y ) = then a particular solution of the original equation is 3 sin t csc t 3 cos t csc t Y (t) = cos t dt + sin t dt = 3t cos t sin t ln sin t, 0 < t < π

19 19 Hence, y(t) = c 1 cos t + c sin t 3t cos t sin t ln sin t, 0 < t < π are the general solutions of the original equation. In each of the problems verify that the given functions y 1 and y satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. 13. t y y = 3t 1, t > 0; y 1 = t, y = t ty + (1 + t)y + 4y = t e t, y 1 = 1 + t, y = e t Answer: 5. It is easy to check that y 1 and y satisfy W (y 1, y ) = 3 and let t y y = 3t 1, t > 0. Y (t) = t t 1 (3t 1) 3 + t 1 t (3t 1) 3 = 3 10 t4 1 3 t ln t t, t > 0. then a particular solution of the original equation is y(t) = 3 10 t4 1 3 t ln t, t > It is easy to check that y 1 and y satisfy ty + (1 + t)y + 4y = t e t W (y 1, y ) = te t and let e t (t e t ) Y (t) = (1 + t) te t = ( 1 t 5 4 t )et. (1 + t)(t + e t e t ) te t

20 0 then a particular solution of the original equation is y(t) = ( 1 t 5 4 t )et.. By choosing the lower limit of the integration in Equ.(8) in the text as the initial point, show that Y (t) becomes Y (t) = t y 1 (s)y (t) y 1 (t)y (s) y 1 (s)y (s) y 1(s)y (s) g(s)ds. Show that Y (t) is a solution of the initial value problem L[y] = g(t), y( ) = 0, y ( ) = 0. Thus Y can be identified with v in problem 1. Answer: Y (t) = y 1 (t) y (s)g(s) W (y 1, y )(s) ds + y (t) y 1 (s)g(s) W (y 1, y )(s) ds = t y 1 (t)y (s)g(s) t y 1 (s)y (s) y (s)y 1(s) ds + y (t)y 1 (s)g(s) y 1 (s)y (s) y (s)y 1(s) ds Hence, = t Y (0) = y 1 (s)y (t) y (s)y 1 (t) y 1 (s)y (s) y (s)y 1(s) g(s)ds t0 y 1 (s)y (t) y (s)y 1 (t) y 1 (s)y (s) y (s)y 1(s) g(s)ds From Y (t) = y 1 (t) t y (s)g(s) t W (y 1, y )(s) ds + y y 1 (s)g(s) (t) W (y 1, y )(s) ds,

21 we get then Y (t) = y 1(t) Y ( ) = y 1( ) t0 t y (s)g(s) t W (y 1, y )(s) ds + y 1 (s)g(s) y (t) W (y 1, y )(s) ds y (s)g(s) t0 W (y 1, y )(s) ds + y ( ) t y Y (t) = y (s)g(s) 1 (t) ds + y (t) W (y 1, y )(s) Hence, Y (t) + p(t)y (t) + q(t)y (t) t 1 y 1 (s)g(s) W (y 1, y )(s) ds = 0 y 1 (s)g(s) ds + g(t) W (y 1, y )(s) t y = [y (t) + p(t)y (t) 1 (s)g(s) + q(t)y (t)] W (y 1, y )(s) ds t y [y 1 (t) + p(t)y 1(t) (s)g(s) + q(t)y 1 (t)] ds + g(t) g(t). W (y 1, y )(s) Therefore, Y (t) is a solution of the initial value problem L[y] = g(t), y( ) = 0, y ( ) = (a) Use the result of Problem to show that the solution of the initial value problem y + y = g(t), y( ) = 0, y ( ) = 0 is y = t sin (t s)g(s)ds (b) Find the solution of the initial value problem y + y = g(t), y(0) = y 0, y (0) = y 0.

22 Answer: The characteristic equation is r + 1 = 0 Thus the possible values of r are r 1 = i and r = i, and the general solution of the homogeneous equation is y(t) = c 1 cos t + c sin t. (a) From the result in Problem, we get the solution of the initial problem Y (t) = t y 1 (s)y (t) y 1 (t)y (s) y 1 (s)y (s) y 1(s)y (s) g(s)ds (b) Because = = t t cos s sin t cos t sin s cos s cos s + sin s sin s g(s)ds (cos s sin t cos t sin s)g(s)ds = t sin (t s)g(s)ds y(t) = c 1 cos t + c sin t satisfy the corresponding homogenous equation, let y(0) = y 0 and y (0) = y 0, then we get c 1 = y 0 and c = y 0. Hence, the solution of the initial problem is y(t) = y 0 cos t + y 0 sin t + t sin (t s)g(s)ds. 8. The method of reduction of order (Section 3.5) can also be used for the nonhomogeneous equation y (t) + p(t)y (t) + q(t)y(t) = g(t), (i)

23 provided one solution y 1 of the corresponding homogeneous equation is known. Let y = v(t)y 1 (t) and show that y satisfies Equ.(i) if and only if y 1 v + [y 1(t) + p(t)y 1 (t)]v = g(t), (ii) Equation (ii) is a first order linear equation for v. Solving this equation, integrating the result, and then multiplying by y 1 (t) lead to the general solution of Eq.(i). 3 Answer: Let y = v(t)y 1 (t), then So, y (t) = v (t)y 1 (t) + v(t)y 1(t) y (t) = v (t)y 1 (t) + v (t)y 1(t) + v(t)y 1 (t) y (t) + p(t)y (t) + q(t)y(t) = y 1 v + [y 1(t) + p(t)y 1 (t)]v + v(y 1 (t) + p(t)y 1(t) + q(t)y 1 (t)) = g(t) if y 1 is the solution of the corresponding homogeneous equation and y 1 v + [y 1(t) + p(t)y 1 (t)]v = g(t). In each of the problems use the method outlined in Problem 8 to solve the given differential equation. 9. t y ty + y = 4t, t > 0; y 1 (t) = t 30. 4y + 7ty + 5y = t, t > 0; y 1 (t) = t Answer: 9. Use the method in Problem 8, the original equation can be written as y (t) t y (t) + y(t) = 4. t

24 4 Let v satisfies then tv = 4 and v = 4t ln t + c 1 t. y 1 v + [y 1(t) + p(t)y 1 (t)]v = g(t) So y = y 1 v = 4t ln t+c 1 t is the solution of t y ty +y = 4t, t > 0. Hence, the general solution are y = c 1 t + c t + 4t ln t 9. Use the method in Problem 8, the original equation can be written as Let v satisfies y (t) + 7 t y (t) + 5 t y(t) = 1 t. y 1 v + [y 1(t) + p(t)y 1 (t)]v = g(t) then v + 5 t v = 1 and v = 1 1 t + c 1 t 4. So y = y 1 v = 1 1 t + c 1t 5 is the solution of 4y + 7ty + 5y = t, t > 0. Hence, the general solution are y = c 1 t 5 + c t t. Section 3.8 In the following problem, determine ω 0, R and δ so as to write the given expression in the form u = R cos ω 0 t δ. 3. u = 4 cos 3t sin 3t

25 Answer: R = 4 + ( ) = 5 and δ = arctan( 4 ) == Hence u = 5 cos 3t δ with δ = arctan( 4 ) == A mass weighing lb stretches a spring 6 in. If the mass is pulled down an additional 3 in. and then released, and if there is no damping, determine the position u of the mass at time t. Plot u versus t. Find the frequency, period, and amplitude of the motion. Answer: Generally, the motion the mass is described by the following equation mu (t) + γu (t) + ku(t) = F (t). Since nothing is said in the statement of the problem about an external force, we assume that F (t) = 0. Also, γ = 0 because there is no damping. To determine m note that m = w g = lb 3ft sec = 1 16 lb sec. ft The spring constant k is be found from the statement that the mass stretches the spring 6 in, or by 1 k = ft. Thus lb 1/ft = 4 lb ft. The equation of motion of the mass is u + 64u(t) = 0. The initial conditions are u(0) = 1 4 and u (0) = 0. The general solution is u = A cos(8t) + B sin(8t).

26 6 The solution satisfies the initial conditions u(0) = 1 4 and u = 0, so we can get A = 1 and B = 0. 4 Hence, the position u of the mass at time t is 1 cos(8t)ft, tin sec; ω = 8rad/sec, T = π/4sec, R = 1/4ft Assume that the system described by the equation mu +γu +ku = 0 is either critically damped or overdamped. Show that the mass can pass through the equilibrium position at most once, regardless of the initial conditions. Answer: Case A: If the system is critically damped, then γ = km. The characteristic equation of the original differential equation is mr + γr + k = 0 Thus the possible value of r is r = γ, and the general solution of the m homogeneous equation is y(t) = (c 1 + c t)e γ m t. Obviously, y(t) at most have one zero, regardless the coefficients of c 1 and c, because y(t) always nondecreasing or nonincreasing. Hence, the mass can pass through the equilibrium position at most once, regardless of the initial conditions. Case B: If the system is overdamped, then γ > km. The characteristic equation of the original differential equation is mr + γr + k = 0

27 γ 4mk m Thus the possible values of r are r 1 = γ+ γ 4mk < 0 and r m = γ < 0 for γ > km, and the general solution of the homogeneous equation is y(t) = c 1 e γ+ = e γ m (c1 e γ 4mk m γ 4mk m t + c e γ γ 4mk t m t + c e γ 4mk m t ). 7 Obviously, y(t) at most have one zero, regardless the coefficients of c 1 and c, then the mass can pass through the equilibrium position at most once, regardless of the initial conditions. 4. The position of a certain spring-mass system satisfied the initial value problem 3 u + ku = 0, u(0) =, u (0) = v. If the period and amplitude of the resulting motion are observed to be π and 3, respectively, determine the value of k and v. Answer: The period of the motion is T = π( m k ) 1 = π( 3/ k ) 1 = π. So we get k = 6 and the equation can written as u + 4u = 0. Obviously, the general solution of this equation is From u(0) =, then A =. u(t) = A cos t + B sin t. From the amplitude of the resulting motion is 3, R = A + B = 3 and then B = ± 5. Hence, u(t) = cos t + ± 5 sin t

28 8 and v = u (0) = ± 5 Section Consider a vibration system described by the initial value problem u u + u = cos ωt, u(0) = 0, u (0) =. (a.) Determine the steady-state part of the solution of this problem. (b.) Find the amplitude A of the steady-state solution in terms of ω. (c.) Plot A versus ω. (d.) Find the maximum value of A and the frequency ω for which it occurs. Answer: The characteristic equation of the original differential equation is r r + = 0 Thus the possible values of r are r 1 = i r 8 8 = 1 17i, and 8 8 the general solution of the homogeneous equation is u(t) = e t (c 1 cos t + c sin t). 8 8 The motion of this system can be described by u(t) = e t (c 1 cos t + c sin t) + R cos(ωt δ). 8 8 where R = / ( ω ) + ω 16

29 and Figure 1. for problem 17 ω = 0.7 cos δ = sin δ = ω ( ω ) + ω 16 1 ω 4 ( ω ) + ω 16 (a.) The steady-state part of the solution of this problem is u = [3( ω ) cos ωt + 8ω sin ωt] 64 63ω + 16ω 4. (b.)the amplitude A of the steady-state is A = ω + 16ω 4. (c.) For graph of A versus ω, see Figure 1. (d.)the maximum value of A is A = 64 17

30 30 and the corresponding frequency ω = Consider the forced but undamped system described by the initial value problem u + u = 3 cos ωt, u(0) = 0, u (0) = 0. a. Find the solution u(t) for ω 1. b. Plot the solution u(t) versus t for ω = 0.7, ω = 0.8 and ω = 0.9. Describe how the response u(t) changes as ω varies in this interval. What happens as ω takes on values closer and closer to 1? Note that the natural frequency of the unforced system is ω 0 = 1. Answer: (a.) ω 0 = k/m = 1, if ω ω 0, then the general solution is u = c 1 cos t + c sin t + 3 cos ωt. 1 ω From u(0) = 0, u (0) = 0, we get c 1 = 3 1 ω, c = 0. Hence the solution is u(t) = 3 (cos ωt cos t). 1 ω (b.) For the graphs of the solution u(t) versus t for ω = 0.7, ω = 0.8 and ω = 0.9, see Figure., Figure 3. and Figure 4. We can write above solution as: 3 (1 ω)t (1 + ω)t u(t) = ( sin ) sin. 1 ω If 1 ω is small, the 1 + ω is much greater than 1 ω. Consequently, sin (1+ω)t is rapidly oscillating function compared to sin (1 ω)t. Thus the motion is a rapid oscillation with frequency 1+ω but with a

31 Figure. for problem 18 ω = 0.7 slowly varying sinusoidal amplitude 3 1 ω sin (1 ω)t. The amplitude of u(t) gets larger and lager as w varies from ω = 0.7, ω = 0.8 to ω = 0.9, and closer and closer to 1, the natural frequency of the unforced system.

32 Figure 3. for problem 18 ω = 0.8; Figure 4. for problem 18 ω = 0.9