Math Ordinary Differential Equations Sample Test 3 Solutions

Transcription

1 Solve the following Math - Ordinary Differential Equations Sample Test Solutions (i x 2 y xy + 8y y(2 2 y (2 (ii x 2 y + xy + 4y y( 2 y ( (iii x 2 y xy + y y( 2 y ( (i The characteristic equation is m(m m + 8 m 2 6m + 8 (m 2(m 4 m 2 4 The solution is y c x 2 + c 2 x 4 With the boundary conditions y(2 2 and y (2 gives c 6 and c 2 Thus the solution is y 6x 2 2x 4 (ii The characteristic equation is m(m + m + 4 m 2 + 4m + 4 (m m The solution is y c x 2 + c 2 ln x x 2 With the boundary conditions y( 2 and y ( gives c and c 2 Thus the solution is y x 2 + ln x x 2 (iii The characteristic equation is m(m m + m 2 4m + m 2 ± i The solution is y c x 2 sin(ln x + c 2 x 2 cos(ln x

2 With the boundary conditions y( 2 and y ( gives c and c 2 2 Thus the solution is y x 2 sin(ln x + 2x 2 cos(ln x 2 Solve the following using the variation of parameters (i The complementary equation is (i y + y tan x (ii y + y + 2y e x + y( 2 y ( y + y whose solution is y c sin x + c 2 cosx For the variation of parameters we replace c and c 2 with u and v Thus y u sin x + v cosx ( so where we set Therefore and y u sin x + u cos x + v cos x v sin x u sin x + v cos x (2 y u cos x v sin x y u cos x u sin x v sin x v cos x Substituting into the original equation gives u cos x u sin x v sin x v cos x u sin x Solving (2 and ( for u and v gives v cos x tan x ( which upon integrating gives u sin x v sin2 x cos x Substituting these into ( gives u cos x v sin x ln sec x + tan x 2

3 y cos x ln sec x + tan x which in turn gives the solution y c sin x + c 2 cos x cos x ln sec x + tan x (ii The complementary equation is y + y + 2y whose solution is y c e x + c 2 e x For the variation of parameters we replace c and c 2 with u and v Thus y ue x + ve x (4 so where we set so Then y u e x ue x + v e x 2ve x u e x + v e x ( y ue x 2ve x y u e x + ue x 2v e x + 4ve x Substituting into the original equation gives u e x + ue x 2v e x + 4ve x ue x + 2ue x Solving ( and (6 for u and v gives which upon integrating gives u Substituting these into (4 gives 6ve x ex e x + v e2x e x + u ln(e x + v e x + ln(e x + y ( e x + e x ln(e x + + 2ve x e x + (6

4 noting that the term in the particular solution e x was neglected because it appears as part of the complementary solution This in turn gives the solution ( y c e x + c 2 e x + e x + e x ln(e x + A -pound weight attached to a spring stretches it 2 feet The weight is attached to a dashpot damping device that offers resistance numerically equal to β (β > times the instantaneous velocity Determine the values of the damping constant β so that the subsequent motion is (a overdamped (b critically damped and (c underdamped The equation which governs the motion is m d2 x dt 2 + βdx + kx dt Since the lb weight stretches the string 2ft then F kx 2k k Further since F mg then 2m m 6 So or The characteristic equation is 6 d 2 x dt 2 + β dx + x dt d2 x dx + 6β + 8x dt2 dt from which we obtain m 2 + 6βm + 8 m 8β ± 6β 2 The motion will be over damped if 6β 2 > critically damped if 6β 2 and under damped if 6β 2 < or if β > /2 β /2 or β < /2 4 Solve the following systems ( d x (i dt x 2 ( d x (iii dt ( d x 6 (v dt 4 x x (ii (iv (vi ( d x dt x ( d x dt x ( d x dt x The general form is (λi A ū (7 and in order to have nontrivial solutions ū we require that λi A (8 4

5 (i The characteristic equation is λ λ λ2 λ 2 (λ + (λ 2 from which we obtain the eigenvalues λ and λ 2 Case : λ From (7 we have ( ( e ( from which we obtain upon expanding 2e + and we deduce the eigenvector ( ū Cas: λ 2 From (7 we have ( 2 ( e ( from which we obtain upon expanding e and we deduce the eigenvector ( ū The general solution is then given by x c ( e t + c 2 ( t (ii The characteristic equation is λ + 2 λ + 2 λ2 + 4λ + (λ + (λ + from which we obtain the eigenvalues λ and λ Case : λ From (7 we have ( ( e ( from which we obtain upon expanding e and we deduce the eigenvector ( ū Cas: λ From (7 we have ( ( e (

6 from which we obtain upon expanding e and we deduce the eigenvector ū ( The general solution is then given by x c ( e t + c 2 ( e t (iii The characteristic equation is λ λ λ2 4λ + 4 (λ 2 2 from which we obtain the eigenvalues λ 2 2 For λ 2 from (7 we have ( ( e ( from which we obtain upon expanding e + and we deduce the eigenvector ū ( For the second solution we seek a solution of the form Substitution into our system gives x ūtt + vt (9 (2I Aū ( (2I A v ū ( Equation ( gives the eigenvector just found whereas ( gives ( ( ( e from which we obtain upon expanding e + and we deduce the eigenvector ū ( So the second solution is x ( ( tt + t 6

7 and general solution is x c ( t + c 2 [( ( tt + t (iv The characteristic equation is λ 4 λ λ2 6λ + 9 (λ 2 from which we obtain the eigenvalues λ For λ from (7 we have ( 4 2 ( e ( from which we obtain upon expanding e 2 and we deduce the eigenvector ū ( 2 For the second solution we seek a solution of the form Substitution into our system gives x ūtt + vt (2 (I Aū ( (I A v ū (4 Equation ( gives the eigenvector just found whereas (4 gives ( ( ( 4 e 2 2 from which we obtain upon expanding e + 2 and we deduce the eigenvector So the second solution is and general solution is x c ( 2 x ( 2 ū ( ( te t + e t + c 2 [( 2 e t ( te t + e t 7

8 (v The characteristic equation is λ 6 λ 4 λ2 λ + 29 from which we obtain the eigenvalues λ ± 2i Case : λ + 2i From (7 we have ( + 2i + 2i ( e ( from which we obtain upon expanding ( 2ie + and we deduce the eigenvector ( ū 2i The second eigenvector would just be the complex conjugate Thus ( ( E E 2 The two solutions are x x 2 [( ( cos 2t [( ( sin 2t + sin 2t cos 2t e t e t and the general solution x c [( + c 2 [( ( cos 2t ( sin 2t sin 2t cos 2t e t e t (vi The characteristic equation is λ λ + λ2 + 2λ + from which we obtain the eigenvalues λ ± i Case : λ + i From (7 we have ( + i 4 + i ( e from which we obtain upon expanding ( + ie and we deduce the eigenvector ( ū + i ( 8

9 The second eigenvector would just be the complex conjugate Thus ( E ( E 2 The two solutions are x x 2 [( ( cos t [( ( sin t + sin t cos t e t e t and the general solution x c [( + c 2 [( ( cos t ( sin t + sin t cos t e t e t 9