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# HW5 - MA341 Homework 5 4.1 3 The derivatives of y are y = 6...

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MA341 Homework 5 4.1 3. The derivatives of y are: y = 6 cos 3 t - 3 sin 3 t, y = - 18 sin 3 t - 9 cos 3 t. Thus, 2 y + 18 y = 2( - 18 sin 3 t - 9 cos 3 t ) + 18(2 sin 3 t + cos 3 t ) = - 36 sin 3 t - 18 cos 3 t + 36 sin 3 t + 18 cos 3 t = 0 . Substitution into the initial conditions yields: y (0) = 2 sin(0) + cos(0) = 1 , y (0) = 6 cos(0) - 3 sin(0) = 6 . Hence, y = 2 sin 3 t + cos 3 t is a solution to the initial value problem. Moreover, y ( t ) = 2 sin 3 t + cos 3 t = A sin(3 t + φ ) = A sin 3 t cos φ + A cos 3 t sin φ . Then, A cos φ = 2 and A sin φ = 1. Consequently, φ = arctan(1 / 2) and A = 5. We conclude that max -∞ <t< | y ( t ) | = A = 5. 5. The derivatives of y = e - 2 t sin( 2 t ) are: y ( t ) = - 2 e - 2 t sin( 2 t ) + 2 e - 2 t cos( 2 t ) , y ( t ) = 2 e - 2 t sin( 2 t ) - 4 2 e - 2 t cos( 2 t ) . Substitution into (3) gives: my + by + ky = y + 4 y + 6 y = 2 e - 2 t sin( 2 t ) - 4 2 e - 2 t cos( 2 t ) + 4 {- 2 e - 2 t sin( 2 t ) + 2 e - 2 t cos( 2 t ) } + 6 e - 2 t sin( 2 t ) = 0 . We also have lim t →∞ e - 2 t sin( 2 t ) = 0. 7. Let y ( t ) = A cos Ω t + B sin Ω t , where Ω = 5. Then, y ( t ) = - 5 A sin 5 t + 5 B cos 5 t, y ( t ) = - 25 A cos 5 t - 25 B sin 5 t. Next, we substitute these expressions into y + 2 y + 4 y = 3 sin 5 t , collect the terms and match coefficients: 3 sin 5 t = y + 2 y + 4 y = ( - 21 B - 10 A ) sin 5 t + ( - 21 A + 10 B ) cos 5 t.

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MA341 Homework 5 Therefore, 3 = - 21 B - 10 A, 0 = - 21 A + 10 B. Hence, we obtain A = - 30 541 , B = - 63 541 , and y ( t ) = - 30 541 cos 5 t - 63 541 sin 5 t . 4.2 3. The characteristic equation is r 2 + 8 r + 16 = ( r + 4) 2 = 0 , which has one root r = - 4.
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HW5 - MA341 Homework 5 4.1 3 The derivatives of y are y = 6...

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