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exam2sampleans

# exam2sampleans - Answers to sample questions for Exam 2 1...

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Answers to sample questions for Exam 2 1. Find the general solution to the following equations: a. y + 2 y + 10 y = 0 r 2 + 2 r + 10 has roots r = - 1 ± 3 i . The general solution is y ( t ) = C 1 e - t cos(3 t ) + C 2 e - t sin(3 t ) . b. y - 4 y - 5 y = 0 r 2 - 4 r - 5 has roots r = 5 and r = - 1. The general solution is y ( t ) = C 1 e 5 t + C 2 e - t . c. y + 6 y + 2 y = 0 r 2 + 6 r + 2 has roots r = - 3 ± 7. The general solution is y ( t ) = C 1 e ( - 3+ 7) t + C 2 e ( - 3 - 7) t d. y + y + 0 . 25 y = 0 r 2 + r + 0 . 25 has a repeated root of r = - 1 / 2. The general solution is y ( t ) = C 1 e - t/ 2 + C 2 te - t/ 2 e. y - 6 y + 25 y = 0 r 2 - 6 r + 25 has roots r = 3 ± 4 i . The general solution is y ( t ) = C 1 e 3 t cos(4 t ) + C 2 e 3 t sin(4 t ) . f. y + 8 y + 15 y = 0 r 2 + 8 r + 15 has roots r = - 3 and r = - 5. The general solution is y ( t ) = C 1 e - 3 t + C 2 e - 5 t . 1

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g. y + 2 y - y - 2 y = 0 The characteristic equation r 3 + 2 r 2 - r - 2 factors as ( r - 1)( r + 1)( r + 2). The general solution is y ( t ) = C 1 e t + C 2 e - t + C 3 e - 2 t . h. y - 2 y + y = 0 r 3 - 2 r 2 + r factors as r ( r - 1) 2 . The general solution is y ( t ) = C 1 + C 2 e t + C 2 te t . i. y - 3 y + 3 y - y = 0. r 3 - 3 r 2 + 3 r - 1 factors as ( r - 1) 3 . The general solution is y ( t ) = C 1 e t + C 2 te t + C 3 t 2 e t . j. y - 3 y - 4 y = 0. The characteristic polynomial r 4 - 3 r 2 - 4 is a quadratic in r 2 ; we can factor the polynomial as ( r 2 - 4)( r 2 + 3). It further factors as ( r - 2)( r + 2)( r 2 + 3). The general solution is y ( t ) = C 1 e 2 t + C 2 e - 2 t + C 3 cos( 3 t ) + C 4 sin( 3 t ) . 2. Solve the initial value problem y + 4 y + 4 y = 0 , y (0) = 3 , y (0) = 1 . The general solution is y ( t ) = C 1 e - 2 t + C 2 te - 2 t . Applying the initial values: y ( t ) = C 1 e - 2 t + C 2 te - 2 t C 1 = 3 y ( t ) = - 2 C 1 e - 2 t + C 2 ( - 2 t + 1) e - 2 t - 2 C 1 + C 2 = 1 . We get C 1 immediately; substituting C 1 = 3 into the second equation gives C 2 = 7. Therefore y ( t ) = 3 e - 2 t + 7 te - 2 t .
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exam2sampleans - Answers to sample questions for Exam 2 1...

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