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Unformatted text preview: Answers to sample questions for Exam 2 1. Find the general solution to the following equations: a. y 00 + 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 00 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 00 + 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 00 + 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 00 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 00 + 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 g. y 000 + 2 y 00 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 000 2 y 00 + 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 000 3 y 00 + 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 0000 3 y 00 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 00 + 4 y + 4 y = 0 , y (0) = 3 , y (0) = 1 ....
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This note was uploaded on 04/23/2008 for the course MA 232 taught by Professor Toland during the Fall '08 term at Clarkson University .
 Fall '08
 Toland
 Equations

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