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Unformatted text preview: r = 3 , ( r + 2) 3 = 0 r = 2 of multiplicity 3 , ( r 2 + 4 r + 5) 2 = 0 r = 2 i of multiplicity 2 , r 5 = 0 r = 0 of multiplicity 5 . Therefore, a general solution is given by y ( x ) = c 1 e 4 x + c 2 e 3 x + ( c 3 + c 4 x + c 5 x 2 ) e 2 x + ( c 6 + c 7 x ) e 2 x cos x + ( c 8 + c 9 x ) e 2 x sin x + c 10 + c 11 x + c 12 x 2 + c 13 x 3 + c 14 x 4 . 19. irst, we nd a general solution to the given equation. Solving the auxiliary equation, r 3 r 2 4 r + 4 = ( r 3 r 2 ) (4 r 4) = ( r 1)( r 2 4) = ( r 1)( r + 2)( r 2) = 0 , 353...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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