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Unformatted text preview: . Using the quadratic formula, we nd that the other two roots are r = 2 4 12 2 = 1 2 i. A general solution is, therefore, y ( t ) = c 1 e t + c 2 e t cos 2 t + c 3 e t sin 2 t . (b) By inspection, r = 2 is a root of the auxiliary equation, r 3 + 2 r 2 + 5 r 26 = 0. Since r 3 + 2 r 2 + 5 r 26 = ( r 2) ( r 2 + 4 r + 13 ) , the other two roots are the roots of r 2 + 4 r + 13 = 0, that is, r = 2 3 i . Therefore, a general solution to the given equation is y ( t ) = c 1 e 2 t + c 2 e 2 t cos 3 t + c 3 e 2 t sin 3 t . 182...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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