Unformatted text preview: r 3 − 2 r 2 − 5 r + 6 = ( r − 1)( r 2 − r − 6) = ( r − 1)( r − 3)( r + 2) = 0 . Thus, the roots to the auxiliary equation are given by r = 1, 3, and − 2, and a general solution to the homogeneous equation is y h ( x ) = c 1 e x + c 2 e 3 x + c 3 e − 2 x . The nonhomogeneous term, g ( x ) = e x + x 2 , is the sum of an exponential term and a polynomial term. Therefore, according to Section 4.5, this equation has a particular solution of the form y p ( x ) = x s 1 C 1 e x + x s 2 ( C 2 + C 3 x + C 4 x 2 ) . 361...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Hyperbolic function, C1 cosh

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