Unformatted text preview: p = Ae2 t , but this term is already included as part of the homogeneous solution. Hence we try a solution of the form y p = Ate2 t . We thus want to ﬁnd A such that y 00 p3 y p10 y p = e2 t . We compute y p = Ae2 t2 Ate2 t and y 00 p =4 Ae2 t + 4 Ate2 t . Crunching everything through, we get (4 Ae2 t + 4 At 32 t ) + (3 Ae2 t + 6 Ate2 t ) + (10 Ate2 t ) = e2 t7 Ae2 t = e2 t A =1 7 Hence y = y h + y p = C 1 e2 t + C 2 e 5 t1 7 te2 t . 1...
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This note was uploaded on 04/02/2008 for the course MATH 33B taught by Professor Staff during the Spring '07 term at UCLA.
 Spring '07
 staff
 Math

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