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Unformatted text preview: equation.) DiFerentiating, we nd y p ( t ) = Ate 2 t + Bte t y p ( t ) = A (1 2 t ) e 2 t + B ( t + 1) e t y p ( t ) = A (4 t 4) e 2 t + B ( t + 2) e t . Substitution into (4.15) yields 3 Ae 2 t + 3 Be t = 6 e 2 t + 3 e t A = 2 , B = 1 . Thus a general solution to (4.15) is given by y ( t ) = y h ( t ) + y p ( t ) = c 1 e t + c 2 e 2 t 2 te 2 t + te t . 252...
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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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