256_pdfsam_math 54 differential equation solutions odd

256_pdfsam_math 54 differential equation solutions odd -...

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Chapter 4 =4 9 Ae 5 t +9 Be 3 t = e 5 t + e 3 t A =1 / 49 ,B =1 / 9 . Therefore, y p ( t )=(1 / 49) e 5 t +(1 / 9) e 3 t and a general solution to the original equation is y ( t )= c 1 e 3 t/ 2 + c 2 te 3 t/ 2 + 1 49 e 5 t + 1 9 e 3 t . 27. This is a Cauchy-Euler equation. Thus we make the substitution x = e t and get x 2 d 2 y dx 2 +2 x dy dx 2 y =6 x 2 +3 x ± d 2 y dt 2 dy dt ² +2 dy dt 2 y =6( e t ) 2 +3( e t ) d 2 y dt 2 + dy dt 2 y =6 e 2 t +3 e t . (4.15) The auxiliary equation, r 2 + r 2 = 0, has the roots r = 2, 1. Therefore, a general solution to the corresponding homogeneous equation is y h ( t )= c 1 e t + c 2 e 2 t . A particular solution to (4.15) has the form y p ( t )= Ate 2 t + Bte t . (The factor t appeared in both terms because e t and e 2 t are both solutions to the homogeneous
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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.

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