340_pdfsam_math 54 differential equation solutions odd

340_pdfsam_math 54 differential equation solutions odd -...

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Chapter 5 A general solution to the corresponding homogeneous equation is given in (3) on page 243 of the text: y h ( t )= c 1 e t/ 2 + c 2 e t/ 6 . A particular solution has the form y p ( t ) C ,wh ichresu lts 3( C ) 0 +2( C ) 0 + 1 4 C =1 . 2 C =4 . 8 . Therefore, y p ( t ) 4 . 8, and a general solution to the system is y ( t y h ( t )+ y p ( t c 1 e t/ 2 + c 2 e t/ 6 +4 . 8 , x ( t )=3 y 0 ( t y ( t c 1 2 e t/ 2 + c 2 2 e t/ 6 . 8 . We Fnd constants c 1 and c 2 from the initial conditions, x (0) = 0 . 1and y (0) = 0 . 3 . Substitu- tion yields the system c 1 2 + c 2 2 . 8=0 . 1 , c 1 + c 2 . . 3 . Solving, we obtain c 1 =49 / 20, c 2 = 139 / 20, and so x ( t 49 40 e t/ 2 139 40 e t/ 6 . 8 , y ( t 49 20 e t/ 2
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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 Berkeley.

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