320_pdfsam_math 54 differential equation solutions odd

320_pdfsam_math 54 differential equation solutions odd -...

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Chapter 5 A =0 ,B = 1 x h ( t )= sin 2 t and x ( t x h ( t )+ x p ( t c 1 cos t + c 2 sin t + c 3 cos 6 t + c 4 sin 6 t sin 2 t. From the frst equation in the original system, we have y ( t 1 2 ( x 0 +5 x ) =2 c 1 cos t +2 c 2 sin t 1 2 c 3 cos 6 t 1 2 c 4 sin 6 t 1 2 sin 2 We determine constants c 1 and c 3 using the initial conditions x (0) = 0 and y (0) = 1. 0= x (0) = c 1 + c 3 , 1= y (0) = 2 c 1 c 3 / 2 c 3 = c 1 , 2 c 1 ( c 1 ) / 2=1 c 3 = 2 / 5 , c 1 / 5 . To fnd c 2 and c 4 , compute x 0 ( t )and y 0 ( t ), evaluate these ±unctions at t = 0, and use the other two initial conditions, x 0 (0) = y 0 (0) = 0. This yields x 0 (0) = c 2 + 6 c 4 2 , y 0 (0) = 2 c 2 6 c 4 / 2 1 c 4 = 6 / 5 , c 2 =4 / 5 . There±ore, the required solution is x ( t 2 5 cos t + 4 5 sin t 2 5 cos 6 t + 6 5 sin 6 t sin 2 t, y ( t 4 5 cos t + 8 5 sin t + 1 5 cos 6 t
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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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