313_pdfsam_math 54 differential equation solutions odd

313_pdfsam_math 54 differential equation solutions odd -...

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Exercises 5.5 which is a quadratic in r 2 .So r 2 = 23 ± 529 240 6 = 23 ± 17 6 . Since 20 / 3and 1 are negative, the roots of the characteristic equation are ± 1 and ± 2 , where β 1 = r 20 3 2 =1 . Hence x ( t )= c 1 cos β 1 t + c 2 sin β 1 t + c 3 cos β 2 t + c 4 sin β 2 t. Solving the Frst equation of the system given in (5.28) for y ,weget y ( t 1 4 ( D 2 +4 ) [ x ]= 1 4 ±( β 2 1 ) c 1 cos β 1 t + ( β 2 1 ) c 2 sin β 1 t + ( β 2 2 ) c 3 cos β 2 t + ( β 2 2 ) c 4 sin β 2 t ² . Next we substitute into the initial conditions. Setting x (0) = 1, x 0 (0) = 0 yields 1= c 1 + c 3 , 0= c 2 β 1 + c 4 β 2 . ±rom the initial conditions y (0) = 0, y 0 (0) = 0, we get
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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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