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271_pdfsam_math 54 differential equation solutions odd

# 271_pdfsam_math 54 differential equation solutions odd -...

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Exercises 5.2 19. From the first equation, we conclude that y = x 4 x . Substitution into the second equation yields ( x 4 x ) = 2 x + ( x 4 x ) x 5 x + 6 x = 0 . The characteristic equation, r 2 5 r + 6 = 0, has roots r = 2 , 3, and so a general solution is x ( t ) = c 1 e 2 t + c 2 e 3 t y ( t ) = ( c 1 e 2 t + c 2 e 3 t ) 4 ( c 1 e 2 t + c 2 e 3 t ) = 2 c 1 e 2 t c 2 e 3 t . We find constants c 1 and c 2 from the initial condition. 1 = x (0) = c 1 e 2(0) + c 2 e 3(0) = c 1 + c 2 , 0 = y (0) = 2 c 1 e 2(0) c 2 e 3(0) = 2 c 1 c 2 c 1 = 1 , c 2 = 2 . Therefore, the answer to this problem is x ( t ) = e 2 t + 2 e 3 t , y ( t ) = 2 e 2 t 2 e 3 t . 21. To apply the elimination method, we write the system using operator notation: D 2 [ x ] y = 0 , x + D 2 [ y ] = 0 . (5.5) Eliminating y by applying D 2 to the first equation and adding to the second equation gives
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