287_pdfsam_math 54 differential equation solutions odd

287_pdfsam_math 54 differential equation solutions odd -...

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Exercises 5.3 x 0 2 =( y 0 ) 0 = y 0 = x 3 , x 0 3 y 0 ) 0 = y (3) = x 4 , x 0 4 = ( y (3) ) 0 = y (4) = y (3) 7 y +cos t = x 4 7 x 1 t. Hence, the required initial value problem for a system in normal form is x 0 1 = x 2 , x 0 2 = x 3 , x 0 3 = x 4 , x 0 4 = x 4 7 x 1 t, x 1 (0) = x 2 (0) = 1 ,x 3 (0) = 0 4 (0) = 2 . 5. First we express the given system as x 0 = x 0 y +2 t, y 0 = x y 1 . Setting x 1 = x , x 2 = x 0 , x 3 = y , x 4 = y 0 we obtain x 0 1 = x 0 = x 2 , x 0 2 = x 0 = x 2 x 3 t, x 0 3 = y 0 = x 4 , x 0 4 = y 0 = x 1 x 3 1 x 0 1 = x 2 , x 0 2 = x 2 x 3 x 0 3 = x 4 , x 0 4 = x 1 x 3 1 with initial conditions x 1 (3) = 5, x 2 (3) = 2, x 3 (3) = 1, and
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