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291_pdfsam_math 54 differential equation solutions odd

# 291_pdfsam_math 54 differential equation solutions odd -...

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Exercises 5.3 k 2 , 2 = h 2 x 1;0 + k 1 , 1 2 3 x 2;0 + k 2 , 1 2 = 2 1 + 1 2 3 1 + 1 2 = 1 2 , k 1 , 3 = h 3 x 1;0 + k 1 , 2 2 4 x 2;0 + k 2 , 2 2 = 3 1 + 1 / 2 2 4 1 + 1 / 2 2 = 3 4 , k 2 , 3 = h 2 x 1;0 + k 1 , 2 2 3 x 2;0 + k 2 , 2 2 = 2 1 + 1 / 2 2 3 1 + 1 / 2 2 = 3 4 , k 1 , 4 = h [3 ( x 1;0 + k 1 , 3 ) 4 ( x 2;0 + k 2 , 3 )] = 3 1 + 3 4 4 1 + 3 4 = 1 4 , k 2 , 4 = h [2 ( x 1;0 + k 1 , 3 ) 3 ( x 2;0 + k 2 , 3 )] = 2 1 + 3 4 3 1 + 3 4 = 1 4 . Using the recursive formulas, we find t 1 = t 0 + h = 0 + 1 = 1 and x 1;1 = x 1;0 + 1 6 ( k 1 , 1 + 2 k 1 , 2 + 2 k 1 , 3 + k 1 , 4 ) = 1 + ( 1) + 2( 1 / 2) + 2( 3 / 4) + ( 1 / 4) 6 = 3 8 , x 2;1 = x 2;0 + 1 6 ( k 2 , 1 + 2 k 2 , 2 + 2 k 2 , 3 + k 2 , 4 ) = 1 + ( 1) + 2( 1 / 2) + 2( 3 / 4) + ( 1 / 4) 6 = 3 8 as approximations to x 1 (1) and x 2 (1) with step h = 1. We repeat the algorithm with h = 2 m , m = 1 , 2 , . . . . The results of these computations are listed in Table 5-A. Table 5–A : Approximations of the solution to Problem 17.
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