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159_pdfsam_math 54 differential equation solutions odd

# 159_pdfsam_math 54 differential equation solutions odd -...

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Exercises 3.7 For the Taylor method of order 4, we need to find f 3 and f 4 . Thus, we have f 3 ( x, y ) = ∂f 2 ( x, y ) ∂x + ∂f 2 ( x, y ) ∂y f ( x, y ) = 1 + 1 · ( x + 1 y ) = x y, f 4 ( x, y ) = ∂f 3 ( x, y ) ∂x + ∂f 3 ( x, y ) ∂y f ( x, y ) = 1 + ( 1) · ( x + 1 y ) = y x. Hence, by equation (6) on page 135 of the text, we see that the recursive formula for y n +1 for the Taylor method of order 4 with h = 0 . 25 is given by y n +1 = y n + 0 . 25 ( x n + 1 y n ) + (0 . 25) 2 2 ( y n x n ) + (0 . 25) 3 6 ( x n y n ) + (0 . 25) 4 24 ( y n x n ) . By starting with x 0 = 0 and y 0 = 1, we can fill in the fourth column of Table 3-H. Table 3–H : Taylor approximations of order 2 and 4 for the equation y = x + 1 y . n x n x n n y n n n (order 2) y n n y (order 4) 0 0 1 1 1 0.25 1.03125 1.02881 2 0.50 1.11035 1.10654 3 0.75 1.22684 1.22238 4 1.00 1.37253 1.36789 Thus, the approximation (rounded to 4 decimal places) of the solution by the Taylor method at the point x = 1 is given by
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