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Unformatted text preview: Exercises 3.7
For the Taylor method of order 4, we need to ﬁnd f3 and f4 . Thus, we have ∂ f2 (x, y ) ∂f2 (x, y ) + f (x, y ) = −1 + 1 · (x + 1 − y ) = x − y, ∂x ∂y ∂ f3 (x, y ) ∂f3 (x, y ) + f4 (x, y ) = f (x, y ) = 1 + (−1) · (x + 1 − y ) = y − x. ∂x ∂y f3 (x, y ) = Hence, by equation (6) on page 135 of the text, we see that the recursive formula for yn+1 for the Taylor method of order 4 with h = 0.25 is given by yn+1 = yn + 0.25 (xn + 1 − yn ) + (0.25)2 (0.25)3 (0.25)4 (yn − xn ) + (xn − yn ) + (yn − xn ) . 2 6 24 By starting with x0 = 0 and y0 = 1, we can ﬁll in the fourth column of Table 3H. Table 3–H: Taylor approximations of order 2 and 4 for the equation y = x + 1 − y .
n 0 1 2 3 4 xn 0 0.25 0.50 0.75 1.00 yn (order 2) 1 1.03125 1.11035 1.22684 1.37253 yn (order 4) 1 1.02881 1.10654 1.22238 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 φ2 (1) = 1.3725 if we use order 2 and by φ4 (1) = 1.3679 if we use order 4. The actual solution is y = x + e−x and so has the value y (1) = 1 + e−1 ≈ 1.3678794 at x = 1. Comparing these results, we see that y (1) − φ2 (1) = 0.00462 and y (1) − φ4 (1) = 0.00002 . 7. We will use the 4th order RungeKutta subroutine described on page 138 of the text. Since x0 = 0 and h = 0.25, we need N = 4 steps to approximate the solution at x = 1. With f (x, y ) = 2y − 6, we set x = x0 = 0, y = y0 = 1 and go to Step 3 to compute kj ’s. k1 = hf (x, y ) = 0.25[2(1) − 6] = −1 ; 155 ...
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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.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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