158_pdfsam_math 54 differential equation solutions odd

158_pdfsam_math 54 differential equation solutions odd - ....

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Chapter 3 Therefore, the recursive formulas of order 4 for the Taylor method are x n +1 = x n + h, y n +1 = y n + h ( x n y n )+ h 2 2 (1 x n + y n )+ h 3 3! ( 1+ x n y n )+ h 4 4! (1 x n + y n ) = y n + h ( x n y n )+ h 2 2 (1 x n + y n ) h 3 6 (1 x n + y n )+ h 4 24 (1 x n + y n ) = y n + h +(1 x n + y n ) ± h + h 2 2 h 3 6 + h 4 24 ² = y n + h ( x n y n )+(1 x n y n ) ± h 2 2 h 3 6 + h 4 24 ² . 5. For the Taylor method of order 2, we need to ±nd (see equation (4) on page 135 of the text) f 2 ( x, y )= ∂f ( x, y ) ∂x + ³ ∂f ( x, y ) ∂y ´ f ( x, y ) for f ( x, y )= x +1 y .Thu s ,w ehav e f 2 ( x, y )=1+( 1)( x +1 y )= y x. Therefore, by equations (5) and (6) on page 135 of the text, we see that the recursive formulas with h =0 . 25 become x n +1 = x n +0 . 25 , y n +1
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Unformatted text preview: . 25 ( x n + 1 − y n ) + (0 . 25) 2 2 ( y n − x n ) . By starting with x = 0 and y = 1 (the initial values for the problem), we ±nd y 1 = 1 + . 0625 2 ≈ 1 . 03125 . Plugging this value into the recursive formulas yields y 2 = 1 . 03125 + 0 . 25(0 . 25 + 1 − 1 . 03125) + ± . 0625 2 ² (1 . 03125 − . 25) ≈ 1 . 11035 . By continuing in this way, we can ±ll in the ±rst three columns in Table 3-H. 154...
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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.

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