147_pdfsam_math 54 differential equation solutions odd

147_pdfsam_math 54 differential equation solutions odd - 2...

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Exercises 3.6 Thus the approximate value of the solution at 1 . 1i s0 . 1045. Next we repeat Step 3 with x =1 . 1and y =0 . 1045 to obtain F =1 . 1+(0 . 1045) 2 1 . 0891 , G =(1 . 1+0 . 1) ± 0 . 1045 + 0 . 1 ( 1 . 1 (0 . 1045) 2 2 1 . 1545 . Hence in Step 4 we compute x =1 . 1+0 . 1=1 . 2 , y =0 . 1045 + 0 . 05(1 . 0891 + 1 . 1545) 0 . 21668 . Thus the approximate value of the solution at 1 . 2is0 . 21668. By continuing in this way, we Fll in Table 3-B. (The reader can also use the software provided free with the text.) Table 3–B : Improved Euler’s method to approximate the solution of y 0 = x y 2 , y (1) = 0, with h =0 . 1. i xy 01 0 1 1.1 0.10450 2
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Unformatted text preview: 2 1.2 0.21668 3 1.3 0.33382 4 1.4 0.45300 5 1.5 0.57135 9. In this initial value problem, f ( x, y ) = x + 3 cos( xy ), x = 0, and y = 0. To approximate the solution on [0 , 2] with a step size h = 0 . 2, we need N = 10 steps. The functions F and G in the improved Eulers method subroutine are F = f ( x, y ) = x + 3 cos( xy ); G = f ( x + h, y + hF ) = x + h + 3 cos[( x + h )( y + hF )] = x + 0 . 2 + 3 cos[( x + 0 . 2)( y + 0 . 2 { x + 3 cos( xy ) } )] . 143...
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