Unformatted text preview: x = 0 + 0 . 5 = 0 . 5 , y = 0 + 0 . 25(1 + 0 . 625) = 0 . 40625 . Thus the approximate value of the solution at 0 . 5 is 0 . 40625. Next we repeat Step 3 with x = 0 . 5 and y = 0 . 40625 to obtain F = 1 − . 40625 + (0 . 40625) 3 = 0 . 6607971 , G = 1 − [0 . 40625 + 0 . 5(0 . 6607971)] + [0 . 40625 + 0 . 5(0 . 6607971)] 3 ≈ . 6630946 . In Step 4 we compute x = 0 . 5 + 0 . 5 = 1 , 146...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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