# pp1a_3 - x 1 , y 1 ) = f (1 , 0) =-2 x 1 + h = 2 y 1 + h f...

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3 3. Use Euler’s method to approximate the differential equation [10] y = xy 2 - x - 1 , y (0) = 1 on the interval x = [0 , 4] . Use a step size of h = 1 . SOLUTION: The update rules for Euler’s method are x n +1 = x n + h y n +1 = y n + h f ( x n , y n ) We create the following table for the solution: n x n y n f ( x n , y n ) = xy 2 - x - 1 x n +1 = x n + h y n +1 = y n + h f ( x n , y n ) 0 0 1 f ( x 0 , y 0 ) = f (0 , 1) = - 1 x 0 + h = 1 y 0 + h f ( x 0 , y 0 ) = 0 1 1 0 f (
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Unformatted text preview: x 1 , y 1 ) = f (1 , 0) =-2 x 1 + h = 2 y 1 + h f ( x 1 , y 1 ) =-2 2 2-2 f ( x 2 , y 2 ) = f (2 ,-2) = 5 x 2 + h = 3 y 2 + h f ( x 2 , y 2 ) = 3 3 3 3 f ( x 3 , y 3 ) = f (3 , 3) = 23 x 3 + h = 4 y 3 + h f ( x 3 , y 3 ) = 26 4 4 26...
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## This note was uploaded on 04/15/2008 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell University (Engineering School).

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