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Unformatted text preview: ) k 2 = hf ( t i + 1 2 h,x i + 1 2 k 1 ) k 3 = hf ( t i + 1 2 h,x i + 1 2 k 2 ) k 4 = hf ( t i + h,x i + k 3 ) x i +1 = x i + 1 6 ( k 1 + 2 k 2 + 2 k 3 + k 4 ) In general RungeKutta method gives more accurate result than Euler’s method at the same step length. However, RungeKutta method is more expensive to implement, that is, at each step, RungeKutta method requires more computation than Euler’s method( four function evaluations compare one in Euler’s method). Example 3.1 . Example 3.2 . Find approximate to x (1) if x ( t ) = t 2e x sin( t ) ,x (0) = 1 with h = 0 . 25 . Solution Here the interval is [0, 1], so a = 0 ,b = 1 . Since h = ba n = 0 . 25 we have n = 4 and we need to compute x 1 ,x 2 ,x 3 ,x 4 starting with x = x (0) = 1 , t = 0...
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This note was uploaded on 10/04/2011 for the course MAP 4231 taught by Professor Dr.han during the Spring '11 term at UNF.
 Spring '11
 Dr.Han

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