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167_pdfsam_math 54 differential equation solutions odd

167_pdfsam_math 54 differential equation solutions odd -...

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Exercises 3.7 solution to the differential equation at x = 1. Since the actual solution is y = e x , this means that y n e . To find the equation we are looking for, we see that y 1 = 1 + 1 n + 1 2 n 2 y 0 = 1 + 1 n + 1 2 n 2 , y 2 = 1 + 1 n + 1 2 n 2 y 1 = 1 + 1 n + 1 2 n 2 2 , y 3 = 1 + 1 n + 1 2 n 2 y 2 = 1 + 1 n + 1 2 n 2 3 , y 4 = 1 + 1 n + 1 2 n 2 y 3 = 1 + 1 n + 1 2 n 2 4 , . . . y n = 1 + 1 n + 1 2 n 2 y n 1 = 1 + 1 n + 1 2 n 2 n . (This can be proved rigorously by mathematical induction.) As we observed above, y n e , and so we have e 1 + 1 n + 1 2 n 2 n . 19. In this initial value problem, the independent variable is u , the dependent variable is v , u 0 = 2, v 0 = 0 . 1, and f ( u, v ) = u u 2 + 1 v 3 + u + 5 2 v 2 . We will use the classical 4th order Runge-Kutta algorithm with tolerance given on page 139 of the text but, since the stopping criteria should be based on the relative error, we will replace
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