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Unformatted text preview: Exercises 3.7
solution to the diﬀerential equation at x = 1. Since the actual solution is y = ex , this means that yn ≈ e. To ﬁnd the equation we are looking for, we see that y1 = y2 = y3 = y4 = . . . yn = 1 1 1+ + 2 n 2n y n−1 = 1 1 1+ + 2 n 2n
n 1+ 1 1 +2 n 2n y0 = y1 = y2 = y3 = 1+ 1 1 +2 n 2n ,
2 1 1 1+ + 2 n 2n 1 1 1+ + 2 n 2n 1+ 1 1 +2 n 2n 1 1 1+ + 2 n 2n 1 1 1+ + 2 n 2n 1+ 1 1 +2 n 2n ,
3 ,
4 , . (This can be proved rigorously by mathematical induction.) As we observed above, yn ≈ e, and so we have e≈ 1 1 1+ + 2 n 2n
n . 19. In this initial value problem, the independent variable is u, the dependent variable is v , u0 = 2, v0 = 0.1, and u 5 + 1 v3 + u + v2 . 2 2 We will use the classical 4th order RungeKutta algorithm with tolerance given on page 139 of f (u, v ) = u the text but, since the stopping criteria should be based on the relative error, we will replace the condition z − v  < ε in Step 6 by (z − v )/v  < ε (see Step 6 on page 138). We start with m = 0, N = 2m = 1, and a step size h = (3 − 2)/N = 1. Setting u = u0 = 2, v = v0 = 0.1, on Step 4 we compute 2 5 + 1 (0.1)3 + 2 + (0.1)2 = 0.049; 2 2 2 + 1/ 2 k2 = hf (u + h/2, v + k1 /2) = (1) (2 + 1/2) + 1 (0.1 + 0.049/2)3 2 5 + (2 + 1/2) + (0.1 + 0.049/2)2 = 0.088356 ; 2 k1 = hf (u, v ) = (1) 2 163 ...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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