Unformatted text preview: ~ v = y ~ i + x ~ j using 2 steps with a time interval of Δ t = . 1. Solution Euler’s method is given by ~ r n + 1 = ~ r n + Δ t ~ F ( ~ r n ) , which in component form is ( x n + 1 = x n + Δ tF 1 ( x n , y n ) y n + 1 = y n + Δ tF 2 ( x n , y n ) . Using Δ t = . 1, F 1 ( x n , y n ) = y n and F 2 ( x n , y n ) = x n yields the equations ( x n + 1 = x n + . 1 y n y n + 1 = y n + . 1 x n . Therefore, ( x 1 = x + . 1 y = 1 + . 1 · 2 = 1 . 2 y 1 = y + . 1 x = 2 + . 1 · 1 = 2 . 1 and ( x 2 = x 1 + . 1 y 1 = 1 . 2 + . 1 · 2 . 1 = 1 . 41 y 2 = y 1 + . 1 x 1 = 2 . 1 + . 1 · 1 . 2 = 2 . 22 . Therefore, the ﬂow through (1, 2) of the vector ﬁeld ~ v = y ~ i + x ~ j at t = . 2 is approximately x (0 . 2) ≈ x 2 = 1 . 41 y (0 . 2) ≈ y 2 = 2 . 22...
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 Spring '08
 Staff
 Multivariable Calculus, yn, Constant of integration, Parametric equation, initial condition, flow line, vector differential equation

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