Differential Equations Solutions 116

# Differential Equations Solutions 116 - 126 Chapter 20...

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Unformatted text preview: 126 Chapter 20. Solutions: Solution of Ordinary Diﬀerential Equations % y(2) is to b_desired at t=1. b_desired = 1; [t,y] = ode45(@yprime,[0,1],[1,z]); f = y(end,1)-b_desired; % end of evalshoot %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function yp = yprime(t,y) yp = [ y(2); -(pi/2)^2*y(1)+(pi/2)^2*t.^2 + 2]; % end of yprime %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% CHALLENGE 20.18. No answer provided. CHALLENGE 20.19. (Partial Solution.) We use Taylor series expansions to derive the second formula: h2 h3 h4 u (t) + u (t) + u (η1 ), η1 ∈ [t, t + h], 2 6 24 h2 h3 h4 u(t − h) = u(t) − hu (t) + u (t) − u (t) + u (η2 ), η2 ∈ [t − h, t]. 2 6 24 u(t + h) = u(t) + hu (t) + Adding, we obtain u(t + h) + u(t − h) = 2u(t) + h2 u (t) + h4 [u (η1 ) + u (η2 )] . 24 Using the Mean Value Theorm on the last term and solving for u (t) gives ⇒ u (t) = u(t − h) − 2u(t) + u(t + h) h4 2u (η ), η ∈ [t − h, t + h]. −2 h2 h · 24 O (h2 ) ...
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