Differential Equations Solutions 110

# Differential Equations Solutions 110 - 120 Chapter 20...

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Unformatted text preview: 120 Chapter 20. Solutions: Solution of Ordinary Diﬀerential Equations CHALLENGE 20.5. Since f (t, y ) = −y , the backward Euler formula is yn+1 = yn + hn f (tn+1 , yn+1 ) = yn − hn yn+1 . Therefore, (1 + hn )yn+1 = yn , so yn+1 = 1 yn . 1 + hn We compute: tn 0 0.1 0.2 0.3 yn 1 1/1.1 = 0.9091 (1/1.1)2 = 0.8264 (1/1.1)3 = 0.7513 y (tn ) 1 0.9048 0.8187 0.7408 CHALLENGE 20.6. Rearranging, we get (1 + ha/2)yn+1 = (1 − ha/2)yn , so yn+1 = 1 − ha/2 yn . 1 + ha/2 Apply Taylor series expansion to the diﬀerential equation to get h2 y (ξ ) 2 h2 = y (tn ) − hay (tn ) + y (ξ ) 2 h2 = (1 − ha)y (tn ) + y (ξ ), 2 y (tn+1 ) = y (tn ) + hy (tn ) + where ξ is a point between tn and tn+1 . Let en = yn − y (tn ), and subtract our two expressions to obtain en+1 = h2 1 − ha/2 1 − ha/2 en − (1 − ha − )y (t) − y (ξ ) 1 + ha/2 1 + ha/2 2 Now, since (1 − ha − 1 − ha/2 h2 a2 h2 )y (t) = − y (t) = − y (t), 1 + ha/2 2 + ha 2 + ha we see that en+1 = 1 − ha/2 h2 h2 en + y (t) + y (ξ ). 1 + ha/2 2 + ha 2 ...
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## This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

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