# Thus the rst block of 3310c is solved for y1 knowing

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Unformatted text preview: bleau c1 a11 0 c2 a21 a22 b1 b2 and it could be of third order. According to Theorem 3.2.1, the conditions for secondorder accuracy are (3.2.5c) with l = 1 2 when (3.2.5b) is satis ed, i.e., b1 + b2 = 1 c1 = a11 c2 = a21 + a22 b1 c1 + b2c2 = 1=2: (As noted earlier, satisfaction of (3.2.5b) is not necessary, but it simpli es the algebraic manipulations.) We might guess that the remaining conditions necessary for third order accuracy are (3.2.5c) with l = 3 and (3.2.7) with k = 3, i.e., b1 c2 + b2 c2 = 1=3 1 2 and bT A2l = bAc = b1 a11 c1 + b2 (a21 c1 + a22 c2 ) = 1=6 27 where (3.2.5b) was used to simplify the last expression. After some e ort, this system of six equations in seven unknowns can be solved to yield 2 ; 3c c2 = 3 ; 6c1 2 b1 = cc ; 1c=2 ; 1 2 ; b2 = 1c=2; cc1 1 2 2 1 a22 = b =(6 ;;b1cc1) c a11 = c1 2 2 1 a21 = c2 ; a22 : 1 As written, the solution is parameterized by c1 . Choosing c1 = 1=3 gives 1/3 1/3 0 1 1/2 1/2 3/4 1/4 Using (3.2.3), the method is Y1 = yn 1 + h f (tn 1 + h Y...
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## This document was uploaded on 03/16/2014 for the course CSCI 6820 at Rensselaer Polytechnic Institute.

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