Chem Differential Eq HW Solutions Fall 2011 198

Chem Differential Eq HW Solutions Fall 2011 198 - A198...

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Unformatted text preview: A198 Appendix A Ordinary Differential Equations: Review of Concepts and Methods Then 4 x2y − 14 x y + (20 − x)y ∞ ∞ 4(m + 2)(m + 1)am xm+2 − 14 = m=0 ∞ (m + 2)am xm+2 + (20 − x) m=1 m=0 ∞ ∞ [4(m + 1) − 14](m + 2)am xm+2 + 20 = am xm+2 m=0 ∞ am xm+2 − m=0 ∞ am xm+3 m=0 ∞ [(4m − 10)(m + 2) + 20]am xm+2 − = m=0 am xm+3 m=0 ∞ ∞ [4m2 − 2m]am xm+2 − = m=0 am−1xm+2 m=1 ∞ [4m2 − 2m]am − am−1 ]xm+2 = m=1 This gives the recurrence relation: For all m ≥ 0, am−1 am−1 am = = . 2 − 2m 4m 2m(2m − 1) Since a0 is arbitrary, take a0 = 1. Then a1 = 1 ; 2 a2 = 1 1 =; 2(12) 4! a3 = 11 1 =; 4! 6 · 5 6! . . . y1 = a0 x2 1 + 1 1 1 x + x2 + x3 + · · · 2! 4! 6! We now turn to the second solution: ∞ ∞ 5 bm xm+ 2 ; y= m=0 y= 3 5 (m + )bm xm+ 2 ; 2 m=0 ∞ y= 1 5 3 (m + )(m + )bm xm+ 2 . 2 2 m=0 So 4 x2 y − 14 x y + (20 − x) y ∞ = m=0 ∞ = m=0 ∞ 5 5 5 3 5 4(m + )(m + ) − 14(m + ) bm xm+ 2 + (20 − x) bm xm+ 2 2 2 2 m=0 ∞ 5 5 7 3 5 4(m + )(m + ) − 14(m + ) + 20 bm xm+ 2 − bm xm+ 2 2 2 2 m=0 ∞ ∞ 5 5 4m2 + 2m bm xm+ 2 − = m=0 bm−1 xm+ 2 m=1 ∞ 5 (4m2 + 2m)bm − bm−1 xm+ 2 = m=1 ...
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