Chem Differential Eq HW Solutions Fall 2011 198

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online