Chem Differential Eq HW Solutions Fall 2011 199

Chem Differential Eq HW Solutions Fall 2011 199 - . We are...

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

View Full Document Right Arrow Icon
Section A.6 The Method of Frobenius A199 This gives b 0 arbitrary and the recurrence relation: For all m 1, b m = b m - 1 2 m (2 m +1) . Since b 0 is arbitrary, take b 0 = 1. Then b 1 = 1 3! ; b 2 = 1 3! 1 4 · 5 = 1 5! ; b 3 = 1 5! 1 6 · 7 = 1 7! ; . . . y 2 = b 0 x 5 / 2 ± 1+ 1 3! x + 1 5! x 2 + 1 7! x 3 + ··· ² 13. For the equation xy ±± +(1 - x ) y ± + y =0 , p ( x )= 1 - x x ,x p ( x )=1 - x, p 0 =1 ; q ( x )= 1 x ,x 2 q ( x )= x, q 0 =0 . p ( x ) and q ( x ) are not analytic at 0. So a = 0 is a singular point. Since xp ( x ) and x 2 q ( x ) are analytic at a = 0, the point a = 0 is a regular singular point. Indicial equation r ( r - 1) + r =0 r = 0 (double root)
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . We are in Case II. The solutions are of the form y 1 = m =0 a m x m and y 2 = y 1 ln x + m =0 b m x m , with a = 0. Let us determine y 1 . We use y instead of y to simplify the notation. We have y = m =0 a m x m ; y = m =0 ma m x m-1 ; y = m =0 m ( m-1) a m x m-2 ....
View Full Document

Ask a homework question - tutors are online