Chem Differential Eq HW Solutions Fall 2011 197

Chem Differential Eq HW Solutions Fall 2011 197 - Section...

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 A197 Solutions to Exercises A.6 1. For the equation y ±± +(1 - x 2 ) y ± + xy =0, p ( x )=1 - x 2 and q ( x )= x are both analytic at a = 0. So a = 0 is a ordinary point. 5. For the equation x 2 y ±± - e x ) y ± + xy =0 , p ( x 1 - e x x 2 ,x p ( x 1 - e x x ; q ( x 1 x 2 q ( x x. 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. To see that xp ( x ) is analytic at 0, derive its Taylor series as follows: for all x , e x =1 + x + x 2 2! + x 3 3! + ··· 1 - e x = - x - x 2 2! - x 3 3! + = x ± - 1 - x 2! - x 2 3! + ² 1 - e x x = - 1 - x 2! - x 2 3! + . Since 1 - e x x has a Taylor series expansion about 0 (valid for all x ), it is analytic at 0. 9. For the equation 4 x 2 y ±± - 14 xy ± + (20 - x ) y , p ( x - 7 2 x p ( x - 7 2 ,p 0 = - 7 2 ; q ( x 20 - x 4 x 2 2 q ( x )=5 - x 4 ,q 0 =5 . 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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

Ask a homework question - tutors are online