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Chem Differential Eq HW Solutions Fall 2011 197

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

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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
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