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Quiz-5-Math-20D-W09solutions

# Quiz-5-Math-20D-W09solutions - Quiz#5 Math 20D March 6 Two...

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Quiz #5 Math 20D March 6 Two problems ( § 5.4, #5 and § 5.6, #3). SOLUTIONS. 1. ( § 5.4, #5. 10 points) Find all singular points of the given equation and determine whether each one is regular or irregular: ( 1 - x 2 ) 2 y 00 + x (1 - x ) y 0 + (1 + x ) y = 0 . Solution : We rewrite this as (since ( 1 - x 2 ) 2 = (1 - x ) 2 (1 + x ) 2 ) 0 = y 00 + x (1 - x ) (1 - x 2 ) 2 y 0 + 1 + x (1 - x 2 ) 2 y = y 00 + x (1 - x ) (1 + x ) 2 y 0 + 1 (1 - x ) 2 (1 + x ) y. The singular points are x = - 1 and x = 1 . (You may write p ( x ) = x (1 - x )(1+ x ) 2 and q ( x ) = 1 (1 - x ) 2 (1+ x ) . ) (a) For x = - 1 we have ( x + 1) p ( x ) = ( x + 1) x (1 - x ) (1 + x ) 2 = x (1 - x ) (1 + x ) is not analytic near x = - 1 . (We do not need to compute ( x + 1) 2 q ( x ) . ) So x = - 1 is an irregular singular point. (b) For x = 1 we have ( x - 1) p ( x ) = ( x - 1) x (1 - x ) (1 + x ) 2 = - x (1 + x ) 2 is analytic near x = 1 . And we have ( x - 1) 2 q ( x ) = ( x - 1) 2 1 (1 - x ) 2 (1 + x ) = 1 1 + x which is also analytic near x = 1 . So x = 1 is a regular singular point. 2. § 5.6, #3. y = X n =0 a n x r + n y 0 = X n =0 ( r + n ) a n x r + n - 1 y 00 = X n =0 ( r + n - 1) ( r + n ) a n x r + n - 2 . Then xy 00 = X n =0 ( r + n - 1) ( r + n ) a n x r + n - 1 .

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Relabel xy 00 = X n = - 1 ( r + n ) ( r + n + 1) a n +1 x r + n . Then 0 = xy 00 + y = X n = - 1 ( r + n ) ( r + n + 1) a n +1 x r + n + X n =0 a n x r + n = ( r - 1) r · a 0 x r - 1 + X n =0 (( r + n ) ( r + n + 1) a n +1 + a n ) x r + n . Indicial equation (from n = - 1 case): ( r - 1) r = 0 . Roots: r 1 = 0 r 2 = 1 . The recurrence relation ( r + n ) ( r + n + 1) a n
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Quiz-5-Math-20D-W09solutions - Quiz#5 Math 20D March 6 Two...

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