de-t3-a

# The equation below has a regular singular point at x

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The equation below has a regular singular point at x 0 = 0. x 2 y xy ( x 2 1) y 0 (a) Obtain the indicial equation for the ODE at x 0 = 0 and its two roots. (b) Then use all the information available and Theorem 6.3 to say what the two nontrivial linearly independent solutions given by theorem look like without attempting to obtain the coefficients of the power series involved. To determine r , you need the indicial equation at x 0 = 0 and its roots. Now the indicial equation is r ( r -1) + p 0 r + q 0 = 0 where p 0 lim x 0 x x x 2 1 and q 0 lim x 0 x 2 x 2 1 x 2 1. Thus, the indicial equation is r 2 - 2 r + 1 = 0, with a single root r 1 = 1 with multiplicity two. Consequently the two linearly independent solutions provided by Theorem 6.3 look like the following: and y 1 ( x ) x 1 n 0 c n x n y 2 ( x ) x 2 n 0 d n x n y 1 ( x )ln x ______________________________________________________________________ 6. (10 pts.) Compute when f ( t ) 1 { F ( s )}( t ) (a) F ( s ) 7 (2 s 1) 3 ( Work ) 7 8( s ( 1/2)) 3 1 { F ( s )}( t ) 7 8 1 2 t 2 e (1/2) t 7 16 t 2 e (1/2) t (b) F ( s ) 2 s 12 s 2 6 s 13 ( Work ) 2 s 12 ( s 3) 2 4 2( s 3) 6 ( s 3) 2 2 2 1 { F ( s )}( t ) 2 e 3 t cos(2 t ) 3 e 3 t sin(2 t ) ’Tis the usual prestidigitation of multiplication by ’1’ in the correct form or the addition of ’0’ suitably transmogrified. I move that we table the motion. Do I hear a second? ______________________________________________________________________ 7. (5 pts.) Circle the letter corresponding to the correct response: If ,then F ( s ) = F ( s ) { t 2 sin( bt )}( s ) (a) (b) 2 s 3 b ( s 2 b 2 ) d ds 2 bs ( s 2 b 2 ) 2 (c) (d) d 2 ds 2 b s 2 b 2 2 bs s 2 ( s 2 b 2 ) 2 (e) None of the above. Obviously (a) and (d) are utter nonsense since the transform of a product is NOT the product of the transforms. (c) is a near miss since the sign is wrong. It turns out that (b) provides the correct answer.

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The equation below has a regular singular point at x 0 x 2...

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