3 to say what the two nontrivial linearly independent

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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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TEST3/MAP2302 Page 4 of 4 ______________________________________________________________________ 8. (15 pts.)
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