Ordinary Diff Eq Exam Review Solutions 70

Ordinary Diff Eq Exam Review Solutions 70 - 72 1 Solutions...

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72 1 Solutions s + 1 ( s 2 + 4 s + 5) 2 ( s 2 + 4 s + 6) 2 = s + 1 ( s 2 + 4 s + 6) 2 + 2 s + 2 s 2 + 4 s + 6 + s + 1 ( s 2 + 4 s + 5) 2 2 s + 2 s 2 + 4 s + 5 14. Using the hint, let u = s 2 . Then the rational function becomes u ( u + 5) 3 ( u + 6) 2 , which can be put in partial fraction form by the technique of Section 2.2. Use Theorem 1 of Section 2.2 to compute the ( u + 5) -chain: u ( u + 5) 3 ( u + 6) 2 = A 1 ( u + 5) 3 + p 1 ( u ) ( u + 5) 2 ( u + 6) 2 where A 1 = u ( u + 6) 2 u = 5 = 5 and p 1 ( u ) = 1 u + 5 ( u ( 5)( u + 6) 2 ) = 5 u 2 + 61 u + 180 u + 6 = (5 u + 36)( u + 5) u + 5 = 5 u + 36 : Continuing gives 5 u + 36 ( u + 5) 2 )( u + 6) 2 = A 2 ( u + 5) 2 + p 2 ( u ) ( u + 5)( u + 6) 2 where A 2 = 5 u + 36 ( u + 6) 2 u = 5 = 11 and p 2 ( u ) = 1 u + 5 (5 u
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