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Math 14 - 8 Let C(y = Y Then £0" — 2y =...

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Unformatted text preview: 8. Let C(y) = Y. Then £0" — 2y) = £(e"’ cost). Note the transform pair. s cost <—> s2+1 Then by Proposition 2.12, “ cost 4+ S +1 e -+—————~. (5 +1)2 +1 Thus, 5 +1 L ’ — 2L? — (y) (y) (5+1)2+1 3 +1 ~ 0 ‘— 2Y — s Y<s> y< ) (s) (S 1)2 +1 But y(0) = —2, so 3 +1 ~ 2 Y 2 = —-————~— (s )(S)+ (S+1)2+1 s +1 ,2 Y“) = s—2+ ((s+1)2+1)(s—2)' Find a partial fraction decomposition. s+1 _ A + Bs+C (3—2)((s+1)2+1)_s-—2 (s+1)2+1 Equating numerators, 5+1=A((s+1)2+1)+(Bs+C)(s—2) =(A+B)s2+(2A—-ZB+C)S +(2A—2C) Thus, A+B=0 2A—2B+C=1 2A—2C=1, and A = 3/10, B = —3/10, and C 2 ~1/5. Thus, ”2 + 3/10 (3/10)s —1/5_ 5—2 5—2 (s+1)2+1 —17/10 —(_3/10)s — 1/5 [5—2 + (s+1)2+1 Y(s) = Note the transform pairs 5 s2+1 1 32+1 COSI 6 sint <—> $3: ,9? and 29 29 2 t —_ —t 2t 2; YU—Sle -~81e +-27te 1 2 2: 1 3 2 _ ”t _ t 9 e + 92‘ e 1 Proposition 2.12 provides 1 e" cost <-> L (s + 1)2 + 1 1 e“ sint <—> ————~———. (s +1)2 +1 __:_11 1 ”:1 s+2/3 m)- 10 ( —2) 10((s+1)2+1) 1 _3__( 3+1 ) s—-2 10 (s+1)2+1 ...
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