**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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