Using the input integral principle below ccL[int_0^t f(u)du](s)=1/s ccL[f(t)](s), s>c Find ccL^(-1)[(1)/(s(s^2+1))](t) without using partial fractions.

easternerjx 2022-09-25 Answered
Using the input integral principle below
L [ 0 t f ( u ) d u ] ( s ) = 1 s L [ f ( t ) ] ( s ) ,   s > c
Find L 1 [ 1 s ( s 2 + 1 ) ] ( t )   without using partial fractions.
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Answers (2)

Katelyn Chapman
Answered 2022-09-26 Author has 13 answers
Compare these
L [ 0 t f ( u ) d u ] ( s ) = 1 s L [ f ( t ) ] ( s )
L [ 0 t sin u   d u ] ( s ) = 1 s 1 ( s 2 + 1 )
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GepGreeloCesyjk
Answered 2022-09-27 Author has 1 answers
Hint: The implied way is
L [ 0 t f ( u ) d u ] ( s ) = 1 s L [ f ( t ) ] ( s ) = 1 s 1 s 2 + 1 L [ f ( t ) ] ( s ) = 1 s 2 + 1 .
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