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inverse of laplace s/(s^2+4s+5)

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inverse laplace ss²+4s+5

Solution

e^−2tcos(t)−2e^−2tsin(t)

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L⁻¹{ss²+4s+5 }

Expand ss²+4s+5 : s+2(s+2)²+1 −2· 1(s+2)²+1

=L⁻¹{s+2(s+2)²+1 −2· 1(s+2)²+1 }

Use the linearity property of Inverse Laplace Transform:
For functions f(s), g(s) and constants a, b: L⁻¹{a·f(s)+b·g(s)}=a·L⁻¹{f(s)}+b·L⁻¹{g(s)}

=L⁻¹{s+2(s+2)²+1 }−2L⁻¹{1(s+2)²+1 }

L⁻¹{s+2(s+2)²+1 }: e^−2tcos(t)

L⁻¹{1(s+2)²+1 }: e^−2tsin(t)

=e^−2tcos(t)−2e^−2tsin(t)

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

inverse of laplace s/(s^2+4s+5)

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