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inverse of laplace (x+2)/(x^3+8)

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inverse laplace x+2x³+8

Solution

1√3 e^tsin(√3t)

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L⁻¹{x+2x³+8 }

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=L⁻¹{xx³+8 +2x³+8 }

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{xx³+8 }+2L⁻¹{1x³+8 }

L⁻¹{xx³+8 }: −16 e^−2t+16 e^tcos(√3t)+12√3 e^tsin(√3t)

L⁻¹{1x³+8 }: 112 e^−2t−112 e^tcos(√3t)+14√3 e^tsin(√3t)

=−16 e^−2t+16 e^tcos(√3t)+12√3 e^tsin(√3t)+2(112 e^−2t−112 e^tcos(√3t)+14√3 e^tsin(√3t))

Simplify −16 e^−2t+16 e^tcos(√3t)+12√3 e^tsin(√3t)+2· (112 e^−2t−112 e^tcos(√3t)+14√3 e^tsin(√3t)): 1√3 e^tsin(√3t)

=1√3 e^tsin(√3t)

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

inverse of laplace (x+2)/(x^3+8)

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