inverse of laplace ((x^3+9))/(x^2+3x+2)

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inverse laplace (x³+9)x²+3x+2

Solution

δ′(t)−3δ(t)+8e^−t−e^−2t

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L⁻¹{(x³+9)x²+3x+2 }

Take the partial fraction of x³+9x²+3x+2 : x−3+8x+1 −1x+2

=L⁻¹{x−3+8x+1 −1x+2 }

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{x}−L⁻¹{3}+L⁻¹{8x+1 }−L⁻¹{1x+2 }

L⁻¹{x}: δ′ (t)

L⁻¹{3}: 3δ(t)

L⁻¹{8x+1 }: 8e^−t

L⁻¹{1x+2 }: e^−2t

=δ′ (t)−3δ(t)+8e^−t−e^−2t

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

inverse of laplace ((x^3+9))/(x^2+3x+2)

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