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laplace ngược (x^3-x)/(x^2-1)

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inverse laplace x³−xx²−1

Solution

δ′(t)+12 e^−t+12 e^t−cosh(t)

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L⁻¹{x³−xx²−1 }

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=L⁻¹{x³x²−1 −xx²−1 }

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{x³x²−1 }−L⁻¹{xx²−1 }

L⁻¹{x³x²−1 }: δ′ (t)+12 e^−t+12 e^t

L⁻¹{xx²−1 }: cosh(t)

=δ′ (t)+12 e^−t+12 e^t−cosh(t)

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

laplace ngược (x^3-x)/(x^2-1)

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