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laplace ngược (4x)/(2-8x^2)

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inverse laplace 4x2−8x²

Solution

−14 e^−t2−14 e^t2

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L⁻¹{4x2−8x² }

Take the partial fraction of 4x2−8x² : −12(2x+1) −12(2x−1)

=L⁻¹{−12(2x+1) −12(2x−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{12(2x+1) }−L⁻¹{12(2x−1) }

L⁻¹{12(2x+1) }: 14 e^−t2

L⁻¹{12(2x−1) }: 14 e^t2

=−14 e^−t2−14 e^t2

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

laplace ngược (4x)/(2-8x^2)

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