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inversa de laplace ((x^3-21x))/((5+4x-x^2))

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inverse laplace (x³−21x)(5+4x−x²)

Solution

−δ′(t)−4δ(t)+103 e^−t−103 e^5t

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L⁻¹{(x³−21x)(5+4x−x²) }

Take the partial fraction of x³−21x5+4x−x² : −x−4+103(x+1) −103(x−5)

=L⁻¹{−x−4+103(x+1) −103(x−5) }

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{4}+L⁻¹{103(x+1) }−L⁻¹{103(x−5) }

L⁻¹{x}: δ′ (t)

L⁻¹{4}: 4δ(t)

L⁻¹{103(x+1) }: 103 e^−t

L⁻¹{103(x−5) }: 103 e^5t

=−δ′ (t)−4δ(t)+103 e^−t−103 e^5t

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

inversa de laplace ((x^3-21x))/((5+4x-x^2))

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