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domain ofe indéfini cos(2t)

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inverser laplace 10(x−1)(x²+9)

Solution

e^t−cos(3t)−13 sin(3t)

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L⁻¹{10(x−1)(x²+9) }

Prendre la fraction partielle de 10(x−1)(x²+9) : 1x−1 +−x−1x²+9

=L⁻¹{1x−1 +−x−1x²+9 }

Développer

=L⁻¹{1x−1 −xx²+9 −1x²+9 }

Utiliser la propriété linéaire de la transformée inverse de Laplace :
Pour les fonctions f(s), g(s) et les constantes a, b: L⁻¹{a·f(s)+b·g(s)}=a·L⁻¹{f(s)}+b·L⁻¹{g(s)}

=L⁻¹{1x−1 }−L⁻¹{xx²+9 }−L⁻¹{1x²+9 }

L⁻¹{1x−1 }: e^t

L⁻¹{xx²+9 }: cos(3t)

L⁻¹{1x²+9 }: 13 sin(3t)

=e^t−cos(3t)−13 sin(3t)

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

domain ofe indéfini cos(2t)

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