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-13^{-1} \mod83

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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inverser laplace (2x−5)e^x

Solution

H(t+1)(2δ′(t+1)−5δ(t+1))

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L⁻¹{e^x(2x−5)}

Appliquer la règle de la transformée inverse : si L⁻¹{F(s)}=f(t) alors L⁻¹{e^−asF(s)}=H(t−a)f(t−a)

Où H(t) est la fonction échelon de Heaviside

Pour (2x−5)e^x: F(x)=(2x−5), a=−1

=H(t+1)L⁻¹{(2x−5)}(t+1)

L⁻¹{(2x−5)}: 2δ′ (t)−5δ(t)

=H(t+1)(2δ′ (t+1)−5δ(t+1))

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

-13^{-1} \mod83

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