inverse of laplace (x^2-5x)/(x-5)

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

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log_□

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ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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inverse laplace x²−5xx−5

Solution

δ′(t)+5δ(t)−5

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L⁻¹{x²−5xx−5 }

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=L⁻¹{x²x−5 −5xx−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{x²x−5 }−5L⁻¹{xx−5 }

L⁻¹{x²x−5 }: δ′ (t)+5δ(t)+25e^5t

L⁻¹{xx−5 }: e^5t· 5+1

=δ′ (t)+5δ(t)+25e^5t−5(e^5t· 5+1)

Simplify δ′ (t)+5δ(t)+25e^5t−5· (e^5t· 5+1): δ′ (t)+5δ(t)−5

=δ′ (t)+5δ(t)−5

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

inverse of laplace (x^2-5x)/(x-5)

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