inversa de laplace (x-3)/(5x-15)

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

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(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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inverse laplace x−35x−15

Solution

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0.2

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L⁻¹{x−35x−15 }

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=L⁻¹{x5x−15 −35x−15 }

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{x5x−15 }−3L⁻¹{15x−15 }

L⁻¹{x5x−15 }: 35 e^3t+15

L⁻¹{15x−15 }: 15 e^3t

=35 e^3t+15 −3· 15 e^3t

Refine 35 e^3t+15 −3· 15 e^3t: 15

=15

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

inversa de laplace (x-3)/(5x-15)

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