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

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inverse laplace 5x²−8x−13x²−5

Solution

5δ(t)−2(2√5+3)√5 e^−√5t+2(3−2√5)√5 e^√5t

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L⁻¹{5x²−8x−13x²−5 }

5x²−8x−13x²−5 خذ الكسر الجزئي لـ: 5−2(2√5+3)√5(x+√5) +2(3−2√5)√5(x−√5)

=L⁻¹{5−2(2√5+3)√5(x+√5) +2(3−2√5)√5(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{5}−L⁻¹{2(2√5+3)√5(x+√5) }+L⁻¹{2(3−2√5)√5(x−√5) }

L⁻¹{5}: 5δ(t)

L⁻¹{2(2√5+3)√5(x+√5) }: 2(2√5+3)√5 e^−√5t

L⁻¹{2(3−2√5)√5(x−√5) }: 2(3−2√5)√5 e^√5t

=5δ(t)−2(2√5+3)√5 e^−√5t+2(3−2√5)√5 e^√5t

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

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