的拉普拉斯逆变换 (26)/((x-1)(x^2+25))

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

☐²

x^☐

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^□√☐

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log_□

θ

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ddx

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

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inverse laplace 26(x−1)(x²+25)

Solution

e^t−cos(5t)−15 sin(5t)

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L⁻¹{26(x−1)(x²+25) }

Take the partial fraction of 26(x−1)(x²+25) : 1x−1 +−x−1x²+25

=L⁻¹{1x−1 +−x−1x²+25 }

Expand

=L⁻¹{1x−1 −xx²+25 −1x²+25 }

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{1x−1 }−L⁻¹{xx²+25 }−L⁻¹{1x²+25 }

L⁻¹{1x−1 }: e^t

L⁻¹{xx²+25 }: cos(5t)

L⁻¹{1x²+25 }: 15 sin(5t)

=e^t−cos(5t)−15 sin(5t)

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

的拉普拉斯逆变换 (26)/((x-1)(x^2+25))

Solution

Solution steps

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