的拉普拉斯逆变换 1/(x^{3/2)}

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

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 1x³²

Solution

2√t√π

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Solution steps

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L⁻¹{1x³² }

=L⁻¹{2√π · (2· 1−1)!!√π2¹x¹⁺¹² }

Use the constant multiplication property of Inverse Laplace Transform:
For function f(t) and constant a: L⁻¹{a·f(t)}=a·L⁻¹{f(t)}

=2√π L⁻¹{(2· 1−1)!!√π2¹x¹⁺¹² }

Use Inverse Laplace Transform table: L⁻¹{(2n−1)!!√π2ⁿsⁿ⁺¹² }=tⁿ⁻¹²

L⁻¹{(2· 1−1)!!√π2¹x¹⁺¹² }=√t

=2√π √t

=2√t√π

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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 1/(x^{3/2)}

Solution

Solution steps

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