的拉普拉斯逆变换 5/(4x^2+1)+3/(x^3)-5 3/(2x)

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

☐²

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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 54x²+1 +3x³ −532x

Solution

52 sin(t2 )+3t²2 −152 H(t)

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L⁻¹{54x²+1 +3x³ −5· 32x }

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{54x²+1 }+L⁻¹{3x³ }−5L⁻¹{32x }

L⁻¹{54x²+1 }: 52 sin(t2 )

L⁻¹{3x³ }: 3t²2

L⁻¹{32x }: 32 H(t)

=52 sin(t2 )+3t²2 −5· 32 H(t)

Refine 52 sin(t2 )+3t²2 −5· 32 H(t): 52 sin(t2 )+3t²2 −152 H(t)

=52 sin(t2 )+3t²2 −152 H(t)

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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 5/(4x^2+1)+3/(x^3)-5 3/(2x)

Solution

Solution steps

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