的拉普拉斯逆变换 (2x-5)e^x

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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 (2x−5)e^x

Solution

H(t+1)(2δ′(t+1)−5δ(t+1))

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Solution steps

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L⁻¹{e^x(2x−5)}

Apply inverse transform rule: if L⁻¹{F(s)}=f(t) then L⁻¹{e^−asF(s)}=H(t−a)f(t−a)

Where H(t) is Heaviside step function

For (2x−5)e^x: F(x)=(2x−5), a=−1

=H(t+1)L⁻¹{(2x−5)}(t+1)

L⁻¹{(2x−5)}: 2δ′ (t)−5δ(t)

=H(t+1)(2δ′ (t+1)−5δ(t+1))

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Notebook

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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 (2x-5)e^x

Solution

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