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axis y=x^2+2

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inverse laplace (1−x)e^x

Solution

H(t+1)(δ(t+1)−δ′(t+1))

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Solution steps

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L⁻¹{e^x(−x+1)}

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 (1−x)e^x: F(x)=(1−x), a=−1

=H(t+1)L⁻¹{(1−x)}(t+1)

L⁻¹{(1−x)}: δ(t)−δ′ (t)

=H(t+1)(δ(t+1)−δ′ (t+1))

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

axis y=x^2+2

Solution

Solution steps

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