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xy=10, 2x+y=1

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

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taylor xe^(−x)

Solution

x−x²+12 x³−16 x⁴+124 x⁵+…

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Solution steps

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L⁻¹{x−32x }

x−32x קח את השבר החלקי של: 12 −32x

=L⁻¹{12 −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{12 }−L⁻¹{32x }

L⁻¹{12 }: 12 δ(t)

L⁻¹{32x }: 32 H(t)

=12 δ(t)−32 H(t)

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

xy=10, 2x+y=1

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