d/(dn)(2^{n^{11}}-2)

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

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≥	≤	·	÷	`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 10x−1914x−10

Solution

57 δ(t)−8398 e^5t7

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Solution steps

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L⁻¹{10x−1914x−10 }

Take the partial fraction of 10x−1914x−10 : 57 −837(14x−10)

=L⁻¹{57 −837(14x−10) }

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{57 }−L⁻¹{837(14x−10) }

L⁻¹{57 }: 57 δ(t)

L⁻¹{837(14x−10) }: 8398 e^5t7

=57 δ(t)−8398 e^5t7

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

d/(dn)(2^{n^{11}}-2)

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Solution steps

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