inverse of laplace (7x^2-9)/(4x+3)

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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 7x²−94x+3

Solution

Undefined−2116 δ(t)−8164 e^−3t4

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L⁻¹{7x²−94x+3 }

Take the partial fraction of 7x²−94x+3 : 7x4 −2116 −8116(4x+3)

=L⁻¹{7x4 −2116 −8116(4x+3) }

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{7x4 }−L⁻¹{2116 }−L⁻¹{8116(4x+3) }

L⁻¹{7x4 }: Undefined

L⁻¹{2116 }: 2116 δ(t)

L⁻¹{8116(4x+3) }: 8164 e^−3t4

=Undefined−2116 δ(t)−8164 e^−3t4

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

inverse of laplace (7x^2-9)/(4x+3)

Solution

Solution steps

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