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axis y=6x^2

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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foci y=x²+4

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L⁻¹{x³4−x² }

x³4−x² خذ الكسر الجزئي لـ: −x−2x+2 −2x−2

=L⁻¹{−x−2x+2 −2x−2 }

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{x}−L⁻¹{2x+2 }−L⁻¹{2x−2 }

L⁻¹{x}: δ′ (t)

L⁻¹{2x+2 }: 2e^−2t

L⁻¹{2x−2 }: 2e^2t

=−δ′ (t)−2e^−2t−2e^2t

Order of operations multi-step without parentheses Examples

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

axis y=6x^2

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