преобразование Лапласа xcos^2(x)

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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

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laplace transform xcos²(x)

Solution

12s² −−s²+42(s²+4)²

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Solution steps

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L{xcos²(x)}

Rewrite using trig identities

L{12 x+12 xcos(2x)}

Use the linearity property of Laplace Transform:
For functions f(t), g(t) and constants a, b: L{a·f(t)+b·g(t)}=a·L{f(t)}+b·L{g(t)}

=12 L{x}+12 L{xcos(2x)}

L{x}: 1s²

L{xcos(2x)}: −−s²+4(s²+4)²

=12 · 1s² +12 (−−s²+4(s²+4)² )

Refine 12 · 1s² +12 · (−−s²+4(s²+4)² ): 12s² −−s²+42(s²+4)²

=12s² −−s²+42(s²+4)²

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inverse laplace xcos²(x)

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

преобразование Лапласа xcos^2(x)

Solution

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