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taylor sin^2(x)

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

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laplace transform 15xcos(x)

Solution

−15(−s²+1)(s²+1)²

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L{15xcos(x)}

Use the constant multiplication property of Laplace Transform:
For function f(t) and constant a: L{a·f(t)}=a·L{f(t)}

=15L{xcos(x)}

L{xcos(x)}: −−s²+1(s²+1)²

=15(−−s²+1(s²+1)² )

15· (−−s²+1(s²+1)² ) פשט את: −15(−s²+1)(s²+1)²

=−15(−s²+1)(s²+1)²

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inverse laplace 15xcos(x)

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

taylor sin^2(x)

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