laplace transform 15xcos(x)

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(☐)^′	`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 )

Refine 15· (−−s²+1(s²+1)² ): −15(−s²+1)(s²+1)²

=−15(−s²+1)(s²+1)²

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inverse laplace 15xcos(x)

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×▭▭	(	1	2	3	−	`x`
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laplace transform 15xcos(x)

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