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transformadora de laplace sin^2(u)

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laplace transform sin²(u)

Solution

2s(s²+4)

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L{sin²(u)}

Use the following identity: sin²(x)=12 −cos(2x)12

=L{12 −cos(2u)12 }

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)}

=L{12 }−12 L{cos(2u)}

L{12 }: 12s

L{cos(2u)}: ss²+4

=12s −12 · ss²+4

Simplify 12s −12 · ss²+4 : 2s(s²+4)

=2s(s²+4)

Solve instead

inverse laplace sin²(u)

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+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

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transformadora de laplace sin^2(u)

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transformadora de laplace sin^2(u)

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