Determine a transformada de Laplace da função f(t) = e2t(sen(3t) - sen(2t)).

F(s)=3s3−4s+13+2s2−4s+8F(s)=3s3−4s+13+2s2−4s+8

F(s)=3s3−4s+13−2s2−4s+8F(s)=3s3−4s+13−2s2−4s+8

F(s)=3s2−4s+9+2s2−4s+8F(s)=3s2−4s+9+2s2−4s+8

F(s)=3s2−4s+4−2s2−4s+8F(s)=3s2−4s+4−2s2−4s+8

F(s)=3s2−4s+13−2s2−4s+8

Determine a transformada de Laplace da função f(t) = e2t(sen(3t) - sen(2t)).

		F(s)=3s3−4s+13+2s2−4s+8F(s)=3s3−4s+13+2s2−4s+8
		F(s)=3s3−4s+13−2s2−4s+8F(s)=3s3−4s+13−2s2−4s+8
		F(s)=3s2−4s+9+2s2−4s+8F(s)=3s2−4s+9+2s2−4s+8
		F(s)=3s2−4s+4−2s2−4s+8F(s)=3s2−4s+4−2s2−4s+8
		F(s)=3s2−4s+13−2s2−4s+8

Cálculo III

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ESTÁCIO

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5

Givaldosamba Givaldo Sambadoval

02.06.2019

💡 5 Respostas

Gabriel Bisson

04.06.2019

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0

Andre Smaira

30.09.2019

Temos a função

f(t)=e^{2t}(sen(3t)-sen(2t))=e^{2t}sen(3t)-e^{2t}sen{2t}

. Assim, como a transformada de Laplace é linear,

\mathcal{L}(f(t))=\mathcal{L}(e^{2t}sen(3t))-\mathcal{L}(e^{2t}sen(2t))

. Além disso, temos que

\mathcal{L}(e^{ct}f(t))=F(s-c)

e que

\mathcal{L}(sen(at))=\dfrac{a}{s^2+a^2}

.

Utilizando essas duas propriedades, obtemos $\mathcal{L}(e^{ct}sen(at))=\dfrac{a}{(s-c)^2+a^2}$ . Portanto:

$\mathcal{L}(f(t))=\mathcal{L}(e^{2t}sen(3t))-\mathcal{L}(e^{2t}sen(2t))=\dfrac{3}{(s-2)^2+9}-\dfrac{2}{(s-2)^2+4}=\dfrac{3(s^2-4s+8)-2(s^2-4s+13)}{(s^2-4s+13)(s^2-4s+8)}=\dfrac{s^2-4s-2}{(s^2-4s+13)(s^2-4s+8)}$ .

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