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anonymous
 4 years ago
laplace transform of u"2u'+2u=e^(t)
anonymous
 4 years ago
laplace transform of u"2u'+2u=e^(t)

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across
 4 years ago
Best ResponseYou've already chosen the best response.0Are you given initial conditions?

across
 4 years ago
Best ResponseYou've already chosen the best response.0Let\[u''2u'+2u=e^{t}.\]Then,\[\mathcal{L}\{u''\}2\mathcal{L}\{u'\}+2\mathcal{L}\{u\}=\mathcal{L}\{e^{t}\},\\s^2U(s)su(0)u'(0)2sU(s)+2u(0)+2U(s)=\frac1{s+1},\\U(s)=\frac{u'(0)}{s^22s+2}+\frac{su(0)}{s^22s+2}\frac{2u(0)}{s^22s+2}+\frac{3s}{5(s^22s+2)}+\frac{1}{5(s+1)}.\]Here, \(u'(0)\) and \(u(0)\) are given. Can you find the inverse Laplace transform of this?
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