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auwally

  • 3 years ago

laplace transform of u"-2u'+2u=e^(-t)

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  1. across
    • 3 years ago
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    Are you given initial conditions?

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

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