## auwally Group Title laplace transform of u"-2u'+2u=e^(-t) 2 years ago 2 years ago

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1. across

Are you given initial conditions?

2. across

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?