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Math  /  Calculus

QuestionCompute the inverse Laplace transform: L−1{−s−6s2+4s+20e−2s}=\mathcal{L}^{-1}\left\{\frac{-s-6}{s^{2}+4 s+20} e^{-2 s}\right\}= □\square (Notation: write u(t−c)\mathbf{u}(\mathbf{t}-\mathbf{c}) for the Heaviside step function uc(t)u_{c}(t) with step at t=ct=c.)

Studdy Solution
Our **final answer** is: (−e−2(t−2)cos⁡(4(t−2))−e−2(t−2)sin⁡(4(t−2)))u(t−2) \left( -e^{-2(t-2)}\cos(4(t-2)) - e^{-2(t-2)}\sin(4(t-2)) \right) u(t-2)

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