Apply the Laplace transform to the differential equation, and solve for $Y(s)$ $y'' + 9y = \begin{cases} 5e^{2t} & 0 < t < 6 \\ 0 & t > 6 \end{cases}$, $y(0) = 4, y'(0) = 0$ $Y(s) = $
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Step 1: Take the Laplace transform of the given differential equation: $\mathcal{L}\{y'' + 9y\} = \mathcal{L}\left\{ \begin{cases} 5e^{2t} & 0 < t < 6 \\ 0 & t > 6 \end{cases} \right\}$ Using the linearity property of the Laplace transform, we Show more…
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