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We will solve the initial value problem using the Laplace transform method numerically. ### Given Differential Equation: \[ y'' - 6y' + 5y = 0 \] with initial conditions: \[ y(0) = 1, \quad y'(0) = -3 \] ### Step 1: Take the Laplace Transform Taking the Laplace transform of both sides and using the properties: \[ \mathcal{L} \{ y'' \} = s^2 Y(s) - sy(0) - y'(0) \] \[ \mathcal{L} \{ y' \} = s Y(s) - y(0) \] \[ \mathcal{L} \{ y \} = Y(s) \] Applying these, \[ (s^2 Y(s) - s(1) - (-3)) - 6(s Y(s) - 1) + 5Y(s) __________ ___ _____ _____ _____.
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