Let's solve for \( Y[s] \), the Laplace transform of the solution \( y(t) \): Given: \[ y'' + 6y = t^2 - 1; \quad y(0) = 0,\, y'(0) = -1 \] ### 1. **Take the Laplace Transform of Both Sides** Recall: - \( \mathcal{L}\{y''\} = s^2 Y(s) - s y(0) - y'(0) \) - \( \mathcal{L}\{y\} = Y(s) \) - \( \mathcal{L}\{t^2\} = \frac{2}{s^3} \) - \( \mathcal{L}\{1\} = \frac{1}{s} \) Apply Laplace transform: \[ \mathcal{L}\{y'' + 6y\} = \mathcal{L}\{t^2 - 1\} \] \[ (s^2 Y(s) - s y(0) - y'(0)) + 6Y(s) = \frac{2}{s^3} - \frac{1}{s} \] Plug in initial conditions (\( y(0) = 0 \), \( y'(0) = -1 \)): \[ (s^2 Y(s) - 0 - (-1)) + 6Y(s) = \frac{2}{s^3} - \frac{1}{s} \] \[ (s^2 Y(s) + 1) + 6Y(s) = \frac{2}{s^3} - \frac{1}{s} \] \[ (s^2 + 6) Y(s) + 1 = \frac{2}{s^3} - \frac{1}{s} \] ### 2. **Solve for \( Y(s) \)** \[ (s^2 + 6)Y(s) = \frac{2}{s^3} - \frac{1}{s} - 1 \] \[ Y(s) = \frac{2}{s^3} \cdot \frac{1}{s^2 + 6} - \frac{1}{s} \cdot \frac{1}{s^2 + 6} - \frac{1}{s^2 + 6} \] Or, combine as a single fraction: \[ Y(s) = \frac{2}{s^3(s^2 + 6)} - \frac{1}{s(s^2 + 6)} - \frac{1}{s^2 + 6} \] --- **Final Answer:** \[ \boxed{ Y(s) = \frac{2}{s^3(s^2 + 6)} - \frac{1}{s(s^2 + 6)} - \frac{1}{s^2 + 6} } \] This is the Laplace transform of the solution to your initial value problem.