Solution: Let $F(s) = \mathcal{L} (f)$ denotes the Laplace transformation of the function $f(t)$. We recall that \[ \mathcal{L} (f^\prime (t)) = s F(s) - f(0). \] Using the above, \begin{align*} \mathcal{L} \left( y^\prime (t) \right) = \mathcal{L} (3 - 2t) & \implies s Y(s) - y(0) = \mathcal{L} (3) - 2 \mathcal{L} (t) \\ & \implies s Y(s) = \frac{3}{s} - \frac{2}{s^2} \\ & \implies Y(s) = \frac{3}{s^2} - \frac{2}{s^3} \\ & \implies y(t) = \mathcal{L} ^{-1} \left( \frac{3}{s^2} - \frac{2}{s^3} \right) \\ & \implies y(t) = \mathcal{L} ^{-1} \left( \frac{3}{s^2} \right) - \mathcal{L} ^{-1} \left( \frac{2}{s^3} \right) \\ & \implies y(t) = 3t - t^2. \end{align*} Thus, the solution the differential equation \eqref{eq:24Nov2023-1} \[ y(t) = 3t - t^2. \]