Solution: At first let us find the complementary function, that is, the solution corresponding to the homogeneous equation. The corresponding homogeneous equation will be \[ y'' - y' - 2y = 0. \] The characteristic equation is \begin{align*} m^2 - m - 2 = 0 \implies (m-2)(m+1) = 0. \end{align*} Thus, the solution will be \begin{equation}\label{eq:30Aug2024-1} y_{\mathrm{CF} } = c_1 e^{2x} + c_2 e^{-x}. \end{equation}

Now we will use the method of undetermined coefficient to find the particular integral. Since the right hand side of the given differential equation is a second order polynomial, we assume that \begin{equation}\label{eq:30Aug2024-2} y_{\mathrm{PI} } = a_0 + a_1 x + a_2 x^2, \quad a_0, a_1, a_2 \in \mathbb{R} . \end{equation} Thus, \begin{align*} & y_{\mathrm{PI} }' = a_1 + 2a_2 x \\ & y_{\mathrm{PI} }'' = 2a_2. \end{align*} Substituting these values in the given differential equation, we get \begin{align*} & 2a_2 - \left( a_1 + 2a_2 x \right) - 2 \left( a_0 + a_1 x + a_2 x^2 \right) = 4x^2 \\ \implies & -2a_2 x^2 + \left( -2a_2 - 2a_1 \right) x + 2a_2 - a_1 - 2a_0 = 4x^2 . \end{align*} Equation the coefficients of same power of $x$, we get \[ -2a_2 = 4, \quad -2a_2 - 2a_1 = 0,\quad 2a_2 - a_1 - 2a_0 = 0. \] Solving the above system we obtain \begin{align*} a_2 = -2, \quad a_1 = 2 \quad \text{ and } \quad a_0 = -3. \end{align*} Hence, \eqref{eq:30Aug2024-2} becomes \[ y_{\mathrm{PI} } = -2x^2 + 2x - 3, \] and the general solution will be \[ y(x) = c_1 e^{2x} + c_2 e^{-x} -2x^2 + 2x - 3. \]