Solution: The given differential equation is \begin{equation}\label{eq:07June2024-1} (3x + 2)^2 \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 5(3x + 2) \frac{\mathrm{d}y}{\mathrm{d}x} -3y = x^2 + x + 1. \end{equation} In order to solve \eqref{eq:07June2024-1}, we first substitute \begin{align*} z = \ln (3x + 2) & \implies 3x + 2 = e^z \\ & \implies \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}z} \frac{\mathrm{d}z}{\mathrm{d}x} \\ & \implies \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{3x+2} \frac{\mathrm{d}y}{\mathrm{d}z}. \end{align*}

Similarly, \begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} & = \frac{\mathrm{d}^2y}{\mathrm{d}z^2} \left( \frac{\mathrm{d}z}{\mathrm{d}x} ^2 \right) + \frac{\mathrm{d}^2z}{\mathrm{d}x^2} \frac{\mathrm{d}y}{\mathrm{d}z} \\ & = \frac{9}{(3x+2)^2} \frac{\mathrm{d}^2y}{\mathrm{d}z^2} - \frac{9}{(3x+2)^2}\frac{\mathrm{d}y}{\mathrm{d}z}. \end{align*} We will write $\frac{d}{dz} \eqqcolon D$. Therefore, \eqref{eq:07June2024-1} can be rewritten as \begin{align*} & \left[ 9\left( D^2 - D \right) + 15D - 3 \right] y = \left( \frac{e^z - 2}{3} \right) ^2 + \left( \frac{e^z - 2}{3} \right) + 1 \\[2ex] \implies & \left( 9D^2 +6D - 3 \right) y = \left( \frac{e^z - 2}{3} \right) ^2 + \left( \frac{e^z - 2}{3} \right) + 1. \end{align*} Therefore, we get a simplified equation as \begin{equation}\label{eq:07June2024-2} \left( 9D^2 + 6D - 3 \right) y = \left( \frac{e^z - 2}{3} \right) ^2 + \left( \frac{e^z - 2}{3} \right) + 1. \end{equation}

The auxiliary equation corresponding to \eqref{eq:07June2024-2} is \[ 9m^2 + 6m - 3 = 0 \implies 3(3m - 1)(m+1) = 0 \implies m = \frac{1}{3},-1. \] So, the complimentary function, which is the solution to the corresponding homogeneous equation, will be \begin{align*} y_{\mathsf{cf} } & = c_1e^{-z} + c_2 e^{\frac{z}{3}} \\ & = c_1 (3x + 2)^{-1} + c_2 (3x + 2)^\frac{1}{3}. \end{align*} Thus, \begin{equation}\label{eq:07June2024-3} y_{\mathsf{cf} } = \frac{c_1}{3x + 2} + c_2 \sqrt[3]{3x+2}. \end{equation} Here one can use the method of undetermined coefficient to find the particular solution. The particular solution will be \[ y_{\mathsf{PI} } = \frac{A}{9}e^{2z} + \frac{B}{9} e^z + 7C. \] Now we can substitute this into the differential equation, to obtain the value of $A,B$ and $C$.

The other way to solve the particular integral. To find the particular integral let us first simply the right hand side of \eqref{eq:07June2024-2}. \begin{align*} \left( \frac{e^z - 2}{3} \right) ^2 + \left( \frac{e^z - 2}{3} \right) + 1 & = \frac{e^{2z} - e^z + 7}{9}. \end{align*} Now the particular integral will be \begin{align*} y_{\mathsf{PI} } & = \frac{e^{2z} - e^z + 7}{9\left( 9D^2 +6D - 3 \right)} \\[1ex] & = \frac{1}{27(3D-1)(D+1)}\left( e^{2z} - e^z + 7 \right). \end{align*} We recall that if the particular integral has the form $ \frac{1}{f(D)}e^{ax}$, then \[ \frac{1}{f(D)}e^{ax} = \frac{1}{f(a)}e^{ax}, \text{ when } f(a) \neq 0. \] If $f(D) = 27(3D-1)(D+1)$, then \begin{align*} y_{\mathsf{PI} } & = \frac{1}{f(2)}e^{2z} - \frac{1}{f(1)}e^z + \frac{7}{f(0)} \\ & = \frac{e^{2z}}{27\times 15} - \frac{e^z}{27\times 4} + \frac{7}{-27} \\ & = \frac{e^{2z}}{405} - \frac{e^z}{108} - \frac{7}{27} \\ & = \frac{(3x + 2)^2}{405} - \frac{3x + 2}{108} - \frac{7}{27}. \end{align*} Hence, the final solution will be \[ y(x) = y_{\mathsf{cf} } + y_{\mathsf{PI} }. \]