Solution: The given differential equation can be written in terms of $D \equiv \frac{\mathrm{d}}{\mathrm{d}x} $ as \begin{equation}\label{eq:17Nov2023-1} \left( x^3 D^3 + 2x^2 D^2 + 2 \right)y = 10 \left( x + \frac{1}{x} \right). \end{equation} We substitute $x = e^z$ in to the given differential equation \eqref{eq:17Nov2023-1}. Let us simplify the derivatives. Note that \begin{align*} & z = \ln x \implies \frac{\mathrm{d}z}{\mathrm{d}x} = \frac{1}{x}, \\ & \frac{1}{x} = e^{-z} \implies \frac{\mathrm{d}}{\mathrm{d}z} \left( \frac{1}{x} \right) = - e^{-z} \end{align*} Using chain rule, we have \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{\mathrm{d}y}{\mathrm{d}z} \cdot \frac{\mathrm{d}z}{\mathrm{d}x} = \frac{1}{x} \frac{\mathrm{d}y}{\mathrm{d}z} \\[1ex] \frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}} & = \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) \\ & = \frac{\mathrm{d}}{\mathrm{d}z} \left( \frac{1}{x} \frac{\mathrm{d}y}{\mathrm{d}z} \right) \cdot \frac{\mathrm{d}z}{\mathrm{d}x} \\ & = \left( \frac{\mathrm{d}}{\mathrm{d}z} \left( \frac{1}{x} \right) \frac{\mathrm{d}y}{\mathrm{d}z} + \frac{1}{x} \frac{\mathrm{d}^{2} y}{\mathrm{d} z^{2}} \right) \cdot \frac{1}{x} \\ & = \left( -\frac{1}{x} \frac{\mathrm{d}y}{\mathrm{d}z} + \frac{1}{x}\frac{\mathrm{d}^{2} y}{\mathrm{d} z^{2}} \right) \cdot \frac{1}{x} \\ & = \frac{1}{x^2}\left( \frac{\mathrm{d}^{2} y}{\mathrm{d} z^{2}} - \frac{\mathrm{d}y}{\mathrm{d}z} \right) \\[1ex] \frac{\mathrm{d}^{3} y}{\mathrm{d} x^{3}} & = \frac{1}{x^3} \left( \frac{\mathrm{d}^{3} y}{\mathrm{d} z^{3}} - 3 \frac{\mathrm{d}^{2} y}{\mathrm{d} z^{2}} - 2 \frac{\mathrm{d}y}{\mathrm{d}z} \right) . \end{align*}

If we write $D_1 = \frac{\mathrm{d}}{\mathrm{d}z} $, then \eqref{eq:17Nov2023-1} becomes \begin{align*} & x^3 \cdot \left[ \frac{1}{x^3} \left( D_1^3 - 3D_1^2 - 2D_1 \right) y \right] + 2x^2 \left[ \frac{1}{x^2} \left( D_1^2 - D_1 \right) y \right] + 2y \\ & \kern 2cm = 10 \left( e^z + e^{-z} \right) \\ \implies & \left( D_1^3 - 3D_1^2 -2 D_1 + 2 D_1^2 - 2 D_1 + 2 \right) y = \left( e^z + e^{-z} \right) \\ \implies & \left( D_1^3 - D_1^2 + 2 \right) y = 10 \left( e^z + e^{-z} \right). \end{align*} Corresponding to the above differential equation, the auxiliary equation will be \[ m^3 - m^2 + 2 = 0 \implies (m+1)(m^2 - 2m + 2) = 0. \] Thus, the roots of the auxiliary equation will be $-1, 1 \pm \iota $. Hence, the complementary function (the solution corresponding to the homogeneous differential equation) will be \begin{equation}\label{eq:17Nov2023-2} y_h(x) = c_1 e^{-z} + e^z \left( c_2 \cos z + c_3 \sin z \right) = \frac{c_1}{x} + \frac{c_2 \cos \ln x + c_3 \sin \ln x}{x}. \end{equation}

Now we need to find the particular integral. For the function $10 e^z$, the particular integral will be \begin{align*} \frac{1}{\left( D_1 + 1 \right) \left( D_1^2 - 2D_1 + 2 \right) }(10 e^z) & = \frac{10}{(1 + 1)(1^2 - 2 + 2)}e^z \\ & = 5e^z = 5x. \end{align*} Similarly, for $-10e^{-z}$, the particular integral will be \begin{align*} \frac{1}{\left( D_1 + 1 \right) \left( D_1^2 - 2D_1 + 2 \right) }(10 e^{-z}) & = \frac{1}{D_1 + 1}\cdot \frac{1}{(-1)^2 - 2(-1) +2 }(10e^{-z}) \\ & = \frac{1}{D_1 + 1}(2 e^{-z}) \\ & = 2e^{-z} \frac{1}{D_1 -1 + 1} \cdot 1 \\ & = 2e^{-z} \frac{1}{D_1}(1) \\ & = 2e^{-z} z = \frac{2}{x}\cdot \ln x \end{align*} Thus, the solution will be \[ y(x) = \frac{c_1}{x} + \frac{c_2 \cos \ln x + c_3 \sin \ln x}{x} + 5x + \frac{2 \ln x}{x}. \]