Solution: We want to solve the equation \begin{equation}\label{eq:19July2024-1} x^2 \frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 2x \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2 \frac{y}{x} = 10 \left( 1 + \frac{1}{x^2} \right). \end{equation} Multiplying \eqref{eq:19July2024-1} by $x$, we get \[ x^3 \frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 2x^2 \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2 y = 10 \left( x + \frac{1}{x} \right). \] Using the $D \equiv \frac{\mathrm{d}}{\mathrm{d}x} $ operator, we can rewrite the above equation as \begin{equation}\label{eq:19July2024-2} \left( x^3 D^3 + 2x^2 D^2 + 2 \right) y = 10 \left( x + \frac{1}{x}. \right) \end{equation} Let $x = e^z$. So, $z = \log x$. Let $D_1 \equiv \frac{\mathrm{d}}{\mathrm{d}z} $. Then 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} = e^{-z} \frac{\mathrm{d}y}{\mathrm{d}z} \\ & = e^{-z}D_1 y \\[2ex] \frac{\mathrm{d}^2y}{\mathrm{d}x^2} & = \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{1}{x}\frac{\mathrm{d}y}{\mathrm{d}z} \right) = \frac{\mathrm{d}}{\mathrm{d}z} \left( e^{-z} \frac{\mathrm{d}y}{\mathrm{d}z} \right) \cdot \frac{\mathrm{d}z}{\mathrm{d}x} \\ & = \left( -e^{-z} \frac{\mathrm{d}y}{\mathrm{d}z} + e^{-z}\frac{\mathrm{d}^2y}{\mathrm{d}z^2} \right) e^{-z} \\ & = e^{-2z} \left( \frac{\mathrm{d}^2y}{\mathrm{d}z^2} - \frac{\mathrm{d}y}{\mathrm{d}z} \right) \\ & = e^{-2z}\left( D_1^2 - D_1 \right) y\\[2ex] \frac{\mathrm{d}^3y}{\mathrm{d}x^3} & = \frac{\mathrm{d}}{\mathrm{d}x} \left[ e^{-2z} \left( \frac{\mathrm{d}^2y}{\mathrm{d}z^2} - \frac{\mathrm{d}y}{\mathrm{d}z} \right) \right] \\ & = e^{-3z} \left[ -2 \left( \frac{\mathrm{d}^2y}{\mathrm{d}z^2} - \frac{\mathrm{d}y}{\mathrm{d}z} \right) + \frac{\mathrm{d}^3y}{\mathrm{d}z^3} - \frac{\mathrm{d}^2y}{\mathrm{d}z^2} \right] \\ & = e^{-3z} \left( D_1^3 - 3D_1^2 + 2D_1 \right) y. \end{align*}

Thus, the given differential equation \eqref{eq:19July2024-2} will be \begin{align*} & \left( D_1^3 - 3D_1^2 + 2D_1 \right) y + 2\left( D_1^2 - D_1 \right)y + 2y = 10 \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*} The auxiliary equation is \begin{align*} m^3 - m^2 + 2 = 0 & \implies (m + 1)(m^2 -2m + 2) = 0 \\ & \implies m = -1 \text{ or } m = 1 \pm \iota . \end{align*} Therefore, \begin{align*} y_{\mathsf{CF} } & = c_1 e^{-z} + e^z \left( c_2 \cos z + c_3 \sin z \right) \\ & = \frac{c_1}{x} + x \left( c_2 \cos \log x + c_{3\sin \log x} \right) . \end{align*}

Now we will find the particular integral. The particular integral corresponding to $10e^z$ will be given by \begin{align*} 10\cdot \frac{1}{(D_1 + 1)(D_1^2 - 2D_1 + 2)}e^ z & = \frac{10}{(1 + 1)(1 - 2 + 2)}e^z \\ & = 5e^z = 5x. \end{align*} Similarly, the particular integral corresponding to $10e^{-z}$ will be \begin{align*} & 10 \cdot \frac{1}{(D_1 + 1)(D_1^2 - 2D_1 + 2)}e^ {-z} \\ = & \frac{10}{D_1 + 1}\cdot \frac{1}{1 + 2 + 2}e^{-z}\\ = & \frac{2}{D_1 + 1}e^{-z} \\ = & 2e^{-z} \frac{1}{D_1 - 1 + 1}\cdot 1 \\ = & 2e^{-z} \frac{1}{D_1}\cdot 1 \\ = & 2e^{-z} z = \frac{2}{x}\log x. \end{align*} Therefore, \[ y(x) = \frac{c_1}{x} + x \left( c_2 \cos \log x + c_{3\sin \log x} \right) + 5x + \frac{2}{x}\log x. \]