\( \left. \begin{array} { c } { x ^ { 2 } y ^ { \prime \prime } + x y ^ { \prime \prime } + y = \sec ( \ln x } \\ { y _ { 1 } = \cos x } \\ { y _ { 2 } = \sin x } \\ { y _ { p } = 2 ? } \end{array} \right. \)

To solve the differential equation given by \[ x^2 y'' + x y' + y = \sec(\ln x), \] we can start by analyzing the structure of the equation. This is a second-order linear ordinary differential equation (ODE) with variable coefficients. The general form of such an equation is: \[ p(x) y'' + q(x) y' + r(x) y = g(x), \] where \( p(x) = x^2 \), \( q(x) = x \), \( r(x) = 1 \), and \( g(x) = \sec(\ln x) \). ### Step 1: Find the Complementary Solution First, we need to find the complementary solution \( y_c \) of the associated homogeneous equation: \[ x^2 y'' + x y' + y = 0. \] This is a Cauchy-Euler equation, which can be solved by assuming a solution of the form \( y = x^m \). We compute the derivatives: \[ y' = m x^{m-1}, \quad y'' = m(m-1) x^{m-2}. \] Substituting these into the homogeneous equation gives: \[ x^2 (m(m-1) x^{m-2}) + x (m x^{m-1}) + x^m = 0, \] which simplifies to: \[ m(m-1) x^m + m x^m + x^m = 0. \] Factoring out \( x^m \) (assuming \( x \neq 0 \)) leads to: \[ (m(m-1) + m + 1) = 0 \implies m^2 = 0 \implies m = 0. \] Thus, the roots of the characteristic equation are \( m = 0 \) (with multiplicity 2). Therefore, the complementary solution is: \[ y_c = C_1 + C_2 \ln x, \] where \( C_1 \) and \( C_2 \) are constants. ### Step 2: Find the Particular Solution Next, we need to find a particular solution \( y_p \) to the non-homogeneous equation. We can use the method of undetermined coefficients or variation of parameters. Given that \( g(x) = \sec(\ln x) \) is not a polynomial, we will use variation of parameters. Assuming a particular solution of the form: \[ y_p = u_1(x) + u_2(x) \ln x, \] where \( u_1 \) and \( u_2 \) are functions to be determined. The formulas for \( u_1 \) and \( u_2 \) in variation of parameters are: \[ u_1' = -\frac{y_2 g}{W}, \quad u_2' = \frac{y_1 g}{W}, \] where \( W \) is the Wronskian of \( y_1 \) and \( y_2 \). Calculating the Wronskian \( W \): \[ W = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{vmatrix} = \cos^2 x + \sin^2 x = 1. \] Now, substituting \( g(x) = \sec(\ln x) \): \[ u_1' = -\frac{\sin x \sec(\ln x)}{1} = -\sin x \sec(\ln x), \] \[ u_2' = \frac{\cos x \sec(\ln x)}{1} = \cos x \sec(\ln x). \] Integrating \( u_1' \) and \( u_2' \) will give us \( u_1 \) and \( u_2 \). However, the integration may be complex, and we may need to use numerical methods or special functions depending on the context. ### Final Solution The general solution to the ODE is: \[ y = y_c + y_p = C_1 + C_2 \ln x + y_p, \] where \( y_p \) is the particular solution found through the method of variation of parameters. ### Conclusion The final form of the solution will depend on the specific form of \( y_p \) obtained from the integration of \( u_1' \) and \( u_2' \). If you have specific values for \( C_1 \) and \(