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The auxiliary equation is a crucial component in solving linear differential equations, especially those with constant coefficients. It is derived from the homogeneous part of the differential equation. To construct the auxiliary equation, you replace the derivative terms with powers of a variable, usually denoted by \( m \). For example, an equation like \( y'' + y = 0 \) becomes \( m^2 + 1 = 0 \). The solutions to this polynomial equation, known as roots, determine the form of the complementary function. In this exercise, the roots are \( m = i \) and \( m = -i \), which indicate complex roots. This suggests that the complementary function will be a combination of sine and cosine functions, specifically \( c_1 \cos x + c_2 \sin x \). This is because exponential terms with imaginary exponents translate into trigonometric functions, reflecting the oscillatory nature of solutions to such differential equations.

The Wronskian is a determinant used to verify the linear independence of solutions to a differential equation. It's particularly helpful in confirmatory checks when forming a general solution. In this exercise, the functions \( y_1 = \cos x \) and \( y_2 = \sin x \) serve as solutions. To compute the Wronskian \( W \), you create a matrix from these solutions and their derivatives and then calculate its determinant:

• \( y_1 = \cos x \), \( y_1' = -\sin x \)
• \( y_2 = \sin x \), \( y_2' = \cos x \)

The Wronskian is given by:\[W = \begin{vmatrix} \cos x & \sin x \ -\sin x & \cos x \end{vmatrix} = \cos^2 x + \sin^2 x = 1\]A non-zero Wronskian, such as 1, confirms that the functions \( \cos x \) and \( \sin x \) are linearly independent, solidifying them as valid components of the complementary function.

Finding a particular solution involves addressing the non-homogeneous part of the differential equation, typically represented by a function \( f(x) \). In this case, \( f(x) = \sec x \). The method of undetermined coefficients isn't feasible here, so instead we employ the method of variation of parameters. First, we calculate the derivatives \( u_1' \) and \( u_2' \):

• \( u_1' = -\frac{y_2 f(x)}{W} = -\frac{\sin x \sec x}{1} = -\tan x \)
• \( u_2' = \frac{y_1 f(x)}{W} = \frac{\cos x \sec x}{1} = 1 \)

Next, we integrate these derivatives to find \( u_1 \) and \( u_2 \):

• \( u_1 = \int (-\tan x) \, dx = \ln |\cos x| \)
• \( u_2 = \int 1 \, dx = x \)

These functions are then used to construct the particular solution: \( y_p = \cos x \ln |\cos x| + x \sin x \). This component aligns with the non-homogeneous part of the equation to complete the general solution.

The complementary function is a solution to the homogeneous equation, formed solely from the homogeneous part, meaning it involves no external function \( f(x) \). It lays the groundwork for the overall solution to the differential equation. With complex roots from the auxiliary equation, the form of the complementary function follows a specific pattern involving sine and cosine functions. Here, the complementary function is given by:\[ y_c = c_1 \cos x + c_2 \sin x \]This function represents all potential solutions arising from the homogeneous equation. The role of the complementary function is vital as it represents the family of curves that the differential equation can produce, assuming no external influence is applied. Once you have determined \( y_c \), it's combined with the particular solution to provide the complete general solution to the differential equation, giving insight into its full range of behavior.