91Ó°ÊÓ

Nonhomogeneous differential equations involve two parts: a homogeneous equation and a nonhomogeneous part. In the equation \( y'' + y = \tan x \), the \( y'' + y \) constitutes the homogeneous part, while \( \tan x \) is the nonhomogeneous part. These equations are characterized by one side being a function that is not zero (the nonhomogeneous part). Being able to identify these parts is crucial for solving the equation.

### Characteristics of Nonhomogeneous Differential Equations

• They include both a differential operator and an external function.
• The solution is composed of a complementary solution from the homogeneous part and a particular solution from the external function.
• Methods such as undetermined coefficients and variation of parameters are often used to solve these equations.

Solving a nonhomogeneous differential equation requires addressing each part separately. First, solve the homogeneous equation, then find a suitable particular solution for the nonhomogeneous part, and finally combine both solutions.

The complementary solution, often denoted as \(y_c\), comes from solving the associated homogeneous differential equation. For our equation \( y'' + y = \tan x \), the homogeneous part \( y'' + y = 0 \) gives rise to the complementary solution.

### Finding the Complementary Solution

• Set up the characteristic equation by replacing the differential operator with a polynomial. Here it is \( r^2 + 1 = 0 \).
• Solve this polynomial, resulting in roots that determine the form of the solution.
• If roots are imaginary, as in our example \( r = \pm i \), the complementary solution involves sine and cosine functions: \( y_c = C_1 \cos x + C_2 \sin x \).

This step involves understanding the behavior of the associated homogeneous function, which influences the overall solution strategy.

The Wronskian is a determinant used in the context of solving differential equations. It helps determine if a set of solutions is linearly independent and is crucial when using the method of variation of parameters. In our exercise involving \( \cos x \) and \( \sin x \), the Wronskian is calculated as follows:

### Calculating the Wronskian

• Take the determinant of a matrix formed by the functions and their derivatives. For \( W(\cos x, \sin x) \), the matrix is \( \begin{pmatrix} \cos x & \sin x \ -\sin x & \cos x \end{pmatrix} \).
• The determinant is computed as \( \cos^2 x + \sin^2 x = 1 \).
• A non-zero Wronskian indicates the functions are linearly independent. Here \( W = 1 \), confirming independence.

Understanding the Wronskian is fundamental when seeking particular solutions, especially in variation of parameters, as it confirms the validity of the functions used in forming the particular solution.