91Ó°ÊÓ

A second-order linear differential equation involves the second derivative of an unknown function and can be written in the general form: \[ a(x) y'' + b(x) y' + c(x) y = 0. \] In this equation, \( y'' \) denotes the second derivative of \( y \) with respect to \( x \), \( y' \) represents the first derivative, and \( y \) is the original function.

• Second-order: Since the highest derivative is the second derivative.
• Linear: It has a linear combination of the function and its derivatives.

For example, the differential equation provided in the problem, \[ x^2 y'' - x y' + 2 y = 0, \] is linear and of the second order.In dealing with second-order linear equations, solutions provide insight into the behavior of related physical, biological, and engineering systems. Recognizing their structure helps in applying appropriate methods of solving them, such as the reduction of order or characteristic equations.

A homogeneous differential equation is a particular type where every term is a function of the unknown variable and its derivatives, without any additional constants or functions. This equation can generally be expressed as: \[ a(x) y'' + b(x) y' + c(x) y = 0. \] The concept of homogeneity implies the solution can be scaled; if a function \( y(x) \) is a solution, then \( c \, y(x) \) for any constant \( c \) is also a solution. In our example: \[ x^2 y'' - x y' + 2 y = 0, \] all terms involve either \( y \) or its derivatives solely, making it homogeneous.Homogeneous differential equations are crucial because their solutions can often be found in a complete form that characterizes the behavior of systems described by these equations thoroughly. A known solution can assist in deriving other possible solutions, typically using methods such as reduction of order.

Finding a second solution to a second-order linear homogeneous differential equation requires a methodical approach when one solution is already known. When given a solution \( y_1(x) \), the reduction of order technique is particularly useful for finding another linearly independent solution.In this method, the second solution \( y_2(x) \) is assumed to have the form \[ y_2(x) = v(x) y_1(x). \] Here, \( v(x) \) is an unknown function to be determined.

• The known solution \( y_1(x) \) acts as a base.
• Substitute \( y_2(x) \) into the original differential equation.
• Through simplification and solving, \( v(x) \) can be resolved, leading to the full form of \( y_2(x) \).

For the given exercise, substitution and calculations reveal the second solution as \( y_2(x) = x \cos(\ln x) \). Ensuring this function is truly a solution involves substituting it back into the original equation to verify.

In the context of differential equations, linearly independent solutions are fundamental in forming the general solution. Two functions, \( y_1(x) \) and \( y_2(x) \), are termed linearly independent if there is no constant \( c \) such that \( y_2(x) = c \, y_1(x) \).The importance of linear independence lies in creating a complete solution set:

• For homogeneous equations, the general solution is a linear combination of linearly independent solutions.
• The Wronskian determinant can be employed to test for linear independence. If it is non-zero, the solutions are independent.

In our exercise, the solutions \( y_1(x) = x \sin(\ln x) \) and the second solution \( y_2(x) = x \cos(\ln x) \) are linearly independent. Together, they construct the general solution to the differential equation:\[ y(x) = C_1 \cdot y_1(x) + C_2 \cdot y_2(x), \]where \( C_1 \) and \( C_2 \) are arbitrary constants representing the family of solutions.