91Ó°ÊÓ

When dealing with solutions to differential equations, it's important to ensure that these solutions are linearly independent. This means that no function in the set can be written as a combination of the others. If they were dependent, one function could be "built" from another, which would mean the solutions aren't genuinely unique or original.
To verify linear independence, we often rely on the Wronskian, a kind of "determinant check" for functions. In our case, the functions given are: \( y_1 = x \), \( y_2 = x^{-2} \), and \( y_3 = x^{-2} \ln x \).
We compute the derivatives of these functions:

• \( y_1' = 1 \) and \( y_1'' = 0 \)
• \( y_2' = -2x^{-3} \) and \( y_2'' = 6x^{-4} \)
• \( y_3' = -2x^{-3} \ln x - x^{-3} \) and \( y_3'' = 6x^{-4} \ln x + 4x^{-4} \)

By plugging these back into the Wronskian determinant and ensuring that the resulting determinant is non-zero over our interval \((0, \infty)\), we confirm that these functions are indeed linearly independent.

Once we establish that the set of functions is linearly independent, we can call them a "fundamental set of solutions" for our differential equation on a given interval. A fundamental set is crucial because it forms the basis for all other solutions to the differential equation.
In simpler terms, any solution of the differential equation can be expressed as a combination of the functions in the fundamental set. This combination is weighted by constants, which adjust the "mix" of how much of each function is in our solution, depending on initial or boundary conditions.
In our exercise, the fundamental set of solutions is \( x, x^{-2}, \text{and} \ x^{-2} \ln x \). Since these functions solve our differential equation and are linearly independent, they create a solid foundation to construct the general solution. This general solution satisfies the differential equation for any partition of \((0, \infty)\).

The Wronskian is a powerful tool used in differential equations to test whether a set of functions is linearly independent. It's a determinant formed by the functions and their derivatives, structured in a matrix.
For three functions like \( y_1 = x \), \( y_2 = x^{-2} \), and \( y_3 = x^{-2} \ln x \), the Wronskian \( W(y_1, y_2, y_3) \) would be:
\[W(y_1, y_2, y_3) = \begin{vmatrix}y_1 & y_2 & y_3 \y_1' & y_2' & y_3' \y_1'' & y_2'' & y_3''\end{vmatrix}\]
Here, by calculating this determinant and finding it non-zero in the interval \((0, \infty)\), we verify linear independence. Hence, these functions are suitable to form a fundamental set of general solutions. It's essential to remember that a zero Wronskian isn't the only determinant of dependence — it hints towards potential dependence, which requires further analysis.