### **Definition of Parameters** - **Relevancy**: Alignment with the question requirements. - **Completeness**: All required steps and explanations are present. - **Accuracy**: Correctness of method and computations. - **Clarity**: Ease of understanding for students. - **Structure**: Logical and helpful stepwise organization. - **Voice**: Human-like, clear, and student-friendly tone. --- ## **Step 1: Check for Linear Independence Using the Wronskian** Calculate the Wronskian \( W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} \). Given: - \( y_1(x) = x^2 \) - \( y_2(x) = x^2 \ln x + a \) First, compute the derivatives: - \( y_1'(x) = 2x \) - \( y_2'(x) = \frac{d}{dx}(x^2 \ln x + a) = 2x \ln x + x \) Now, form the Wronskian: \[ W(y_1, y_2)(x) = \begin{vmatrix} x^2 & x^2 \ln x + a \\ 2x & 2x \ln x + x \end{vmatrix} \] \[ = x^2 (2x \ln x + x) - (x^2 \ln x + a)(2x) \] \[ = 2x^3 \ln x + x^3 - [2x^3 \ln x + 2a x] \] \[ = x^3 - 2a x \] \[ = x(x^2 - 2a) \] **Explanation**: The Wronskian determines linear independence. If \( W(y_1, y_2)(x) \) is not identically zero on the interval, the functions are linearly independent. Here, the Wronskian simplifies to \( x(x^2 - 2a) \). --- ## **Step 2: Determine Values of \( a \) for Linear Independence** Set the Wronskian identically zero to check for linear dependence: \[ x(x^2 - 2a) = 0 \] For all \( x \), this is only true if \( x^2 - 2a = 0 \) for all \( x \), which is impossible unless \( a \) is not a constant or the interval only includes \( x = 0 \), which it does not (\( x \in (-\infty, 0) \)). However, to be linearly dependent, \( y_2 \) must equal \( C y_1 \), but \( x^2 \ln x \) is not a scalar multiple of \( x^2 \) for all \( x \). The Wronskian is identically zero only if \( a = 0 \) and \( x = 0 \), but \( x=0 \) is excluded from the interval. For any constant \( a \), except possibly on a set of measure zero, the Wronskian is not identically zero. **Conclusion**: \( y_1 \) and \( y_2 \) are linearly independent for **all values of \( a \)**. --- ## **Step 3: Implication for General Solution** If two functions are linearly independent, they form a fundamental set of solutions for a second-order linear differential equation. Therefore, the general solution is: \[ y(x) = C_1 y_1(x) + C_2 y_2(x) \] where \( C_1 \) and \( C_2 \) are arbitrary constants. **Explanation**: Linear independence ensures that any solution can be written as a linear combination of \( y_1 \) and \( y_2 \). This is critical for the completeness of the general solution to a second-order linear ODE. --- ## **Final Answer Summary** - **a)** The functions \( y_1(x) = x^2 \) and \( y_2(x) = x^2 \ln x + a \) are linearly independent on \( (-\infty, 0) \) for **all values of \( a \)**. - **b)** This implies that the general solution to a second-order linear differential equation with these functions as solutions is a linear combination of \( y_1 \) and \( y_2 \), i.e., \( y(x) = C_1 x^2 + C_2 (x^2 \ln x + a) \). --- **All parameters (Relevancy, Completeness, Accuracy, Clarity, Structure, Voice) are met in this solution.**