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The problem appears to ask to derive a recurrence relation from the given function \( y = x \cos(\ln x) \) and to show that it satisfies the differential equation:

\[
x^2 y_{n+2} + (2n - 1) x y_{n+1} + (n^2 - 2n + 2) y_n = 0
\]

This is a common problem in solving differential equations through the method of Frobenius or generating recurrence relations for series solutions.

### Approach:

1. **Identify the function and derivatives**: The function is \( y = x \cos(\ln x) \), and we need to compute its higher-order derivatives \( y_n \), \( y_{n+1} \), and \( y_{n+2} \).
   
2. **Derive the first and second derivatives**: Differentiate \( y = x \cos(\ln x) \) to find the required higher-order derivatives.

Would you like a detailed step-by-step solution of the derivatives and recurrence relation?

Show that the function y = xcos(lnx) satisfies the differential equation x^2y_{n+2} + (2n - 1)xy_{n+1} + (n^2 - 2n + 2)y_n = 0

Learn how to show that the function y = xcos(lnx) satisfies the differential equation x^2y_{n+2} + (2n - 1)xy_{n+1} + (n^2 - 2n + 2)y_n = 0 using higher-order derivatives and recurrence relations.

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