Word Count : 448

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We are asked to solve the differential equation: [ x^2 y'' - 2x y' - 4y = x^2 + 2 \ln x ] Step 1: Solve the corresponding homogeneous equation [ x^2 y'' - 2x y' - 4y = 0 ] This is a Cauchy-Euler (equidimensional) equation. We assume a solution of the form (y = x^m). Then [ y' = m x^{m-1}, \quad y'' = m(m-1)x^{m-2} ] Substitute into the homogeneous equation: [ x^2[m(m-1)x^{m-2}] - 2x[m x^{m-1}] - 4 x^m = 0 ] Simplify each term: [ m(m-1)x^m - 2 m x^m - 4 x^m = 0 ] Combine terms: [ [m(m-1) - 2m - 4] x^m = 0 ] [ ____ ____ ___ ___ ___ _____ __________ _______ _______.
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