Solve x²y''−xy'+y=log x

This is a Cauchy–Euler equation. Put z = log x (x = eᶻ) and D = d/dz; then x dy/dx = Dy and x² d²y/dx² = D(D − 1)y.

Transformed equation

[D(D − 1) − D + 1]y = z

(D² − 2D + 1)y = z, i.e. (D − 1)²y = z

Complementary function

Auxiliary equation (m − 1)² = 0, so m = 1, 1.

CF = (c₁ + c₂z)eᶻ

Particular integral

PI = 1/(D − 1)² · z = (1 − D)⁻² z = (1 + 2D + 3D² + …) z = z + 2

General solution

y = (c₁ + c₂z)eᶻ + z + 2

Replacing z = log x:

y = (c₁ + c₂ log x) x + log x + 2

(Check: y = log x + 2 gives x²y'' − xy' + y = −1 − 1 + log x + 2 = log x ✓.)