Solve x²y''−2xy'+2y=1/x

x²y'' − 2xy' + 2y = 1/x is a Cauchy–Euler equation.

Transformation

Put z = log x and D = d/dz, so x²y'' = D(D − 1)y and xy' = Dy:

[D(D − 1) − 2D + 2]y = e^(−z) (D² − 3D + 2)y = e^(−z), i.e. (D − 1)(D − 2)y = e^(−z)

Complementary function

m = 1, 2: CF = c₁eᶻ + c₂e^(2z) = c₁x + c₂x²

Particular integral

Put D = −1: PI = e^(−z)/[(−1)² − 3(−1) + 2] = e^(−z)/6 = 1/(6x)

General solution

y = c₁x + c₂x² + 1/(6x)