Solve x²y''−2xy'−4y=x⁴

x²y'' − 2xy' − 4y = x⁴ is a Cauchy–Euler equation.

Transformation

Put z = log x, so x = eᶻ, and D = d/dz. Then x y' = Dy and x² y'' = D(D − 1)y:

[D(D − 1) − 2D − 4]y = e^(4z) (D² − 3D − 4)y = e^(4z), i.e. (D − 4)(D + 1)y = e^(4z)

Complementary function

m = 4, −1: CF = c₁e^(4z) + c₂e^(−z) = c₁x⁴ + c₂/x

Particular integral

The right side e^(4z) has the same exponent as a root, so use the resonance rule:

PI = z e^(4z) / (2·4 − 3) = z e^(4z)/5 = (x⁴ log x)/5

General solution

y = c₁x⁴ + c₂/x + (x⁴ log x)/5