Solve (x²D²+xD+1)y=sin(log x²)

(x²D² + xD + 1)y = sin(log x²), where D = d/dx. This is a Cauchy–Euler equation.

Transformation

Put z = log x. With θ = d/dz: x²D² = θ(θ − 1) and xD = θ. Also log x² = 2z. So

[θ(θ − 1) + θ + 1]y = sin 2z (θ² + 1)y = sin 2z

Complementary function

m² + 1 = 0, m = ±i: CF = c₁ cos z + c₂ sin z = c₁ cos(log x) + c₂ sin(log x)

Particular integral

Put θ² = −4:

PI = 1/(θ² + 1) · sin 2z = sin 2z/(−4 + 1) = −(1/3) sin 2z

General solution

y = c₁ cos(log x) + c₂ sin(log x) − (1/3) sin(2 log x)