Linear independence of solutions of x²y'''−3xy''+3y'=0

Equation: x²y‴ − 3xy″ + 3y′ = 0 (x > 0). It is an equidimensional (Cauchy–Euler) equation, so try solutions of the form y = xᵐ.

Find the solutions

Because y itself does not appear, y = 1 (a constant) is a solution (m = 0). For the others, y′ = mxᵐ⁻¹, y″ = m(m−1)xᵐ⁻², y‴ = m(m−1)(m−2)xᵐ⁻³:

m(m−1)(m−2) − 3m(m−1) + 3m = 0
m[(m−1)(m−2) − 3(m−1) + 3] = 0
m[m² − 6m + 8] = 0
m(m − 2)(m − 4) = 0   →   m = 0, 2, 4

Three solutions: y₁ = 1, y₂ = x², y₃ = x⁴.

Check, e.g. y = x⁴: x²(24x) − 3x(12x²) + 3(4x³) = 24x³ − 36x³ + 12x³ = 0 ✓.

Wronskian

        | 1   x²   x⁴   |
W(x) =  | 0   2x   4x³  | = 1·(2x·12x² − 4x³·2) = 24x³ − 8x³ = 16x³
        | 0   2    12x² |

For x > 0, W(x) = 16x³ ≠ 0. A non-zero Wronskian means the solutions 1, x², x⁴ are linearly independent.

General solution: y = c₁ + c₂x² + c₃x⁴.