Series solution of x²y''+xy'+(x²−4)y=0

Equation: x²y″ + xy′ + (x² − 4)y = 0. This is Bessel's equation of order 2. x = 0 is a regular singular point, so use the Frobenius method: y = Σ aₖ x^(k+m), a₀ ≠ 0.

Indicial equation and recurrence

Substituting, the coefficient of x^(k+m) gives

[(k+m)² − 4] aₖ + aₖ₋₂ = 0.

• Lowest power (k = 0): m² − 4 = 0, so m = 2 or −2.
• Recurrence: aₖ = − aₖ₋₂ / ((k + m)² − 4), and a₁ = a₃ = … = 0.

Root m = 2

aₖ = − aₖ₋₂ / (k(k + 4)):

a₂ = −a₀/12,   a₄ = a₀/(12·32) = a₀/384,   a₆ = −a₀/(384·60) …

Taking a₀ = 1/8 (i.e. 1/(2²·2!)):

y₁ = J₂(x) = x²/8 − x⁴/96 + x⁶/3072 − …
   = Σ (−1)ᵏ / (k!(k+2)!) · (x/2)^(2k+2)

Root m = −2

The roots differ by an integer (4). With m = −2 the recurrence is aₖ = −aₖ₋₂/(k(k − 4)), which fails at k = 4 (division by zero, while a₂ = a₀/4 ≠ 0). So the second solution contains a logarithm and is not a pure power series; it is the Bessel function of the second kind Y₂(x) (J₋₂ = J₂ for integer order, so it is not independent).

Answer: y = A J₂(x) + B Y₂(x), where J₂(x) = Σ (−1)ᵏ/(k!(k+2)!)(x/2)^(2k+2) is the series solution above.