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Power series solution of y''−y=x

Short answer

Substituting y = Σ a_n x^n gives recurrence a_{n+2} = a_n/[(n+1)(n+2)] for n≠1, with 6a_3 - a_1 = 1. This yields y = a_0 cosh x + (a_1+1) sinh x - x, i.e., y = C_1 e^x + C_2 e^{-x} - x.

Solve by power series method y'' − y = x.

Equation: y″ − y = x. Take y = Σ aₙxⁿ, so y″ = Σ (n+2)(n+1)aₙ₊₂ xⁿ.

Equate coefficients of xⁿ

(n+2)(n+1)aₙ₊₂ − aₙ = 0     for n ≠ 1
 6a₃ − a₁ = 1               for n = 1  (coefficient of x on the right is 1)

So aₙ₊₂ = aₙ/((n+1)(n+2)) except a₃ = (a₁ + 1)/6.

Coefficients

Even:  a₂ = a₀/2!,  a₄ = a₀/4!,  a₆ = a₀/6! …
Odd:   a₃ = (a₁+1)/3!,  a₅ = a₃/20 = (a₁+1)/5!,  a₇ = (a₁+1)/7! …

Solution

y = a₀(1 + x²/2! + x⁴/4! + …) + a₁x + (a₁+1)(x³/3! + x⁵/5! + …)
  = a₀ cosh x + (a₁+1)(x + x³/3! + x⁵/5! + …) − x
  = a₀ cosh x + (a₁+1) sinh x − x

Writing the constants as C₁ and C₂:

y = C₁eˣ + C₂e⁻ˣ − x

Check: for y = −x, y″ − y = 0 + x = x ✓ (this is the particular solution).

✓ Answer checked by Claude Opus 5.5 (high effort)

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