Skip to main content

Power series solution of y''−y=0

Short answer

Assume y = Σ a_n x^n. Recurrence a_{n+2}=a_n/((n+1)(n+2)) gives a_{2n}=a_0/(2n)!, a_{2n+1}=a_1/(2n+1)!. Summing yields y = a_0 cosh x + a_1 sinh x = C_1 e^x + C_2 e^{-x}.

Solve the differential equation by power series method: y'' − y = 0.

Equation: y″ − y = 0. x = 0 is an ordinary point, so assume y = Σ aₙxⁿ (n = 0, 1, 2, …).

Substitution

y″ = Σ (n+2)(n+1)aₙ₊₂ xⁿ. Putting in the equation and equating the coefficient of xⁿ to zero:

(n+2)(n+1)aₙ₊₂ − aₙ = 0, so aₙ₊₂ = aₙ / ((n+1)(n+2))

Coefficients

a₀ and a₁ are arbitrary.

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

Solution

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

Writing cosh x = (eˣ + e⁻ˣ)/2 and sinh x = (eˣ − e⁻ˣ)/2, this is the same as

y = C₁eˣ + C₂e⁻ˣ,

which agrees with the direct solution of the auxiliary equation m² − 1 = 0.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics chapter 4 questions →