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Differential Equations Unit 9: Series solutions (optional)
Power-series solutions handle equations with variable coefficients where other methods fail.
Covers: Power-series solutions near ordinary points
Study guide
Some differential equations, especially those with variable coefficients like y″ + xy = 0, have no elementary solutions. Power series solutions assume y = Σ aₙxⁿ and solve for the coefficients.
The method: write y, y′ = Σ naₙxⁿ⁻¹ and y″ = Σ n(n − 1)aₙxⁿ⁻² as series. Substitute them into the equation, multiply through by any powers of x, and shift indices so every sum uses the same power xⁿ. Combine the sums and set the coefficient of each power of x equal to zero. This gives a recurrence relation that expresses later coefficients in terms of earlier ones.
For a second-order equation, a₀ and a₁ are arbitrary, since they're y(0) and y′(0). The recurrence then determines all the others, and the solution splits into two independent series, one multiplied by a₀ and one by a₁. Write out the first several terms and, if possible, find a general pattern.
A point x₀ is an ordinary point if, after dividing by the coefficient of y″, the remaining coefficient functions are analytic (have convergent power series) there. Near an ordinary point, two independent power series solutions exist, and they converge at least out to the nearest singular point. At singular points, the Frobenius method (series with a power xʳ in front) is needed. Famous equations solved this way include Airy's, Legendre's and Bessel's equations.
The radius of convergence of Σ aₙxⁿ is R = lim |aₙ/aₙ₊₁| when that limit exists, which follows from the ratio test. For a series in x² (only even powers), apply the ratio test to consecutive nonzero terms. A singular point x₀ of y″ + P(x)y′ + Q(x)y = 0 is regular if (x − x₀)P(x) and (x − x₀)²Q(x) are analytic there. Only at regular singular points does the Frobenius method's guess y = Σ aₙ(x − x₀)ⁿ⁺ʳ work, and r comes from an indicial equation.
Key ideas
- Assume y = Σ aₙxⁿ near an ordinary point.
- Substitute y, y′, y″, shift indices so powers match, and set each coefficient to zero.
- That gives a recurrence relation for the aₙ.
- a₀ and a₁ stay arbitrary and come from initial conditions.
Formulas and rules
Common mistakes
- Index-shifting errors when matching powers.
- Forgetting the first few terms that don't fit the general sum.
Words to know
- Assumed form of a power-series solution
- y = Σ aₙxⁿ
- Recurrence relation (series solutions)
- The rule linking coefficients, found by matching powers of x.
- Arbitrary coefficients (series)
- a₀ and a₁ for a second-order equation.
- Ordinary point
- Where the coefficient functions (after dividing by y″'s) are analytic.
- Index shifting
- Rewriting sums so they share the same power of x.
- Power series solution
- y = Σ aₙxⁿ substituted into the DE.
- Singular point
- Where a coefficient function isn't analytic after dividing by y″'s coefficient.
- Radius of convergence (series solution)
- At least the distance to the nearest singular point.
- Frobenius method
- A series with a factor xʳ, used near regular singular points.
- Airy's equation
- y″ − xy = 0, solved by power series.
Sample practice questions
15 more questions like these, from level 1 to 5, are in the practice set.
Assumed form of a power-series solution.
- y = Σ aₙxⁿ
- y = Σ aₙe^(nx)
- y = Σ aₙ sin(nx)
- y = Σ aₙ/xⁿ
Show the answer and steps
Answer: A) y = Σ aₙxⁿ
- A power-series solution is a series in powers of x.
- y = Σ aₙxⁿ.
If y = Σ aₙxⁿ, then y′ =
- Σ (n + 1)aₙxⁿ⁺¹
- Σ aₙxⁿ⁻¹
- Σ n·aₙxⁿ
- Σ n·aₙxⁿ⁻¹
Show the answer and steps
Answer: D) Σ n·aₙxⁿ⁻¹
- Differentiate term by term.
- y′ = Σ n·aₙxⁿ⁻¹.
Recurrence for y′ = y.
- aₙ₊₁ = (n + 1)aₙ
- aₙ₊₁ = aₙ
- aₙ₊₁ = aₙ/(n + 1)
- aₙ₊₁ = −aₙ
Show the answer and steps
Answer: C) aₙ₊₁ = aₙ/(n + 1)
- Substitute into y′ = y and match the xⁿ coefficients.
- (n + 1)aₙ₊₁ = aₙ.
- aₙ₊₁ = aₙ/(n + 1).