Series Solutions

Study Sheet

Series Solutions

Power series where formulas fail

The Power Series Method

Tip
Substitute, shift, match

Write y=∑anxny = \sum a_nx^n, substitute into the equation, shift indices so all powers align, and set each coefficient to zero — a RECURRENCE for the ana_n. For y′=yy' = y: (n+1)an+1=an(n+1)a_{n+1} = a_n, giving an=1n!a_n = \tfrac{1}{n!} — the series discovers exe^x on its own.

Concept
Equations beyond elementary functions

The Airy-type equation y′′=xyy'' = xy has no closed-form solution, but its recurrence (n+2)(n+1)an+2=an−1(n+2)(n+1)a_{n+2} = a_{n-1} generates the answer to any accuracy: with y(0)=1y(0)=1, y′(0)=0y'(0)=0 the series starts 1+x36+⋯1 + \tfrac{x^3}{6} + \cdots. Series ARE the solution.

Example
Recognizing an old friend

y′=2xyy' = 2xy, y(0)=1y(0) = 1: the recurrence nan=2an−2na_n = 2a_{n-2} gives y=1+x2+x42+x66+⋯=ex2y = 1 + x^2 + \tfrac{x^4}{2} + \tfrac{x^6}{6} + \cdots = e^{x^2} — matching the separable-method answer, coefficient for coefficient.

Side note
Significance: when formulas run out

Most differential equations of physics — Airy (optics near caustics), Bessel (drumheads), Hermite (quantum oscillators) — have NO elementary solutions. Series are not a fallback; they are how these functions are DEFINED, tabulated, and computed inside every scientific library you have ever called.

Try it
Try it: two terms by recurrence

For y′=x+yy' = x + y, y(0)=1y(0) = 1, find a2a_2 and a3a_3. Work: matching gives (n+1)an+1=an(n+1)a_{n+1} = a_n plus the extra xx: a1=a0=1a_1 = a_0 = 1; 2a2=a1+1=22a_2 = a_1 + 1 = 2 so a2=1a_2 = 1; 3a3=a23a_3 = a_2 so a3=13a_3 = \tfrac13. Series: 1+x+x2+x33+⋯1 + x + x^2 + \tfrac{x^3}{3}+\cdots (indeed the exact solution is 2ex−x−12e^x - x - 1 — expand it and check both coefficients ✓).

Proofs & Why It Matters

Tip
Proof: coefficient matching is valid

A power series that sums to zero on an interval has ALL coefficients zero (differentiate repeatedly and evaluate at the center: an=f(n)(0)n!=0a_n = \tfrac{f^{(n)}(0)}{n!} = 0).

So when substituting y=∑anxny = \sum a_nx^n into an equation produces ∑cnxn=0\sum c_nx^n = 0, each cnc_n must individually vanish — turning the differential equation into an exact recurrence, not an approximation. ■\blacksquare

Tip
Proof: y′ = y forces aₙ = 1/n!

Substitute: ∑n≥1nanxn−1=∑n≥0anxn\sum_{n\ge1} na_nx^{n-1} = \sum_{n\ge0} a_nx^n. Shift the left index and match coefficients of xnx^n: (n+1)an+1=an(n+1)a_{n+1} = a_n. With a0=y(0)=1a_0 = y(0) = 1, induction gives an=1n!a_n = \tfrac{1}{n!}: a1=1a_1 = 1, and an+1=ann+1=1(n+1)!a_{n+1} = \tfrac{a_n}{n+1} = \tfrac{1}{(n+1)!}.

The series ∑xnn!\sum\tfrac{x^n}{n!} — the definition of exe^x — emerges with no prior knowledge of exponentials. ■\blacksquare

Going Deeper: Explanations & Worked Problems

Concept
The recurrence method, slowly

Write y=∑n≥0anxny = \sum_{n\ge0} a_nx^n and differentiate term by term: y′=∑n≥1nanxn−1y' = \sum_{n\ge1} na_nx^{n-1}, y′′=∑n≥2n(n−1)anxn−2y'' = \sum_{n\ge2} n(n-1)a_nx^{n-2}.

Substitute into the equation; now every sum must be re-indexed so all run over the SAME power xnx^n — the shift m=n−2m = n - 2 turns ∑n(n−1)anxn−2\sum n(n-1)a_nx^{n-2} into ∑(m+2)(m+1)am+2xm\sum (m+2)(m+1)a_{m+2}x^m, and multiplying by xx shifts the other way. Collect the coefficient of each xnx^n and set it to zero (legal because a power series vanishing identically has all coefficients zero — the proof slide): out drops a RECURRENCE expressing high coefficients in terms of low ones. Initial conditions seed it: a0=y(0)a_0 = y(0), a1=y′(0)a_1 = y'(0). Then compute coefficients one at a time, as many as accuracy demands. The equation is never "solved" in closed form — and never needs to be.

Example
Worked: deriving a recurrence with the index shift shown

Solve y′′=−yy'' = -y, y(0)=1y(0) = 1, y′(0)=0y'(0) = 0, by series.

Step 1 — substitute: ∑n≥2n(n−1)anxn−2=−∑n≥0anxn\sum_{n\ge2} n(n-1)a_nx^{n-2} = -\sum_{n\ge0} a_nx^n.

Step 2 — shift the left index by 22: ∑n≥0(n+2)(n+1)an+2xn=−∑n≥0anxn\sum_{n\ge0} (n+2)(n+1)a_{n+2}x^{n} = -\sum_{n\ge0} a_nx^n.

Step 3 — equate the coefficient of every xnx^n: (n+2)(n+1)an+2=−an(n+2)(n+1)a_{n+2} = -a_n.

Step 4 — seed and run: a0=1a_0 = 1, a1=0a_1 = 0; then a2=−12⋅1=−12a_2 = \tfrac{-1}{2\cdot1} = -\tfrac12; a3=0a_3 = 0 (all odd coefficients inherit a1=0a_1 = 0); a4=−a24⋅3=124a_4 = \tfrac{-a_2}{4\cdot3} = \tfrac{1}{24}; a6=−a46⋅5=−1720a_6 = \tfrac{-a_4}{6\cdot5} = -\tfrac{1}{720}.

Step 5 — recognize: 1−x22!+x44!−x66!+⋯=cos⁡x1 - \tfrac{x^2}{2!} + \tfrac{x^4}{4!} - \tfrac{x^6}{6!} + \cdots = \cos x — the series method has DERIVED the cosine series from the oscillator equation, not assumed it.

Example
Worked: the Airy-type start, where no closed form exists

Begin y′′=xyy'' = xy, y(0)=1y(0) = 1, y′(0)=0y'(0) = 0.

Step 1 — substitute and align powers: the left side is ∑n≥0(n+2)(n+1)an+2xn\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n; the right side xy=∑n≥1an−1xnxy = \sum_{n\ge1} a_{n-1}x^{n}.

Step 2 — match: for n=0n = 0: 2a2=02a_2 = 0, so a2=0a_2 = 0; for n≥1n \ge 1: (n+2)(n+1)an+2=an−1(n+2)(n+1)a_{n+2} = a_{n-1}.

Step 3 — run it: a3=a03⋅2=16a_3 = \tfrac{a_0}{3\cdot2} = \tfrac16; a4=a14⋅3=0a_4 = \tfrac{a_1}{4\cdot3} = 0; a5=a25⋅4=0a_5 = \tfrac{a_2}{5\cdot4} = 0; a6=a36⋅5=1180a_6 = \tfrac{a_3}{6\cdot5} = \tfrac{1}{180}.

Step 4 — the solution so far: y=1+x36+x6180+⋯y = 1 + \tfrac{x^3}{6} + \tfrac{x^6}{180} + \cdots, with nonzero coefficients only at powers 0,3,6,9,…0, 3, 6, 9, \ldots — a three-step lattice the recurrence forces. No elementary function has this series; the series itself is the honest, computable answer, good to any accuracy on any bounded interval.