Power series solutions — ordinary points

Chapter: Ordinary Differential Equations Subtopic: Power series solutions at an ordinary point

HOW the method runs (the recipe)

Formula

Goal: solve $y'' + P y' + Q y = 0$ near an ordinary point $x_0$ (shift so $x_0=0$ ).

Step 1. Assume $y=\sum_{n=0}^{\infty} a_n x^n.$ Step 2. Differentiate term by term: $y'=\sum_{n=1}^{\infty} n\,a_n x^{n-1},\qquad y''=\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}.$ Step 3. Substitute into the ODE. Step 4. Re-index every sum so the power of $x$ is the same ( $x^k$ ), then combine into one series. Step 5. Set the coefficient of each $x^k$ to $0$ ⇒ recurrence relation for $a_{k+2}$ in terms of earlier $a$ 's. Step 6. $a_0,a_1$ stay free (the two integration constants). Generate all others; collect $a_0$ -terms and $a_1$ -terms ⇒ two independent solutions $y_1,y_2$ .

Worked Example 1 — the simplest possible

Solve $y'' + y = 0$ about $x_0=0$ . (We know the answer is $\cos x,\sin x$ — perfect for Forecast-then-Verify.)

Forecast: I expect $a_0$ -series $=\cos x$ and $a_1$ -series $=\sin x$ .

Let $y=\sum a_n x^n$ , so $y''=\sum_{n\ge2}n(n-1)a_nx^{n-2}$ .

Substitute: $\sum_{n=2}^\infty n(n-1)a_n x^{n-2} + \sum_{n=0}^\infty a_n x^n = 0.$

Re-index first sum with $k=n-2$ (so $n=k+2$ ): Why? — to make both sums carry $x^k$ . $\sum_{k=0}^\infty (k+2)(k+1)a_{k+2}x^k + \sum_{k=0}^\infty a_k x^k = 0.$

Coefficient of $x^k$ must vanish: $(k+2)(k+1)a_{k+2} + a_k = 0 \;\Rightarrow\; \boxed{a_{k+2} = -\frac{a_k}{(k+2)(k+1)}}.$ Why? — a single power series is zero iff every coefficient is zero.

Now crank it:

Even: $a_2=-\frac{a_0}{2!}$ , $a_4=-\frac{a_2}{4\cdot3}=\frac{a_0}{4!}$ , $a_6=-\frac{a_0}{6!}$ …
Odd: $a_3=-\frac{a_1}{3!}$ , $a_5=\frac{a_1}{5!}$ …

So $y=a_0\underbrace{\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\right)}_{\cos x}+a_1\underbrace{\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)}_{\sin x}.$ Verified ✔ — the forecast was exactly right.

Worked Example 2 — Legendre-type, $(1-x^2)y''-2xy'+6y=0$ about $x_0=0$

Why $x_0=0$ is legal: shown above to be an ordinary point.

Plug $y=\sum a_nx^n$ into each piece. Keep the polynomial coefficients multiplied inside:

$\underbrace{\sum n(n-1)a_nx^{n-2}}_{y''}-\underbrace{\sum n(n-1)a_nx^{n}}_{x^2y''}-\underbrace{\sum 2n\,a_nx^{n}}_{2xy'}+\underbrace{6\sum a_nx^n}_{6y}=0.$

Re-index the first sum ( $k=n-2$ ); the rest already carry $x^n$ (rename $n\to k$ ):

$\sum_k\Big[(k+2)(k+1)a_{k+2} - k(k-1)a_k - 2k\,a_k + 6a_k\Big]x^k=0.$

Why combine the last three? They all multiply $a_k x^k$ , so: $-k(k-1)-2k+6 = -(k^2-k+2k-6)=-(k^2+k-6)=-(k+3)(k-2).$

Recurrence: $(k+2)(k+1)a_{k+2}=(k+3)(k-2)a_k\;\Rightarrow\; a_{k+2}=\frac{(k+3)(k-2)}{(k+2)(k+1)}a_k.$

Notice: $k=2$ gives $a_4=0$ , so all higher even terms vanish — the $a_0$ branch terminates:

$a_2=\frac{(3)(-2)}{2\cdot1}a_0=-3a_0$ , $a_4=0$ .
$y_{\text{even}}=a_0(1-3x^2)$ — a polynomial solution (this is why Legendre equations give Legendre polynomials!).

Odd branch ( $a_1$ ) does not terminate (numerator $(k+3)(k-2)$ never hits $0$ for odd $k\ge1$ ):

$a_3=\frac{(4)(-1)}{3\cdot2}a_1=-\frac{2}{3}a_1$ .
$a_5=\frac{(5)(1)}{5\cdot4}a_3=\frac{6}{20}a_3=\frac{3}{10}a_3=\frac{3}{10}\!\left(-\tfrac23\right)a_1=-\frac15a_1$ . Why this step? — apply the recurrence with $k=3$ : numerator $(k+3)(k-2)=6\cdot1=6$ , denominator $(k+2)(k+1)=5\cdot4=20$ , so factor $=6/20=3/10$ ; then substitute $a_3=-\tfrac23a_1$ .

$y=a_0(1-3x^2)+a_1\left(x-\tfrac23x^3-\tfrac15x^5-\cdots\right).$

Figure — Power series solutions — ordinary points

Common mistakes (Steel-man + fix)

Active recall

Recall Self-test (try before peeking)

Define an ordinary point.
Why are $a_0,a_1$ free?
What does Fuchs' theorem guarantee about the radius of convergence?
In Example 2, why did the even series terminate?

Answers: 1. $P,Q$ in standard form are analytic at $x_0$ . 2. Recurrence only gives $a_{k+2}$ from $a_k$ ; first two unconstrained = two integration constants. 3. ≥ distance from $x_0$ to nearest singular point. 4. Recurrence numerator $(k+3)(k-2)$ hit $0$ at $k=2$ , killing $a_4$ and all higher even terms.

Recall Feynman: explain to a 12-year-old

Some equations connect a curve to how steep it is and how curvy it is. We can't write their answer with normal functions. So we guess the answer is a giant polynomial — $a_0 + a_1x + a_2x^2+\dots$ — but we don't know the numbers. When we put this guess into the equation, the equation becomes a chain of simple sums that say: "if you tell me the first two numbers, I'll compute all the rest." So we pick the first two freely (those are our two free choices), and the machine spits out the whole infinite list. As long as the equation isn't "broken" (no dividing-by-zero spot) near our starting point, this guessed polynomial actually works.

Flashcards

What makes

x_0

an ordinary point of

y''+Py'+Qy=0

Both

P

and

Q

are analytic (have convergent power series) at

x_0

Why do exactly two free constants

a_0,a_1

appear?

The recurrence determines

a_{k+2}

from

a_k

only, leaving

a_0,a_1

undetermined = the two integration constants of a 2nd-order ODE.

What does Fuchs' theorem guarantee at an ordinary point?

Two linearly independent power-series solutions exist, convergent at least up to the nearest singular point.

First step before classifying the point?

Put the ODE in standard form

y''+Py'+Qy=0

(divide by leading coefficient).

Recurrence for

y''+y=0

a_{k+2}=-\dfrac{a_k}{(k+2)(k+1)}

, giving

\cos x

and

\sin x

Why re-index the sums after substituting?

To make every sum carry the same power

x^k

so coefficients can be combined and set to zero.

When does a power-series solution terminate into a polynomial?

When the recurrence's numerator vanishes for some

k

, zeroing all further coefficients in that branch (e.g. Legendre polynomials).

Radius of convergence relation?

R \ge

distance from

x_0

to the nearest singular point in the complex plane.

Connections

Singular points and Frobenius method — what to do when $P,Q$ are not analytic.
Legendre's equation and Legendre polynomials — terminating power series.
Taylor series and analyticity — meaning of "analytic".
Existence and uniqueness for linear ODEs — Fuchs' theorem context.
Recurrence relations — the algebraic engine here.
Hermite and Airy equations — classic ordinary-point examples.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, kuch differential equations ka solution normal functions (jaise $\sin$ , $e^x$ ) se nahi likh paate. To trick yeh hai: hum maan lete hain ki answer ek infinite polynomial hai, $y=a_0+a_1x+a_2x^2+\dots$ , par numbers $a_n$ humein nahi pata. Phir is guess ko equation mein daal dete hain. Kyunki series ko differentiate karna bas coefficients ko shift-scale karta hai, poori differential equation ek simple recurrence relation ban jaati hai — yaani "agar pehle do number bata do, baaki saare main nikaal dunga".

Lekin yeh sab tabhi guarantee se kaam karta hai jab $x_0$ ek ordinary point ho. Iska matlab: equation ko standard form $y''+Py'+Qy=0$ mein likho, aur dekho ki $P$ aur $Q$ us point par analytic hain (koi divide-by-zero nahi). Jaise $(1-x^2)y''-\dots$ mein $x=\pm1$ par dikkat hai, par $x=0$ bilkul theek — ordinary point. Fuchs ka theorem kehta hai: ordinary point par hamesha do independent power-series solutions milenge.

Sabse important baat: $a_0$ aur $a_1$ free rehte hain — yeh hamare do integration constants hain, kyunki recurrence sirf $a_{k+2}$ ko $a_k$ se nikalta hai. Toh $a_0$ wale terms se ek solution $y_1$ banta hai aur $a_1$ wale terms se doosra $y_2$ . Kabhi-kabhi (jaise Legendre equation) series beech mein hi terminate ho jaati hai aur clean polynomial ban jaata hai. Bas dhyan rakho: pehle standard form, phir re-index karke same power $x^k$ pe lao, tabhi coefficients add karo — yeh do common galtiyan hain.

Power series solutions — ordinary points

The big picture

What is an ordinary point?

HOW the method runs (the recipe)

Worked Example 1 — the simplest possible

Worked Example 2 — Legendre-type, $(1-x^2)y''-2xy'+6y=0$ about $x_0=0$

Common mistakes (Steel-man + fix)

Active recall

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Ordinary Differential Equations

Connections