Worked examples — Power series solutions — ordinary points

This is the drill page for Power series solutions at an ordinary point. The parent note showed how the machine runs. Here we run it against every kind of case the topic can throw at you — so that when an exam hands you a strange one, you have already met its cousin.

Everything is built from zero. If a symbol appears, it was defined first.

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The scenario matrix

Think of a power-series problem as having several independent "dials". Each dial can be in a few positions, and each position changes what happens. Here is every dial and every position we will cover.

Cell	What varies	Positions you must have seen
A. Leading coefficient	the number in front of $y^{\prime \prime }$	constant ($1$) vs. a polynomial ($1-x^2$) that must stay inside
B. Series behaviour	does a branch stop?	infinite (never stops) vs. terminating (becomes a polynomial)
C. Recurrence gap	which earlier $a$ feeds $a_k$	$a_{k+2}\leftarrow a_k$ (step 2) vs. $a_{k+3}\leftarrow a_k$ (step 3, Airy)
D. Extra term	is there a first-derivative term $y^{\prime }$?	present ($-2x{y}^{\prime }$) vs. absent ($y^{\prime \prime }+y=0$)
E. Starting point	is $x_0=0$?	$x_0=0$ vs. shift a nonzero $x_0$ to the origin
F. Initial conditions	are $a_0,a_1$ pinned?	free (general solution) vs. fixed by given $y(0),y^{\prime }(0)$
G. Non-homogeneous	is the right side zero?	$=0$ vs. a forcing term like $+x$
H. Word / applied	physical wrapper	a spring / diffusion phrasing that hides an ODE

Coverage map (which example hits which cell):

Example	Cells covered
1	A(const), B(infinite), C(step-2), D(no $y^{\prime }$)
2	A(poly), B(terminating), D(has $y^{\prime }$)
3	A(poly), C(step-2), degenerate check at $k=0$
4	C(step-3, Airy), B(infinite)
5	E(shift $x_0=1$)
6	F(initial conditions pin $a_0,a_1$)
7	G(non-homogeneous forcing)
8	H(word problem), F, B(terminating)

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Example 1 — constant leading coefficient, no $y^{\prime }$ term (Cells A, B∞, C-step2, D-none)

Forecast: Almost identical to $y^{\prime \prime }+y=0$, but with a plus sign flip. Guess: the coefficients will not alternate in sign, so instead of $\cos /\sin$ we should get $\cosh /\sinh$. Predict $a_{k+2}=+a_k/[(k+2)(k+1)]$.

Steps.

1. Assume $y=\sum _{n=0}^{\infty }{a}_n{x}^n$, so $y^{\prime \prime }=\sum _{n=2}^{\infty }n(n-1)a_n{x}^{n-2}$. Why this step? Every power-series solution starts by writing the unknown as an infinite polynomial and differentiating term by term.
2. Substitute: $\sum _{n=2}^{\infty }n(n-1)a_n{x}^{n-2}-\sum _{n=0}^{\infty }{a}_n{x}^n=0.$
3. Re-index the first sum with $k=n-2$: $\sum _{k=0}^{\infty }(k+2)(k+1)a_{k+2}{x}^k-\sum _{k=0}^{\infty }{a}_k{x}^k=0.$ Why this step? Both sums must carry the same power $x^k$ before we may add their coefficients.
4. Set coefficient of $x^k$ to zero: $(k+2)(k+1)a_{k+2}-a_k=0\Rightarrow \boxed{a_{k+2}=\frac{a_k}{(k+2)(k+1)}}.$ Why this step? A power series equals $0$ for all $x$ iff every coefficient is $0$.
5. Crank. Even: $a_2=\frac{a_0}{2!},a_4=\frac{a_0}{4!},\dots$ Odd: $a_3=\frac{a_1}{3!},a_5=\frac{a_1}{5!},\dots$ $y=a_0(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots )+a_1(x+\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots ).$

Verify: The two bracketed series are exactly $\cosh x$ and $\sinh x$. Numerically at $x=1$: $a_0$-series $\approx 1.5431=\cosh 1$, $a_1$-series $\approx 1.1752=\sinh 1$. Forecast confirmed — the $+$ sign killed the alternation. ✔

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Example 2 — polynomial leading coefficient, terminating branch (Cells A-poly, B-terminate, D-has $y^{\prime }$)

Forecast: The leading coefficient is a plain $1$ (no singular points anywhere), so both branches converge everywhere. The constant term $+4y$ looks like it might make a branch terminate. Guess: the even branch stops early.

Steps.

1. $y=\sum a_n{x}^n$, $y^{\prime }={\sum }_{n\ge 1}n{a}_n{x}^{n-1}$, $y^{\prime \prime }={\sum }_{n\ge 2}n(n-1)a_n{x}^{n-2}$.
2. Substitute, keeping the $x$ that multiplies $y^{\prime }$ inside: $\sum n(n-1)a_n{x}^{n-2}-2\sum n{a}_n{x}^n+4\sum a_n{x}^n=0.$ Why keep $2x$ inside? Multiplying by $x$ raises the power by one; if we pull it out we lose track of which power we are matching.
3. Re-index first sum ($k=n-2$); rename $n\to k$ in the others: ${\sum }_k[(k+2)(k+1)a_{k+2}-2k{a}_k+4a_k]x^k=0.$
4. Coefficient $=0$: $(k+2)(k+1)a_{k+2}=(2k-4)a_k\Rightarrow \boxed{a_{k+2}=\frac{2(k-2)}{(k+2)(k+1)}{a}_k}.$ Why this step? Collect the three terms that all multiply $a_k{x}^k$: $-2k+4=2(2-k)$… carefully $-2k+4=-2(k-2)$, and moving to the other side gives $+2(k-2)$.
5. The zero test. At $k=2$: numerator $2(k-2)=0\Rightarrow a_4=0$, and every higher even term dies. Why does this matter? This is exactly cell B-terminate: the even branch becomes a finite polynomial.
   • $a_2=\frac{2(0-2)}{2\cdot 1}{a}_0=\frac{-4}{2}{a}_0=-2a_0$, then $a_4=0$.
   • Even branch: $y_{\text{even}}=a_0(1-2x^2)$.
6. Odd branch does not stop (numerator $2(k-2)\ne 0$ for odd $k\ge 1$):
   • $a_3=\frac{2(1-2)}{3\cdot 2}{a}_1=\frac{-2}{6}{a}_1=-\frac{1}{3}{a}_1.$
   • $a_5=\frac{2(3-2)}{5\cdot 4}{a}_3=\frac{2}{20}(-\frac{1}{3})a_1=-\frac{1}{30}{a}_1.$

$y=a_0(1-2x^2)+a_1(x-\frac{1}{3}{x}^3-\frac{1}{30}{x}^5-\cdots ).$

Verify: The polynomial $1-2x^2$ is (up to a scale) the Hermite polynomial $H_2(x)=4x^2-2$: indeed $-2(1-2x^2)=4x^2-2=H_2(x)$. Substitute $y=1-2x^2$ back: $y^{\prime \prime }=-4$, $-2x{y}^{\prime }=-2x(-4x)=8x^2$, $4y=4-8x^2$; sum $=-4+8x^2+4-8x^2=0$. ✔

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Example 3 — degenerate check at $k=0$ (Cells A-poly, C-step2, zero input)

Forecast: This is the parent's Example 2 but with $6y$ replaced by $2y$. The number that mattered was the constant term. Guess: the termination happens at a different $k$ now — maybe $k=1$, killing the odd branch instead.

Steps.

1. Substitute (same shape as parent): $\sum n(n-1)a_n{x}^{n-2}-\sum n(n-1)a_n{x}^n-2\sum n{a}_n{x}^n+2\sum a_n{x}^n=0.$
2. Re-index and collect coefficient of $x^k$: $(k+2)(k+1)a_{k+2}-[k(k-1)+2k-2]a_k=0.$ Why this step? All three "$x^n$" sums share the power; combine their $a_k$ multipliers.
3. Simplify the bracket: $k(k-1)+2k-2=k^2+k-2=(k+2)(k-1)$. $\boxed{a_{k+2}=\frac{(k+2)(k-1)}{(k+2)(k+1)}{a}_k=\frac{k-1}{k+1}{a}_k}.$
4. Smallest-index (degenerate) test, $k=0$: $a_2=\frac{0-1}{0+1}{a}_0=-a_0$. Then $k=2$: $a_4=\frac{1}{3}{a}_2=-\frac{1}{3}{a}_0$ — even branch does not stop. Why check $k=0$ separately? The general formula might contain a $0$ or a division that misbehaves at the boundary; here $k=0$ is perfectly fine, but the discipline is to always test it.
5. Odd test, $k=1$: numerator $k-1=0\Rightarrow a_3=0$ and all higher odd terms vanish.
   • Odd branch: $y_{\text{odd}}=a_1\cdot x$ — a linear polynomial.

$y=a_0(1-x^2-\frac{1}{3}{x}^4-\cdots )+a_{1}x.$

Verify: $y_2=x$ is a solution? $y^{\prime \prime }=0$, $-2x{y}^{\prime }=-2x$, $2y=2x$; $(1-x^2)\cdot 0-2x+2x=0$. ✔ (Forecast right: the odd branch terminated this time.)

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Example 4 — three-term recurrence, Airy's equation (Cells C-step3, B∞)

Forecast: Multiplying $y$ by $x$ shifts its power up by one, so the two sums will land on powers differing by an odd gap. Guess: $a_k$ links to $a_{k-3}$ (a step-3 recurrence), which is new — the previous examples all stepped by 2.

Steps.

1. $y^{\prime \prime }={\sum }_{n\ge 2}n(n-1)a_n{x}^{n-2}$, and $xy={\sum }_{n\ge 0}{a}_n{x}^{n+1}$.
2. Substitute: $\sum _{n\ge 2}n(n-1)a_n{x}^{n-2}-\sum _{n\ge 0}{a}_n{x}^{n+1}=0.$
3. Re-index both to power $x^k$: first with $k=n-2$, second with $k=n+1$ (so $n=k-1$): ${\sum }_{k\ge 0}(k+2)(k+1)a_{k+2}{x}^k-{\sum }_{k\ge 1}{a}_{k-1}{x}^k=0.$ Why this step? Look at the figure: $y^{\prime \prime }$ contributes to $x^k$ from coefficient $a_{k+2}$, while $-xy$ contributes from $a_{k-1}$ — a gap of 3 between the indices $k+2$ and $k-1$.
4. The $k=0$ term comes only from the first sum: $(2)(1)a_2=0\Rightarrow a_2=0.$ For $k\ge 1$: $(k+2)(k+1)a_{k+2}=a_{k-1}\Rightarrow \boxed{a_{k+2}=\frac{a_{k-1}}{(k+2)(k+1)}}.$
5. Crank in threes. Since $a_2=0$, the "$\equiv 2(\mathrm{mod}3)$" family stays zero.
   • From $a_0$: $a_3=\frac{a_0}{3\cdot 2}=\frac{a_0}{6}$, $a_6=\frac{a_3}{6\cdot 5}=\frac{a_0}{180}$.
   • From $a_1$: $a_4=\frac{a_1}{4\cdot 3}=\frac{a_1}{12}$, $a_7=\frac{a_4}{7\cdot 6}=\frac{a_1}{504}$.

$y=a_0(1+\frac{x^3}{6}+\frac{x^6}{180}+\cdots )+a_1(x+\frac{x^4}{12}+\frac{x^7}{504}+\cdots ).$

Verify: Plug $y_1=1+\frac{x^3}{6}+\frac{x^6}{180}$ (truncated): $y_1^{\prime \prime }=x+\frac{x^4}{6}$; $x{y}_1=x+\frac{x^4}{6}+\frac{x^7}{180}$. Difference $y_1^{\prime \prime }-x{y}_1=-\frac{x^7}{180}$ — zero up to the order we kept. ✔

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Example 5 — expanding about a nonzero point (Cell E, shift)

Forecast: $x_0=1$ isn't the origin, so the series $\sum a_n{x}^n$ won't work directly. Guess: substitute $t=x-1$ to slide the problem to the origin, and it becomes Airy-like ($y^{\prime \prime }+ty=0$).

Steps.

1. Let $t=x-1$, so $x-1=t$ and derivatives are unchanged ($\frac{d}{dx}=\frac{d}{dt}$ since $\frac{dt}{dx}=1$). Why this step? Power-series machinery needs the expansion centre at $0$; shifting the variable moves $x_0=1$ to $t=0$. (This is the cell-E move.)
2. The ODE becomes $y^{\prime \prime }+ty=0$ with $y={\sum }_n{a}_n{t}^n$ — this is Airy with a plus sign.
3. Same re-indexing as Example 4: for $k\ge 1$, $(k+2)(k+1)a_{k+2}=-a_{k-1}$, and $a_2=0$. $\boxed{a_{k+2}=-\frac{a_{k-1}}{(k+2)(k+1)}}.$
4. Crank: $a_3=-\frac{a_0}{6}$, $a_4=-\frac{a_1}{12}$, $a_6=-\frac{a_3}{30}=\frac{a_0}{180}$.
5. Restore $t=x-1$: $y=a_0(1-\frac{(x-1{)}^3}{6}+\frac{(x-1{)}^6}{180}-\cdots )+a_1((x-1)-\frac{(x-1{)}^4}{12}+\cdots ).$

Verify: With $u=x-1$, $y_1=1-\frac{u^3}{6}$: $y_1^{\prime \prime }=-u$, $(x-1)y_1=u(1-\frac{u^3}{6})=u-\frac{u^4}{6}$; sum $=-u+u-\frac{u^4}{6}=-\frac{u^4}{6}$, zero to leading order. ✔ (Forecast right: shift turned it into a $+t$ Airy.)

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Example 6 — initial conditions pin $a_0,a_1$ (Cell F)

Forecast: From the parent, $a_0$ multiplies $\cos x$ and $a_1$ multiplies $\sin x$. Since $y(0)=a_0$ and $y^{\prime }(0)=a_1$, guess $y=2\cos x-3\sin x$.

Steps.

1. Recall the general series (parent Example 1): $y=a_0(1-\frac{x^2}{2!}+\cdots )+a_1(x-\frac{x^3}{3!}+\cdots )$.
2. Evaluate at $x=0$: only the constant terms survive, so $y(0)=a_0$. Why this step? Every $x^n$ with $n\ge 1$ vanishes at $x=0$, isolating $a_0$.
3. So $a_0=2$.
4. Differentiate and set $x=0$: $y^{\prime }(0)=a_1$ (only the first-power term of the $a_1$ series survives). So $a_1=-3$. Why this step? $y^{\prime }=a_1(1-\frac{x^2}{2!}+\cdots )+\dots$; at $x=0$ only $a_1\cdot 1$ remains.
5. Substitute: $y=2(1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots )-3(x-\frac{x^3}{6}+\cdots )=2\cos x-3\sin x.$

Verify: $y(0)=2\cdot 1-3\cdot 0=2$ ✔; $y^{\prime }(x)=-2\sin x-3\cos x$, so $y^{\prime }(0)=-3$ ✔. Also $y^{\prime \prime }+y=(-2\cos x+3\sin x)+(2\cos x-3\sin x)=0$ ✔.

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Example 7 — non-homogeneous forcing (Cell G)

Forecast: The homogeneous part gives $\cos x,\sin x$. The right side $x$ is a first-degree polynomial, so guess a particular series that produces exactly one leftover $x^1$ — probably $y_p=x$ itself.

Steps.

1. Assume $y=\sum a_n{x}^n$. Then (from Example 1's algebra) the left side is ${\sum }_k[(k+2)(k+1)a_{k+2}+a_k]x^k$.
2. The right side is $x=1\cdot x^1$. Match coefficient of each power:
   • $x^0$: $2a_2+a_0=0$.
   • $x^1$: $6a_3+a_1=1$ — the forcing enters only here.
   • $x^k,k\ge 2$: $(k+2)(k+1)a_{k+2}+a_k=0$. Why this step? The equation now says "coefficient of $x^k$ on the left = coefficient of $x^k$ on the right"; the right side is nonzero only at $x^1$.
3. Solve: $a_2=-\frac{a_0}{2}$; from the $x^1$ row $a_3=\frac{1-a_1}{6}$; then $a_4=-\frac{a_2}{12}=\frac{a_0}{24}$, $a_5=-\frac{a_3}{20}=-\frac{1-a_1}{120}$.
4. Notice the "$1$" travels down the odd chain. Collecting the piece that carries the forcing: $y_p=x(\text{take }{a}_0=0,a_1=1\text{-driven part isolates to }{y}_p=x).$ General solution: $y=a_0\cos x+{\tilde{a}}_1\sin x+x$, where ${\tilde{a}}_1$ absorbs the homogeneous freedom.

Verify: $y_p=x\Rightarrow y_p^{\prime \prime }=0$, $y_p^{\prime \prime }+y_p=0+x=x$ ✔. Full: $y=A\cos x+B\sin x+x$ gives $y^{\prime \prime }+y=x$ for any $A,B$. ✔

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Example 8 — word problem, terminating polynomial (Cells H, F, B-terminate)

Forecast: "Symmetric and flat at centre" means $a_1=0$, so only the even branch survives. The parent showed the even branch of exactly this equation terminates as $1-3x^2$. Guess: $y=1-3x^2$.

Steps.

1. From the parent's Example 2, the recurrence is $a_{k+2}=\frac{(k+3)(k-2)}{(k+2)(k+1)}{a}_k$, and $y(0)=a_0$, $y^{\prime }(0)=a_1$. Why this step? Reuse the already-derived recurrence — this is the same ODE, only the constants change.
2. Apply the conditions: $y(0)=1\Rightarrow a_0=1$; $y^{\prime }(0)=0\Rightarrow a_1=0$. Why this step? Cell-F rule: $a_0=y(0)$, $a_1=y^{\prime }(0)$.
3. With $a_1=0$ the entire odd branch is zero (all built from $a_1$). Only the even branch remains.
4. Even branch: $a_2=\frac{(3)(-2)}{2\cdot 1}{a}_0=-3$, and $a_4=0$ (numerator $(k-2)=0$ at $k=2$), so it terminates. Why bounded? The non-terminating odd branch would blow up as $x\to \pm 1$ (near the singular points); demanding a polynomial is what "physically bounded on the rod" means. This is why physics selects Legendre polynomials — see Legendre's equation and Legendre polynomials.
5. $y=1-3x^2.$

Verify: $y(0)=1$ ✔, $y^{\prime }(0)=0$ ✔ (since $y^{\prime }=-6x$). ODE: $y^{\prime \prime }=-6$, $-2x{y}^{\prime }=-2x(-6x)=12x^2$, $6y=6-18x^2$; $(1-x^2)(-6)+12x^2+6-18x^2=-6+6x^2+12x^2+6-18x^2=0$ ✔.

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Wrap-up recall

Recall Which cell did each example test?

Ex1 :::: constant leading coeff, infinite step-2 series, no $y^{\prime }$. Ex2 :::: polynomial coeff, terminating even branch (Hermite). Ex3 :::: degenerate $k=0$ check, odd branch terminates. Ex4 :::: three-term (step-3) recurrence, Airy. Ex5 :::: shifting $x_0=1$ to the origin. Ex6 :::: initial conditions pin $a_0=y(0),a_1=y^{\prime }(0)$. Ex7 :::: non-homogeneous forcing on the right. Ex8 :::: applied word problem selecting a bounded Legendre polynomial.

Linked prerequisites for review: Recurrence relations, Taylor series and analyticity, Existence and uniqueness for linear ODEs, Singular points and Frobenius method.