Series Solutions

Why this is legitimate: a power series can be differentiated term by term inside its interval of convergence, so every term of the equation becomes a power series, and two power series that agree on an interval agree coefficient by coefficient (proved below). Matching the coefficient of each $x^n$ therefore loses no information — the recurrence is the equation, rewritten. For $y' = y$ the coefficient of $x^n$ on the left is $(n+1)a_{n+1}$ and on the right $a_n$; with $a_0 = 1$ the partial sums $1 + x$, $1 + x + \tfrac{x^2}{2}$, $1 + x + \tfrac{x^2}{2} + \tfrac{x^3}{6}$ close in on $e^x$.

Why the recurrence reaches back three places: in $y'' = xy$ the second derivative lowers every power by two, while multiplying by $x$ raises it by one, so the coefficient of $x^n$ links $a_{n+2}$ to $a_{n-1}$. With $a_1 = 0$ and $a_2 = 0$ only the powers $0, 3, 6, \ldots$ survive. The series converges for every $x$, but each partial sum is accurate only near $0$; more terms reach farther out, as the figure shows on the oscillating side $x < 0$.

Where the recurrence comes from: substituting gives $\sum_{n\ge1} na_nx^{n-1} = \sum_{n\ge0} 2a_nx^{n+1}$. The coefficient of $x^{n-1}$ is $na_n$ on the left and $2a_{n-2}$ on the right (for $n \ge 2$); for $n = 1$ the right side has no constant term, so $a_1 = 0$ and every odd coefficient vanishes. From $a_0 = 1$: $a_2 = a_0 = 1$, $a_4 = \tfrac{2a_2}{4} = \tfrac12$, $a_6 = \tfrac{2a_4}{6} = \tfrac16$, and in general $a_{2k} = \tfrac{1}{k!}$ — the series of $e^u$ with $u = x^2$.

Why exactly two are free: the coefficient of $x^n$ in $y''$ is $(n+2)(n+1)a_{n+2}$, and the other terms involve only $a_0, \ldots, a_{n+1}$, so each matching equation determines $a_{n+2}$ from earlier coefficients. Nothing ever determines $a_0$ or $a_1$ — the series version of the fact that a second-order initial-value problem needs two numbers. The two solutions are independent because their Wronskian at $x_0$ is $y_1y_2' - y_1'y_2 = 1\cdot1 - 0\cdot0 = 1$.

Step 1 — match the coefficient of $x^n$: $(n+2)(n+1)a_{n+2} + na_n + a_n = 0$, and since $na_n + a_n = (n+1)a_n$, the factor $n + 1$ cancels: $a_{n+2} = -\dfrac{a_n}{n+2}$.

Step 2 — seed with $a_0 = 1$, $a_1 = 0$. Every odd coefficient inherits $a_1 = 0$; the even ones are $a_2 = -\tfrac12$, $a_4 = \tfrac18$, $a_6 = -\tfrac1{48}$.

Step 3 — recognize the pattern $a_{2k} = \dfrac{(-1)^k}{2^kk!}$: the series is $\sum \dfrac{(-x^2/2)^k}{k!} = e^{-x^2/2}$.

Step 4 — check without series: $y'' + xy' + y = (y' + xy)'$, and $y = e^{-x^2/2}$ has $y' = -xy$, so $y' + xy = 0$ and the equation holds. The recurrence found a closed form nobody had to guess.

Step 1 — even solution, $a_0 = 1$, $a_1 = 0$: $a_2 = \dfrac{2(-2)}{2\cdot1} = -2$, and then $a_4 = \dfrac{2\cdot0}{4\cdot3}a_2 = 0$. The factor $n - 2$ vanishes at $n = 2$, so every later even coefficient is $0$ as well: $y_1 = 1 - 2x^2$, a polynomial. Check: $y_1'' = -4$ and $-2x\,y_1' = 8x^2$, so $-4 + 8x^2 + 4(1 - 2x^2) = 0$.

Step 2 — odd solution, $a_0 = 0$, $a_1 = 1$: $a_3 = \dfrac{2(-1)}{3\cdot2} = -\tfrac13$, $a_5 = \dfrac{2\cdot1}{5\cdot4}a_3 = -\tfrac1{30}$, $a_7 = \dfrac{2\cdot3}{7\cdot6}a_5 = -\tfrac1{210}$; the factor $n - 2$ is never $0$ for odd $n$, so $y_2 = x - \tfrac{x^3}{3} - \tfrac{x^5}{30} - \tfrac{x^7}{210} - \cdots$ never stops.

Step 3 — the general solution is $a_0y_1 + a_1y_2$. This is Hermite's equation $y'' - 2xy' + 2\lambda y = 0$ with $\lambda = 2$; whenever $\lambda$ is a whole number, one of the two series terminates, giving the Hermite polynomials of the quantum oscillator.

Why complex zeros count: a power series converges on a disk in the complex plane, and the disk cannot contain a point where the solution may blow up or branch. The real line can look perfectly clear while a singular point sits just off it at $\pm i$ — the series still feels it, exactly as the geometric series $1 - x^2 + x^4 - \cdots$ for $\dfrac{1}{1 + x^2}$ stops at $|x| = 1$.

Step 1 — cancel the common factor $n + 1$: $a_{n+2} = -\dfrac{n}{n+2}\,a_n$.

Step 2 — with $a_0 = 0$ all even coefficients vanish; from $a_1 = 1$: $a_3 = -\tfrac13$, $a_5 = -\tfrac35a_3 = \tfrac15$, $a_7 = -\tfrac57a_5 = -\tfrac17$. So $y = x - \tfrac{x^3}{3} + \tfrac{x^5}{5} - \tfrac{x^7}{7} + \cdots = \arctan x$.

Step 3 — confirm directly: the equation says $\big((1 + x^2)y'\big)' = 0$, so $y' = \dfrac{1}{1 + x^2}$ and $y = \arctan x$.

Step 4 — the radius: the leading coefficient $1 + x^2$ vanishes at $x = \pm i$, at distance $1$ from the centre, and indeed the terms $\tfrac{x^{2k+1}}{2k+1}$ grow without bound when $|x| > 1$. The function $\arctan x$ is smooth on the whole real line, yet its series about $0$ works only for $|x| \le 1$.

Step 1 — no real singular points, but the complex ones still limit the disk of convergence.

Step 2 — about $x_0 = 0$ the distance to $1 + 2i$ is $\sqrt{1^2 + 2^2} = \sqrt5 \approx 2.24$ (the same for $1 - 2i$), so both series solutions converge at least for $|x| < \sqrt5$.

Step 3 — about $x_0 = 1$ the singular points are straight above and below, at distance $2$: the guaranteed interval is $(-1, 3)$.

Step 4 — the guarantee is a lower bound; a particular solution may converge farther (here $y = x$ is a solution, a polynomial, converging everywhere), but the general one cannot be promised more.

Step 1 — sum through $x^9$ at $x = 1$: $1 + \tfrac16 + \tfrac1{180} + \tfrac1{12960} = \tfrac{15193}{12960} \approx 1.1722994$.

Step 2 — bound the tail. From $a_{12}$ on, each coefficient is at most $\tfrac{1}{15\cdot14} < \tfrac1{200}$ of the previous one, so the tail is less than $a_{12}\left(1 + \tfrac1{200} + \tfrac1{200^2} + \cdots\right) < 2a_{12} \approx 1.2\times10^{-6}$.

Step 3 — so $1.1722994 < y(1) < 1.1723006$, and $y(1) = 1.17230$ to five decimals.

Step 4 — the lesson: near the centre a handful of terms suffices, and the recurrence itself supplies the error bound. At $x = -6$ the same accuracy needs far more terms, because the powers $6^n$ fight the factorials much longer.

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

The series $\sum\tfrac{x^n}{n!}$ — the definition of $e^x$ — emerges with no prior knowledge of exponentials. $\blacksquare$

Substitute into the equation; now every sum must be re-indexed so all run over the same power $x^n$ — the shift $m = n - 2$ turns $\sum n(n-1)a_nx^{n-2}$ into $\sum (m+2)(m+1)a_{m+2}x^m$, and multiplying by $x$ shifts the other way. Collect the coefficient of each $x^n$ 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: $a_0 = y(0)$, $a_1 = y'(0)$. 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.

Step 1 — substitute: $\sum_{n\ge2} n(n-1)a_nx^{n-2} = -\sum_{n\ge0} a_nx^n$.

Step 2 — shift the left index by $2$: $\sum_{n\ge0} (n+2)(n+1)a_{n+2}x^{n} = -\sum_{n\ge0} a_nx^n$.

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

Step 4 — seed and run: $a_0 = 1$, $a_1 = 0$; then $a_2 = \tfrac{-1}{2\cdot1} = -\tfrac12$; $a_3 = 0$ (all odd coefficients inherit $a_1 = 0$); $a_4 = \tfrac{-a_2}{4\cdot3} = \tfrac{1}{24}$; $a_6 = \tfrac{-a_4}{6\cdot5} = -\tfrac{1}{720}$.

Step 5 — recognize: $1 - \tfrac{x^2}{2!} + \tfrac{x^4}{4!} - \tfrac{x^6}{6!} + \cdots = \cos x$ — the series method has derived the cosine series from the oscillator equation, not assumed it.

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

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

Step 3 — run it: $a_3 = \tfrac{a_0}{3\cdot2} = \tfrac16$; $a_4 = \tfrac{a_1}{4\cdot3} = 0$; $a_5 = \tfrac{a_2}{5\cdot4} = 0$; $a_6 = \tfrac{a_3}{6\cdot5} = \tfrac{1}{180}$.

Step 4 — the solution so far: $y = 1 + \tfrac{x^3}{6} + \tfrac{x^6}{180} + \cdots$, with nonzero coefficients only at powers $0, 3, 6, 9, \ldots$ — 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.