Power-Series Solutions at Ordinary Points, differential equations lesson

Take the equation $y'' + 2xy' = 0$ , where $x$ is the independent variable and $y' = dy/dx$ , with the data $y(0) = 0$ and $y'(0) = 1$ . The coefficient $2x$ changes with $x$ , so the exponential trial $e^{rx}$ that works for constant coefficients is of no use here. Which function solves this problem? Writing $y$ as a power series about $0$ turns the equation into a rule that produces the coefficients one at a time. This lesson explains when that approach is justified, how to line up the powers, and how the two starting values pick out one solution, which here turns out to be $\int_0^x e^{-u^2}\,du$ .

Key terms

Ordinary point

Definition Write a second-order linear homogeneous equation in the independent variable $x$ in normalized form,

y'' + p(x)\,y' + q(x)\,y = 0.

A function is analytic at $x_0$ if it equals a power series in $x - x_0$ that converges on some open interval around $x_0$ . The point $x_0$ is an ordinary point of the equation if $p$ and $q$ are both analytic at $x_0$ , and a singular point otherwise. For an equation $P(x)\,y'' + Q(x)\,y' + R(x)\,y = 0$ whose coefficients are polynomials with no common factor, the normalized coefficients are $p = Q/P$ and $q = R/P$ , and $x_0$ is an ordinary point exactly when $P(x_0) \ne 0$ .

In plain words Near an ordinary point, the coefficients of the normalized equation are themselves given by power series. Nothing in the equation breaks down there, and its solutions can be built as power series too.

Example In $y'' + 2xy' = 0$ the coefficients $p(x) = 2x$ and $q(x) = 0$ are polynomials, so every real number is an ordinary point. For $(1 - x^2)\,y'' - 2x\,y' + 6y = 0$ the leading coefficient vanishes only at $x = 1$ and $x = -1$ . So $x_0 = 0$ is an ordinary point, while $1$ and $-1$ are singular points, where the normalized coefficients

\begin{gathered} p(x) = \frac{-2x}{1 - x^2}, \\[0.4em] q(x) = \frac{6}{1 - x^2} \end{gathered}

are not even defined.

Power-series solution

Definition A power-series solution about $x_0$ is a solution of the form

y = \sum_{n=0}^{\infty} a_n (x - x_0)^n

whose series converges on an open interval $\lvert x - x_0 \rvert < \rho$ with $\rho > 0$ and satisfies the equation there. Inside that interval the series can be differentiated term by term as often as needed, and the differentiated series converge on the same interval. The first derivative is

y' = \sum_{n=1}^{\infty} n\,a_n (x - x_0)^{n-1},

and differentiating once more gives a series for $y''$ that starts at $n = 2$ . It follows that the coefficients are the Taylor coefficients $a_n = y^{(n)}(x_0)/n!$ . Substituting the trial series into the equation and matching powers is a formal computation, and its result is a solution once the series is known to converge. At an ordinary point this is guaranteed. For every choice of $a_0$ and $a_1$ there is exactly one power-series solution about $x_0$ with these first two coefficients, and its series converges on some interval around $x_0$ . The next lesson shows how large that interval is guaranteed to be.

In plain words Instead of guessing a formula for $y$ , treat its Taylor coefficients as the unknowns and let the equation decide them.

Example For $y'' + 2xy' = 0$ with $y(0) = 0$ and $y'(0) = 1$ , the steps in the cards below give

\begin{aligned} y &= \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{k!\,(2k+1)} \\[0.4em] &= x - \frac{x^3}{3} + \frac{x^5}{10} - \cdots. \end{aligned}

Differentiating term by term cancels each factor $2k + 1$ and leaves the exponential series with $-x^2$ in place of the variable,

y' = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{k!} = e^{-x^2},

which converges for every real $x$ . Together with $y(0) = 0$ this identifies the solution as

y(x) = \int_0^x e^{-u^2}\,du,

which also equals $\tfrac{\sqrt{\pi}}{2}\operatorname{erf}(x)$ in terms of the error function. A direct check confirms it. The fundamental theorem of calculus gives $y' = e^{-x^2}$ , then $y'' = -2xe^{-x^2}$ , so $y'' + 2xy' = 0$ for every real $x$ , and the integral has value $0$ and slope $1$ at $x = 0$ .

Index shift

Definition An index shift relabels the terms of a sum so that it shows a chosen power. Substituting $n = m + k$ in a sum that starts at $n = k$ gives

\begin{aligned} &\sum_{n=k}^{\infty} c_n (x - x_0)^{n-k} \\[0.4em] &\qquad = \sum_{m=0}^{\infty} c_{m+k} (x - x_0)^{m}. \end{aligned}

Both sides list the same terms in the same order. The index of the coefficient and the lower limit change together with the exponent, and afterward the index may be called $n$ again. Terms that are zero, such as a term with a factor $n$ at $n = 0$ , may be added or dropped so that all sums start at the same index.

In plain words Shifting the index is bookkeeping. It changes how the terms are labeled, never which terms are there, and it lets every sum in the equation talk about the same power.

Example With $y = \sum a_n x^n$ about $0$ , the second derivative is

y'' = \sum_{n=2}^{\infty} n(n-1)\,a_n x^{n-2}.

Replacing $n$ by $n + 2$ throughout writes the same sum as

\sum_{n=0}^{\infty} (n+2)(n+1)\,a_{n+2}\,x^{n}.

In the other term, the factor $x$ raises each power of $y'$ by one, so no shift is needed,

\begin{aligned} 2xy' &= \sum_{n=1}^{\infty} 2n\,a_n x^{n} \\[0.4em] &= \sum_{n=0}^{\infty} 2n\,a_n x^{n}, \end{aligned}

where the second line adds the term for $n = 0$ , which is zero. A common slip changes the exponent but not the coefficient, writing $\sum n(n-1)\,a_n x^n$ for $y''$ . That sum has constant term $0$ , while the true constant term of $y''$ is $2a_2$ .

Coefficient recurrence

Definition Once every sum in the equation is written with the power $(x - x_0)^n$ and starts at the same index, the equation reads $\sum c_n (x - x_0)^n = 0$ on an interval around $x_0$ , where each $c_n$ is a combination of the unknown coefficients. A power series that is zero at every point of an open interval around its center has all its coefficients zero, so $c_n = 0$ for every $n$ . Solving $c_n = 0$ for the coefficient with the highest index gives the coefficient recurrence, a formula for a later coefficient in terms of earlier ones. Low powers that do not fit the general pattern give separate equations, which must be solved one by one.

In plain words A power series is zero on an interval only if every one of its coefficients is zero. Setting each coefficient of the equation to zero gives a rule that builds the solution one coefficient at a time.

Example Adding the two shifted sums for $y'' + 2xy' = 0$ gives $(n+2)(n+1)\,a_{n+2} + 2n\,a_n = 0$ for every $n \ge 0$ , so

a_{n+2} = \frac{-2n\,a_n}{(n+2)(n+1)}.

At $n = 0$ the recurrence gives $a_2 = 0$ , whatever $a_0$ is. At $n = 1$ it gives $a_3 = -\tfrac{2}{6}a_1 = -\tfrac13 a_1$ , and at $n = 3$ it gives $a_5 = -\tfrac{6}{20}a_3$ .

Initial coefficients

Definition At an ordinary point $x_0$ of a second-order equation, the recurrence leaves $a_0$ and $a_1$ free and determines every later coefficient from them. These two initial coefficients carry the initial data,

a_0 = y(x_0), \quad a_1 = y'(x_0),

so the initial-value problem with $y(x_0) = k_0$ and $y'(x_0) = k_1$ is solved by the series with $a_0 = k_0$ and $a_1 = k_1$ . Let $y_1$ be the series solution with $(a_0, a_1) = (1, 0)$ and $y_2$ the one with $(a_0, a_1) = (0, 1)$ . On an interval around $x_0$ where both series converge, every solution is

y = a_0\,y_1 + a_1\,y_2,

and the pair is a fundamental set, since its Wronskian at $x_0$ is $y_1(x_0)\,y_2'(x_0) - y_1'(x_0)\,y_2(x_0) = 1$ .

In plain words A second-order equation needs two numbers to fix a solution. In the series method those two numbers are simply the first two coefficients, the value and the slope at the center.

Example For $y'' + 2xy' = 0$ , the data $y(0) = 0$ and $y'(0) = 1$ give $a_0 = 0$ and $a_1 = 1$ . The recurrence then produces $a_3 = -\tfrac13$ , $a_5 = \tfrac{1}{10}$ and $a_7 = -\tfrac{1}{42}$ , while every even coefficient is zero. The two basic choices give $y_1 = 1$ , because $(a_0, a_1) = (1, 0)$ makes every coefficient after $a_0$ zero, and $y_2 = \int_0^x e^{-u^2}\,du$ , the solution just found. Both series converge for every real $x$ , so the general solution on the whole real line is

y = C_0 + C_1\int_0^x e^{-u^2}\,du,

and the data above select $C_0 = 0$ and $C_1 = 1$ .

Even and odd coefficient chains

Definition When the recurrence expresses $a_{n+2}$ through $a_n$ alone, the coefficients split into two chains that never interact. The even chain $a_0, a_2, a_4, \dots$ consists of multiples of $a_0$ , and the odd chain $a_1, a_3, a_5, \dots$ consists of multiples of $a_1$ . The solution then splits as $y = a_0\,y_1 + a_1\,y_2$ , where $y_1$ contains only even powers of $x - x_0$ and $y_2$ only odd powers. If the factor in front of $a_n$ vanishes at some index, every later coefficient of that chain is zero and the chain sums to a polynomial. A recurrence that links indices three apart, or one that involves more than two coefficients, splits differently or not at all, so the chains have to be read off the recurrence of the equation at hand.

In plain words A two-step recurrence hops over every other coefficient. One chain of hops starts at $a_0$ and the other at $a_1$ , so the value and the slope at the center each control their own half of the series.

Example For $y'' + 2xy' = 0$ the even chain starts with $a_2 = 0 \cdot a_0 = 0$ , so $a_4$ , $a_6$ and all later even coefficients vanish, and the even part of every solution is the constant $a_0$ . The odd chain with $a_1 = 1$ follows the formula

a_{2k+1} = \frac{(-1)^k}{k!\,(2k+1)}

for every $k \ge 0$ . A direct check confirms it. With $n = 2k + 1$ the recurrence multiplies this coefficient by $-2(2k + 1)/((2k + 3)(2k + 2))$ , and the product simplifies to $(-1)^{k+1}/((k+1)!\,(2k + 3))$ , which is the same formula with $k + 1$ in place of $k$ .

Check yourself

Use a power series about $0$ to solve $y'' - y = 0$ with $y(0) = 1$ and $y'(0) = 1$ . Which function is it?

Show the answer

Write $y = \sum a_n x^n$ . The shifted second derivative gives $(n + 2)(n + 1)\,a_{n+2} - a_n = 0$ for every $n \ge 0$ , so

a_{n+2} = \frac{a_n}{(n+2)(n+1)}.

The data give $a_0 = 1$ and $a_1 = 1$ . If $a_n = 1/n!$ , the recurrence gives $a_{n+2} = 1/(n+2)!$ , so by induction along both chains $a_n = 1/n!$ for every $n$ and

y = \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x.

The even chain alone sums to $\cosh x$ and the odd chain alone to $\sinh x$ , so the solution is $\cosh x + \sinh x = e^x$ , and indeed $(e^x)'' - e^x = 0$ .

For $y'' + 2xy' = 0$ with $y(0) = 0$ and $y'(0) = 1$ , a classmate keeps the first two terms of the series and claims that $y = x - x^3/3$ solves the problem. Is the claim right?

Show the answer

No. For this polynomial $y' = 1 - x^2$ and $y'' = -2x$ , so $y'' + 2xy' = -2x + 2x - 2x^3$ . The terms in $x$ cancel and leave

y'' + 2xy' = -2x^3.

The polynomial has the right value and slope at $0$ , and the equation is satisfied through degree two, but the residual $-2x^3$ is not zero, so it is only an approximation near $0$ . Adding terms pushes the residual to higher powers. With $x^5/10$ added it becomes $x^5$ , and with $-x^7/42$ also added it becomes $-x^7/3$ . Only the full series, whose sum is $\int_0^x e^{-u^2}\,du$ , makes the residual vanish.

Explain why $x_0 = 0$ is an ordinary point of Airy’s equation $y'' - xy = 0$ . Find its coefficient recurrence and the solution with $y(0) = 1$ and $y'(0) = 0$ through the term in $x^6$ . Do the coefficients split into even and odd chains?

Show the answer

The normalized coefficients $p(x) = 0$ and $q(x) = -x$ are polynomials, so every point is ordinary. With $y = \sum a_n x^n$ , a shift gives

\begin{aligned} xy &= \sum_{n=0}^{\infty} a_n x^{n+1} \\[0.4em] &= \sum_{n=1}^{\infty} a_{n-1} x^{n}. \end{aligned}

The constant term of the equation is $2a_2 = 0$ , so $a_2 = 0$ . For $n \ge 1$ the coefficient of $x^n$ gives

a_{n+2} = \frac{a_{n-1}}{(n+2)(n+1)}.

This recurrence links indices three apart, so the chains are $a_0, a_3, a_6, \dots$ and $a_1, a_4, a_7, \dots$ and $a_2, a_5, a_8, \dots$ , and the last one is zero because $a_2 = 0$ . With $a_0 = 1$ and $a_1 = 0$ the recurrence gives $a_3 = \tfrac16$ , $a_6 = \tfrac{1}{6 \cdot 30} = \tfrac{1}{180}$ , and $a_4 = a_5 = 0$ , so

y = 1 + \frac{x^3}{6} + \frac{x^6}{180} + \cdots.

The chains are not even and odd, since this one solution already mixes the odd power $x^3$ with the even power $x^6$ .

Power-Series Solutions at Ordinary Points

Key terms

Ordinary point

Power-series solution

Index shift

Coefficient recurrence

Initial coefficients

Even and odd coefficient chains

Check yourself

How this lesson connects

Builds on

Used later in