Power Series Solutions Of First And Second Order Equations

1. Foundational Concepts, Definitions, and Axioms

1.1 Why a new method is needed

The methods of the earlier lessons - separation of variables, exact equations, integrating factors, and the auxiliary equation for constant coefficients - all end with a formula built from elementary functions. That class of methods is now exhausted. The equation

$$
\frac{d^{2}y}{dx^{2}}-xy=0
$$

has variable coefficients, and neither substitution nor an integrating factor reduces it. Its solutions exist, are smooth, and are perfectly well behaved, but they are not expressible in terms of powers, roots, exponentials, logarithms, trigonometric functions or their inverses. The same is true of Legendre's equation

$$
\left(1-x^{2}\right)\frac{d^{2}y}{dx^{2}}-2x\frac{dy}{dx}+n(n+1)y=0 ,
$$

whose solutions are the Legendre polynomials and the Legendre functions of the second kind, and of Bessel's equation in its many forms.

A power series solution abandons the search for a closed formula and instead represents the unknown function as an infinite sum of powers of the independent variable. The representation is exact, not approximate: a convergent power series is the function. The method replaces a differential equation by an infinite system of algebraic equations, one for each coefficient, and that system is solved recursively. Nothing about the equation needs to be guessed except the shape of the answer, and the shape is dictated by the classification of the point about which the expansion is made.

1.2 Power series, convergence, and analytic functions

> Definition (power series about a point). A power series about the point $x_{0}$ is an expression of the form

$$
> \sum_{n=0}^{\infty}a_{n}\left(x-x_{0}\right)^{n}=a_{0}+a_{1}\left(x-x_{0}\right)+a_{2}\left(x-x_{0}\right)^{2}+\cdots ,
$$

> where the coefficients $a_{0},a_{1},a_{2},\dots$ are fixed numbers and $x$ is the variable.

> Definition (radius of convergence). The series has radius of convergence $R$, where $0\le R\le\infty$, if it converges absolutely for every $x$ with $\left|x-x_{0}\right|<R$ and diverges for every $x$ with $\left|x-x_{0}\right|>R$. The open interval $\left|x-x_{0}\right|<R$ is the interval of convergence.

Three familiar series show the range of possibilities. The exponential series

$$
e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\cdots
$$

converges for every real $x$, so its radius is infinite. The geometric series

$$
\frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}=1+x+x^{2}+x^{3}+\cdots
$$

converges only for $\left|x\right|<1$, so its radius is one, and the obstruction is the pole of the function at $x=1$. The series $\sum n!\,x^{n}$ converges only at $x=0$, so its radius is zero, and such a series is useless for representing a function on an interval.

> Definition (analytic function). A function $f$ is analytic at $x_{0}$ if it equals its own power series about $x_{0}$ on some interval: there are numbers $a_{n}$ and a radius $R>0$ such that

$$
> f(x)=\sum_{n=0}^{\infty}a_{n}\left(x-x_{0}\right)^{n}\qquad\text{for }\left|x-x_{0}\right|<R .
$$

Polynomials, exponentials, sines, cosines and rational functions are analytic at every point where their denominator does not vanish. The quotient of two analytic functions is analytic wherever the denominator is non-zero, which is the fact used to classify the points of a differential equation in the next subsection.

1.3 The normalised form and the classification of points

Every second-order linear equation can be written with the second derivative carrying coefficient one, by dividing by the coefficient of that derivative.

> Definition (normalised form). Let

$$
> a_{0}(x)\frac{d^{2}y}{dx^{2}}+a_{1}(x)\frac{dy}{dx}+a_{2}(x)y=0
$$

> be a second-order linear homogeneous equation with $a_{0}$ not identically zero. On any interval where $a_{0}$ does not vanish, division by $a_{0}$ produces the normalised form

$$
> \frac{d^{2}y}{dx^{2}}+p(x)\frac{dy}{dx}+q(x)y=0 ,
$$

> with $p$ equal to $a_{1}$ divided by $a_{0}$ and $q$ equal to $a_{2}$ divided by $a_{0}$.

> Definition (ordinary and singular points). A point $x_{0}$ is an ordinary point of the normalised equation if both $p$ and $q$ are analytic at $x_{0}$. A point that is not ordinary is a singular point.

The classification is made on the normalised form and not on the original equation, and this is the first place where candidates lose marks: the coefficient $a_{0}$ may vanish at a point where $p$ and $q$ are nevertheless perfectly regular, or the reverse. Three examples show the mechanics.

For the equation

$$
\left(x^{2}-1\right)\frac{d^{2}y}{dx^{2}}+3x\frac{dy}{dx}+xy=0 ,
$$

normalising gives

$$
\frac{d^{2}y}{dx^{2}}+\frac{3x}{x^{2}-1}\frac{dy}{dx}+\frac{x}{x^{2}-1}y=0 .
$$

The coefficient functions are $p$ equal to $3x$ divided by $x^{2}-1$ and $q$ equal to $x$ divided by $x^{2}-1$. Both are rational with denominator $\left(x-1\right)\left(x+1\right)$, so both are analytic except at $x=1$ and $x=-1$. Every other point, and in particular $x=0$, is ordinary. The distance from $0$ to the nearer singular point is one, so a series about $0$ is guaranteed to converge at least for $\left|x\right|<1$.

For the equation

$$
2x^{2}\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}+\left(1-x^{2}\right)y=0 ,
$$

normalising gives

$$
\frac{d^{2}y}{dx^{2}}-\frac{1}{2x}\frac{dy}{dx}+\frac{1-x^{2}}{2x^{2}}y=0 .
$$

Here $p$ is $-1$ divided by $2x$, which is not analytic at $0$, so $0$ is singular. Nevertheless the two products $xp(x)$ and $x^{2}q(x)$ are

$$
xp(x)=-\frac{1}{2},\qquad x^{2}q(x)=\frac{1-x^{2}}{2},
$$

and both are analytic at $0$, indeed analytic everywhere. The singularity is of the mild kind that the Frobenius method handles.

For the equation

$$
x^{3}\frac{d^{2}y}{dx^{2}}+xy=0 ,
$$

normalising gives $p$ equal to zero and $q$ equal to $1$ divided by $x^{2}$, so

$$
xp(x)=0,\qquad x^{2}q(x)=\frac{1}{x} .
$$

The second product is not analytic at $0$ and tends to infinity there, so the singularity is severe. No series of either template describes the solution near $0$.

1.4 Regular and irregular singular points

> Definition (regular singular point). A singular point $x_{0}$ is regular if the two products $\left(x-x_{0}\right)p(x)$ and $\left(x-x_{0}\right)^{2}q(x)$ are analytic at $x_{0}$. Equivalently, writing

$$
> p_{0}=\lim_{x\to x_{0}}\left(x-x_{0}\right)p(x),\qquad q_{0}=\lim_{x\to x_{0}}\left(x-x_{0}\right)^{2}q(x),
$$

> both limits exist and are finite, and the products are analytic.

> Definition (irregular singular point). A singular point that is not regular is irregular.

The two numbers $p_{0}$ and $q_{0}$ are not decoration. They are the only data about the coefficients that enter the indicial equation, and therefore they alone decide the powers $x^{r}$ that can appear in a solution.

The classification is a table worth memorising in the form of a test rather than a definition.

In the first row the products are polynomials, which is the strongest possible form of regularity, and the point is ordinary rather than singular. In the last row the equation is Euler's equation, whose solutions are pure powers; the products are constants, so the regularity is also maximal in its class.

1.5 The first-order equation and its three existence conditions

For a first-order equation written in the solved form

$$
\frac{dy}{dx}=f(x,y),
$$

the classical existence and uniqueness statement is local and needs three hypotheses.

> Definition (the three conditions). The equation $\frac{dy}{dx}=f(x,y)$ has a unique solution through each point of a domain $D$ of the $xy$-plane when the following hold:

> (i) $f(x,y)$ is continuous and single valued on $D$;

> (ii) $f(x,y)$ is bounded on $D$, that is $\left|f(x,y)\right|\le M$ for a fixed number $M$;

> (iii) the partial derivative $\frac{\partial f}{\partial y}$ exists and is continuous at every point of $D$.

The first condition makes the slope well defined at every point, so that a direction field exists. The second prevents the slope from becoming infinite anywhere in the domain, which would allow a solution to turn vertically. The third is the Lipschitz-type condition that rules out the branching behaviour seen in $y'$ equal to three times $y$ to the power two thirds, where two solutions pass through the same point. When all three hold, exactly one solution curve passes through each point of $D$.

A polynomial $f$ satisfies all three conditions on the whole plane, so every equation whose right-hand side is a polynomial in $x$ and $y$ has a unique local solution through every point. The interesting case for this topic is the linear one, where the right-hand side is $Q(x)-P(x)y$ with $P$ and $Q$ analytic; then the solution is itself analytic and its series can be computed coefficient by coefficient.

1.6 The two templates for a series solution

There are exactly two shapes of answer in this topic, and choosing the wrong one is fatal.

The first template belongs to an ordinary point:

$$
y=\sum_{n=0}^{\infty}c_{n}x^{n}=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+\cdots .
$$

Here every power present is a non-negative integer and the two lowest coefficients $c_{0}$ and $c_{1}$ are free. They are the values of $y$ and of $y'$ at the expansion point, so the general solution appears as $c_{0}$ times one series plus $c_{1}$ times another.

The second template belongs to a regular singular point and carries the name of Frobenius:

$$
y=x^{r}\sum_{n=0}^{\infty}c_{n}x^{n}=c_{0}x^{r}+c_{1}x^{r+1}+c_{2}x^{r+2}+\cdots ,
$$

where $c_{0}$ is non-zero and $r$ is a number still to be found. The factor $x^{r}$ allows the first power to be fractional or negative, which is exactly what the mild singularity requires. The condition $c_{0}\neq 0$ is a normalisation: if $c_{0}$ were zero the whole series could be divided by a higher power of $x$ and rewritten with a different, larger $r$.

1.7 What the method actually does

Every calculation in this topic is the same five moves in the same order.

1. Classify the expansion point, using the normalised form, and choose the template accordingly.
2. Differentiate the template term by term as many times as the equation requires.
3. Substitute into the equation and multiply out, so that the equation becomes a single sum of powers of $x$ equal to a known right-hand side.
4. Align the indices, so that every term is written as a coefficient times the same power $x^{m}$; then equate coefficients.
5. Solve the resulting recurrence from the bottom upwards, and write the answer both as a list of terms and as a sum.

Only one step is conceptually delicate, and it is step four. The alignment of indices is pure bookkeeping with no mathematics beyond a change of summation letter, yet almost every error in the topic is an alignment error. Step five is mechanical once the recurrence is right. The two roles of the free coefficients are decided at the very first alignment: the coefficient of the lowest power in the equation produces either a determination of $c_{2}$ in the ordinary case, or the indicial equation in the Frobenius case.

2. Theorems, Lemmas, and Analytical Properties

2.1 Term-by-term differentiation

> Lemma (differentiation of a power series). If $f(x)=\sum c_{n}\left(x-x_{0}\right)^{n}$ on the interval $\left|x-x_{0}\right|<R$ with $R>0$, then $f$ is differentiable there and

$$
> f'(x)=\sum_{n=1}^{\infty}n\,c_{n}\left(x-x_{0}\right)^{n-1},
$$

> the differentiated series having the same radius of convergence $R$.

The lemma is what allows the template to be substituted into a differential equation at all. Differentiating $k$ times multiplies the coefficient $c_{n}$ by the falling product $n(n-1)\cdots(n-k+1)$ and lowers the exponent by $k$; that product vanishes for $n<k$, which is why the differentiated sum may start at $n=k$ instead of $n=0$.

> Corollary. The $k$-th derivative of the template at the expansion point is $k!$ times the coefficient of $\left(x-x_{0}\right)^{k}$, so

$$
> c_{k}=\frac{f^{(k)}(x_{0})}{k!} .
$$

The corollary identifies the coefficients of a series solution with the Taylor coefficients of the solution, so a power series solution and a Taylor expansion are the same object; the differential equation is merely a device for computing the Taylor coefficients without knowing the function.

2.2 Uniqueness of the coefficients

> Lemma (identity theorem for power series). Suppose $\sum_{n=0}^{\infty}a_{n}\left(x-x_{0}\right)^{n}=0$ for every $x$ in some interval $\left|x-x_{0}\right|<R$ with $R>0$. Then $a_{n}=0$ for every $n$.

> Corollary (matching of coefficients). If two power series agree on an interval, then their coefficients agree term by term.

This is the licence to equate coefficients, and it is the step that converts one differential equation into infinitely many algebraic equations. It fails for ordinary functions: two different functions can agree on an interval without being equal, and two different trigonometric sums can agree at infinitely many points. For power series, agreement on any interval is agreement everywhere the series converge.

2.3 The index-shifting lemma

> Lemma (shift of index). Let $k$ be an integer. Then, as formal series,

$$
> \sum_{n=N}^{\infty}A_{n}x^{n+k}=\sum_{m=N+k}^{\infty}A_{m-k}x^{m},
$$

> and

$$
> \sum_{n=N}^{\infty}A_{n}x^{n-k}=\sum_{m=N-k}^{\infty}A_{m+k}x^{m}.
$$

The lemma says only that relabelling the summation letter changes the range of summation in a dictated way. Its content is entirely in the range: the lower limit moves with the exponent, and a reader who shifts the exponent but not the lower limit silently deletes or invents terms. The proof is given in Section 3.1.

> Corollary (alignment rule). A term of the form $x^{k}y$ contributes to the coefficient of $x^{m}$ the value $c_{m-k}$, and a term of the form $x^{k}y'$ contributes $\left(m-k+1\right)c_{m-k+1}$, with the understanding that a coefficient with negative index is zero.

The corollary is the working form of the lemma, and it is worth deriving once by hand for the two cases $x^{2}y''$ and $xy$, which are the two that appear in almost every question.

2.4 Existence of a series solution for a first-order equation

> Theorem (first-order linear equation with analytic coefficients). Let $P$ and $Q$ be analytic at $x_{0}$, with power series convergent for $\left|x-x_{0}\right|<R$. Then the equation

$$
> \frac{dy}{dx}+P(x)y=Q(x)
$$

> has exactly one solution analytic at $x_{0}$ with any prescribed value $y(x_{0})=y_{0}$, and its power series converges for $\left|x-x_{0}\right|<R$.

The theorem converts the local existence statement of Section 1.5 into a statement about series, and it adds the information that the radius is at least as large as the smaller radius of the two coefficient series. The proof is the recurrence of Section 3.3, together with an estimate showing that the coefficients so produced do not grow too fast; the estimate is not needed for computation.

2.5 Existence of two series solutions at an ordinary point

> Theorem (ordinary point). Let $x_{0}$ be an ordinary point of the normalised equation $y''+p(x)y'+q(x)y=0$, so that $p$ and $q$ are analytic at $x_{0}$ with series convergent for $\left|x-x_{0}\right|<R$. Then there are two solutions

$$
> y_{1}=\sum_{n=0}^{\infty}a_{n}\left(x-x_{0}\right)^{n},\qquad y_{2}=\sum_{n=0}^{\infty}b_{n}\left(x-x_{0}\right)^{n},
$$

> analytic at $x_{0}$ and linearly independent there, and every solution on the interval is a combination $A y_{1}+B y_{2}$. Both series converge at least for $\left|x-x_{0}\right|<R$.

The freedom is exactly two-dimensional because the recurrence determines $c_{2},c_{3},c_{4},\dots$ from $c_{0}$ and $c_{1}$ and never determines $c_{0}$ and $c_{1}$ themselves. The pair is conveniently normalised by taking $\left(c_{0},c_{1}\right)=\left(1,0\right)$ for $y_{1}$ and $\left(0,1\right)$ for $y_{2}$.

> Corollary (radius bound). If the only singular points of the equation are at distances $d_{1},d_{2},\dots$ from $x_{0}$, then $R$ is at least the smallest of these distances. The radius is a property of the coefficients, not of the solution.

The corollary explains why the series for the equation with singular points at $1$ and $-1$ is guaranteed only for $\left|x\right|<1$ even though the solution may be perfectly smooth beyond that point. The guarantee is local, and the true radius may be larger.

2.6 Existence of a Frobenius solution

> Theorem (Frobenius, regular singular point). Let $x_{0}$ be a regular singular point of $y''+p(x)y'+q(x)y=0$, and let $p_{0}$ and $q_{0}$ be the limits of $\left(x-x_{0}\right)p$ and $\left(x-x_{0}\right)^{2}q$ at $x_{0}$. Then the equation has at least one solution of the form

$$
> y=x^{r}\sum_{n=0}^{\infty}c_{n}x^{n},\qquad c_{0}\neq 0,
$$

> where $r$ is a root of the indicial equation of Section 2.7 and the series converges for $0<\left|x\right|<R$ with $R$ the distance to the next singular point.

The theorem promises one solution, not two. The second solution may require a logarithmic factor, and that is the content of Section 2.8. At an irregular singular point the theorem fails outright: no power of $x$ times an analytic function can reproduce the behaviour of a solution there.

2.7 The indicial equation

> Lemma (indicial equation). Let $x_{0}$ be a regular singular point and write the coefficients in the form

$$
> \left(x-x_{0}\right)p(x)=\sum_{m=0}^{\infty}p_{m}\left(x-x_{0}\right)^{m},\qquad \left(x-x_{0}\right)^{2}q(x)=\sum_{m=0}^{\infty}q_{m}\left(x-x_{0}\right)^{m}.
$$

> Then the admissible exponents $r$ in the Frobenius template are the roots of

$$
> r\left(r-1\right)+p_{0}r+q_{0}=0 .
$$

The polynomial on the left is called the indicial polynomial, and its two roots are the indicial roots. For the equation $2x^{2}y''-xy'+\left(1-x^{2}\right)y=0$ the coefficients are $p_{0}$ equal to $-\frac{1}{2}$ and $q_{0}$ equal to $\frac{1}{2}$, so the indicial equation is

$$
r\left(r-1\right)-\frac{r}{2}+\frac{1}{2}=0 ,
$$

which is $2r^{2}-3r+1=0$, that is $\left(2r-1\right)\left(r-1\right)=0$. The roots are $r=1$ and $r=\frac{1}{2}$.

2.8 The three cases for the second solution

> Theorem (form of the second solution). Let $r_{1}$ and $r_{2}$ be the indicial roots with $\operatorname{Re}r_{1}\ge\operatorname{Re}r_{2}$.

> (a) If $r_{1}-r_{2}$ is not an integer, there are two independent solutions of Frobenius type,

$$
> y_{1}=x^{r_{1}}\sum_{n=0}^{\infty}a_{n}x^{n},\qquad y_{2}=x^{r_{2}}\sum_{n=0}^{\infty}b_{n}x^{n}.
$$

> (b) If $r_{1}=r_{2}=r$, the first solution is of Frobenius type and the second has the form

$$
> y_{2}=y_{1}\ln x+x^{r}\sum_{n=0}^{\infty}b_{n}x^{n}.
$$

> (c) If $r_{1}-r_{2}=N$ is a positive integer, the second solution has the form

$$
> y_{2}=C\,y_{1}\ln x+x^{r_{2}}\sum_{n=0}^{\infty}b_{n}x^{n},
$$

> where $C$ is a constant that may turn out to be zero.

Case (c) is the one that costs marks, because the constant $C$ is not always non-zero and because the recurrence for the coefficients $b_{n}$ breaks down at the step $n=N$. The appearance of $\ln x$ is not an accident of algebra; it reflects the fact that between two roots differing by an integer the two solution families merge, in the same way that a repeated root of the auxiliary equation in the constant-coefficient case produces a factor $x$.

2.9 Superposition and the non-homogeneous problem

> Theorem (superposition). If $L$ denotes the operator $L[y]=y''+p y'+q y$, if $L[y_{h}]=0$ and $L[y_{p}]=F$, then $L[y_{p}+y_{h}]=F$. Consequently the general solution of $L[y]=F$ is $y_{p}$ plus the general solution of $L[y]=0$.

> Corollary (series forcing). If $F$ is analytic at $x_{0}$, so that $F=\sum f_{n}\left(x-x_{0}\right)^{n}$, then a series of the ordinary template may be substituted into $L[y]=F$ and coefficients matched against the coefficients $f_{n}$. The homogeneous coefficients remain free, and the particular solution is fixed once the free coefficients are assigned any convenient values.

The corollary is the practical form of the theorem: the presence of a right-hand side changes the equation for finitely or infinitely many coefficients, but it never changes the shape of the recurrence for large $n$. That is why the same bookkeeping, and the same control of indices, serves both the homogeneous and the non-homogeneous problem.

2.10 Identification with a known solution

> Corollary (identification). If a series solution has been computed and the resulting series is recognised as the expansion of an elementary function, then that function is the solution, by uniqueness.

This corollary is the cheapest possible check on a long computation. A first-order calculation that produces $1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\cdots$ must be the expansion of $e^{x}$, and a second-order calculation on $y''-y=0$ that produces $1+\frac{x^{2}}{2}+\frac{x^{4}}{24}+\cdots$ must be the expansion of $\cosh x$. If the computed terms do not match a series that is known to solve the equation, the recurrence is wrong.