8.3: Series Solutions About an Ordinary Point

Many physical applications give rise to second order linear differential equations of the form

\[\label{eq:7.2.1} P_0(x)y''+P_1(x)y'+P_2(x)y=F(x).\]

Often the solutions of these equations can’t be expressed in terms of familiar elementary functions. Therefore we’ll consider the problem of representing solutions of Equation \ref{eq:7.2.1} with series.

We'll divide \ref{eq:7.2.1} by \(P_0(x)\) to get the form we'll use throughout the remainder of this chapter:

\[\label{eq:7.2.2} y''+p(x)y'+q(x)y=f(x).\]

Definition 8.3.1

A function \(g(x)\) is analytic at \(x_0\) if it can be expressed as a convergent power series at \(x_0\) (with a positive or infinite radius of convergence) that is infinitely differentiable:

\[g(x)=\sum_{n=0}^\infty a_n(x-x_0)^n\nonumber\]

Homogeneous Equations 

Definition 8.3.2

Let \[ y''+p(x)y'+q(x)y=0.\nonumber\]

We say that \(x_0\) is an ordinary point of Equation \ref{eq:7.2.2} if \(p(x)\) and \(q(x)\) are both analytic at \(x_0\). Otherwise, \(x_0\) is a singular point.

Note: Since polynomials are continuous everywhere, if \ref{eq:7.2.1} has polynomial coefficients, then \(p(x)={P_1(x)\over P_0(x)}\), \(q(x)={P_2(x)\over P_0(x)}\), and \(f(x)={F(x)\over P_0(x)}\) from \ref{eq:7.2.2} are analytic at any point \(x_0\) that isn’t a zero of \(P_0(x)\). Therefore, if the polynomial \(P_0(x_0)\ne 0\), then \(x_0\) is an ordinary point of Equation \ref{eq:7.2.2} when \(P_0(x)\), \(P_1(x)\), and \(P_2(x)\) are all polynomials.

For Legendre’s equation,

\[(1-x^2)y''-2xy'+\alpha(\alpha+1)y=0,\nonumber\]

\(x_0=1\) and \(x_0=-1\) are singular points and all other points are ordinary points. For Bessel’s equation,

\[x^2y''+xy'+(x^2-\nu^2)y=0, \nonumber \]

\(x_0=0\) is a singular point and all other points are ordinary points. If \(P_0\) is a nonzero constant as in Airy’s equation,

\[ y''-xy=0,\nonumber\]

then every point is an ordinary point.

Since polynomials are continuous everywhere, \(P_1/P_0\) and \(P_2/P_0\) are continuous at any point \(x_0\) that isn’t a zero of \(P_0\). Therefore, if \(x_0\) is an ordinary point of Equation \ref{eq:7.2.1} and \(a_0\) and \(a_1\) are arbitrary real numbers, then the initial value problem

\[\label{eq:7.2.4} P_0(x)y''+P_1(x)y'+P_2(x)y=0, \quad y(x_0)=a_0,\quad y'(x_0)=a_1\]

has a unique solution on the largest open interval that contains \(x_0\) and does not contain any zeros of \(P_0\). To see this, we rewrite the differential equation in Equation \ref{eq:7.2.4} as

\[y''+{P_1(x)\over P_0(x)}y'+{P_2(x)\over P_0(x)}y=0 \nonumber\]

and apply Theorem 5.1.1 with \(p=P_1/P_0\) and \(q=P_2/P_0\).

In the remainder of this section we consider the problem of representing solutions of Equation \ref{eq:7.2.1} by power series that converge for values of \(x\) near an ordinary point \(x_0\).

We state the next theorem without proof.

Theorem 8.3.1

Let \[\label{eq:7.2.5} y''+p(x)y'+q(x)y=0\]

Let \(x_0\) be an ordinary point of \ref{eq:7.2.5} and \(\rho\) be the distance from \(x_0\) to the nearest singular point of \ref{eq:7.2.5}. Then every solution of \ref{eq:7.2.5}

can be represented by a power series

\[\label{eq:7.2.6} y=\sum_{n=0}^\infty a_n(x-x_0)^n\]

that converges at least on the open interval \((x_0-\rho,x_0+\rho)\).

Furthermore, \ref{eq:7.2.6} will generate the two linearly independent solutions of \ref{eq:7.2.5}.

We call Equation \ref{eq:7.2.6} a power series solution in \(x-x_0\) of Equation \ref{eq:7.2.5}. We’ll now develop a method for finding power series solutions of Equation \ref{eq:7.2.5}.

Theorem 8.3.1 implies that every solution of

\[\label{eq:7.2.7}y''+p(x)y'+q(x)y=0\]

 on \((x_0-\rho,x_0+\rho)\) can be written as

\[y=\sum_{n=0}^\infty a_n(x-x_0)^n=a_0+a_1(x-x_0)+a_2(x-x_0)^2+\cdots. \nonumber\]

Setting \(x=x_0\) in this series and in the series

\[y'=\sum_{n=1}^\infty na_n(x-x_0)^{n-1}=a_1+2a_2(x-x_0)+3a_3(x-x_0)^2+\cdots \nonumber\]

shows that \(y(x_0)=a_0\) and \(y'(x_0)=a_1\). Since every initial value problem Equation \ref{eq:7.2.4} has a unique solution, this means that \(a_0\) and \(a_1\) can be chosen arbitrarily, and \(a_2\), \(a_3\), …are uniquely determined by them.

To find \(a_2, a_3, a_4,\cdots\), we write \(P_0\), \(P_1\), and \(P_2\) in powers of \(x-x_0\), substitute

\[ \begin{align*} y &=\sum^\infty_{n=0}a_n(x-x_0)^n, \\[4pt] y' &=\sum^\infty_{n=1}na_n(x-x_0)^{n-1}, \\[4pt] y''&=\sum^\infty_{n=2}n(n-1)a_n(x-x_0)^{n-2} \end{align*}\]

into Equation \ref{eq:7.2.7}, and collect the coefficients of like powers of \(x-x_0\). This yields

\[\label{eq:7.2.8} \sum^\infty_{n=0}b_n(x-x_0)^n=0,\]

where \(\{b_0, b_1, \dots, b_n, \dots\}\) are expressed in terms of \(\{a_0, a_1, \dots,a_n, \dots\}\) and the coefficients of \(P_0\), \(P_1\), and \(P_2\), written in powers of \(x-x_0\). Equation \ref{eq:7.2.8} is satisfied if and only if \(b_n=0\) for \(n\ge0\), and all power series solutions in \(x-x_0\) of \ref{eq:7.2.7} can be obtained by choosing \(a_0\) and \(a_1\) arbitrarily and computing \(a_2\), \(a_3\), …, successively so that \(b_n=0\) for \(n\ge0\). For simplicity, we call the power series obtained this way the power series in \(x-x_0\) for the general solution of \ref{eq:7.2.7}.

Example 8.3.1

Let \(x_0\) be an arbitrary real number. Find the power series in \(x-x_0\) for the general solution of

\[\label{eq:7.2.9} y''+ y=0.\]

Solution

Here

\[y''+y=0. \nonumber\]

Since \(x_0\) is an ordinary point (for all \(x_0\)) of \ref{eq:7.2.9}, \(y=\sum_{n=0}^\infty a_n(x-x_0)^n,\) will generate two linearly independent solutions to \ref{eq:7.2.9}.

If

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

then

\[y''=\sum_{n=2}^\infty n(n-1)a_n(x-x_0)^{n-2}, \nonumber\]

so

\[\sum_{n=2}^\infty n(n-1)a_n(x-x_0)^{n-2}+\sum_{n=0}^\infty a_n(x-x_0)^n=0. \nonumber\]

To collect coefficients of like powers of \(x-x_0\), we shift the summation index in the first sum. This yields

\[\sum^\infty_{n=0}(n+2)(n+1)a_{n+2}(x-x_0)^n + \sum^\infty_{n=0}a_n(x-x_0)^n =\sum^\infty_{n=0}[(n+2)(n+1)a_{n+2}+a_n](x-x_0)^n=0, \nonumber\]

Therefore

\[\label{eq:7.2.10} a_{n+2}={-a_n\over(n+2)(n+1)},\quad n\ge0,\]

where \(a_0\) and \(a_1\) are arbitrary.

Computing the coefficients of \(x-x_0\) from Equation \ref{eq:7.2.10} yields

\[\begin{aligned} a_2 &= -{a_0\over2\cdot1},\\[4pt]a_3 &= -{a_1\over3\cdot2},\\ a_4 &= -{a_2\over4\cdot3}=-{1\over4\cdot3} \left(-{a_0\over2\cdot1}\right)= {a_0\over4\cdot3\cdot2\cdot1},\\ a_5 &= -{a_3\over5\cdot4}=-{1\over5\cdot4} \left(-{a_1\over3\cdot2}\right)= {a_1\over5\cdot4\cdot3\cdot2},\cdots\end{aligned}\nonumber \]

Thus, the general solution of Equation \ref{eq:7.2.9} can be written as

\[\begin{aligned} y&=a_0+a_1(x-x_0)-{a_0\over2\cdot1}(x-x_0)^2-{a_1\over3\cdot2}(x-x_0)^3+{a_0\over4\cdot3\cdot2\cdot1}(x-x_0)^4+{a_1\over5\cdot4\cdot3\cdot2}(x-x_0)^5+\cdots\nonumber\\

&= a_0(1-{(x-x_0)^2\over2\cdot1}+{(x-x_0)^4\over4\cdot3\cdot2\cdot1}+\cdots)+a_1((x-x_0)-{(x-x_0)^3\over3\cdot2}+{(x-x_0)^5\over5\cdot4\cdot3\cdot2}+\cdots).\end{aligned}\nonumber \]

Writing these series in closed form we get

\[\label{eq:7.2.11}y=a_0\sum_{m=0}^\infty(-1)^m{(x-x_0)^{2m}\over(2m)!} +a_1\sum_{m=0}^\infty(-1)^m{(x-x_0)^{2m+1}\over(2m+1)!}.\]

If we recall from calculus that

\[\sum_{m=0}^\infty(-1)^m{(x-x_0)^{2m}\over(2m)!}=\cos(x-x_0) \quad \text{and} \quad \sum_{m=0}^\infty(-1)^m{(x-x_0)^{2m+1}\over(2m+1)!}=\sin(x-x_0), \nonumber\]

then Equation \ref{eq:7.2.11} becomes

\[y=a_0\cos(x-x_0)+a_1\sin(x-x_0), \nonumber\]

Equations like Equation \ref{eq:7.2.10}, which define a given coefficient in the sequence \(\{a_n\}\) in terms of one or more coefficients with lesser indices are called recurrence relations. When we use a recurrence relation to compute terms of a sequence we are computing recursively.

In the example above we were able to obtain a closed formula for coefficients in the power series solution. In many cases this is very difficult, or impossible, and we must settle for computing a finite number of terms in the series, as the following example illustrates.

Example 8.3.2

Find the power series in \(x\) for the general solution of

\[\label{eq:7.2.17} (1+2x^2)y''+6xy'+2y=0.\]

Solution

Here

\[(1+2x^2)y''+6xy'+2y=y''+2x^2y''+6xy'+2y=0. \nonumber\]

Since \(0\) is an ordinary point of \ref{eq:7.2.17}, \(y=\sum_{n=0}^\infty a_nx^n,\) will generate two linearly independent solutions to \ref{eq:7.2.17}.

If

\[y=\sum_{n=0}^\infty a_nx^n \nonumber\]

then

\[y'=\sum_{n=1}^\infty na_nx^{n-1}\quad\mbox{ and }\quad y''=\sum_{n=2}^\infty n(n-1)a_nx^{n-2},\nonumber \]

so, we get

\[\begin{aligned}\sum^\infty_{n=2}n(n-1)a_nx^{n-2}+2x^2\sum^\infty_{n=2}n(n-1)a_nx^{n-2}+ 6x \sum^\infty_{n=1}na_nx^{n-1} +2 \sum^\infty_{n=0}a_nx^n= \sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum^\infty_{n=2}2n(n-1)a_nx^n+\sum^\infty_{n=1}6na_nx^n + \sum^\infty_{n=0}2a_nx^n=0.\end{aligned}\nonumber \]

Running out terms we get

\[2a_2+6a_3x+\sum_{n=4}^\infty n(n-1)a_nx^{n-2}+\sum^\infty_{n=2}2n(n-1)a_nx^n+6a_1x+\sum^\infty_{n=2}6na_nx^n +2a_0+2a_1x+ \sum^\infty_{n=2}2a_nx^n=0.\nonumber \]

Reindexing we get

\[2a_2+6a_3x+\sum_{n=2}^\infty (n+2)(n+1)a_{n+2}x^n+\sum^\infty_{n=2}2n(n-1)a_nx^n+6a_1x+\sum^\infty_{n=2}6na_nx^n +2a_0+2a_1x+ \sum^\infty_{n=2}2a_nx^n=0.\nonumber \]

We now collect like terms to obtain

\[(2a_2+2a_0)+(6a_3+8a_1)x+\sum_{n=2}^\infty [(n+2)(n+1)a_{n+2}+(2n^2+4n+2)a_n]x^n=0\nonumber\]

Equating coefficients we get

\[\begin{aligned}a_2&= -a_0,\\a_3&= -{4\over 3}a_1,\\a_{n+2}&=-{2(n+1)^2\over (n+2)(n+1)}a_n=-{2(n+1)\over n+2}a_n, \quad n=2,3,4,\cdots\end{aligned}\nonumber\]

Now substituting \(n=2,3,4,\cdots\) into the last equation yields

\[\begin{aligned} a_4 &= -{3\over2}a_2={3\over2}a_0, \\[4pt]a_5 &= -{8\over 5}a_3={32\over 15}a_1,\\[4pt]a_6&=-{5\over 3}a_4=-{5\over 2}a_0,\\[4pt]a_7&=-{12\over 7}a_5=-{128\over 35}a_1,\cdots \end{aligned}\nonumber \] 

Thus,

\[\begin{aligned}y&=a_0+a_1x-a_0x^2-{4\over 3}a_1x^3+{3\over 2}a_0x^4+{32\over 15}a_1x^5-{5\over 2}a_0x^6-{128\over 35}a_1x^7+\cdots\\&=a_0(1-x^2+{3\over 2}x^4-{5\over 2}x^6+\cdots)+a_1(x-{4\over 3}x^3+{32\over 15}x^5-{128\over 35}x^7+\cdots)\end{aligned}\nonumber\]

is the power series in \(x\) for the general solution of Equation \ref{eq:7.2.17}. Since \(P_0(x)=1+2x^2\) has no real zeros, Theorem 5.1.1 implies that every solution of Equation \ref{eq:7.2.17} is defined on \((-\infty,\infty)\).

Example 8.3.3

Find the power series in \(x-1\) for the general solution of

\[\label{eq:7.2.28} (2+4x-2x^2)y''-12(x-1)y'-12y=0.\]

Solution

We must first write the coefficient \(P_0(x)=2+4x-x^2\) in powers of \(x-1\). To do this, we write \(x=(x-1)+1\) in \(P_0(x)\) and then expand the terms, collecting powers of \(x-1\); thus,

\[\begin{aligned} 2+4x-2x^2 &= 2+4[(x-1)+1]-2[(x-1)+1]^2\\ &= 4-2(x-1)^2.\end{aligned}\nonumber \]

Therefore we can rewrite Equation \ref{eq:7.2.28} as

\[\left(4-2(x-1)^2\right)y''-12(x-1)y'-12y=0, \nonumber\]

Since \(x=1\) is an ordinary point, we will generate two linearly independent solutions to \ref{eq:7.2.28} from

 \[ \begin{align*} y &=\sum^\infty_{n=0}a_n(x-1)^n, \\[4pt] y' &=\sum^\infty_{n=1}na_n(x-1)^{n-1}, \\[4pt] y''&=\sum^\infty_{n=2}n(n-1)a_n(x-1)^{n-2} \end{align*}\]

So, we get

\[\begin{aligned}y&=4\sum^\infty_{n=2}n(n-1)a_n(x-1)^{n-2}-2(x-1)^2\sum^\infty_{n=2}n(n-1)a_n(x-1)^{n-2}-12(x-1)\sum^\infty_{n=1}na_n(x-1)^{n-1}-12\sum^\infty_{n=0}a_n(x-1)^n\\&=\sum^\infty_{n=2}4n(n-1)a_n(x-1)^{n-2}-\sum^\infty_{n=2}2n(n-1)a_n(x-1)^n-\sum^\infty_{n=1}12na_n(x-1)^n-\sum^\infty_{n=0}12a_n(x-1)^n\end{aligned}\nonumber\]

Running out terms we get

\(8a_2+24a_3(x-1)+\sum^\infty_{n=4}4n(n-1)a_n(x-1)^{n-2}-\sum^\infty_{n=2}2n(n-1)a_n(x-1)^n-12a_1(x-1)-\sum^\infty_{n=2}12na_n(x-1)^n-12a_0-12a_1(x-1)-\sum^\infty_{n=2}12a_n(x-1)^n=0.\)

Reindexing we get

\(8a_2+24a_3(x-1)+\sum^\infty_{n=2}4(n+2)(n+1)a_{n+2}(x-1)^n-\sum^\infty_{n=2}2n(n-1)a_n(x-1)^n-12a_1(x-1)-\sum^\infty_{n=2}12na_n(x-1)^n-12a_0-12a_1(x-1)-\sum^\infty_{n=2}12a_n(x-1)^n=0.\)

We now collect like terms to obtain

\[(8a_2-12a_0)+(24a_3-24a_1)(x-1)+\sum_{n=2}^\infty [4(n+2)(n+1)a_{n+2}-(2n^2+10n+12)a_n](x-1)^n=0\nonumber\]

Equating coefficients we get

\[\begin{aligned}a_2&= {3\over 2}a_0,\\a_3&= a_1,\\a_{n+2}&={2(n+2)(n+3)\over 4(n+2)(n+1)}a_n={n+3\over 2(n+1)}a_n, \quad n=2,3,4,\cdots\end{aligned}\nonumber\]

Now substituting \(n=2,3,4,\cdots\) into the last equation yields

\[\begin{aligned} a_4 &= {5\over 6}a_2={5\over 4}a_0, \\[4pt]a_5 &= {3\over 4}a_3={3\over 4}a_1,\\[4pt]a_6&={7\over 10}a_4={7\over 8}a_0,\\[4pt]a_7&={2\over 3}a_5={1\over 2}a_1,\cdots \end{aligned}\nonumber \] 

Thus,

\[\begin{aligned}y&=a_0+a_1(x-1)+{3\over 2}a_0(x-1)^2+a_1(x-1)^3+{5\over 4}a_0(x-1)^4+{3\over 4}a_1(x-1)^5+{7\over 8}a_0(x-1)^6+{1\over 2}a_1(x-1)^7+\cdots\\&=a_0(1+{3\over 2}(x-1)^2+{5\over 4}(x-1)^4+{7\over 8}(x-1)^6+\cdots)+a_1((x-1)+(x-1)^3+{3\over 4}(x-1)^5+{1\over 2}(x-1)^7+\cdots)\end{aligned}\nonumber\]

which implies that the power series in \(x-1\) for the general solution of Equation \ref{eq:7.2.28} is

\[y=a_0\sum_{m=0}^\infty{2m+1\over2^m}(x-1)^{2m}+a_1\sum_{m=0}^\infty {m+1\over2^m}(x-1)^{2m+1}.\nonumber \]

Since \(P_0(x)=2+4x-2x^2\) has real zeros at \(1\pm\sqrt2\), Theorem 5.1.1 implies that every solution of Equation \ref{eq:7.2.28} is defined on \((-\infty,1-\sqrt2)\), \((1-\sqrt2,1+\sqrt2)\), or \((1+\sqrt2,\infty)\).

The next example illustrates an initial value problem.

Example 8.3.4

Find the power series in \(x\) for the general solution of

\[\label{eq:7.2.29} (1+2x^2)y''+6xy'+2y=0,\quad y(0)=2,\quad y'(0)=-3.\]

Solution

From Example 6 we know the solution to the differential equation is

\[y=a_0(1-x^2+{3\over 2}x^4-{5\over 2}x^6+\cdots)+a_1(x-{4\over 3}x^3+{32\over 15}x^5-{128\over 35}x^7+\cdots)\nonumber\]

The condition \(y(0)=2\) yields \(a_0=2.\)

Differentiating our solution gives us

\[y'=a_0(-2x+6x^3-15x^5+\cdots)+a_1(1-4x^2+{32\over 3}x^4-{128\over 5}x^6+\cdots)\nonumber\]

The condition \(y'(0)=-3\) yields \(a_1=-3.\)

So the solution to \ref{eq:7.2.29} is

\[y=2(1-x^2+{3\over 2}x^4-{5\over 2}x^6+\cdots)-3(x-{4\over 3}x^3+{32\over 15}x^5-{128\over 35}x^7+\cdots)\nonumber\]

Nonhomogeneous Equations 

Definition: 8.3.3

Let \[ y''+p(x)y'+q(x)y=f(x).\nonumber\]

We say that \(x_0\) is an ordinary point of Equation \ref{eq:7.2.2} if \(p(x)\), \(q(x)\), and \(f(x)\) are all analytic at \(x_0\). Otherwise, \(x_0\) is a singular point.

Theorem \(\PageIndex{2}\)

Let \[\label{eq:7.2.30} y''+p(x)y'+q(x)y=f(x)\]

Let \(x_0\) be an ordinary point of \ref{eq:7.2.30} and \(\rho\) be the distance from \(x_0\) to the nearest singular point of \ref{eq:7.2.30}. Then every solution of \ref{eq:7.2.30}

can be represented by a power series

\[\label{eq:7.2.31} y=\sum_{n=0}^\infty a_n(x-x_0)^n\]

that converges at least on the open interval \((x_0-\rho,x_0+\rho)\).

Furthermore, \ref{eq:7.2.31} will generate the two linearly independent solutions of the homogeneous part of \ref{eq:7.2.30} AND a particular solution to the nonhomogeneous part of \ref{eq:7.2.30}.

Example \(\PageIndex{5}\)

Find the power series in \(x\) for the general solution of

\[\label{eq:7.2.32} y''+xy'+y={1\over 1-x}\]

Solution

The series for \({1\over 1-x}\) is given by

\[{1\over 1-x}=\sum_{n=0}^\infty x^n\nonumber\]

which converges on \((-1,1)\).

Since \(x=0\) is an ordinary point, Theorem 2 says we we will generate both the homogeneous and the nonhomogeneous solutions of \ref{eq:7.2.32} from

\[ \begin{align*} y &=\sum^\infty_{n=0}a_n(x-1)^n, \\[4pt] y' &=\sum^\infty_{n=1}na_n(x-1)^{n-1}, \\[4pt] y''&=\sum^\infty_{n=2}n(n-1)a_n(x-1)^{n-2} \end{align*}\nonumber\]

so, we get

\[\begin{aligned}\sum^\infty_{n=2}n(n-1)a_nx^{n-2}+x \sum^\infty_{n=1}na_nx^{n-1} +\sum^\infty_{n=0}a_nx^n= \sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum^\infty_{n=1}na_nx^n + \sum^\infty_{n=0}a_nx^n=\sum_{n=0}^\infty x^n.\end{aligned}\nonumber \]

Running out terms we get

\[2a_2+\sum_{n=3}^\infty n(n-1)a_nx^{n-2}+\sum^\infty_{n=1}na_nx^n +a_0 +\sum^\infty_{n=1}a_nx^n=1+\sum_{n=1}^\infty x^n.\nonumber \]

Reindexing we get

\[2a_2+\sum_{n=1}^\infty (n+2)(n+1)a_{n+2}x^n+\sum^\infty_{n=1}na_nx^n +a_0+ \sum^\infty_{n=1}a_nx^n=1+\sum_{n=1}^\infty x^n.\nonumber \]

We now collect like terms to obtain

\[(2a_2+a_0)+\sum_{n=1}^\infty [(n+2)(n+1)a_{n+2}+(n+1)a_n]x^n=1+\sum_{n=1}^\infty x^n.\nonumber \]

Equating coefficients we get

\[\begin{aligned}a_2&= -{1\over 2}a_0+{1\over 2},\\a_{n+2}&={-a_n\over n+2}+{1\over (n+2)(n+1)}, \quad n=1,2,3,\cdots\end{aligned}\nonumber\]

Now substituting \(n=1,2,3,\cdots\) into the last equation yields

\[\begin{aligned} a_3 &= -{1\over3}a_1+{1\over 6}, \\[4pt]a_4 &= -{1\over 4}a_2+{1\over 12}={1\over 8}a_0-{1\over 24},\\[4pt]a_5&=-{1\over 5}a_3+{1\over 20}={1\over 15}a_1+{1\over 60},\cdots \end{aligned}\nonumber \] 

Thus,

\[\begin{aligned}y&=a_0+a_1x-({1\over 2}a_0-{1\over 2})x^2-({1\over 3}a_1-{1\over 6})x^3+({1\over 8}a_0-{1\over 24})x^4+({1\over 15}a_1+{1\over 60})x^5+\cdots\\&=a_0(1-{1\over 2}x^2+{1\over 8}x^4+\cdots)+a_1(x-{1\over 3}x^3+{1\over 15}x^5+\cdots)+({1\over 2}x^2+{1\over 6}x^3-{1\over 24}x^4+{1\over 60}x^5+\cdots)\end{aligned}\nonumber\]

is the power series in \(x\) for the general solution of Equation \ref{eq:7.2.32} defined on \((-1,1)\).