7.2 Series Solutions Near an Ordinary Point I

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

where \(P_0\), \(P_1\), and \(P_2\) are polynomials. Usually 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 (7.2.1) with series.

We assume throughout that \(P_0\), \(P_1\) and \(P_2\) have no common factors. Then we say that \(x_0\) is an ordinary point of (7.2.1) if \(P_0(x_0)\ne0\), or a singular point if \(P_0(x_0)=0\). For Legendre’s equation,

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

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

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 (7.2.1) and \(a_0\) and \(a_1\) are arbitrary real numbers, then the initial value problem

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 (7.2.4) as

and apply Theorem 5.1.1 with \(p=P_1/P_0\) and \(q=P_2/P_0\). In this section and the next we consider the problem of representing solutions of (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.

We call (7.2.6) a power series solution in \(x-x_0\) of (7.2.5). We’ll now develop a method for finding power series solutions of (7.2.5). For this purpose we write (7.2.5) as \(Ly=0\), where

Theorem 7.2.1 implies that every solution of \(Ly=0\) on \((x_0-\rho,x_0+\rho)\) can be written as

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

shows that \(y(x_0)=a_0\) and \(y'(x_0)=a_1\). Since every initial value problem (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 \(a_2\), \(a_3\), …, we write \(P_0\), \(P_1\), and \(P_2\) in powers of \(x-x_0\), substitute

into (7.2.7), and collect the coefficients of like powers of \(x-x_0\). This yields

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 \(x-x_0\). Since (7.2.8) and (a) of Theorem 7.1.6 imply that \(Ly=0\) if and only if \(b_n=0\) for \(n\ge0\), all power series solutions in \(x-x_0\) of \(Ly=0\) 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 \(Ly=0\), without explicitly identifying the open interval of convergence of the series.

Solution Here

If

then

so

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

with

Therefore \(Ly=0\) if and only if

where \(a_0\) and \(a_1\) are arbitrary. Since the indices on the left and right sides of (7.2.10) differ by two, we write (7.2.10) separately for \(n\) even \((n=2m)\) and \(n\) odd \((n=2m+1)\). This yields

Computing the coefficients of the even powers of \(x-x_0\) from (7.2.11) yields

and, in general,

Computing the coefficients of the odd powers of \(x-x_0\) from (7.2.12) yields

and, in general,

Thus, the general solution of (7.2.9) can be written as

or, from (7.2.13) and (7.2.14), as

If we recall from calculus that

then (7.2.15) becomes

which should look familiar. ∎

Equations like (7.2.10), (7.2.11), and (7.2.12), 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’re computing recursively.

In the remainder of this section we consider the problem of finding power series solutions in \(x-x_0\) for equations of the form

Many important equations that arise in applications are of this form with \(x_0=0\), including Legendre’s equation (7.2.2), Airy’s equation (7.2.3), Chebyshev’s equation,

and Hermite’s equation,

Since

in (7.2.16), the point \(x_0\) is an ordinary point of (7.2.16), and Theorem 7.2.1 implies that the solutions of (7.2.16) can be written as power series in \(x-x_0\) that converge on the interval \((x_0-1/\sqrt|\alpha|,x_0+1/\sqrt|\alpha|)\) if \(\alpha\ne0\), or on \((-\infty,\infty)\) if \(\alpha=0\). We’ll see that the coefficients in these power series can be obtained by methods similar to the one used in Example 7.2.1.

To simplify finding the coefficients, we introduce some notation for products:

Thus,

and

We define

no matter what the form of \(b_j\).

Solution Here

If

then

so

To collect coefficients of \(x^n\), we shift the summation index in the first sum. This yields

with

To obtain solutions of (7.2.17), we set \(b_n=0\) for \(n\ge0\). This is equivalent to the recurrence relation

Since the indices on the left and right differ by two, we write (7.2.18) separately for \(n=2m\) and \(n=2m+1\), as in Example 7.2.1. This yields

Computing the coefficients of even powers of \(x\) from (7.2.19) yields

In general,

(Note that (7.2.21) is correct for \(m=0\) because we defined \(\prod_{j=1}^0b_j=1\) for any \(b_j\).)

Computing the coefficients of odd powers of \(x\) from (7.2.20) yields

In general,

From (7.2.21) and (7.2.22),

is the power series in \(x\) for the general solution of (7.2.17). Since \(P_0(x)=1+2x^2\) has no real zeros, Theorem 5.1.1 implies that every solution of (7.2.17) is defined on \((-\infty,\infty)\). However, since \(P_0(\pm i/\sqrt2)=0\), Theorem 7.2.1 implies only that the power series converges in \((-1/\sqrt2,1/\sqrt2)\) for any choice of \(a_0\) and \(a_1\).

The results in Examples 7.2.1 and 7.2.2 are consequences of the following general theorem.

Proof Here

If

then

Hence,

from (7.2.25). To collect coefficients of powers of \(x-x_0\), we shift the summation index in the first sum. This yields

Thus, \(Ly=0\) if and only if

which is equivalent to (7.2.24). Writing (7.2.24) separately for the cases where \(n=2m\) and \(n=2m+1\) yields (7.2.26) and (7.2.27).

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,

Therefore we can rewrite (7.2.28) as

or, equivalently,

This is of the form (7.2.23) with \(\alpha=-1/2\), \(\beta=-3\), and \(\gamma=-3\). Therefore, from (7.2.25)

Hence, Theorem 7.2.2 implies that

We leave it to you to show that

which implies that the power series in \(x-1\) for the general solution of (7.2.28) is

In the examples considered so far we were able to obtain closed formulas for coefficients in the power series solutions. In some cases this is impossible, and we must settle for computing a finite number of terms in the series. The next example illustrates this with an initial value problem.

Solution Since \(\alpha=2\), \(\beta=10\), and \(\gamma=8\) in (7.2.29),

Therefore

Writing this equation separately for \(n=2m\) and \(n=2m+1\) yields

Starting with \(a_0=y(0)=2\), we compute \(a_2, a_4\), and \(a_6\) from (7.2.30):

Starting with \(a_1=y'(0)=-3\), we compute \(a_3,a_5\) and \(a_7\) from (7.2.31):

Therefore the solution of (7.2.29) is

Using Technology

Computing coefficients recursively as in Example 7.2.4 is tedious. We recommend that you do this kind of computation by writing a short program to implement the appropriate recurrence relation on a calculator or computer. You may wish to do this in verifying examples and doing exercises (identified by the symbol C ) in this chapter that call for numerical computation of the coefficients in series solutions. We obtained the answers to these exercises by using software that can produce answers in the form of rational numbers. However, it’s perfectly acceptable - and more practical - to get your answers in decimal form. You can always check them by converting our fractions to decimals.

If you’re interested in actually using series to compute numerical approximations to solutions of a differential equation, then whether or not there’s a simple closed form for the coefficents is essentially irrelevant. For computational purposes it’s usually more efficient to start with the given coefficients \(a_0=y(x_0)\) and \(a_1=y'(x_0)\), compute \(a_2\), …, \(a_N\) recursively, and then compute approximate values of the solution from the Taylor polynomial

The trick is to decide how to choose \(N\) so the approximation \(y(x)\approx T_N(x)\) is sufficiently accurate on the subinterval of the interval of convergence that you’re interested in. In the computational exercises in this and the next two sections, you will often be asked to obtain the solution of a given problem by numerical integration with software of your choice (see Section 10.1 for a brief discussion of one such method), and to compare the solution obtained in this way with the approximations obtained with \(T_N\) for various values of \(N\). This is a typical textbook kind of exercise, designed to give you insight into how the accuracy of the approximation \(y(x)\approx T_N(x)\) behaves as a function of \(N\) and the interval that you’re working on. In real life, you would choose one or the other of the two methods (numerical integration or series solution). If you choose the method of series solution, then a practical procedure for determining a suitable value of \(N\) is to continue increasing \(N\) until the maximum of \(|T_N-T_{N-1}|\) on the interval of interest is within the margin of error that you’re willing to accept.

In doing computational problems that call for numerical solution of differential equations you should choose the most accurate numerical integration procedure your software supports, and experiment with the step size until you’re confident that the numerical results are sufficiently accurate for the problem at hand.

7.2 Exercises

In Exercises 1 –8 find the power series in \(x\) for the general solution.

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

   Show answer

   \(y=\dst{a_0\sum^\infty_{m=0}(-1)^m(2m+1)x^{2m}+ a_1\sum^\infty_{m=0}(-1)^m(m+1)x^{2m+1}}\)

2. \((1+x^2)y''+2xy'-2y=0\)

   Show answer

   \(y=\dst{a_0\sum_{m=0}^\infty(-1)^{m+1}{x^{2m}\over2m-1}+a_1x}\)

3. \((1+x^2)y''-8xy'+20y=0\)

   Show answer

   \(y=\dst{a_0(1-10x^2+5x^4)+a_1\left(x-2x^3+{1 \over5}x^5\right)}\)

4. \((1-x^2)y''-8xy'-12y=0\)

   Show answer

   \(y=\dst{a_0\sum_{m=0}^\infty(m+1)(2m+1)x^{2m}+{a_1\over3}\sum_{m=0}^\infty (m+1)(2m+3)x^{2m+1}}\)

5. \((1+2x^2)y''+7xy'+2y=0\)

   Show answer

   \(y=\dst{a_0\sum_{m=0}^\infty (-1)^m\left[\prod_{j=0}^{m-1}{4j+1\over2j+1}\right]x^{2m} +a_1\sum_{m=0}^\infty (-1)^m\left[\prod_{j=0}^{m-1}(4j+3)\right]{x^{2m+1}\over2^mm!}}\)

6. \(\dst{(1+x^2)y''+2xy'+{1\over4}y=0}\)

   Show answer

   \(y=\dst{a_0\sum_{m=0}^\infty (-1)^m\left[\prod_{j=0}^{m-1}{(4j+1)^2\over2j+1}\right]{x^{2m}\over8^mm!} +a_1\sum_{m=0}^\infty (-1)^m\left[\prod_{j=0}^{m-1}{(4j+3)^2\over2j+3}\right] {x^{2m+1}\over8^mm!}}\)

7. \((1-x^2)y''-5xy'-4y=0\)

   Show answer

   \(y=\dst{a_0\sum_{m=0}^\infty{2^mm!\over\prod_{j=0}^{m-1}(2j+1)}x^{2m} +a_1\sum_{m=0}^\infty{\prod_{j=0}^{m-1}(2j+3)\over2^mm!}x^{2m+1}}\)

8. \((1+x^2)y''-10xy'+28y=0\)

   Show answer

   \(y=\dst{a_0\left(1-14x^2+{35\over3}x^4\right)+a_1\left(x-3x^3+{3\over5}x^5 +{1\over35}x^7\right)}\)

9. L

   1. Find the power series in \(x\) for the general solution of \(y''+xy'+2y=0\).

   2. For several choices of \(a_0\) and \(a_1\), use differential equations software to solve the initial value problem

      \[ y''+xy'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \tag*{\rm(A)} \]

      numerically on \((-5,5)\).

   3. For fixed \(r\) in \(\{1,2,3,4,5\}\) graph

      \[ T_N(x)=\sum_{n=0}^N a_nx^n \]

      and the solution obtained in (a) on \((-r,r)\). Continue increasing \(N\) until there’s no perceptible difference between the two graphs.

   Show answer

   (a) \(y=\dst{a_0\sum_{m=0}^\infty(-1)^m {x^{2m}\over\prod_{j=0}^{m-1}(2j+1)} +a_1\sum_{m=0}^\infty(-1)^m{x^{2m+1}\over2^mm!}}\)

10. L Follow the directions of Exercise 9 for the differential equation

    \[ y''+2xy'+3y=0. \]
    Show answer

    (a) \(y=\dst{a_0\sum_{m=0}^\infty (-1)^m\left[\prod_{j=0}^{m-1}{4j+3\over2j+1}\right]{x^{2m}\over2^mm!} +a_1\sum_{m=0}^\infty (-1)^m\left[\prod_{j=0}^{m-1}{4j+5\over2j+3}\right]{x^{2m+1}\over 2^mm!}}\)

11. C \((1+x^2)y''+xy'+y=0,\quad y(0)=2,\quad y'(0)=-1\)

    Show answer

    \(y=\dst{2-x-x^2+{1\over3}x^3+{5\over12}x^4-{1\over6}x^5-{17\over72}x^6 +{13\over126}x^7+\cdots}\)

12. C \((1+2x^2)y''-9xy'-6y=0,\quad y(0)=1,\quad y'(0)=-1\)

    Show answer

    \(y=\dst{1-x+3x^2-{5\over2}x^3+5x^4-{21\over8}x^5+3x^6 -{11\over16}x^7+\cdots}\)

13. C \((1+8x^2)y''+2y=0,\quad y(0)=2,\quad y'(0)=-1\)

    Show answer

    \(y=\dst{2-x-2x^2+{1\over3}x^3+3x^4-{5\over6}x^5-{49\over5}x^6 +{45\over14}x^7+\cdots}\)

14. L Do the next experiment for various choices of real numbers \(a_0\), \(a_1\), and \(r\), with \(0<r<1/\sqrt2\).

    1. Use differential equations software to solve the initial value problem

       \[ (1-2x^2)y''-xy'+3y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \tag*{\rm(A)} \]

       numerically on \((-r,r)\).

    2. For \(N=2\), \(3\), \(4\), …, compute \(a_2\), …, \(a_N\) in the power series solution \(y=\sum_{n=0}^\infty a_nx^n\) of (A), and graph

       \[ T_N(x)=\sum_{n=0}^N a_nx^n \]

       and the solution obtained in (a) on \((-r,r)\). Continue increasing \(N\) until there’s no perceptible difference between the two graphs.

15. L Do (a) and (b) for several values of \(r\) in \((0,1)\):

    1. Use differential equations software to solve the initial value problem

       \[ (1+x^2)y''+10xy'+14y=0,\quad y(0)=5,\quad y'(0)=1, \tag*{\rm(A)} \]

       numerically on \((-r,r)\).

    2. For \(N=2\), \(3\), \(4\), …, compute \(a_2\), …, \(a_N\) in the power series solution \(y=\sum_{n=0}^\infty a_nx^n\) of (A) , and graph

       \[ T_N(x)=\sum_{n=0}^N a_nx^n \]

       and the solution obtained in (a) on \((-r,r)\). Continue increasing \(N\) until there’s no perceptible difference between the two graphs. What happens to the required \(N\) as \(r\to1\)?

    3. Try (a) and (b) with \(r=1.2\). Explain your results.

16. \(y''-y=0;\quad x_0=3\)

    Show answer

    \(y=\dst{a_0\sum_{m=0}^\infty{(x-3)^{2m}\over(2m)!}+a_1\sum_{m=0}^\infty {(x-3)^{2m+1}\over(2m+1)!}}\)

17. \(y''-(x-3)y'-y=0;\quad x_0=3\)

    Show answer

    \(y=\dst{a_0\sum_{m=0}^\infty{(x-3)^{2m}\over2^mm!}+a_1\sum_{m=0}^\infty {(x-3)^{2m+1}\over\prod_{j=0}^{m-1}(2j+3)}}\)

18. \((1-4x+2x^2)y''+10(x-1)y'+6y=0;\quad x_0=1\)

    Show answer

    \(y=\dst{a_0\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j+3)\right]{ (x-1)^{2m}\over m!}+a_1\sum_{m=0}^\infty {4^m(m+1)!\over\prod_{j=0}^{m-1}(2j+3)}(x-1)^{2m+1}}\)

19. \((11-8x+2x^2)y''-16(x-2)y'+36y=0;\quad x_0=2\)

    Show answer

    \(y=\dst{a_0\left(1-6(x-2)^2+{4\over3}(x-2)^4+{8\over135}(x-2)^6\right) +a_1\left((x-2)-{10\over9}(x-2)^3\right)}\)

20. \((5+6x+3x^2)y''+9(x+1)y'+3y=0;\quad x_0=-1\)

    Show answer

    \(y=\dst{a_0\sum_{m=0}^\infty(-1)^m\left[\prod_{j=0}^{m-1}(2j+1)\right] {3^m\over4^mm!}(x+1)^{2m}+a_1\sum_{m=0}^\infty(-1)^m{3^mm!\over \prod_{j=0}^{m-1}(2j+3)}(x+1)^{2m+1}}\)

21. C \((x^2-4)y''-xy'-3y=0,\quad y(0)=-1,\quad y'(0)=2\)

    Show answer

    \(y=\dst{-1+2x+{3\over8}x^2-{1\over3}x^3-{3\over128}x^4-{1\over1024}x^6+\cdots}\)

22. C \(y''+(x-3)y'+3y=0,\quad y(3)=-2,\quad y'(3)=3\)

    Show answer

    \(y=\dst{-2+3(x-3)+3(x-3)^2-2(x-3)^3-{5\over4}(x-3)^4+{3\over5}(x-3)^5 +{7\over24}(x-3)^6-{4\over35}(x-3)^7+\cdots}\)

23. C \((5-6x+3x^2)y''+(x-1)y'+12y=0,\quad y(1)=-1,\quad y'(1)=1\)

    Show answer

    \(y=\dst{-1+(x-1)+3(x-1)^2-{5\over2}(x-1)^3-{27\over4}(x-1)^4+{21\over4}(x-1)^5 +{27\over2}(x-1)^6-{81\over8}(x-1)^7+\cdots}\)

24. C \((4x^2-24x+37)y''+y=0,\quad y(3)=4,\quad y'(3)=-6\)

    Show answer

    \(y=\dst{4-6(x-3)-2(x-3)^2+(x-3)^3+{3\over2}(x-3)^4-{5\over4}(x-3)^5- {49\over20}(x-3)^6+{135\over56}(x-3)^7+\cdots}\)

25. C \((x^2-8x+14)y''-8(x-4)y'+20y=0,\quad y(4)=3,\quad y'(4)=-4\)

    Show answer

    \(y=\dst{3-4(x-4)+15(x-4)^2-4(x-4)^3+{15\over4}(x-4)^4-{1\over5}(x-4)^5}\)

26. C \((2x^2+4x+5)y''-20(x+1)y'+60y=0,\quad y(-1)=3,\quad y'(-1)=-3\)

    Show answer

    \(y=\dst{3-3(x+1)-30(x+1)^2+{20\over3}(x+1)^3+20(x+1)^4-{4\over3}(x+1)^5-{8\over 9}(x+1)^6}\)

27. 1. Find a power series in \(x\) for the general solution of

       \[ (1+x^2)y''+4xy'+2y=0. \tag*{\rm (A)} \]

    2. Use (a) and the formula

       \[ {1\over1-r}=1+r+r^2+\cdots+r^n+\cdots \quad(-1<r<1) \]

       for the sum of a geometric series to find a closed form expression for the general solution of (A) on \((-1,1)\).

    3. Show that the expression obtained in (b) is actually the general solution of of (A) on \((-\infty,\infty)\).

    Show answer

    (a)\(y=\dst{a_0\sum_{m=0}^\infty(-1)^m x^{2m}+a_1\sum_{m=0}^\infty(-1)^mx^{2m+1}}\) (b)\(y=\dst{a_0+a_1x\over1+x^2}\)

28. Use Theorem 7.2.2 to show that the power series in \(x\) for the general solution of

    \[ (1+\alpha x^2)y''+\beta xy'+\gamma y=0 \]

    is

    \[ y=a_0\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0} p(2j)\right] {x^{2m}\over(2m)!} + a_1\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0}p(2j+1)\right] {x^{2m+1}\over(2m+1)!}. \]

29. Use Exercise 28 to show that all solutions of

    \[ (1+\alpha x^2)y''+\beta xy'+\gamma y=0 \]

    are polynomials if and only if

    \[ \alpha n(n-1)+\beta n+\gamma=\alpha(n-2r)(n-2s-1), \]

    where \(r\) and \(s\) are nonnegative integers.

30. 1. Use Exercise 28 to show that the power series in \(x\) for the general solution of

       \[ (1-x^2)y''-2bxy'+\alpha(\alpha+2b-1)y=0 \]

       is \(y=a_0y_1+a_1y_2\), where

       \begin{eqnarray*} y_1&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j-\alpha)(2j+\alpha+2b-1) \right]{x^{2m}\over(2m)!} \\ \noalign{\hspace*{50pt}\mbox{and}} \\ y_2&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j+1-\alpha)(2j+\alpha+2b) \right]{x^{2m+1}\over(2m+1)!}. \end{eqnarray*}

    2. Suppose \(2b\) isn’t a negative odd integer and \(k\) is a nonnegative integer. Show that \(y_1\) is a polynomial of degree \(2k\) such that \(y_1(-x)=y_1(x)\) if \(\alpha=2k\), while \(y_2\) is a polynomial of degree \(2k+1\) such that \(y_2(-x)=-y_2(-x)\) if \(\alpha=2k+1\). Conclude that if \(n\) is a nonnegative integer, then there’s a polynomial \(P_n\) of degree \(n\) such that \(P_n(-x)=(-1)^nP_n(x)\) and

       \[ (1-x^2)P_n''-2bxP_n'+n(n+2b-1)P_n=0. \tag*{\rm (A)} \]

    3. Show that (A) implies that

       \[ [(1-x^2)^b P_n']'=-n(n+2b-1)(1-x^2)^{b-1}P_n, \]

       and use this to show that if \(m\) and \(n\) are nonnegative integers, then

       \[ \begin{array}{ll} [(1-x^2)^bP_n']'P_m-[(1-x^2)^bP_m']'P_n=&\\[6pt] \hspace*{40pt}\left[m(m+2b-1)-n(n+2b-1)\right](1-x^2)^{b-1}P_mP_n.& \end{array} \tag*{\rm (B)} \]

    4. Now suppose \(b>0\). Use (B) and integration by parts to show that if \(m\ne n\), then

       \[ \int_{-1}^1 (1-x^2)^{b-1}P_m(x)P_n(x)\,dx=0. \]

       (We say that \(P_m\) and \(P_n\) are orthogonal on \((-1,1)\) with respect to the weighting function \((1-x^2)^{b-1}\).)

31. 1. Use Exercise 28 to show that the power series in \(x\) for the general solution of Hermite’s equation

       \[ y''-2xy'+2\alpha y=0 \]

       is \(y=a_0y_1+a_1y_1\), where

       \begin{eqnarray*} y_1&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j-\alpha) \right]{2^mx^{2m}\over(2m)!} \\ \noalign{\hspace*{25pt}\mbox{and}} \\ y_2&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j+1-\alpha) \right]{2^mx^{2m+1}\over(2m+1)!}. \end{eqnarray*}

    2. Suppose \(k\) is a nonnegative integer. Show that \(y_1\) is a polynomial of degree \(2k\) such that \(y_1(-x)=y_1(x)\) if \(\alpha=2k\), while \(y_2\) is a polynomial of degree \(2k+1\) such that \(y_2(-x)=-y_2(-x)\) if \(\alpha=2k+1\). Conclude that if \(n\) is a nonnegative integer then there’s a polynomial \(P_n\) of degree \(n\) such that \(P_n(-x)=(-1)^nP_n(x)\) and

       \[ P_n''-2xP_n'+2nP_n=0. \tag*{\rm (A)} \]

    3. Show that (A) implies that

       \[ [e^{-x^2}P_n']'=-2ne^{-x^2}P_n, \]

       and use this to show that if \(m\) and \(n\) are nonnegative integers, then

       \[ [e^{-x^2}P_n']'P_m-[e^{-x^2}P_m']'P_n= 2(m-n)e^{-x^2}P_mP_n. \tag*{\rm (B)} \]

    4. Use (B) and integration by parts to show that if \(m\ne n\), then

       \[ \int_{-\infty}^\infty e^{-x^2}P_m(x)P_n(x)\,dx=0. \]

       (We say that \(P_m\) and \(P_n\) are orthogonal on \((-\infty,\infty)\) with respect to the weighting function \(e^{-x^2}\).)

32. Consider the equation

    \[ \left(1+\alpha x^3\right)y''+\beta x^2y'+\gamma xy=0, \tag*{\rm (A)} \]

    and let \(p(n)=\alpha n(n-1)+\beta n+\gamma\). (The special case \(y''-xy=0\) of (A) is Airy’s equation.)

    1. Modify the argument used to prove Theorem 7.2.2 to show that

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

       is a solution of (A) if and only if \(a_2=0\) and

       \[ a_{n+3}=-{p(n)\over(n+3)(n+2)}a_n,\quad n\ge0. \]

    2. Show from (a) that \(a_n=0\) unless \(n=3m\) or \(n=3m+1\) for some nonnegative integer \(m\), and that

       \begin{eqnarray*} a_{3m+3}&=&-{p(3m)\over(3m+3)(3m+2)}a_{3m},\quad m\ge 0, \\ \noalign{\hspace*{50pt}\mbox{and}} \\ a_{3m+4}&=&-{p(3m+1)\over(3m+4)(3m+3)} a_{3m+1},\quad m\ge0, \end{eqnarray*}

       where \(a_0\) and \(a_1\) may be specified arbitrarily.

    3. Conclude from (b) that the power series in \(x\) for the general solution of (A) is

       \[ \begin{array}{l} y=\dst{a_0\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0} {p(3j)\over3j+2}\right] {x^{3m}\over3^m m!}}\\[6pt] \qquad\dst{+a_1\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0}{p(3j+1)\over3j+4}\right] {x^{3m+1}\over3^mm!}}. \end{array} \]

33. \(y''-xy=0\)

    Show answer

    \(y=\dst{a_0\sum_{m=0}^\infty {x^{3m}\over3^mm!\prod_{j=0}^{m-1}(3j+2)}+a_1\sum_{m=0}^\infty {x^{3m+1}\over3^mm!\prod_{j=0}^{m-1}(3j+4)}}\)

34. \((1-2x^3)y''-10x^2y'-8xy=0\)

    Show answer

    \(y=\dst{a_0\sum_{m=0}^\infty \left(2\over3\right)^m\left[\prod_{j=0}^{m-1}(3j+2)\right]{x^{3m}\over m!} +a_1\sum_{m=0}^\infty{6^mm!\over\prod_{j=0}^{m-1}(3j+4)}x^{3m+1}}\)

35. \((1+x^3)y''+7x^2y'+9xy=0\)

    Show answer

    \(y=\dst{a_0\sum_{m=0}^\infty (-1)^m{3^mm!\over\prod_{j=0}^{m-1}(3j+2)}x^{3m} +a_1\sum_{m=0}^\infty (-1)^m\left[\prod_{j=0}^{m-1}(3j+4)\right]{x^{3m+1}\over 3^mm!}}\)

36. \((1-2x^3)y''+6x^2y'+24xy=0\)

    Show answer

    \(y=\dst{a_0(1-4x^3+4x^6)+a_1\sum_{m=0}^\infty 2^m\left[\prod_{j=0}^{m-1}{3j-5\over3j+4}\right]x^{3m+1}}\)

37. \((1-x^3)y''+15x^2y'-63xy=0\)

    Show answer

    \(y=\dst{a_0\left(1+{21\over2}x^3+{42\over5}x^6+{7\over20}x^9\right) +a_1\left(x+4x^4+{10\over7}x^7\right)}\)

38. Consider the equation

    \[ \left(1+\alpha x^{k+2}\right)y''+\beta x^{k+1}y'+\gamma x^ky=0, \tag*{\rm (A)} \]

    where \(k\) is a positive integer, and let \(p(n)=\alpha n(n-1)+\beta n+\gamma\).

    1. Modify the argument used to prove Theorem 7.2.2 to show that

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

       is a solution of (A) if and only if \(a_n=0\) for \(2\le n\le k+1\) and

       \[ a_{n+k+2}=-{p(n)\over(n+k+2)(n+k+1)}a_n,\quad n\ge0. \]

    2. Show from (a) that \(a_n=0\) unless \(n=(k+2)m\) or \(n=(k+2)m+1\) for some nonnegative integer \(m\), and that

       \begin{eqnarray*} a_{(k+2)(m+1)}&=&-{p\left((k+2)m\right)\over (k+2)(m+1)[(k+2)(m+1)-1]}a_{(k+2)m},\quad m\ge 0, \\ \noalign{\hspace*{50pt}\mbox{and}} \\ a_{(k+2)(m+1)+1}&=&-{p\left((k+2)m+1\right)\over[(k+2)(m+1)+1](k+2)(m+1)} a_{(k+2)m+1},\quad m\ge0, \end{eqnarray*}

       where \(a_0\) and \(a_1\) may be specified arbitrarily.

    3. Conclude from (b) that the power series in \(x\) for the general solution of (A) is

       \[ \begin{array}{l} y=a_0\dst{\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0} {p\left((k+2)j\right)\over(k+2)(j+1)-1}\right] {x^{(k+2)m}\over(k+2)^m m!}}\\ [15pt] \qquad+a_1\dst{\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0}{p\left((k+2)j+1\right)\over(k+2)(j+1)+1}\right] {x^{(k+2)m+1}\over(k+2)^mm!}}. \end{array} \]

39. \((1+2x^5)y''+14x^4y'+10x^3y=0\)

    Show answer

    \(y=\dst{a_0\sum_{m=0}^\infty (-2)^m\left[\prod_{j=0}^{m-1}{5j+1\over5j+4}\right]x^{5m}+ a_1\sum_{m=0}^\infty \left(-{2\over5}\right)^m\left[\prod_{j=0}^{m-1}(5j+2)\right]{x^{5m+1}\over m!}}\)

40. \(y''+x^2y=0\)

    Show answer

    \(y=\dst{a_0\sum_{m=0}^\infty (-1)^m{x^{4m}\over4^mm!\prod_{j=0}^{m-1}(4j+3)}+a_1\sum_{m=0}^\infty (-1)^m{x^{4m+1}\over4^mm!\prod_{j=0}^{m-1}(4j+5)}}\)

41. \(y''+x^6y'+7x^5y=0\)

    Show answer

    \(y=\dst{a_0\sum_{m=0}^\infty(-1)^m{x^{7m}\over\prod_{j=0}^{m-1}(7j+6)}+ a_1\sum_{m=0}^\infty(-1)^m{x^{7m+1}\over7^mm!}}\)

42. \((1+x^8)y''-16x^7y'+72x^6y=0\)

    Show answer

    \(y=\dst{a_0\left(1-{9\over7}x^8\right)+a_1\left(x-{7\over9}x^9\right)}\)

43. \((1-x^6)y''-12x^5y'-30x^4y=0\)

    Show answer

    \(y=\dst{a_0\sum_{m=0}^\infty x^{6m}+a_1\sum_{m=0}^\infty x^{6m+1}}\)

44. \(y''+x^5y'+6x^4y=0\)

    Show answer

    \(y=\dst{a_0\sum_{m=0}^\infty(-1)^m{x^{6m}\over\prod_{j=0}^{m-1}(6j+5)}+ a_1\sum_{m=0}^\infty(-1)^m{x^{6m+1}\over6^mm!}}\)