7.2E: Series Solutions Near an Ordinary Point I (Exercises)

Q7.2.1

In Exercises 7.2.1-7.2.8 find the power series in $x$ for the general solution.

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

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

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

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

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

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

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

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

Q7.2.2

9.

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 $$y''+xy'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \tag{A}$$ numerically on $(-5,5)$.
3. For fixed $r$ in $\{1,2,3,4,5\}$ graph $$T_N(x)=\sum_{n=0}^N a_nx^n\nonumber$$ and the solution obtained in (a) on $(-r,r)$. Continue increasing $N$ until there’s no perceptible difference between the two graphs.

10. Follow the directions of Exercise [exer:7.2.9} for the differential equation $$y''+2xy'+3y=0.\nonumber$$

Q7.2.3

In Exercises 7.2.11-7.2.13 find $a_{0}, ..., a_{N}$ for $N$ at least $7$ in the power series solution $y=\sum _{n=0}^{\infty} a_{n}x^{n}$ of the initial value problem.

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

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

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

Q7.2.4

14. 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 $$(1-2x^2)y''-xy'+3y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \tag{A}$$ 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 $$T_N(x)=\sum_{n=0}^N a_nx^n\nonumber$$ and the solution obtained in (a) on $(-r,r)$. Continue increasing $N$ until there’s no perceptible difference between the two graphs.

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

1. Use differential equations software to solve the initial value problem $$(1+x^2)y''+10xy'+14y=0,\quad y(0)=5,\quad y'(0)=1, \tag{A}$$ 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 $$T_N(x)=\sum_{n=0}^N a_nx^n\nonumber$$ 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.

Q7.2.5

In Exercises 7.2.16-7.2.20 find the power series in $x-x_{0}$ for the general solution.

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

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

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

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

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

Q7.2.6

In Exercises 7.2.21-7.2.26 find $a_{0}, ... a_{N}$ for $N$ at least $7$ in the power series $y=\sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n}$ for the solution of the initial value problem. Take $x_{0}$ to be the point where the initial conditions are imposed.

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

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

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

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

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

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

Q7.2.7

27.

1. Find a power series in $x$ for the general solution of $$(1+x^2)y''+4xy'+2y=0. \tag{A}$$
2. Use (a) and the formula $${1\over1-r}=1+r+r^2+\cdots+r^n+\cdots \quad(-1<r<1)\nonumber$$ 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 $(-\infty,\infty)$.

28. Use Theorem 7.2.2 to show that the power series in $x$ for the general solution of $$(1+\alpha x^2)y''+\beta xy'+\gamma y=0\nonumber$$

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

29. Use Exercise 7.2.28 to show that all solutions of $$(1+\alpha x^2)y''+\beta xy'+\gamma y=0\nonumber$$

are polynomials if and only if $$\alpha n(n-1)+\beta n+\gamma=\alpha(n-2r)(n-2s-1),\nonumber$$

where $r$ and $s$ are nonnegative integers.

30.

1. Use Exercise [exer:7.2.28} to show that the power series in $x$ for the general solution of $$(1-x^2)y''-2bxy'+\alpha(\alpha+2b-1)y=0\nonumber$$ is $y=a_0y_1+a_1y_2$, where $$y_{1}=\sum_{m=0}^{\infty} \left[\prod_{j=0}^{m-1}(2j-\alpha )(2j+\alpha +2b-1) \right]\frac{x^{2m}}{(2m)!}\nonumber$$ and $$y_{2}=\sum_{m=0}^{\infty}\left[\prod_{j=0}^{m-1}(2j+1-\alpha )(2j+\alpha +2b) \right] \frac{x^{2m+1}}{(2m+1)!}\nonumber$$
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 $$(1-x^2)P_n''-2bxP_n'+n(n+2b-1)P_n=0. \tag{A}$$
3. Show that (A) implies that $$[(1-x^2)^b P_n']'=-n(n+2b-1)(1-x^2)^{b-1}P_n,\nonumber$$ and use this to show that if $m$ and $n$ are nonnegative integers, then $$[(1-x^{2})^{b}P_{n}']P_{m}-[(1-x^{2})^{b}P_{m}']P_{n}=[m(m+2b-1)-n(n+2b-1)](1-x^{2})^{b-1}P_{m}P_{n}\tag{B}$$
4. Now suppose $b>0$. Use (B) and integration by parts to show that if $m\ne n$, then $$\int_{-1}^1 (1-x^2)^{b-1}P_m(x)P_n(x)\,dx=0.\nonumber$$ (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 7.2.28 to show that the power series in $x$ for the general solution of Hermite’s equation $$y''-2xy'+2\alpha y=0\nonumber$$ is $y=a_0y_1+a_1y_1$, where $$y_{1}=\sum_{m=0}^{\infty}\left[ \prod_{j=0}^{m-1}(2j-\alpha ) \right]\frac{2^{m}x^{2m}}{(2m)!}\nonumber$$ and $$y_{2}=\sum_{m=0}^{\infty}\left[\prod_{j=0}^{m-1} (2j+1-\alpha ) \right] \frac{2^{m}x^{2m+1}}{(2m+1)!}\nonumber$$
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 $$P_n''-2xP_n'+2nP_n=0. \tag{A}$$
3. Show that (A) implies that $$[e^{-x^2}P_n']'=-2ne^{-x^2}P_n,\nonumber$$ and use this to show that if $m$ and $n$ are nonnegative integers, then $$[e^{-x^2}P_n']'P_m-[e^{-x^2}P_m']'P_n= 2(m-n)e^{-x^2}P_mP_n. \tag{B}$$
4. Use (B) and integration by parts to show that if $m\ne n$, then $$\int_{-\infty}^\infty e^{-x^2}P_m(x)P_n(x)\,dx=0.\nonumber$$ (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 $$\left(1+\alpha x^3\right)y''+\beta x^2y'+\gamma xy=0, \tag{A}$$ 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 [thmtype:7.2.2} to show that $$y=\sum_{n=0}^\infty a_nx^n\nonumber$$ is a solution of (A) if and only if $a_2=0$ and $$a_{n+3}=-{p(n)\over(n+3)(n+2)}a_n,\quad n\ge0.\nonumber$$
2. Show from (a) that $a_n=0$ unless $n=3m$ or $n=3m+1$ for some nonnegative integer $m$, and that $$\begin{aligned} a_{3m+3}&=&-{p(3m)\over(3m+3)(3m+2)}a_{3m},\quad m\ge 0,\\[4pt] \text{and} \\[4pt] a_{3m+4}&=&-{p(3m+1)\over(3m+4)(3m+3)} a_{3m+1},\quad m\ge0,\end{aligned}\nonumber$$ 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 $$\begin{array}{l} y={a_0\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0} {p(3j)\over3j+2}\right] {x^{3m}\over3^m m!}}\\[4pt] \qquad{+a_1\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0}{p(3j+1)\over3j+4}\right] {x^{3m+1}\over3^mm!}}. \end{array}\nonumber$$

Q7.2.8

In Exercises 7.2.33-7.2.37 use the method of Exercise 7.2.32 to find the power series in $x$ for the general solution.

33. $y''-xy=0$

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

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

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

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

Q7.2.9

38. Consider the equation $$\left(1+\alpha x^{k+2}\right)y''+\beta x^{k+1}y'+\gamma x^ky=0, \tag{A}$$ 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 $$y=\sum_{n=0}^\infty a_nx^n\nonumber$$ is a solution of (A) if and only if $a_n=0$ for $2\le n\le k+1$ and $$a_{n+k+2}=-{p(n)\over(n+k+2)(n+k+1)}a_n,\quad n\ge0.\nonumber$$
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 $$\begin{aligned} a_{(k+2)(m+1)}&=&-{p\left((k+2)m\right)\over (k+2)(m+1)[(k+2)(m+1)-1]}a_{(k+2)m},\quad m\ge 0, \\[4pt] \text{and}\\[4pt] a_{(k+2)(m+1)+1}&=&-{p\left((k+2)m+1\right)\over[(k+2)(m+1)+1](k+2)(m+1)} a_{(k+2)m+1},\quad m\ge0,\end{aligned}\nonumber$$ 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 $$\begin{array}{l} y=a_0{\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0} {p\left((k+2)j\right)\over(k+2)(j+1)-1}\right] {x^{(k+2)m}\over(k+2)^m m!}}\\[4pt] \qquad+a_1{\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0}{p\left((k+2)j+1\right)\over(k+2)(j+1)+1}\right] {x^{(k+2)m+1}\over(k+2)^mm!}}. \end{array}\nonumber$$

Q7.2.10

In Exercises 7.2.39-7.2.44 use the method of Exercise 7.2.38 to find the power series in $x$ for the general solution.

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

40. $y''+x^2y=0$

41. $y''+x^6y'+7x^5y=0$

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

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

44. $y''+x^5y'+6x^4y=0$