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.
- Find the power series in \(x\) for the general solution of \(y''+xy'+2y=0\).
- 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 \((-5,5)\).
- 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 \(0<r<1/\sqrt2\).
- 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 \((-r,r)\).
- 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 \((0,1)\):
- 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 \((-r,r)\).
- 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 \(r\to1\)?
- 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.
- Find a power series in \(x\) for the general solution of \[(1+x^2)y''+4xy'+2y=0. \tag{A} \]
- 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 \((-1,1)\).
- 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.
- 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 \[y_{2}=\sum_{m=0}^{\infty}\left[\prod_{j=0}^{m-1}(2j+1-\alpha )(2j+\alpha +2b) \right] \frac{x^{2m+1}}{(2m+1)!}\nonumber \]
- 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 \[(1-x^2)P_n''-2bxP_n'+n(n+2b-1)P_n=0. \tag{A} \]
- 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 \[[(1-x^{2})^{b}P_{n}']P_{m}-[(1-x^{2})^{b}P_{m}']P_{n}=[m(m+2b-1)-n(n+2b-1)](1-x^{2})^{b-1}P_{m}P_{n}\tag{B} \]
- 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.
- 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 \[y_{2}=\sum_{m=0}^{\infty}\left[\prod_{j=0}^{m-1} (2j+1-\alpha ) \right] \frac{2^{m}x^{2m+1}}{(2m+1)!}\nonumber \]
- 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 \[P_n''-2xP_n'+2nP_n=0. \tag{A} \]
- 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 \[[e^{-x^2}P_n']'P_m-[e^{-x^2}P_m']'P_n= 2(m-n)e^{-x^2}P_mP_n. \tag{B} \]
- 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.)
- 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 \[a_{n+3}=-{p(n)\over(n+3)(n+2)}a_n,\quad n\ge0.\nonumber \]
- 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.
- 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 \(p(n)=\alpha n(n-1)+\beta n+\gamma\).
- 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 \[a_{n+k+2}=-{p(n)\over(n+k+2)(n+k+1)}a_n,\quad n\ge0.\nonumber \]
- 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.
- 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. \(y''+x^5y'+6x^4y=0\)