Q7.6.1
In Exercises 7.6.1-7.6.11 find a fundamental set of Frobenius solutions. Compute the terms involving \(x^{n+r_1}\), where \(0\le n\le N\) (\(N\) at least \(7\)) and \(r_1\) is the root of the indicial equation. Optionally, write a computer program to implement the applicable recurrence formulas and take \(N>7\).
1. \(x^2y''-x(1-x)y'+(1-x^2)y=0\)
2. \(x^2(1+x+2x^2)y'+x(3+6x+7x^2)y'+(1+6x-3x^2)y=0\)
3. \(x^2(1+2x+x^2)y''+x(1+3x+4x^2)y'-x(1-2x)y=0\)
4. \(4x^2(1+x+x^2)y''+12x^2(1+x)y'+(1+3x+3x^2)y=0\)
5. \(x^2(1+x+x^2)y''-x(1-4x-2x^2)y'+y=0\)
6. \(9x^2y''+3x(5+3x-2x^2)y'+(1+12x-14x^2)y=0\)
7. \(x^2y''+x(1+x+x^2)y'+x(2-x)y=0\)
8. \(x^2(1+2x)y''+x(5+14x+3x^2)y'+(4+18x+12x^2)y=0\)
9. \(4x^2y''+2x(4+x+x^2)y'+(1+5x+3x^2)y=0\)
10. \(16x^2y''+4x(6+x+2x^2)y'+(1+5x+18x^2)y=0\)
11. \(9x^2(1+x)y''+3x(5+11x-x^2)y'+(1+16x-7x^2)y=0\)
Q7.6.2
In Exercises 7.6.12-7.6.22 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients.
12. \(4x^2y''+(1+4x)y=0\)
13. \(36x^2(1-2x)y''+24x(1-9x)y'+(1-70x)y=0\)
14. \(x^2(1+x)y''-x(3-x)y'+4y=0\)
15. \(x^2(1-2x)y''-x(5-4x)y'+(9-4x)y=0\)
16. \(25x^2y''+x(15+x)y'+(1+x)y=0\)
17. \(2x^2(2+x)y''+x^2y'+(1-x)y=0\)
18. \(x^2(9+4x)y''+3xy'+(1+x)y=0\)
19. \(x^2y''-x(3-2x)y'+(4+3x)y=0\)
20. \(x^2(1-4x)y''+3x(1-6x)y'+(1-12x)y=0\)
21. \(x^2(1+2x)y''+x(3+5x)y'+(1-2x)y=0\)
22. \(2x^2(1+x)y''-x(6-x)y'+(8-x)y=0\)
Q7.6.3
In Exercises 7.6.23-7.6.27 find a fundamental set of Frobenius solutions. Compare the terms involving \(x^{n+r_{1}}\), where \(0\leq n\leq N\) (\(N\) at least \(7\)) and \(r_{1}\) is the root of the indicial equation. Optionally, write a computer program to implement the applicable recurrence formulas and take \(N>7\).
23. \(x^2(1+2x)y''+x(5+9x)y'+(4+3x)y=0\)
24. \(x^2(1-2x)y''-x(5+4x)y'+(9+4x)y=0\)
25. \(x^2(1+4x)y''-x(1-4x)y'+(1+x)y=0\)
26. \(x^2(1+x)y''+x(1+2x)y'+xy=0\)
27. \(x^2(1-x)y''+x(7+x)y'+(9-x)y=0\)
Q7.6.4
In Exercises 7.6.28-7.6.38 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients.
28. \(x^2y''-x(1-x^2)y'+(1+x^2)y=0\)
29. \(x^2(1+x^2)y''-3x(1-x^2)y'+4y=0\)
30. \(4x^2y''+2x^3y'+(1+3x^2)y=0\)
31. \(x^2(1+x^2)y''-x(1-2x^2)y'+y=0\)
32. \(2x^2(2+x^2)y''+7x^3y'+(1+3x^2)y=0\)
33. \(x^2(1+x^2)y''-x(1-4x^2)y'+(1+2x^2)y=0\)
34. \(4x^2(4+x^2)y''+3x(8+3x^2)y'+(1-9x^2)y=0\)
35. \(3x^2(3+x^2)y''+x(3+11x^2)y'+(1+5x^2)y=0\)
36. \(4x^2(1+4x^2)y''+32x^3y'+y=0\)
37. \(9x^2y''-3x(7-2x^2)y'+(25+2x^2)y=0\)
38. \(x^2(1+2x^2)y''+x(3+7x^2)y'+(1-3x^2)y=0\)
Q7.6.5
In Exercises 7.6.39-7.6.43 find a fundamental set of Frobenius solutions. Compute the terms involving \(x^{2m+r_{1}}\), where \(0 ≤ m ≤ M\) (\(M\) at least \(3\)) and \(r_{1}\) is the root of the indicial equation. Optionally, write a computer program to implement the applicable recurrence formulas and take \(M > 3\).
39. \(x^2(1+x^2)y''+x(3+8x^2)y'+(1+12x^2)y\)
40. \(x^2y''-x(1-x^2)y'+(1+x^2)y=0\)
41. \(x^2(1-2x^2)y''+x(5-9x^2)y'+(4-3x^2)y=0\)
42. \(x^2(2+x^2)y''+x(14-x^2)y'+2(9+x^2)y=0\)
43. \(x^2(1+x^2)y''+x(3+7x^2)y'+(1+8x^2)y=0\)
Q7.6.6
In Exercises 7.6.44-7.6.52 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients.
44. \(x^2(1-2x)y''+3xy'+(1+4x)y=0\)
45. \(x(1+x)y''+(1-x)y'+y=0\)
46. \(x^2(1-x)y''+x(3-2x)y'+(1+2x)y=0\)
47. \(4x^2(1+x)y''-4x^2y'+(1-5x)y=0\)
48. \(x^2(1-x)y''-x(3-5x)y'+(4-5x)y=0\)
49. \(x^2(1+x^2)y''-x(1+9x^2)y'+(1+25x^2)y=0\)
50. \(9x^2y''+3x(1-x^2)y'+(1+7x^2)y=0\)
51. \(x(1+x^2)y''+(1-x^2)y'-8xy=0\)
52. \(4x^2y''+2x(4-x^2)y'+(1+7x^2)y=0\)
Q7.6.7
53. Under the assumptions of Theorem 7.6.2, suppose the power series
\[\sum_{n=0}^\infty a_n(r_1)x^n \quad\mbox{ and }\quad \sum_{n=1}^\infty a_n'(r_1)x^n\nonumber \]
converge on \((-\rho,\rho)\).
- Show that \[y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n\quad\mbox{ and }\quad y_2=y_1\ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n\nonumber\] are linearly independent on \((0,\rho)\). HINT: Show that if \(c_{1}\) and \(c_{2}\) are constants such that \(c_{1}y_{1}+c_{2}y_{2}≡0\) on \((0, \rho )\), then \[(c_{1}+c_{2}\ln x)\sum_{n=0}^{\infty}a_{n}(r_{1})x^{n}+c_{2}\sum_{n=1}^{\infty}a_{n}'(r_{1})x^{n}=0,\quad 0<x<\rho \nonumber\] Then let \(x\to 0+\) to conclude that \(c_{2}=0\).
- Use the result of (a) to complete the proof of Theorem 7.6.2.
54. Let
\[Ly=x^2(\alpha_0+\alpha_1x)y''+x(\beta_0+\beta_1x)y'+(\gamma_0+\gamma_1x)y\nonumber\]
and define
\[p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0\quad\mbox{ and }\quad p_1(r)=\alpha_1r(r-1)+\beta_1r+\gamma_1.\nonumber\]
Theorem 7.6.1 and Exercise 7.5.55a
imply that if
\[y(x,r)=x^r\sum_{n=0}^\infty a_n(r)x^n\nonumber\]
where
\[a_n(r)=(-1)^n\prod_{j=1}^n{p_1(j+r-1)\over p_0(j+r)},\nonumber\]
then
\[Ly(x,r)=p_0(r)x^r.\nonumber\]
Now suppose \(p_0(r)=\alpha_0(r-r_1)^2\) and \(p_1(k+r_1)\ne0\) if \(k\) is a nonnegative integer.
- Show that \(Ly=0\) has the solution \[y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n,\nonumber\] where \[a_n(r_1)={(-1)^n\over\alpha_0^n(n!)^2}\prod_{j=1}^np_1(j+r_1-1).\nonumber\]
- Show that \(Ly=0\) has the second solution \[y_2=y_1\ln x+x^{r_1}\sum_{n=1}^\infty a_n(r_1)J_nx^n,\nonumber\] where \[J_n=\sum_{j=1}^n{p_1'(j+r_1-1)\over p_1(j+r_1-1)}-2\sum_{j=1}^n{1\over j}.\nonumber\]
- Conclude from (a) and (b) that if \(\gamma_1\ne0\) then \[y_1=x^{r_1}\sum_{n=0}^\infty {(-1)^n\over(n!)^2}\left(\gamma_1\over\alpha_0\right)^nx^n\nonumber\] and \[y_2=y_1\ln x-2x^{r_1}\sum_{n=1}^\infty {(-1)^n\over(n!)^2}\left(\gamma_1\over\alpha_0\right)^n \left(\sum_{j=1}^n{1\over j}\right)x^n\nonumber\] are solutions of \[\alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_1x)y=0.\nonumber\] (The conclusion is also valid if \(\gamma_1=0\). Why?)
55. Let
\[Ly=x^2(\alpha_0+\alpha_qx^q)y''+x(\beta_0+\beta_qx^q)y'+(\gamma_0+\gamma_qx^q)y\nonumber\]
where \(q\) is a positive integer, and define
\[p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0\quad\mbox{ and }\quad p_q(r)=\alpha_qr(r-1)+\beta_qr+\gamma_q.\nonumber\]
Suppose
\[p_0(r)=\alpha_0(r-r_1)^2 \quad\mbox{ and }\quad p_q(r)\not\equiv0.\nonumber\]
- Recall from Exercise 7.5.59 that \(Ly~=0\) has the solution \[y_1=x^{r_1}\sum_{m=0}^\infty a_{qm}(r_1)x^{qm},\nonumber\] where \[a_{qm}(r_1)={(-1)^m\over (q^2\alpha_0)^m(m!)^2}\prod_{j=1}^mp_q\left(q(j-1)+r_1\right).\nonumber\]
- Show that \(Ly=0\) has the second solution \[y_2=y_1\ln x+x^{r_1}\sum_{m=1}^\infty a_{qm}'(r_1)J_mx^{qm},\nonumber\] where \[J_m=\sum_{j=1}^m{p_q'\left(q(j-1)+r_1\right)\over p_q\left(q(j-1)+r_1\right)}-{2\over q}\sum_{j=1}^m{1\over j}.\nonumber\]
- Conclude from (a) and (b) that if \(\gamma_q\ne0\) then \[y_1=x^{r_1}\sum_{m=0}^\infty {(-1)^m\over(m!)^2}\left(\gamma_q\over q^2\alpha_0\right)^mx^{qm}\nonumber\] and \[y_2=y_1\ln x-{2\over q}x^{r_1}\sum_{m=1}^\infty {(-1)^m\over(m!)^2}\left(\gamma_q\over q^2\alpha_0\right)^m\left(\sum_{j=1}^m{1\over j}\right)x^{qm}\nonumber\] are solutions of \[\alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_qx^q)y=0.\nonumber\]
56. The equation
\[xy''+y'+xy=0\nonumber\]
is Bessel’s equation of order zero. (See Exercise 7.5.53.) Find two linearly independent Frobenius solutions of this equation.
57. Suppose the assumptions of Exercise 7.5.53 hold, except that
\[p_0(r)=\alpha_0(r-r_1)^2.\nonumber\]
Show that
\[y_1={x^{r_1}\over\alpha_0+\alpha_1x+\alpha_2x^2}\quad\mbox{ and }\quad y_2={x^{r_1}\ln x\over\alpha_0+\alpha_1x+\alpha_2x^2}\nonumber\]
are linearly independent Frobenius solutions of
\[x^2(\alpha_0+\alpha_1x+\alpha_2 x^2)y''+x(\beta_0+\beta_1x+\beta_2x^2)y'+ (\gamma_0+\gamma_1x+\gamma_2x^2)y=0\nonumber\]
on any interval \((0,\rho)\) on which \(\alpha_0+\alpha_1x+\alpha_2x^2\) has no zeros.
Q7.6.8
58. \(4x^2(1+x)y''+8x^2y'+(1+x)y=0\)
59. \(9x^2(3+x)y''+3x(3+7x)y'+(3+4x)y=0\)
60. \(x^2(2-x^2)y''-x(2+3x^2)y'+(2-x^2)y=0\)
61. \(16x^2(1+x^2)y''+8x(1+9x^2)y'+(1+49x^2)y=0\)
62. \(x^2(4+3x)y''-x(4-3x)y'+4y=0\)
63. \(4x^2(1+3x+x^2)y''+8x^2(3+2x)y'+(1+3x+9x^2)y=0\)
64. \(x^2(1-x)^2y''-x(1+2x-3x^2)y'+(1+x^2)y=0\)
65. \(9x^2(1+x+x^2)y''+3x(1+7x+13x^2)y'+(1+4x+25x^2)y=0\)
Q7.6.9
66.
- Let \(L\) and \(y(x,r)\) be as in Exercises 7.5.57 and 7.5.58. Extend Theorem 7.6.1 by showing that \[L\left({\partial y\over \partial r}(x,r)\right)=p'_0(r)x^r+x^rp_0(r)\ln x.\nonumber\]
- Show that if \[p_0(r)=\alpha_0(r-r_1)^2\nonumber\] then \[y_1=y(x,r_1) \quad \text{and} \quad y_2={\partial y\over\partial r}(x,r_1)\nonumber\] are solutions of \(Ly=0\).
