7.3.1: Series Solutions Near an Ordinary Point II (Exercises)
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Q7.3.1
In Exercises 7.3.1-7.3.12 find the coefficients a0,…, aN for N at least 7 in the series solution y=∑∞n=0anxn of the initial value problem.
1. (1+3x)y″
2. (1+x+2x^2)y''+(2+8x)y'+4y=0,\quad y(0)=-1,\quad y'(0)=2
3. (1-2x^2)y''+(2-6x)y'-2y=0,\quad y(0)=1,\quad y'(0)=0
4. (1+x+3x^2)y''+(2+15x)y'+12y=0,\quad y(0)=0,\quad y'(0)=1
5. (2+x)y''+(1+x)y'+3y=0,\quad y(0)=4,\quad y'(0)=3
6. (3+3x+x^2)y''+(6+4x)y'+2y=0,\quad y(0)=7,\quad y'(0)=3
7. (4+x)y''+(2+x)y'+2y=0,\quad y(0)=2,\quad y'(0)=5
8. (2-3x+2x^2)y''-(4-6x)y'+2y=0,\quad y(1)=1,\quad y'(1)=-1
9. (3x+2x^2)y''+10(1+x)y'+8y=0,\quad y(-1)=1,\quad y'(-1)=-1
10. (1-x+x^2)y''-(1-4x)y'+2y=0,\quad y(1)=2,\quad y'(1)=-1
11. (2+x)y''+(2+x)y'+y=0,\quad y(-1)=-2,\quad y'(-1)=3
12. x^2y''-(6-7x)y'+8y=0,\quad y(1)=1,\quad y'(1)=-2
Q7.3.2
13. Do the following 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+x+2x^2)y''+(1+7x)y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \tag{A} numerically on (-r,r). (See Example 7.3.1.)
- 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.
14. Do the following experiment for various choices of real numbers a_0, a_1, and r, with 0<r<2.
- Use differential equations software to solve the initial value problem (3+x)y''+(1+2x)y'-(2-x)y=0,\quad y(-1)=a_0,\quad y'(-1)=a_1, \tag{A} numerically on (-1-r,-1+r). (See Example 7.3.2). Why this interval?)
- For N=2, 3, 4, …, compute a_2,\dots,a_N in the power series solution y=\sum_{n=0}^\infty a_n(x+1)^n\nonumber of (A), and graph T_N(x)=\sum_{n=0}^N a_n(x+1)^n\nonumber and the solution obtained in (a) on (-1-r,-1+r). Continue increasing N until there’s no perceptible difference between the two graphs.
15. Do the following experiment for several choices of a_0, a_1, and r, with r>0.
- Use differential equations software to solve the initial value problem y''+3xy'+(4+2x^2)y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \tag{A} numerically on (-r,r). (See Example 7.3.3.)
- Find the coefficients a_0, a_1, …, 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.
16. Do the following experiment for several choices of a_0 and a_1.
- Use differential equations software to solve the initial value problem (1-x)y''-(2-x)y'+y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \tag{A} numerically on (-r,r).
- Find the coefficients a_0, a_1, …, a_N in the power series solution y=\sum_{n=0}^Na_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 as you let r\to 1?
17. Follow the directions of Exercise 7.3.16 for the initial value problem (1+x)y''+3y'+32y=0,\quad y(0)=a_0,\quad y'(0)=a_1.\nonumber
18. Follow the directions of Exercise 7.3.16 for the initial value problem (1+x^2)y''+y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1.\nonumber
Q7.3.3
In Exercises 7.3.19-7.3.28 find the coefficients a_{0},...a_{N} for N at least 7 in the series solution y=\sum_{n=0}^{\infty}a_{n}(x-x_{0})^{n}\nonumber of the initial value problem. Take x_{0} to be the point where the initial conditions are imposed.
19. (2+4x)y''-4y'-(6+4x)y=0,\quad y(0)=2,\quad y'(0)=-7
20. (1+2x)y''-(1-2x)y'-(3-2x)y=0,\quad y(1)=1,\quad y'(1)=-2
21. (5+2x)y''-y'+(5+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1
22. (4+x)y''-(4+2x)y'+(6+x)y=0,\quad y(-3)=2,\quad y'(-3)=-2
23. (2+3x)y''-xy'+2xy=0,\quad y(0)=-1,\quad y'(0)=2
24. (3+2x)y''+3y'-xy=0,\quad y(-1)=2,\quad y'(-1)=-3
25. (3+2x)y''-3y'-(2+x)y=0,\quad y(-2)=-2,\quad y'(-2)=3
26. (10-2x)y''+(1+x)y=0,\quad y(2)=2,\quad y'(2)=-4
27. (7+x)y''+(8+2x)y'+(5+x)y=0,\quad y(-4)=1,\quad y'(-4)=2
28. (6+4x)y''+(1+2x)y=0,\quad y(-1)=-1,\quad y'(-1)=2
Q7.3.4
29. Show that the coefficients in the power series in x for the general solution of (1+\alpha x+\beta x^2)y''+(\gamma+\delta x)y'+\epsilon y=0\nonumber satisfy the recurrrence relation a_{n+2}=-{\gamma+\alpha n\over n+2}\,a_{n+1}-{\beta n(n-1)+\delta n+\epsilon\over(n+2)(n+1)}\, a_n.\nonumber
30.
- Let \alpha and \beta be constants, with \beta\ne0. Show that y=\sum_{n=0}^\infty a_nx^n is a solution of (1+\alpha x+\beta x^2)y''+(2\alpha+4\beta x)y'+2\beta y=0 \tag{A} if and only if a_{n+2}+\alpha a_{n+1}+\beta a_n=0,\quad n\ge0. \tag{B} An equation of this form is called a second order homogeneous linear difference equation. The polynomial p(r)=r^2+\alpha r+\beta is called the characteristic polynomial of (B). If r_1 and r_2 are the zeros of p, then 1/r_1 and 1/r_2 are the zeros of P_{0}(x)=1+\alpha x+\beta x^{2}\nonumber
- Suppose p(r)=(r-r_1)(r-r_2) where r_1 and r_2 are real and distinct, and let \rho be the smaller of the two numbers \{1/|r_1|,1/|r_2|\}. Show that if c_1 and c_2 are constants then the sequence a_n=c_1r_1^n+c_2r_2^n,\quad n\ge0\nonumber satisfies (B). Conclude from this that any function of the form y=\sum_{n=0}^\infty (c_1r_1^n+c_2r_2^n)x^n\nonumber is a solution of (A) on (-\rho,\rho).
- Use (b) and the formula for the sum of a geometric series to show that the functions y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={1\over1-r_2x}\nonumber form a fundamental set of solutions of (A) on (-\rho,\rho).
- Show that \{y_1,y_2\} is a fundamental set of solutions of (A) on any interval that does’nt contain either 1/r_1 or 1/r_2.
- Suppose p(r)=(r-r_1)^2, and let \rho=1/|r_1|. Show that if c_1 and c_2 are constants then the sequence a_n=(c_1+c_2n)r_1^n,\quad n\ge0\nonumber satisfies (B). Conclude from this that any function of the form y=\sum_{n=0}^\infty (c_1+c_2n)r_1^nx^n\nonumber is a solution of (A) on (-\rho,\rho).
- Use (e) and the formula for the sum of a geometric series to show that the functions y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={x\over(1-r_1x)^2}\nonumber form a fundamental set of solutions of (A) on (-\rho,\rho).
- Show that \{y_1,y_2\} is a fundamental set of solutions of (A) on any interval that does not contain 1/r_1.
31. Use the results of Exercise 7.3.30 to find the general solution of the given equation on any interval on which polynomial multiplying y'' has no zeros.
- (1+3x+2x^2)y''+(6+8x)y'+4y=0
- (1-5x+6x^2)y''-(10-24x)y'+12y=0
- (1-4x+4x^2)y''-(8-16x)y'+8y=0
- (4+4x+x^2)y''+(8+4x)y'+2y=0
- (4+8x+3x^2)y''+(16+12x)y'+6y=0
Q7.3.5
In Exercises 7.3.32-7.3.38 find the coefficients a_{0}, ..., a_{N} for N at least 7 in the series solution y=\sum_{n=0}^{\infty} a_{n}x^{n} of the initial value problem.
32. y''+2xy'+(3+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2
33. y''-3xy'+(5+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2
34. y''+5xy'-(3-x^2)y=0,\quad y(0)=6,\quad y'(0)=-2
35. y''-2xy'-(2+3x^2)y=0,\quad y(0)=2,\quad y'(0)=-5
36. y''-3xy'+(2+4x^2)y=0,\quad y(0)=3,\quad y'(0)=6
37. 2y''+5xy'+(4+2x^2)y=0,\quad y(0)=3,\quad y'(0)=-2
38. 3y''+2xy'+(4-x^2)y=0,\quad y(0)=-2,\quad y'(0)=3
Q7.3.6
39. Find power series in x for the solutions y_1 and y_2 of y''+4xy'+(2+4x^2)y=0\nonumber such that y_1(0)=1, y'_1(0)=0, y_2(0)=0, y'_2(0)=1, and identify y_1 and y_2 in terms of familiar elementary functions.
Q7.3.7
In Exercises 7.3.40-7.3.49 find the coefficients a_{0}, ..., a_{N} for N at least 7 in the series solution y=\sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n}\nonumber of the initial value problem. Take x_{0} to be the point where the initial conditions are imposed.
40. (1+x)y''+x^2y'+(1+2x)y=0,\quad y(0)-2,\quad y'(0)=3
41. y''+(1+2x+x^2)y'+2y=0,\quad y(0)=2,\quad y'(0)=3
42. (1+x^2)y''+(2+x^2)y'+xy=0,\quad y(0)=-3,\quad y'(0)=5
43. (1+x)y''+(1-3x+2x^2)y'-(x-4)y=0,\quad y(1)=-2,\quad y'(1)=3
44. y''+(13+12x+3x^2)y'+(5+2x),\quad y(-2)=2,\quad y'(-2)=-3
45. (1+2x+3x^2)y''+(2-x^2)y'+(1+x)y=0,\quad y(0)=1,\quad y'(0)=-2
46. (3+4x+x^2)y''-(5+4x-x^2)y'-(2+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1
47. (1+2x+x^2)y''+(1-x)y=0,\quad y(0)=2,\quad y'(0)=-1
48. (x-2x^2)y''+(1+3x-x^2)y'+(2+x)y=0,\quad y(1)=1,\quad y'(1)=0
49. (16-11x+2x^2)y''+(10-6x+x^2)y'-(2-x)y,\quad y(3)=1,\quad y'(3)=-2