7.2E: Series Solutions Near an Ordinary Point I (Exercises)
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Q7.2.1
In Exercises 7.2.1-7.2.8 find the power series in x for the general solution.
1. (1+x2)y″+6xy′+6y=0
2. (1+x2)y″+2xy′−2y=0
3. (1+x2)y″−8xy′+20y=0
4. (1−x2)y″−8xy′−12y=0
5. (1+2x2)y″+7xy′+2y=0
6. (1+x2)y″+2xy′+14y=0
7. (1−x2)y″−5xy′−4y=0
8. (1+x2)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 a0 and a1, use differential equations software to solve the initial value problem y″+xy′+2y=0,y(0)=a0,y′(0)=a1, numerically on (−5,5).
- For fixed r in {1,2,3,4,5} graph TN(x)=N∑n=0anxn 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.
Q7.2.3
In Exercises 7.2.11-7.2.13 find a0,...,aN for N at least 7 in the power series solution y=∑∞n=0anxn of the initial value problem.
11. (1+x2)y″+xy′+y=0,y(0)=2,y′(0)=−1
12. (1+2x2)y″−9xy′−6y=0,y(0)=1,y′(0)=−1
13. (1+8x2)y″+2y=0,y(0)=2,y′(0)=−1
Q7.2.4
14. Do the next experiment for various choices of real numbers a0, a1, and r, with 0<r<1/√2.
- Use differential equations software to solve the initial value problem (1−2x2)y″−xy′+3y=0,y(0)=a0,y′(0)=a1, numerically on (−r,r).
- For N=2, 3, 4, …, compute a2, …, aN in the power series solution y=∑∞n=0anxn of (A), and graph TN(x)=N∑n=0anxn 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+x2)y″+10xy′+14y=0,y(0)=5,y′(0)=1, numerically on (−r,r).
- For N=2, 3, 4, …, compute a2, …, aN in the power series solution y=∑∞n=0anxn of (A), and graph TN(x)=N∑n=0anxn 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→1?
- 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−x0 for the general solution.
16. y″−y=0;x0=3
17. y″−(x−3)y′−y=0;x0=3
18. (1−4x+2x2)y″+10(x−1)y′+6y=0;x0=1
19. (11−8x+2x2)y″−16(x−2)y′+36y=0;x0=2
20. (5+6x+3x2)y″+9(x+1)y′+3y=0;x0=−1
Q7.2.6
In Exercises 7.2.21-7.2.26 find a0,...aN for N at least 7 in the power series y=∑∞n=0an(x−x0)n for the solution of the initial value problem. Take x0 to be the point where the initial conditions are imposed.
21. (x2−4)y″−xy′−3y=0,y(0)=−1,y′(0)=2
22. y″+(x−3)y′+3y=0,y(3)=−2,y′(3)=3
23. (5−6x+3x2)y″+(x−1)y′+12y=0,y(1)=−1,y′(1)=1
24. (4x2−24x+37)y″+y=0,y(3)=4,y′(3)=−6
25. (x2−8x+14)y″−8(x−4)y′+20y=0,y(4)=3,y′(4)=−4
26. (2x2+4x+5)y″−20(x+1)y′+60y=0,y(−1)=3,y′(−1)=−3
Q7.2.7
27.
- Find a power series in x for the general solution of (1+x2)y″+4xy′+2y=0.
- Use (a) and the formula 11−r=1+r+r2+⋯+rn+⋯(−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).
- Show that the expression obtained in (b) is actually the general solution of of (A) on (−∞,∞).
28. Use Theorem 7.2.2 to show that the power series in x for the general solution of (1+αx2)y″+βxy′+γy=0
is y=a0∞∑m=0(−1)m[m−1∏j=0p(2j)]x2m(2m)!+a1∞∑m=0(−1)m[m−1∏j=0p(2j+1)]x2m+1(2m+1)!.
29. Use Exercise 7.2.28 to show that all solutions of (1+αx2)y″+βxy′+γy=0
are polynomials if and only if αn(n−1)+βn+γ=α(n−2r)(n−2s−1),
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−x2)y″−2bxy′+α(α+2b−1)y=0 is y=a0y1+a1y2, where y1=∞∑m=0[m−1∏j=0(2j−α)(2j+α+2b−1)]x2m(2m)! and y2=∞∑m=0[m−1∏j=0(2j+1−α)(2j+α+2b)]x2m+1(2m+1)!
- Suppose 2b isn’t a negative odd integer and k is a nonnegative integer. Show that y1 is a polynomial of degree 2k such that y1(−x)=y1(x) if α=2k, while y2 is a polynomial of degree 2k+1 such that y2(−x)=−y2(−x) if α=2k+1. Conclude that if n is a nonnegative integer, then there’s a polynomial Pn of degree n such that Pn(−x)=(−1)nPn(x) and (1−x2)P″n−2bxP′n+n(n+2b−1)Pn=0.
- Show that (A) implies that [(1−x2)bP′n]′=−n(n+2b−1)(1−x2)b−1Pn, and use this to show that if m and n are nonnegative integers, then [(1−x2)bP′n]Pm−[(1−x2)bP′m]Pn=[m(m+2b−1)−n(n+2b−1)](1−x2)b−1PmPn
- Now suppose b>0. Use (B) and integration by parts to show that if m≠n, then ∫1−1(1−x2)b−1Pm(x)Pn(x)dx=0. (We say that Pm and Pn are orthogonal on (−1,1) with respect to the weighting function(1−x2)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′+2αy=0 is y=a0y1+a1y1, where y1=∞∑m=0[m−1∏j=0(2j−α)]2mx2m(2m)! and y2=∞∑m=0[m−1∏j=0(2j+1−α)]2mx2m+1(2m+1)!
- Suppose k is a nonnegative integer. Show that y1 is a polynomial of degree 2k such that y1(−x)=y1(x) if α=2k, while y2 is a polynomial of degree 2k+1 such that y2(−x)=−y2(−x) if α=2k+1. Conclude that if n is a nonnegative integer then there’s a polynomial Pn of degree n such that Pn(−x)=(−1)nPn(x) and P″n−2xP′n+2nPn=0.
- Show that (A) implies that [e−x2P′n]′=−2ne−x2Pn, and use this to show that if m and n are nonnegative integers, then [e−x2P′n]′Pm−[e−x2P′m]′Pn=2(m−n)e−x2PmPn.
- Use (B) and integration by parts to show that if m≠n, then ∫∞−∞e−x2Pm(x)Pn(x)dx=0. (We say that Pm and Pn are orthogonal on (−∞,∞) with respect to the weighting function e−x2.)
32. Consider the equation (1+αx3)y″+βx2y′+γxy=0, and let p(n)=αn(n−1)+βn+γ. (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=∞∑n=0anxn is a solution of (A) if and only if a2=0 and an+3=−p(n)(n+3)(n+2)an,n≥0.
- Show from (a) that an=0 unless n=3m or n=3m+1 for some nonnegative integer m, and that a3m+3=−p(3m)(3m+3)(3m+2)a3m,m≥0,anda3m+4=−p(3m+1)(3m+4)(3m+3)a3m+1,m≥0, where a0 and a1 may be specified arbitrarily.
- Conclude from (b) that the power series in x for the general solution of (A) is y=a0∑∞m=0(−1)m[∏m−1j=0p(3j)3j+2]x3m3mm!+a1∑∞m=0(−1)m[∏m−1j=0p(3j+1)3j+4]x3m+13mm!.
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. (1−2x3)y″−10x2y′−8xy=0
35. (1+x3)y″+7x2y′+9xy=0
36. (1−2x3)y″+6x2y′+24xy=0
37. (1−x3)y″+15x2y′−63xy=0
Q7.2.9
38. Consider the equation (1+αxk+2)y″+βxk+1y′+γxky=0, where k is a positive integer, and let p(n)=αn(n−1)+βn+γ.
- Modify the argument used to prove Theorem 7.2.2 to show that y=∞∑n=0anxn is a solution of (A) if and only if an=0 for 2≤n≤k+1 and an+k+2=−p(n)(n+k+2)(n+k+1)an,n≥0.
- Show from (a) that an=0 unless n=(k+2)m or n=(k+2)m+1 for some nonnegative integer m, and that a(k+2)(m+1)=−p((k+2)m)(k+2)(m+1)[(k+2)(m+1)−1]a(k+2)m,m≥0,anda(k+2)(m+1)+1=−p((k+2)m+1)[(k+2)(m+1)+1](k+2)(m+1)a(k+2)m+1,m≥0, where a0 and a1 may be specified arbitrarily.
- Conclude from (b) that the power series in x for the general solution of (A) is y=a0∑∞m=0(−1)m[∏m−1j=0p((k+2)j)(k+2)(j+1)−1]x(k+2)m(k+2)mm!+a1∑∞m=0(−1)m[∏m−1j=0p((k+2)j+1)(k+2)(j+1)+1]x(k+2)m+1(k+2)mm!.
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