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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+2xy2y=0

3. (1+x2)y8xy+20y=0

4. (1x2)y8xy12y=0

5. (1+2x2)y+7xy+2y=0

6. (1+x2)y+2xy+14y=0

7. (1x2)y5xy4y=0

8. (1+x2)y10xy+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 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).
  3. For fixed r in {1,2,3,4,5} graph TN(x)=Nn=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)y9xy6y=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.

  1. Use differential equations software to solve the initial value problem (12x2)yxy+3y=0,y(0)=a0,y(0)=a1, numerically on (r,r).
  2. For N=2, 3, 4, …, compute a2, …, aN in the power series solution y=n=0anxn of (A), and graph TN(x)=Nn=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):

  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).
  2. For N=2, 3, 4, …, compute a2, …, aN in the power series solution y=n=0anxn of (A), and graph TN(x)=Nn=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 r1?
  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 xx0 for the general solution.

16. yy=0;x0=3

17. y(x3)yy=0;x0=3

18. (14x+2x2)y+10(x1)y+6y=0;x0=1

19. (118x+2x2)y16(x2)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(xx0)n for the solution of the initial value problem. Take x0 to be the point where the initial conditions are imposed.

21. (x24)yxy3y=0,y(0)=1,y(0)=2

22. y+(x3)y+3y=0,y(3)=2,y(3)=3

23. (56x+3x2)y+(x1)y+12y=0,y(1)=1,y(1)=1

24. (4x224x+37)y+y=0,y(3)=4,y(3)=6

25. (x28x+14)y8(x4)y+20y=0,y(4)=3,y(4)=4

26. (2x2+4x+5)y20(x+1)y+60y=0,y(1)=3,y(1)=3

Q7.2.7

27.

  1. Find a power series in x for the general solution of (1+x2)y+4xy+2y=0.
  2. Use (a) and the formula 11r=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).
  3. 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=a0m=0(1)m[m1j=0p(2j)]x2m(2m)!+a1m=0(1)m[m1j=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(n1)+βn+γ=α(n2r)(n2s1),

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 (1x2)y2bxy+α(α+2b1)y=0 is y=a0y1+a1y2, where y1=m=0[m1j=0(2jα)(2j+α+2b1)]x2m(2m)! and y2=m=0[m1j=0(2j+1α)(2j+α+2b)]x2m+1(2m+1)!
  2. 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 (1x2)Pn2bxPn+n(n+2b1)Pn=0.
  3. Show that (A) implies that [(1x2)bPn]=n(n+2b1)(1x2)b1Pn, and use this to show that if m and n are nonnegative integers, then [(1x2)bPn]Pm[(1x2)bPm]Pn=[m(m+2b1)n(n+2b1)](1x2)b1PmPn
  4. Now suppose b>0. Use (B) and integration by parts to show that if mn, then 11(1x2)b1Pm(x)Pn(x)dx=0. (We say that Pm and Pn are orthogonal on (1,1) with respect to the weighting function(1x2)b1.)

31.

  1. Use Exercise 7.2.28 to show that the power series in x for the general solution of Hermite’s equation y2xy+2αy=0 is y=a0y1+a1y1, where y1=m=0[m1j=0(2jα)]2mx2m(2m)! and y2=m=0[m1j=0(2j+1α)]2mx2m+1(2m+1)!
  2. 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 Pn2xPn+2nPn=0.
  3. Show that (A) implies that [ex2Pn]=2nex2Pn, and use this to show that if m and n are nonnegative integers, then [ex2Pn]Pm[ex2Pm]Pn=2(mn)ex2PmPn.
  4. Use (B) and integration by parts to show that if mn, then ex2Pm(x)Pn(x)dx=0. (We say that Pm and Pn are orthogonal on (,) with respect to the weighting function ex2.)

32. Consider the equation (1+αx3)y+βx2y+γxy=0, and let p(n)=αn(n1)+βn+γ. (The special case yxy=0 of (A) is Airy’s equation.)

  1. 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,n0.
  2. 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,m0,anda3m+4=p(3m+1)(3m+4)(3m+3)a3m+1,m0, where a0 and a1 may be specified arbitrarily.
  3. Conclude from (b) that the power series in x for the general solution of (A) is y=a0m=0(1)m[m1j=0p(3j)3j+2]x3m3mm!+a1m=0(1)m[m1j=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. yxy=0

34. (12x3)y10x2y8xy=0

35. (1+x3)y+7x2y+9xy=0

36. (12x3)y+6x2y+24xy=0

37. (1x3)y+15x2y63xy=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(n1)+βn+γ.

  1. 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 2nk+1 and an+k+2=p(n)(n+k+2)(n+k+1)an,n0.
  2. 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,m0,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,m0, where a0 and a1 may be specified arbitrarily.
  3. Conclude from (b) that the power series in x for the general solution of (A) is y=a0m=0(1)m[m1j=0p((k+2)j)(k+2)(j+1)1]x(k+2)m(k+2)mm!+a1m=0(1)m[m1j=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


This page titled 7.2E: Series Solutions Near an Ordinary Point I (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.

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