# 6.7.1: The Method of Frobenius III (Exercises)

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## Q7.7.1

In Exercises 7.7.1-7.7.40 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients.

1. $$x^2y''-3xy'+(3+4x)y=0$$

2. $$xy''+y=0$$

3. $$4x^2(1+x)y''+4x(1+2x)y'-(1+3x)y=0$$

4. $$xy''+xy'+y=0$$

5. $$2x^2(2+3x)y''+x(4+21x)y'-(1-9x)y=0$$

6. $$x^2y''+x(2+x)y'-(2-3x)y=0$$

7. $$4x^2y''+4xy'-(9-x)y=0$$

8. $$x^2y''+10xy'+(14+x)y=0$$

9. $$4x^2(1+x)y''+4x(3+8x)y'-(5-49x)y=0$$

10. $$x^2(1+x)y''-x(3+10x)y'+30xy=0$$

11. $$x^2y''+x(1+x)y'-3(3+x)y=0$$

12. $$x^2y''+x(1-2x)y'-(4+x)y=0$$

13. $$x(1+x)y''-4y'-2y=0$$

14. $$x^2(1+2x)y''+x(9+13x)y'+(7+5x)y=0$$

15. $$4x^2y''-2x(4-x)y'-(7+5x)y=0$$

16. $$3x^2(3+x)y''-x(15+x)y'-20y=0$$

17. $$x^2(1+x)y''+x(1-10x)y'-(9-10x)y=0$$

18. $$x^2(1+x)y''+3x^2y'-(6-x)y=0$$

19. $$x^2(1+2x)y''-2x(3+14x)y'+(6+100x)y=0$$

20. $$x^2(1+x)y''-x(6+11x)y'+(6+32x)y=0$$

21. $$4x^2(1+x)y''+4x(1+4x)y'-(49+27x)y=0$$

22. $$x^2(1+2x)y''-x(9+8x)y'-12xy=0$$

23. $$x^2(1+x^2)y''-x(7-2x^2)y'+12y=0$$

24. $$x^2y''-x(7-x^2)y'+12y=0$$

25. $$xy''-5y'+xy=0$$

26. $$x^2y''+x(1+2x^2)y'-(1-10x^2)y=0$$

27. $$x^2y''-xy'-(3-x^2)y=0$$

28. $$4x^2y''+2x(8+x^2)y'+(5+3x^2)y=0$$

29. $$x^2y''+x(1+x^2)y'-(1-3x^2)y=0$$

30. $$x^2y''+x(1-2x^2)y'-4(1+2x^2)y=0$$

31. $$4x^2y''+8xy'-(35-x^2)y=0$$

32. $$9x^2y''-3x(11+2x^2)y'+(13+10x^2)y=0$$

33. $$x^2y''+x(1-2x^2)y'-4(1-x^2)y=0$$

34. $$x^2y''+x(1-3x^2)y'-4(1-3x^2)y=0$$

35. $$x^2(1+x^2)y''+x(5+11x^2)y'+24x^2y=0$$

36. $$4x^2(1+x^2)y''+8xy'-(35-x^2)y=0$$

37. $$x^2(1+x^2)y''-x(5-x^2)y'-(7+25x^2)y=0$$

38. $$x^2(1+x^2)y''+x(5+2x^2)y'-21y=0$$

39. $$x^2(1+2x^2)y''-x(3+x^2)y'-2x^2y=0$$

40. $$4x^2(1+x^2)y''+4x(2+x^2)y'-(15+x^2)y=0$$

## Q7.7.2

41.

1. Under the assumptions of Theorem 7.7.1, show that $y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n\nonumber$ and $y_2=x^{r_2}\sum_{n=0}^{k-1}a_n(r_2)x^n+C\left(y_1\ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n\right)\nonumber$ are linearly independent.
2. Use the result of (a) to complete the proof of Theorem 7.7.1.

42. Find a fundamental set of Frobenius solutions of Bessel’s equation $x^2y''+xy'+(x^2-\nu^2)y=0\nonumber$ in the case where $$\nu$$ is a positive integer.

43. Prove Theorem 7.7.2.

44. Under the assumptions of Theorem 7.7.1, show that $$C=0$$ if and only if $$p_1(r_2+)=0$$ for some integer $$\ell$$ in $$\{0,1,\dots,k-1\}$$.

45. Under the assumptions of Theorem 7.7.2, show that $$C=0$$ if and only if $$p_2(r_2+2)=0$$ for some integer $$\ell$$ in $$\{0,1,\dots,k-1\}$$.

46. Let $Ly=\alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_1x)y\nonumber$ and define $p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0.\nonumber$

Show that if $p_0(r)=\alpha_0(r-r_1)(r-r_2)\nonumber$ where $$r_1-r_2=k$$, a positive integer, then $$Ly=0$$ has the solutions

$y_1=x^{r_1}\sum_{n=0}^\infty {(-1)^n\over n!\prod_{j=1}^n(j+k)}\left(\gamma_1\over\alpha_0\right)^n x^n\nonumber$

and

\begin{aligned} y_2&=x^{r_2}\sum_{n=0}^{k-1} {(-1)^n\over n!\prod_{j=1}^n(j-k)} \left(\gamma_1\over\alpha_0\right)^n x^n\\[4pt]&-{1\over k!(k-1)!}\left(\gamma_1\over\alpha_0\right)^k\left(y_1\ln x- x^{r_1}\sum_{n=1}^\infty {(-1)^n\over n!\prod_{j=1}^n(j+k)}\left(\gamma_1\over\alpha_0\right)^n \left(\sum_{j=1}^n{2j+k\over j(j+k)}\right)x^n\right).\end{aligned}\nonumber

47. Let $Ly=\alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_2x^2)y\nonumber$ and define $p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0.\nonumber$

Show that if $p_0(r)=\alpha_0(r-r_1)(r-r_2)\nonumber$ where $$r_1-r_2=2k$$, an even positive integer, then $$Ly=0$$ has the solutions

$y_1=x^{r_1}\sum_{m=0}^\infty {(-1)^m\over 4^mm!\prod_{j=1}^m(j+k)}\left(\gamma_2\over\alpha_0\right)^m x^{2m}\nonumber$

and

\begin{aligned} y_2&=x^{r_2}\sum_{m=0}^{k-1} {(-1)^m\over4^mm!\prod_{j=1}^m(j-k)} \left(\gamma_2\over\alpha_0\right)^m x^{2m}\\[4pt]&-{2\over 4^kk!(k-1)!}\left(\gamma_2\over\alpha_0\right)^k\left(y_1\ln x- {x^{r_1}\over2}\sum_{m=1}^\infty {(-1)^m\over 4^mm!\prod_{j=1}^m(j+k)}\left(\gamma_2\over\alpha_0\right)^m \left(\sum_{j=1}^m{2j+k\over j(j+k)}\right)x^{2m}\right).\end{aligned}\nonumber

48. Let $$L$$ be as in Exercises 7.5.57 and 7.5.58, and suppose the indicial polynomial of $$Ly=0$$ is

$p_0(r)=\alpha_0(r-r_1)(r-r_2),\nonumber$

with $$k=r_1-r_2$$, where $$k$$ is a positive integer. Define $$a_0(r)=1$$ for all $$r$$. If $$r$$ is a real number such that $$p_0(n+r)$$ is nonzero for all positive integers $$n$$, define

$a_n(r)=-{1\over p_0(n+r)}\sum_{j=1}^n p_j(n+r-j)a_{n-j}(r),\,n\ge1,\nonumber$

and let $y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n.\nonumber$

Define $a_n(r_2)=-{1\over p_0(n+r_2)}\sum_{j=1}^n p_j(n+r_2-j)a_{n-j}(r_2) \, \text{if} n\ge1 \, \text{and}\, n\ne k,\nonumber$ and let $$a_k(r_2)$$ be arbitrary.

1. Conclude from Exercise 7.6.66 that $L\left(y_1\ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n\right)=k\alpha_0x^{r_1}.\nonumber$
2. Conclude from Exercise 7.5.57 that $L\left(x^{r_2}\sum_{n=0}^\infty a_n(r_2)x^n\right)=Ax^{r_1},\nonumber$ where $A=\sum_{j=1}^k p_j(r_1-j)a_{k-j}(r_2).\nonumber$
3. Show that $$y_1$$ and $y_2=x^{r_2}\sum_{n=0}^\infty a_n(r_2)x^n -{A\over k\alpha_0} \left(y_1 \ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n\right)\nonumber$ form a fundamental set of Frobenius solutions of $$Ly=0$$.
4. Show that choosing the arbitrary quantity $$a_k(r_2)$$ to be nonzero merely adds a multiple of $$y_1$$ to $$y_2$$. Conclude that we may as well take $$a_k(r_2)~=~0$$.

This page titled 6.7.1: The Method of Frobenius III (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.