
# 7.6E: The Method of Frobenius II (Exercises)

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In Exercises [exer:7.6.1} –[exer: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 $$N>7$$.

[exer:7.6.1] $$x^2y''-x(1-x)y'+(1-x^2)y=0$$

[exer:7.6.2] $$x^2(1+x+2x^2)y'+x(3+6x+7x^2)y'+(1+6x-3x^2)y=0$$

[exer:7.6.3] $$x^2(1+2x+x^2)y''+x(1+3x+4x^2)y'-x(1-2x)y=0$$

[exer:7.6.4] $$4x^2(1+x+x^2)y''+12x^2(1+x)y'+(1+3x+3x^2)y=0$$

[exer:7.6.5] $$x^2(1+x+x^2)y''-x(1-4x-2x^2)y'+y=0$$

[exer:7.6.6] $$9x^2y''+3x(5+3x-2x^2)y'+(1+12x-14x^2)y=0$$

[exer:7.6.7] $$x^2y''+x(1+x+x^2)y'+x(2-x)y=0$$

[exer:7.6.8] $$x^2(1+2x)y''+x(5+14x+3x^2)y'+(4+18x+12x^2)y=0$$

[exer:7.6.9] $$4x^2y''+2x(4+x+x^2)y'+(1+5x+3x^2)y=0$$

[exer:7.6.10] $$16x^2y''+4x(6+x+2x^2)y'+(1+5x+18x^2)y=0$$

[exer:7.6.11] $$9x^2(1+x)y''+3x(5+11x-x^2)y'+(1+16x-7x^2)y=0$$

[exer:7.6.12] $$4x^2y''+(1+4x)y=0$$

[exer:7.6.13] $$36x^2(1-2x)y''+24x(1-9x)y'+(1-70x)y=0$$

[exer:7.6.14] $$x^2(1+x)y''-x(3-x)y'+4y=0$$

[exer:7.6.15] $$x^2(1-2x)y''-x(5-4x)y'+(9-4x)y=0$$

[exer:7.6.16] $$25x^2y''+x(15+x)y'+(1+x)y=0$$

[exer:7.6.17] $$2x^2(2+x)y''+x^2y'+(1-x)y=0$$

[exer:7.6.18] $$x^2(9+4x)y''+3xy'+(1+x)y=0$$

[exer:7.6.19] $$x^2y''-x(3-2x)y'+(4+3x)y=0$$

[exer:7.6.20] $$x^2(1-4x)y''+3x(1-6x)y'+(1-12x)y=0$$

[exer:7.6.21] $$x^2(1+2x)y''+x(3+5x)y'+(1-2x)y=0$$

[exer:7.6.22] $$2x^2(1+x)y''-x(6-x)y'+(8-x)y=0$$

[exer:7.6.23] $$x^2(1+2x)y''+x(5+9x)y'+(4+3x)y=0$$

[exer:7.6.24] $$x^2(1-2x)y''-x(5+4x)y'+(9+4x)y=0$$

[exer:7.6.25] $$x^2(1+4x)y''-x(1-4x)y'+(1+x)y=0$$

[exer:7.6.26] $$x^2(1+x)y''+x(1+2x)y'+xy=0$$

[exer:7.6.27] $$x^2(1-x)y''+x(7+x)y'+(9-x)y=0$$

[exer:7.6.28] $$x^2y''-x(1-x^2)y'+(1+x^2)y=0$$

[exer:7.6.29] $$x^2(1+x^2)y''-3x(1-x^2)y'+4y=0$$

[exer:7.6.30] $$4x^2y''+2x^3y'+(1+3x^2)y=0$$

[exer:7.6.31] $$x^2(1+x^2)y''-x(1-2x^2)y'+y=0$$

[exer:7.6.32] $$2x^2(2+x^2)y''+7x^3y'+(1+3x^2)y=0$$

[exer:7.6.33] $$x^2(1+x^2)y''-x(1-4x^2)y'+(1+2x^2)y=0$$

[exer:7.6.34] $$4x^2(4+x^2)y''+3x(8+3x^2)y'+(1-9x^2)y=0$$

[exer:7.6.35] $$3x^2(3+x^2)y''+x(3+11x^2)y'+(1+5x^2)y=0$$

[exer:7.6.36] $$4x^2(1+4x^2)y''+32x^3y'+y=0$$

[exer:7.6.37] $$9x^2y''-3x(7-2x^2)y'+(25+2x^2)y=0$$

[exer:7.6.38] $$x^2(1+2x^2)y''+x(3+7x^2)y'+(1-3x^2)y=0$$

[exer:7.6.39] $$x^2(1+x^2)y''+x(3+8x^2)y'+(1+12x^2)y$$

[exer:7.6.40] $$x^2y''-x(1-x^2)y'+(1+x^2)y=0$$

[exer:7.6.41] $$x^2(1-2x^2)y''+x(5-9x^2)y'+(4-3x^2)y=0$$

[exer:7.6.42] $$x^2(2+x^2)y''+x(14-x^2)y'+2(9+x^2)y=0$$

[exer:7.6.43] $$x^2(1+x^2)y''+x(3+7x^2)y'+(1+8x^2)y=0$$

[exer:7.6.44] $$x^2(1-2x)y''+3xy'+(1+4x)y=0$$

[exer:7.6.45] $$x(1+x)y''+(1-x)y'+y=0$$

[exer:7.6.46] $$x^2(1-x)y''+x(3-2x)y'+(1+2x)y=0$$

[exer:7.6.47] $$4x^2(1+x)y''-4x^2y'+(1-5x)y=0$$

[exer:7.6.48] $$x^2(1-x)y''-x(3-5x)y'+(4-5x)y=0$$

[exer:7.6.49] $$x^2(1+x^2)y''-x(1+9x^2)y'+(1+25x^2)y=0$$

[exer:7.6.50] $$9x^2y''+3x(1-x^2)y'+(1+7x^2)y=0$$

[exer:7.6.51] $$x(1+x^2)y''+(1-x^2)y'-8xy=0$$

[exer:7.6.52] $$4x^2y''+2x(4-x^2)y'+(1+7x^2)y=0$$

[exer:7.6.53] Under the assumptions of Theorem [thmtype:7.6.2} , suppose the power series

$\sum_{n=0}^\infty a_n(r_1)x^n \quad\mbox{ and }\quad \sum_{n=1}^\infty a_n'(r_1)x^n$

converge on $$(-\rho,\rho)$$.

Show that

$y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n\quad\mbox{ and }\quad y_2=y_1\ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n$

are linearly independent on $$(0,\rho)$$.

Use the result of

## a

to complete the proof of Theorem [thmtype:7.6.2}.

[exer:7.6.54] Let

$Ly=x^2(\alpha_0+\alpha_1x)y''+x(\beta_0+\beta_1x)y'+(\gamma_0+\gamma_1x)y$

and define

$p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0\quad\mbox{ and }\quad p_1(r)=\alpha_1r(r-1)+\beta_1r+\gamma_1.$

Theorem [thmtype:7.6.1} and Exercise 7.5. [exer:7.5.55}

## a

imply that if

$y(x,r)=x^r\sum_{n=0}^\infty a_n(r)x^n$

where

$a_n(r)=(-1)^n\prod_{j=1}^n{p_1(j+r-1)\over p_0(j+r)},$

then

$Ly(x,r)=p_0(r)x^r.$

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

$y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n,$

where

$a_n(r_1)={(-1)^n\over\alpha_0^n(n!)^2}\prod_{j=1}^np_1(j+r_1-1).$

Show that $$Ly=0$$ has the second solution

$y_2=y_1\ln x+x^{r_1}\sum_{n=1}^\infty a_n(r_1)J_nx^n,$

where

$J_n=\sum_{j=1}^n{p_1'(j+r_1-1)\over p_1(j+r_1-1)}-2\sum_{j=1}^n{1\over j}.$

Conclude from

and

## b

that if $$\gamma_1\ne0$$ then

$y_1=x^{r_1}\sum_{n=0}^\infty {(-1)^n\over(n!)^2}\left(\gamma_1\over\alpha_0\right)^nx^n$

and

$y_2=y_1\ln x-2x^{r_1}\sum_{n=1}^\infty {(-1)^n\over(n!)^2}\left(\gamma_1\over\alpha_0\right)^n \left(\sum_{j=1}^n{1\over j}\right)x^n$

are solutions of

$\alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_1x)y=0.$

(The conclusion is also valid if $$\gamma_1=0$$. Why?)

[exer:7.6.55] Let

$Ly=x^2(\alpha_0+\alpha_qx^q)y''+x(\beta_0+\beta_qx^q)y'+(\gamma_0+\gamma_qx^q)y$

where $$q$$ is a positive integer, and define

$p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0\quad\mbox{ and }\quad p_q(r)=\alpha_qr(r-1)+\beta_qr+\gamma_q.$

Suppose

$p_0(r)=\alpha_0(r-r_1)^2 \quad\mbox{ and }\quad p_q(r)\not\equiv0.$

Recall from Exercise 7.5. [exer:7.5.59} that $$Ly~=0$$ has the solution

$y_1=x^{r_1}\sum_{m=0}^\infty a_{qm}(r_1)x^{qm},$

where

$a_{qm}(r_1)={(-1)^m\over (q^2\alpha_0)^m(m!)^2}\prod_{j=1}^mp_q\left(q(j-1)+r_1\right).$

Show that $$Ly=0$$ has the second solution

$y_2=y_1\ln x+x^{r_1}\sum_{m=1}^\infty a_{qm}'(r_1)J_mx^{qm},$

where

$J_m=\sum_{j=1}^m{p_q'\left(q(j-1)+r_1\right)\over p_q\left(q(j-1)+r_1\right)}-{2\over q}\sum_{j=1}^m{1\over j}.$

Conclude from

and

## b

that if $$\gamma_q\ne0$$ then

$y_1=x^{r_1}\sum_{m=0}^\infty {(-1)^m\over(m!)^2}\left(\gamma_q\over q^2\alpha_0\right)^mx^{qm}$

and

$y_2=y_1\ln x-{2\over q}x^{r_1}\sum_{m=1}^\infty {(-1)^m\over(m!)^2}\left(\gamma_q\over q^2\alpha_0\right)^m\left(\sum_{j=1}^m{1\over j}\right)x^{qm}$

are solutions of

$\alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_qx^q)y=0.$

[exer:7.6.56] The equation

$xy''+y'+xy=0$

is Bessel’s equation of order zero. (See Exercise [exer:7.5.53}.) Find two linearly independent Frobenius solutions of this equation.

[exer:7.6.57] Suppose the assumptions of Exercise 7.5. [exer:7.5.53} hold, except that

$p_0(r)=\alpha_0(r-r_1)^2.$

Show that

$y_1={x^{r_1}\over\alpha_0+\alpha_1x+\alpha_2x^2}\quad\mbox{ and }\quad y_2={x^{r_1}\ln x\over\alpha_0+\alpha_1x+\alpha_2x^2}$

are linearly independent Frobenius solutions of

$x^2(\alpha_0+\alpha_1x+\alpha_2 x^2)y''+x(\beta_0+\beta_1x+\beta_2x^2)y'+ (\gamma_0+\gamma_1x+\gamma_2x^2)y=0$

on any interval $$(0,\rho)$$ on which $$\alpha_0+\alpha_1x+\alpha_2x^2$$ has no zeros.

[exer:7.6.58] $$4x^2(1+x)y''+8x^2y'+(1+x)y=0$$

[exer:7.6.59] $$9x^2(3+x)y''+3x(3+7x)y'+(3+4x)y=0$$

[exer:7.6.60] $$x^2(2-x^2)y''-x(2+3x^2)y'+(2-x^2)y=0$$

[exer:7.6.61] $$16x^2(1+x^2)y''+8x(1+9x^2)y'+(1+49x^2)y=0$$

[exer:7.6.62] $$x^2(4+3x)y''-x(4-3x)y'+4y=0$$

[exer:7.6.63] $$4x^2(1+3x+x^2)y''+8x^2(3+2x)y'+(1+3x+9x^2)y=0$$

[exer:7.6.64] $$x^2(1-x)^2y''-x(1+2x-3x^2)y'+(1+x^2)y=0$$

[exer:7.6.65] $$9x^2(1+x+x^2)y''+3x(1+7x+13x^2)y'+(1+4x+25x^2)y=0$$

[exer:7.6.66]

Let $$L$$ and $$y(x,r)$$ be as in Exercises [exer:7.5.57} and [exer:7.5.58}. Extend Theorem [thmtype:7.6.1} by showing that

$L\left({\partial y\over \partial r}(x,r)\right)=p'_0(r)x^r+x^rp_0(r)\ln x.$

Show that if

$p_0(r)=\alpha_0(r-r_1)^2$

then

$y_1=y(x,r_1) \mbox{\quad and \quad} y_2={\partial y\over\partial r}(x,r_1)$

are solutions of $$Ly=0$$.