3.5E: Exercises

This set contains exercises specifically identified by \Lex that ask you to implement the verification procedure. These particular exercises were chosen arbitrarily you can just as well formulate such laboratory problems for any of the equations in Exercises \((3.5E.1)\) to \((3.5E.10)\), \((3.5E.14)\) to \((3.5E.25)\), and \((3.5E.28)\) to \((3.5E.51)\).

In Exercises \((3.5E.1)\) to \((3.5E.10)\), find a fundamental set of Frobenius solutions. Compute \(a_0\), \(a_{1}\) \(\dots\), \(a_N\) for \(N\) at least \(7\) in each solution.

Exercise \(\PageIndex{1}\)

\(2x^2(1+x+x^2)y''+x(3+3x+5x^2)y'-y=0\)

Exercise \(\PageIndex{2}\)

\(3x^2y''+2x(1+x-2x^2)y'+(2x-8x^2)y=0\)

Answer

\(p_0(r)=r(3r-1) \);
\(p_1(r)=2(r+1) \);
\(p_2(r)=-4(r+2) \).

\(a_1(r)=\displaystyle-{2\over3r+2} \);
\(a_n(r)=\displaystyle-{2a_{n-1}(r)-4a_{n-2}(r)\over3n+3r-1} \),
\(n\ge1 \).

\(r_1=1/3 \);
\(a_1(1/3)=-2/3 \);
\(a_n(1/3)=\displaystyle-{2a_{n-1}(1/3)-4a_{n-2}(1/3)\over3n} \),
\(n\ge \)1;

\(y_1=\displaystyle{x^{1/3}\left(1-{2\over3}x+{8\over9}x^2-
{40\over81}x^3+\cdots\right)} \).

 

\(r_2=0 \);
\(a_1(0)=-1 \);
\(a_n(0)=-\displaystyle{2a_{n-1}(0)-4a_{n-2}(0)\over3n-1} \),
\(n\ge1 \);

\(y_2=\displaystyle{1-x+{6\over5}x^2-{4\over5}x^3+\cdots} \).
 

 

Exercise \(\PageIndex{3}\)

\(x^2(3+3x+x^2)y''+x(5+8x+7x^2)y'-(1-2x-9x^2)y=0\)

Exercise \(\PageIndex{4}\)

\(4x^2y''+x(7+2x+4x^2)y'-(1-4x-7x^2)y=0\)

Answer

\(p_0(r)=(r+1)(4r-1) \);
\(p_1(r)=2(r+2) \);
\(p_2(r)=4r+7 \).

\(a_1(r)=\displaystyle-{2\over 4r+3} \);
\(a_n(r)=\displaystyle-{2\over 4n+4r-1}a_{n-1}(r)-{1\over n+r+1}a_{n-2}(r) \),
\(n\ge1 \).

\(r_1=1/4 \);
\(a_1(1/4)=-1/2 \);
\(a_n(1/4)=\displaystyle-{1\over 2n}a_{n-1}(1/4)-{4\over 4n+5}a_{n-2}(1/4) \),
\(n\ge \)1;

\(y_1=\displaystyle{x^{1/4}\left(1-{1\over2}x-{19\over104}x^2+
{1571\over10608}x^3+\cdots\right)} \).

\(r_2=-1 \);
\(a_1(-1)=2 \);
\(a_n(-1)=\displaystyle-{2\over 4n-5}a_{n-1}(-1)-{1\over n}a_{n-2}(-1) \),
\(n\ge1 \);

\(y_2=\displaystyle{x^{-1}\left(1+2x-{11\over6}x^2-{1\over7}x^3+\cdots\right)} \).

Exercise \(\PageIndex{5}\)

\(12x^2(1+x)y''+x(11+35x+3x^2)y'-(1-10x-5x^2)y=0\)

Exercise \(\PageIndex{6}\)

\(x^2(5+x+10x^2)y''+x(4+3x+48x^2)y'+(x+36x^2)y=0\)

Answer

\(p_0(r)=r(5r-1) \);
\(p_1(r)=(r+1)^2 \);
\(p_2(r)=2(r+2)(5r+9) \).

\(a_1(r)=\displaystyle-{r+1\over5r+4}\displaystyle \);
\(a_n(r)=\displaystyle-{n+r\over5n+5r-1}a_{n-1}(r)-2a_{n-2}(r) \),
\(n\ge1 \).

\(r_1=1/5 \);
\(a_1(1/5)=-6/25 \);
\(a_n(1/5)=\displaystyle-{5n+1\over25n}a_{n-1}(1/5)-2a_{n-2}(1/5) \),
\(n\ge \)1;

\(y_1=\displaystyle{x^{1/5}\left(1-{6\over25}x-{1217\over625}x^2+
{41972\over46875}x^3
+\cdots\right)} \).

\(r_2=0 \);
\(a_1(0)=-1/4 \);
\(a_n(0)=\displaystyle-{n\over5n-1}a_{n-1}(0)-2a_{n-2}(0) \),
\(n\ge1 \);

\(y_2=\displaystyle{x-{1\over4}x^2-{35\over18}x^3+{11\over12}x^4+\cdots} \).
 

Exercise \(\PageIndex{7}\)

\(8x^2y''-2x(3-4x-x^2)y'+(3+6x+x^2)y=0\)

Exercise \(\PageIndex{8}\)

\(18x^2(1+x)y''+3x(5+11x+x^2)y'-(1-2x-5x^2)y=0\)

Answer

\(p_0(r)=(3r-1)(6r+1) \);
\(p_1(r)=(3r+2)(6r+1) \);
\(p_2(r)=3r+5 \).

\(a_1(r)=\displaystyle-{6r+1\over6r+7} \);
\(a_n(r)=\displaystyle-{6n+6r-5\over6n+6r+1}a_{n-1}(r)-{1\over6n+6r+1}a_{n-2}(r) \),
\(n\ge1 \).

\(r_1=1/3 \);
\(a_1(1/3)=-1/3 \);
\(a_n(1/3)=\displaystyle-{2n-1\over2n+1}a_{n-1}(1/3)-{1\over6n+3}a_{n-2}(1/3) \),
\(n\ge \)1;

\(y_1=\displaystyle{x^{1/3}\left(1-{1\over3}x+{2\over15}x^2-
{5\over63}x^3+\cdots\right)} \).

\(r_2=-1/6 \);
\(a_1(-1/6)=0 \);
\(a_n(-1/6)=\displaystyle-{n-1\over n}a_{n-1}(-1/6)-{1\over6n}a_{n-2}(-1/6) \),
\(n\ge1 \);

\(y_2=\displaystyle{x^{-1/6}\left(1-{1\over12}x^2+{1\over18}x^3+\cdots\right)} \).

Exercise \(\PageIndex{9}\)

\(x(3+x+x^2)y''+(4+x-x^2)y'+xy=0\)

Exercise \(\PageIndex{10}\)

\(10x^2(1+x+2x^2)y''+x(13+13x+66x^2)y'-(1+4x+10x^2)y=0\)

Answer

\(p_0(r)=(2r+1)(5r-1) \);
\(p_1(r)=(2r-1)(5r+4) \);
\(p_2(r)=2(2r+5)(5r-1) \).

\(a_1(r)=\displaystyle-{2r-1\over2r+3} \);
\(a_n(r)=\displaystyle-{2n+2r-3\over2n+2r+1}a_{n-1}(r)-
{10n+10r-22\over5n+5r-1}a_{n-2}(r) \),
\(n\ge1 \).

\(r_1=1/5 \);
\(a_1(1/5)=3/17 \);
\(a_n(1/5)=\displaystyle-{10n-13\over10n+7}a_{n-1}(1/5)-{2n-4\over n}a_{n-2}(1/5) \),
\(n\ge \)1;

\(y_1=\displaystyle{x^{1/5}\left(1+{3\over17}x-{7\over153}x^2-
{547\over5661}x^3+\cdots\right)} \).

\(r_2=-1/2 \);
\(a_1(-1/2)=1 \);
\(a_n(-1/2)=\displaystyle-{n-2\over n}a_{n-1}(-1/2)-{20n-54\over10n-7}a_{n-2}(-1/2) \),
\(n\ge1 \);

\(y_2=\displaystyle{x^{-1/2}\left(1+x+{14\over13}x^2-{556\over897}x^3+\cdots\right)} \).

Exercise \(\PageIndex{11}\)

\Lex
The Frobenius solutions of

\begin{eqnarray*}
2x^2(1+x+x^2)y''+x(9+11x+11x^2)y'+(6+10x+7x^2)y=0
\end{eqnarray*}

obtained in Example \((3.5.1)\) are defined on \((0,\rho)\), where \(\rho\) is defined in Theorem \((3.5.2)\). Find \(\rho\). Then do the following experiments for each Frobenius solution, with \(M=20\) and \(\delta=.5\rho\), \(.7\rho\), and \(.9\rho\) in the verification procedure described at the end of this section.

(a) Compute \(\sigma_N(\delta)\) (see Equation \((3.5.28)\)) for \(N=5\), \(10\), \(15\), \(\dots\), \(50\).

(b) Find \(N\) such that \(\sigma_N(\delta)<10^{-5}\).

(c) Find \(N\) such that \(\sigma_N(\delta)<10^{-10}\).

Exercise \(\PageIndex{12}\)

\Lex
By Theorem \((3.5.2)\), the Frobenius solutions of the equation in Exercise \((3.5E.4)\) are defined on \((0,\infty)\). Do experiments (a), (b), and (c) of Exercise \((3.5E.11)\) for each Frobenius solution, with \(M=20\) and \(\delta=1\), \(2\), and \(3\) in the verification procedure described at the end of this section.

Exercise \(\PageIndex{13}\)

\Lex
The Frobenius solutions of the equation in Exercise \((3.5E.6)\) are defined on \((0,\rho)\), where \(\rho\) is defined in Theorem \((3.5.2)\). Find \(\rho\) and do experiments (a), (b), and (c), of Exercise \((3.5E.11)\) for each Frobenius solution, with \(M=20\) and \(\delta=.3\rho\), \(.4\rho\), and \(.5\rho\), in the verification procedure described at the end of this section.

In Exercises \9(3.5E.14)\) to \((3.5E.25)\), find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients in each solution.

Exercise \(\PageIndex{14}\)

\(2x^2y''+x(3+2x)y'-(1-x)y=0\)

Answer

\(p_0(r)=(r+1)(2r-1) \);
\(p_1(r)=2r+1 \);
\(a_n(r)=-\displaystyle{1\over n+r+1}a_{n-1}(r) \).

\(r_1=1/2 \); \(a_n(1/2)=\displaystyle-{2\over 2n+3}a_{n-1}(1/2) \);
\(y_1=\displaystyle{x^{1/2}\sum_{n=0}^\infty{(-2)^n\over\prod_{j=1}^n(2j+3)}x^n} \).

\(r_2=-1 \); \(a_n(-1)=\displaystyle-{1\over n}a_{n-1}(-1) \);
\(y_2=\displaystyle{x^{-1}\sum_{n=0}^\infty{(-1)^n\over n!}x^n} \).

Exercise \(\PageIndex{15}\)

\(x^2(3+x)y''+x(5+4x)y'-(1-2x)y=0\)

Exercise \(\PageIndex{16}\)

\(2x^2y''+x(5+x)y'-(2-3x)y=0\)

Answer

\(p_0(r)=(r+2)(2r-1) \);
\(p_1(r)=r+3 \);
\(a_n(r)=-\displaystyle{1\over 2n+2r-1}a_{n-1}(r) \).

\(r_1=1/2 \);
\(a_n(1/2)= -\displaystyle{1\over 2n}a_{n-1}(1/2) \);
\(y_1=\displaystyle{x^{1/2}\sum_{n=0}^\infty {(-1)^n\over2^nn!}x^n} \).

\(r_2=-2 \);
\(a_n(-2)=-\displaystyle{1\over 2n-5}a_{n-1}(-2) \);
\(y_2=\displaystyle{{1\over x^2}\sum_{n=0}^\infty{(-1)^n\over\prod_{j=1}^n(2j-5)}
x^n} \).

Exercise \(\PageIndex{17}\)

\(3x^2y''+x(1+x)y'-y=0\)

Exercise \(\PageIndex{18}\)

\(x^2y''-xy'+(1-2x)y=0\)

Answer

\(p_0(r)=(r-1)(2r-1) \);
\(p_1(r)= -2 \);
\(a_n(r)=\displaystyle{2\over (n+r-1)(2n+2r-1)}a_{n-1}(r) \).

\(r_1=1 \);
\(a_n(1)=\displaystyle{2\over n(2n+1})a_{n-1}(1) \);
\(y_1=\displaystyle{x\sum_{n=0}^\infty{2^n\over
n!\prod_{j=1}^n(2j+1)}x^n} \).

\(r_2=1/2 \);
\(a_n(1/2)=\displaystyle{2\over n(2n-1})a_{n-1}(1/2) \);
\(y_2=\displaystyle{x^{1/2}\sum_{n=0}^\infty{2^n\over
n!\prod_{j=1}^n(2j-1)}x^n} \).

Exercise \(\PageIndex{19}\)

\(9x^2y''+9xy'-(1+3x)y=0\)

Exercise \(\PageIndex{20}\)

\(3x^2y''+x(1+x)y'-(1+3x)y=0\)

Answer

\(p_0(r)=(r-1)(3r+1) \);
\(p_1(r)=r-3 \);
\(a_n(r)=-\displaystyle{n+r-4\over (n+r-1)(3n+3r+1)}a_{n-1}(r) \).

\(r_1= \);
\(a_n(1)=-\displaystyle{n-3\over n(3n+4)}a_{n-1}(1) \);
\(y_1=\displaystyle{x\left(1+{2\over7}x+{1\over70}x^2\right)} \).

\(r_2=-1/3 \);
\(a_n(-1/3)=-\displaystyle{3n-13\over 3n(3n-4)}a_{n-1}(-1/3) \);
\(y_2=\displaystyle{x^{-1/3}\sum_{n=0}^\infty{(-1)^n\over3^nn!}\left(\prod_{j=1}^n
{3j-13\over3j-4}\right)x^n} \).

Exercise \(\PageIndex{21}\)

\(2x^2(3+x)y''+x(1+5x)y'+(1+x)y=0\)

Exercise \(\PageIndex{22}\)

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

Answer

\(p_0(r)=(r-1)(4r-1) \);
\(p_1(r)=r(r+2) \);
\(a_n(r)=-\displaystyle{n+r+1\over 4n+4r-1}a_{n-1}(r) \).

\(r_1=1 \);
\(a_n(1)= -\displaystyle{n+2\over 4n+3}a_{n-1}(1) \);
\(y_1=\displaystyle{x\sum_{n=0}^\infty{(-1)^n(n+2)!\over2\prod_{j=1}^n(4j+3)}
x^n} \).

\(r_2=1/4 \);
\(a_n(1/4)=-\displaystyle{4n+5\over 16n}a_{n-1}(1/4) \);
\(y_2=\displaystyle{x^{1/4}\sum_{n=0}^\infty{(-1)^n\over16^nn!}\prod_{j=1}^n(4j+5)
x^n} \)

Exercise \(\PageIndex{23}\)

\(2x^2y''+5xy'+(1+x)y=0\)

Exercise \(\PageIndex{24}\)

\(x^2(3+4x)y''+x(5+18x)y'-(1-12x)y=0\)

Answer

\(p_0(r)=(r+1)(3r-1) \);
\(p_1(r)=2(r+2)(2r+3) \);
\(a_n(r)=-2\displaystyle{2n+2r+1\over 3n+3r-1}a_{n-1}(r) \).

\(r_1=1/3 \);
\(a_n(1/3)=-2\displaystyle{6n+5\over 9n}a_{n-1}(1/3) \);
\(y_1=\displaystyle{x^{1/3}\sum_{n=0}^\infty{(-1)^n\over
n!}\left(2\over9\right)^n\left(\prod_{j=1}^n(6j+5)\right)
x^n} \);

\(r_2=-1 \);
\(a_n(-1)=-2\displaystyle{2n-1\over 3n-4}a_{n-1}(-1) \);
\(y_2=\displaystyle{x^{-1}\sum_{n=0}^\infty(-1)^n2^n\left(\prod_{j=1}^n
{2j-1\over3j-4}\right) x^n} \)

Exercise \(\PageIndex{25}\)

\(6x^2y''+x(10-x)y'-(2+x)y=0\)

Exercise \(\PageIndex{26}\)

\Lex
By Theorem \((3.5.2)\) the Frobenius solutions of the equation in Exercise \((3.5E.17)\) are defined on \((0,\infty)\). Do experiments (a), (b), and (c) of Exercise \((3.5E.11)\) for each Frobenius solution, with \(M=20\) and \(\delta=3\), \(6\), \(9\), and \(12\) in the verification procedure described at the end of this section.

Exercise \(\PageIndex{27}\)

\Lex
The Frobenius solutions of the equation in Exercise \((3.5E.22)\) are defined on \((0,\rho)\), where \(\rho\) is defined in Theorem \((3.5.2)\). Find \(\rho\) and do experiments (a), (b), and (c) of Exercise \((3.5E.11)\) for each Frobenius solution, with \(M=20\) and \(\delta=.25\rho\), \(.5\rho\), and \(.75\rho\) in the verification procedure described at the end of this section.

In Exercises \((3.5E.28)\) to \((3.5E.32)\), find a fundamental set of Frobenius solutions. Compute coefficients \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in each solution.

Exercise \(\PageIndex{28}\)

\(x^2(8+x)y''+x(2+3x)y'+(1+x)y=0\)

Answer

\(p_0(r)=(2r-1)(4r-1) \);
\(p_1(r)=(r+1)^2 \);
\(a_n(r)=\displaystyle-{(n+r)^2\over(2n+2r-1)(4n+4r-1)} a_{n-1}(r) \).

\(r_1=1/2 \);
\(a_n(1/2)=\displaystyle-{4n^2+4n+1\over8n(4n+1)} a_{n-1}(1/2) \);
\(y_1=\displaystyle{x^{1/2}\left(1-{9\over40}x+{5\over128}x^2-{245\over39936}x^3
+\cdots\right)} \).

\(r_2=1/4 \);
\(a_n(1/4)=\displaystyle-{16n^2+8n+1\over32n(4n-1)} a_{n-1}(1/4) \);
\(y_2=\displaystyle{x^{1/4}\left(1-{25\over96}x+{675\over14336}x^2-
{38025\over5046272}x^3
+\cdots\right)} \).

Exercise \(\PageIndex{29}\)

\(x^2(3+4x)y''+x(11+4x)y'-(3+4x)y=0\)

Exercise \(\PageIndex{30}\)

\(2x^2(2+3x)y''+x(4+11x)y'-(1-x)y=0\)

Answer

\(p_0(r)=(2r-1)(2r+1) \);
\(p_1(r)=(2r+1)(3r+1) \);
\(a_n(r)=\displaystyle-{(3n+3r-2)\over(2n+2r+1)} a_n(r) \).

\(r_1=1/2 \);
\(a_n(1/2)=\displaystyle-{6n-1\over4(n+1)} a_{n-1}(1/2) \);
\(y_1=\displaystyle{x^{1/2}\left(1-{5\over8}x+{55\over96}x^2
-{935\over1536}x^3+\cdots\right)} \).

\(r_2=-1/2 \);
\(a_n(-1/2)=\displaystyle-{6n-7\over4n} a_{n-1}(-1/2) \);
\(y_2=\displaystyle{x^{-1/2}\left(1+{1\over4}x-{5\over32}x^2
-{55\over384}x^3+\cdots\right)} \).

Exercise \(\PageIndex{31}\)

\(x^2(2+x)y''+5x(1-x)y'-(2-8x)y\)

Exercise \(\PageIndex{32}\)

\(x^2(6+x)y''+x(11+4x)y'+(1+2x)y=0\)

Answer

\(p_0(r)=(2r+1)(3r+1) \);
\(p_1(r)=(r+1)(r+2) \);
\(a_n(r)=\displaystyle{(n+r)(n+r+1)\over(2n+2r+1)(3n+3r+1)} a_n(r) \).

\(r_1=-1/3 \);
\(a_n(-1/3)=\displaystyle-{(3n-1)(3n+2)\over9n(6n+1)}a_{n-1}(-1/3) \);
\(y_1=\displaystyle{x^{-1/3}\left(1-{10\over63}x+{200\over7371}x^2-
{17600\over3781323}x^3
+\cdots\right)} \).

\(r_2=-1/2 \);
\(a_n(-1/2)=\displaystyle-{(2n-1)(2n+1)\over4n(6n-1)} a_{n-1}(-1/2) \);
\(y_2=\displaystyle{x^{-1/2}\left(1-{3\over20}x+{9\over352}x^2
-{105\over23936}x^3
+\cdots\right)} \).

In Exercises \((3.5E.33)\) to \((3.5E.46)\), find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients in each solution.

Exercise \(\PageIndex{33}\)

\(8x^2y''+x(2+x^2)y'+y=0\)

Exercise \(\PageIndex{34}\)

\(8x^2(1-x^2)y''+2x(1-13x^2)y'+(1-9x^2)y=0\)

Answer

\(p_0(r)=(2r-1)(4r-1) \);
\(p_2(r)=-(2r+3)(4r+3) \);
\(a_{2m}(r)=\displaystyle{8m+4r-5\over8m+4r-1}a_{2m-2}(r) \).

\(r_1=1/2 \);
\(a_{2m}(1/2)=\displaystyle{8m-3\over8m+1}a_{2m-2}(1/2) \);
\(y_1=\displaystyle{x^{1/2}\sum_{m=0}^\infty\left(\prod_{j=1}^m{8j-3\over8j+
1}\right)x^{2m}} \).

\(r_2=1/4 \);
\(a_{2m}(1/4)=\displaystyle{2m-1\over2m}a_{2m-2}(1/4) \);
\(y_2=\displaystyle{x^{1/4}\sum_{m=0}^\infty{1\over2^mm!}\left(\prod_{j=1}^m(2j-1)
\right)x^{2m}} \)

Exercise \(\PageIndex{35}\)

\(x^2(1+x^2)y''-2x(2-x^2)y'+4y=0\)

Exercise \(\PageIndex{36}\)

\(x(3+x^2)y''+(2-x^2)y'-8xy=0\)

Answer

\(p_0(r)=r(3r-1) \);
\(p_2(r)=(r-4)(r+2) \);
\(a_{2m}(r)=-\displaystyle{2m+r-6\over6m+3r-1}a_{2m-2}(r) \).

\(r_1=1/3 \);
\(a_{2m}(1/3)=-\displaystyle{6m-17\over18m}a_{2m-2}(1/3) \);
\(y_1=\displaystyle{x^{1/3}\sum_{m=0}^\infty{(-1)^m\over18^mm!}\left(\prod_{j=1}^m
(6j-17)\right) x^{2m}} \).

\(r_2=0 \);
\(a_{2m}(0)=-\displaystyle{2m-6\over6m-1}a_{2m-2}(0) \);
\(y_2=\displaystyle{1+{4\over5}x^2+{8\over55}x^4} \)

Exercise \(\PageIndex{37}\)

\(4x^2(1-x^2)y''+x(7-19x^2)y'-(1+14x^2)y=0\)

Exercise \(\PageIndex{38}\)

\(3x^2(2-x^2)y''+x(1-11x^2)y'+(1-5x^2)y=0\)

Answer

\(p_0(r)=(2r-1)(3r-1) \);
\(p_2(r)=-(r+1)(3r+5) \);
\(a_{2m}(r)=\displaystyle{2m+r-1\over4m+2r-1}a_{2m-2}(r) \).

\(r_1=1/2 \);
\(a_{2m}(1/2)=\displaystyle{4m-1\over8m}a_{2m-2}(1/2) \);
\(y_1=\displaystyle{x^{1/2}\sum_{m=0}^\infty{1\over8^mm!}\left(\prod_{j=1}^m(4j-1)
 \right)x^{2m}} \).

\(r_2=1/3 \);
\(a_{2m}(1/3)=\displaystyle{6m-2\over12m-1}a_{2m-2}(1/3) \);
\(y_2=\displaystyle{x^{1/3}\sum_{m=0}^\infty2^m\left(\prod_{j=1}^m{3j-1\over12j-1}
\right) x^{2m}} \).

Exercise \(\PageIndex{39}\)

\(2x^2(2+x^2)y''-x(12-7x^2)y'+(7+3x^2)y=0\)

Exercise \(\PageIndex{40}\)

\(2x^2(2+x^2)y''+x(4+7x^2)y'-(1-3x^2)y=0\)

Answer

\(p_0(r)=(2r-1)(2r+1) \);
\(p_1(r)=(r+1)(2r+3) \);
\(a_{2m}(r)=-\displaystyle{2m+r-1\over4m+2r+1}a_{2m-2}(r) \).

\(r_1=1/2 \);
\(a_{2m}(1/2)=-\displaystyle{4m-1\over4(2m+1)}a_{2m-2}(1/2) \);
\(y_1=\displaystyle{x^{1/2}\sum_{m=0}^\infty{(-1)^m\over4^m}\left(\prod_{j=1}^m
{4j-1\over2j+1}\right)x^{2m}} \).

\(r_2=-1/2 \);
\(a_{2m}(-1/2)=-\displaystyle{4m-3\over8m}a_{2m-2}(-1/2) \);
\(y_2=\displaystyle{x^{-1/2}\sum_{m=0}^\infty{(-1)^m\over8^mm!}\left(\prod_{j=1}^m
(4j-3)\right)x^{2m}} \)

Exercise \(\PageIndex{41}\)

\(2x^2(1+2x^2)y''+5x(1+6x^2)y'-(2-40x^2)y=0\)

Exercise \(\PageIndex{42}\)

\(3x^2(1+x^2)y''+5x(1+x^2)y'-(1+5x^2)y=0\)

Answer

\(p_0(r)=(r+1)(3r-1) \);
\(p_1(r)=(r-1)(3r+5) \);
\(a_{2m}(r)=-\displaystyle{2m+r-3\over2m+r+1}a_{2m-2}(r) \).

\(r_1=1/3 \);
\(a_{2m}(1/3)=-\displaystyle{3m-4\over3m+2}a_{2m-2}(1/3) \);
\(y_1=\displaystyle{x^{1/3}\sum_{m=0}^\infty(-1)^m\left(\prod_{j=1}^m
{3j-4\over3j+2}\right)x^{2m}} \).

\(r_2=-1 \);
\(a_{2m}(-1)=-\displaystyle{m-2\over m}a_{2m-2}(-1) \);
 \(y_2=\displaystyle{x^{-1}(1+x^2)} \)

Exercise \(\PageIndex{43}\)

\(x(1+x^2)y''+(4+7x^2)y'+8xy=0\)

Exercise \(\PageIndex{44}\)

\(x^2(2+x^2)y''+x(3+x^2)y'-y=0\)

Answer

\(p_0(r)=(r+1)(2r-1) \);
\(p_1(r)=r^2 \);
\(a_{2m}(r)=-\displaystyle{(2m+r-2)^2\over(2m+r+1)(4m+2r-1)}a_{2m-2}(r) \).

\(r_1=1/2 \);
\(a_{2m}(1/2)=-\displaystyle{(4m-3)^2\over8m(4m+3)}a_{2m-2}(1/2) \);
\(y_1=\displaystyle{x^{1/2}\sum_{m=0}^\infty{(-1)^m\over8^mm!}\left(\prod_{j=1}^m
{(4j-3)^2\over4j+3}\right)x^{2m}} \).

\(r_2=-1 \);
\(a_{2m}(-1)=-\displaystyle{(2m-3)^2\over2m(4m-3)}a_{2m-2}(-1) \);
\(y_2=\displaystyle{x^{-1}\sum_{m=0}^\infty{(-1)^m\over2^mm!}\left(\prod_{j=1}^m
{(2j-3)^2\over4j-3}\right)x^{2m}} \).

Exercise \(\PageIndex{45}\)

\(2x^2(1+x^2)y''+x(3+8x^2)y'-(3-4x^2)y=0\)

Exercise \(\PageIndex{46}\)

\(9x^2y''+3x(3+x^2)y'-(1-5x^2)y=0\)

Answer

\(p_0(r)=(3r-1)(3r+1) \);
\(p_1(r)=3r+5 \);
\(a_{2m}(r)=-\displaystyle{1\over6m+3r+1}a_{2m-2}(r) \).

\(r_1=1/3 \);
\(a_{2m}(1/3)=-\displaystyle{1\over2(3m+1)}a_{2m-2}(1/3) \);
\(y_1=\displaystyle{x^{1/3}\sum_{m=0}^\infty{(-1)^m\over2^m\prod_{j=1}^m(3j+1)}
x^{2m}} \).

\(r_2=-1/3 \);
\(a_{2m}(-1/3)=-\displaystyle{1\over6m}a_{2m-2}(-1/3) \);
\(y_2=\displaystyle{x^{-1/3}\sum_{m=0}^\infty{(-1)^m\over6^mm!} x^{2m}} \)
 

In Exercises \((3.5E.47)\) to \((3.5E.51)\), find a fundamental set of Frobenius solutions. Compute the coefficients \(a_0\), \(\dots\), \(a_{2M}\) for \(M\) at least \(7\) in each solution.

Exercise \(\PageIndex{47}\)

\(6x^2y''+x(1+6x^2)y'+(1+9x^2)y=0\)

Exercise \(\PageIndex{48}\)

\(x^2(8+x^2)y''+7x(2+x^2)y'-(2-9x^2)y=0\)

Answer

\(p_0(r)=2(r+1)(4r-1) \);
\(p_2(r)=(r+3)^2 \);
\(a_{2m}(r)=\displaystyle-{2m+r+1\over2(8m+4r-1)} a_{2m-2}(r) \).

\(r_1=1/4 \);
\(a_{2m}(1/4)=\displaystyle-{8m+5\over64m} a_{2m-2}(1/4) \);
\(y_1=\displaystyle{x^{1/4}\left(1-{13\over64}x^2+{273\over8192}x^4-
{2639\over524288}x^6 +\cdots\right)} \).

\(r_2=-1 \);
\(a_{2m}(-1)=\displaystyle-{m\over8m-5} a_{2m-2}(-1) \);
\(y_2=\displaystyle{x^{-1}\left(1-{1\over3}x^2+{2\over33}x^4-{2\over209}x^6
+\cdots\right)} \).
 

Exercise \(\PageIndex{49}\)

\(9x^2(1+x^2)y''+3x(3+13x^2)y'-(1-25x^2)y=0\)

Exercise \(\PageIndex{50}\)

\(4x^2(1+x^2)y''+4x(1+6x^2)y'-(1-25x^2)y=0\)

Answer

\(p_0(r)=(2r-1)(2r+1) \);
\(p_2(r)=(2r+5)^2 \);
\(a_{2m}(r)=\displaystyle-{4m+2r+1\over4m+2r-1} a_{2m-2}(r) \).

\(r_1=1/2 \);
\(a_{2m}(1/2)=\displaystyle-{2m+1\over2m} a_{2m-2}(1/2) \);
\(y_1=\displaystyle{x^{1/2}\left(1-{3\over2}x^2+{15\over8}x^4-{35\over16}x^6
+\cdots\right)} \).

\(r_2=-1/2 \);
\(a_{2m}(-1/2)=\displaystyle-{2m\over2m-1} a_{2m-2}(-1/2) \);
\(y_2=\displaystyle{x^{-1/2}\left(1-2x^2+{8\over3}x^4-{16\over5}x^6
+\cdots\right)} \).
 

Exercise \(\PageIndex{51}\)

\(8x^2(1+2x^2)y''+2x(5+34x^2)y'-(1-30x^2)y=0\)

Exercise \(\PageIndex{52}\)

Suppose \(r_1>r_2\), \(a_0=b_0=1\), and the Frobenius series

\begin{eqnarray*}
y_1=x^{r_1}\sum_{n=0}^\infty a_nx^n\quad\mbox{ and } \quad y_2=x^{r_2}\sum_{n=0}^\infty b_nx^n
\end{eqnarray*}

both converge on an interval \((0,\rho)\).

(a) Show that \(y_1\) and \(y_2\) are linearly independent on \((0,\rho)\).

Hint: Show that if \(c_1\) and \(c_2\) are constants such that \(c_1y_1+c_2y_2\equiv0\) on \((0,\rho)\), then

\begin{eqnarray*}
c_1x^{r_1-r_2}\sum_{n=0}^\infty a_nx^n+ c_2\sum_{n=0}^\infty b_nx^n=0,\quad 0<x<\rho.
\end{eqnarray*}

Then let \(x\to0+\) to conclude that \(c_2=0\).

(b) Use the result of part (a) to complete the proof of Theorem \((3.5.3)\).

Answer

\part{a}
Multiplying (A) \)c_1y_1+c_2y_2\equiv0 \) by \)x^{-r_2} \) yields
\)c_1x^{r_1-r_2}\sum_{n=0}^\infty a_nx^n+
c_2\sum_{n=0}^\infty b_nx^n=0 \), \)0<x<\rho \).
Letting \)x\to0+ \) shows that \)c_2=0 \), since \)b_0=1 \). Now (A)
reduces to \)c_1y_1\equiv0 \), so \)c_1=0 \). Therefore,\(y_1 \)
and \(y_2 \) are linearly independent on \)(0,\rho) \).

\part{b} Since
\(y_1=\sum_{n=0}^\infty a_n(r_1)x^n \) and
\(y_2=\sum_{n=0}^\infty a_n(r_2)x^n \) are linearly independent solutions
of \)Ly=0 \)   \)(0,\rho) \), \)\{y_1,y_2\} \) is a fundamental
set of solutions of \)Ly=0 \) on \)(0,\rho) \), by Theorem~5.1.6.
 

Exercise \(\PageIndex{53}\)

The equation

\begin{equation}\label{eq:3.5E.1}
x^2y''+xy'+(x^2-\nu^2)y=0
\end{equation}

is Bessel's equation of order \(\nu\). (Here \(\nu\) is a parameter, and this use of "order'' should not be confused with its usual use as in "the order of the equation.'') The solutions of \eqref{eq:3.5E.1} are Bessel functions of order \(\nu\).

(a) Assuming that \(\nu\) isn't an integer, find a fundamental set of Frobenius solutions of \eqref{eq:3.5E.1}.

(b) If \(\nu=1/2\), the solutions of \eqref{eq:3.5E.1} reduce to familiar elementary functions. Identify these functions.

Exercise \(\PageIndex{54}\)

(a) Verify that

\begin{eqnarray*}
{d\over dx}\left(|x|^rx^n\right)=(n+r)|x|^rx^{n-1} \quad \mbox{and} \quad {d^2\over dx^2}\left(|x|^rx^n\right)=(n+r)(n+r-1)|x|^rx^{n-2}
\end{eqnarray*}

if \(x\ne0\).

(b) Let

\begin{eqnarray*}
Ly= x^2(\alpha_0+\alpha_1x+\alpha_2x^2)y''+x(\beta_0+\beta_1x+\beta_2x^2)y' +(\gamma_0+\gamma_1x+\gamma_2x^2)y=0.
\end{eqnarray*}

Show that if \(x^r\sum_{n=0}^\infty a_nx^n\) is a solution of \(Ly=0\) on \((0,\rho)\) then \(|x|^r\sum_{n=0}^\infty a_nx^n\) is a solution on \((-\rho,0)\) and \((0,\rho)\).

Answer

\part{a}  If \)x>0 \), then \)|x|^rx^n=x^{n+r} \), so the assertions
are obvious. If \)x<0 \), then \)|x|^r=(-x)^r \), so
\(\displaystyle {d\over dx}|x|^r=-r(-x)^{r-1}={r(-x)^r\over x}={r|x|^r\over x} \).
Therefore,(A) \(\displaystyle {d\over dx}(|x|^rx^n)={r|x|^r\over
x}x^n+|x|^r(nx^{n-1})=(n+r)|x|^rx^{n-1} \) and
\(\displaystyle {d^2\over dx^2}(|x|^rx^n)=(n+r){d\over dx}(|x|^rx^{n-1})=
(n+r)(n+r-1)|x|^rx^{n-2} \), from (A) with \(n \) replaced by \(n-1 \).

Exercise \(\PageIndex{55}\)

(a) Deduce from Equation \((3.5.20)\) that

\begin{eqnarray*}
a_n(r)=(-1)^n\prod_{j=1}^n{p_1(j+r-1)\over p_0(j+r)}.
\end{eqnarray*}

(b) Conclude that if \(p_0(r)=\alpha_0(r-r_1)(r-r_2)\) where \(r_1-r_2\) is not an integer, then

\begin{eqnarray*}
y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n\quad\mbox{ and }\quad y_2=x^{r_2}\sum_{n=0}^\infty a_n(r_2)x^n
\end{eqnarray*}

form a fundamental set of Frobenius solutions of

\begin{eqnarray*}
x^2(\alpha_0+\alpha_1x)y''+x(\beta_0+\beta_1x)y'+(\gamma_0+\gamma_1x)y=0.
\end{eqnarray*}

(c) Show that if \(p_0\) satisfies the hypotheses of part (b) then

\begin{eqnarray*}
y_1=x^{r_1}\sum_{n=0}^\infty {(-1)^n\over n!\prod_{j=1}^n(j+r_1-r_2)} \left(\gamma_1\over\alpha_0\right)^nx^n
\end{eqnarray*}

and

\begin{eqnarray*}
y_2=x^{r_2}\sum_{n=0}^\infty {(-1)^n\over n!\prod_{j=1}^n(j+r_2-r_1)} \left(\gamma_1\over\alpha_0\right)^nx^n
\end{eqnarray*}

form a fundamental set of Frobenius solutions of

\begin{eqnarray*}
\alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_1x)y=0.
\end{eqnarray*}

Exercise \(\PageIndex{56}\)

Let

\begin{eqnarray*}
Ly=x^2(\alpha_0+\alpha_2x^2)y''+x(\beta_0+\beta_2x^2)y'+ (\gamma_0+\gamma_2x^2)y=0
\end{eqnarray*}

and define

\begin{eqnarray*}
p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0\quad\mbox{ and }\quad p_2(r)=\alpha_2r(r-1)+\beta_2r+\gamma_2.
\end{eqnarray*}

(a) Use Theorem \((3.5.2)\) to show that if

\begin{equation}\label{eq:3.5E.2}
\begin{array}{rcl}
a_0(r)&=&1,\\
p_0(2m+r)a_{2m}(r)+p_2(2m+r-2)a_{2m-2}(r)&=&0,\quad m\ge1,
\end{array}
\end{equation}

then the Frobenius series \(y(x,r)=x^r\sum_{m=0}^\infty a_{2m}x^{2m}\) satisfies \(Ly(x,r)=p_0(r)x^r\).

(b) Deduce from \eqref{eq:3.5E.2} that if \(p_0(2m+r)\) is nonzero for every positive integer \(m\) then

\begin{eqnarray*}
a_{2m}(r)=(-1)^m\prod_{j=1}^m{p_2(2j+r-2)\over p_0(2j+r)}.
\end{eqnarray*}

(c) Conclude that if \(p_0(r)=\alpha_0(r-r_1)(r-r_2)\) where \(r_1-r_2\) is not an even integer, then

\begin{eqnarray*}
y_1=x^{r_1}\sum_{m=0}^\infty a_{2m}(r_1)x^{2m}\quad\mbox{ and }\quad y_2=x^{r_2}\sum_{m=0}^\infty a_{2m}(r_2)x^{2m}
\end{eqnarray*}

form a fundamental set of Frobenius solutions of \(Ly=0\).

(d) Show that if \(p_0\) satisfies the hypotheses of part (c) then

\begin{eqnarray*}
y_1=x^{r_1}\sum_{m=0}^\infty {(-1)^m\over 2^mm!\prod_{j=1}^m(2j+r_1-r_2)} \left(\gamma_2\over\alpha_0\right)^mx^{2m}
\end{eqnarray*}

and

\begin{eqnarray*}
y_2=x^{r_2}\sum_{m=0}^\infty {(-1)^m\over 2^mm!\prod_{j=1}^m(2j+r_2-r_1)} \left(\gamma_2\over\alpha_0\right)^mx^{2m}
\end{eqnarray*}

form a fundamental set of Frobenius solutions of

\begin{eqnarray*}
\alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_2x^2)y=0.
\end{eqnarray*}

Answer

\part{a} Here \(p_1\equiv0 \), so Eqn.~(7.5.12) reduces to
\(a_0(r)=1 \), \(a_1(r)=0 \), \(a_n(r)=-\displaystyle{p_2(n+r-2)\over
p_0(n+r)}a_{n-2}(r) \), \(r\ge0 \), which implies that \(a_{2m+1}(r)=0 \)
for \)m=1,2,3,\dots \). Therefore,Eqn.~(7.5.12) actually reduces to
\(a_0(r)=1 \), \(a_{2m}(r)=\displaystyle-{p_2(2m+r-2)\over p_0(2m+r)} \), which
holds because of condition (A).

\part{b} Similar to the proof of Exercise~7.5.55\part{a}.

\part{c}
\(p_0(2m+r_1)=2m\alpha_0(2m+r_1-r_2) \),
which is nonzero if \)m>0 \), since \(r_1-r_2\ge0 \). Therefore, the
assumptions of Theorem~7.5.2 hold with \(r=r_1 \), and
\)Ly_1=p_0(r_1)x^{r_1}=0 \).
If \(r_1-r_2 \) is not an even integer, then
\(p_0(2m+r_2)=2m\alpha_0(2m-r_1+r_2)\ne0 \), \)m=1,2,\cdots \).
Hence, the assumptions of Theorem~7.5.2 hold with \(r=r_2 \) and
  \)Ly_2=p_0(r_2)x^{r_2}=0 \).
From Exercise~7.5.52,
\)\{y_1,y_2\} \) is a fundamental set of solutions.

\part{d}
 Similar to the proof of Exercise~7.5.55\part{c}.
 

Exercise \(\PageIndex{57}\)

Let

\begin{eqnarray*}
Ly=x^2q_0(x)y''+xq_1(x)y'+q_2(x)y,
\end{eqnarray*}

where

\begin{eqnarray*}
q_0(x)=\sum_{j=0}^\infty \alpha_jx^j,\quad q_1(x)=\sum_{j=0}^\infty \beta_jx^j,\quad q_2(x)=\sum_{j=0}^\infty \gamma_jx^j,
\end{eqnarray*}

and define

\begin{eqnarray*}
p_j(r)=\alpha_jr(r-1)+\beta_jr+\gamma_j,\quad j=0,1,\dots.
\end{eqnarray*}

Let \(y=x^r\sum_{n=0}^\infty a_nx^n\). Show that

\begin{eqnarray*}
Ly=x^r\sum_{n=0}^\infty b_nx^n,
\end{eqnarray*}

where

\begin{eqnarray*}
b_n=\sum_{j=0}^np_j(n+r-j)a_{n-j}.
\end{eqnarray*}

Exercise \(\PageIndex{58}\)

(a) Let \(L\) be as in Exercise \((3.5E.57)\). Show that if

\begin{eqnarray*}
y(x,r)=x^r\sum_{n=0}^\infty a_n(r)x^n
\end{eqnarray*}

where

\begin{eqnarray*}
a_0(r)&=&1,\\
a_n(r)&=&-\displaystyle{1\over p_0(n+r)}\sum_{j=1}^n p_j(n+r-j)a_{n-j}(r),\quad n\ge1,
\end{eqnarray*}

then

\begin{eqnarray*}
Ly(x,r)=p_0(r)x^r.
\end{eqnarray*}

(b) Conclude that if

\begin{eqnarray*}
p_0(r)=\alpha_0(r-r_1)(r-r_2)
\end{eqnarray*}

where \(r_1-r_2\) isn't an integer then \(y_1=y(x,r_1)\) and \(y_2=y(x,r_2)\) are solutions of \(Ly=0\).

Answer

\part{a} From Exercise~3.5.57, \)b_n=0 \) for \(n\ge1 \).

Exercise \(\PageIndex{59}\)

Let

\begin{eqnarray*}
Ly=x^2(\alpha_0+\alpha_qx^q)y''+x(\beta_0+\beta_qx^q)y'+ (\gamma_0+\gamma_qx^q)y
\end{eqnarray*}

where \(q\) is a positive integer, and define

\begin{eqnarray*}
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.
\end{eqnarray*}

(a) Show that if

\begin{eqnarray*}
y(x,r)=x^{r}\sum_{m=0}^\infty a_{qm}(r)x^{qm}
\end{eqnarray*}

where

\begin{equation}\label{eq:3.5E.3}
\begin{array}{rcl}
a_0(r)&=&1,\\
a_{qm}(r)&=&-\displaystyle{p_q\left(q(m-1)+r\right)\over p_0(qm+r)}a_{q(m-1)}(r),\quad m\ge1,
\end{array}
\end{equation}

then

\begin{eqnarray*}
Ly(x,r)=p_0(r)x^r.
\end{eqnarray*}

(b) Deduce from \eqref{eq:3.5E.3} that

\begin{eqnarray*}
a_{qm}(r)=(-1)^m\prod_{j=1}^m{p_q\left(q(j-1)+r\right)\over p_0(qj+r)}.
\end{eqnarray*}

(c) Conclude that if \(p_0(r)=\alpha_0(r-r_1)(r-r_2)\) where \(r_1-r_2\) is not an integer multiple of \(q\), then

\begin{eqnarray*}
y_1=x^{r_1}\sum_{m=0}^\infty a_{qm}(r_1)x^{qm}\quad\mbox{ and }\quad y_2=x^{r_2}\sum_{m=0}^\infty a_{qm}(r_2)x^{qm}
\end{eqnarray*}

form a fundamental set of Frobenius solutions of \(Ly=0\).

(d) Show that if \(p_0\) satisfies the hypotheses of part (c) then

\begin{eqnarray*}
y_1=x^{r_1}\sum_{m=0}^\infty {(-1)^m\over q^mm!\prod_{j=1}^m(qj+r_1-r_2)} \left(\gamma_q\over\alpha_0\right)^mx^{qm}
\end{eqnarray*}

and

\begin{eqnarray*}
y_2=x^{r_2}\sum_{m=0}^\infty {(-1)^m\over q^mm!\prod_{j=1}^m(qj+r_2-r_1)} \left(\gamma_q\over\alpha_0\right)^mx^{qm}
\end{eqnarray*}

form a fundamental set of Frobenius solutions of

\begin{eqnarray*}
\alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_qx^q)y=0.
\end{eqnarray*}

Exercise \(\PageIndex{60}\)

(a) Suppose \(\alpha_0,\alpha_1\), and \(\alpha_2\) are real numbers with \(\alpha_0\ne0\), and \(\{a_n\}_{n=0}^\infty\) is defined by

\begin{eqnarray*}
\alpha_0a_1+\alpha_1a_0=0
\end{eqnarray*}

and

\begin{eqnarray*}
\alpha_0a_n+\alpha_1a_{n-1}+\alpha_2a_{n-2}=0,\quad n\ge2.
\end{eqnarray*}

Show that

\begin{eqnarray*}
(\alpha_0+\alpha_1x+\alpha_2x^2)\sum_{n=0}^\infty a_nx^n=\alpha_0a_0,
\end{eqnarray*}

and infer that

\begin{eqnarray*}
\sum_{n=0}^\infty a_nx^n={\alpha_0a_0\over\alpha_0+\alpha_1x+\alpha_2x^2}.
\end{eqnarray*}

(b) With \(\alpha_0,\alpha_1\), and \(\alpha_2\) as in part (a), consider the equation

\begin{equation}\label{eq:3.5E.4}
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,
\end{equation}

and define

\begin{eqnarray*}
p_j(r)=\alpha_jr(r-1)+\beta_jr+\gamma_j,\quad j=0,1,2.
\end{eqnarray*}

Suppose

\begin{eqnarray*}
{p_1(r-1)\over p_0(r)}= {\alpha_1\over\alpha_0},\qquad {p_2(r-2)\over p_0(r)}= {\alpha_2\over\alpha_0},
\end{eqnarray*}

and

\begin{eqnarray*}
p_0(r)=\alpha_0(r-r_1)(r-r_2),
\end{eqnarray*}

where \(r_1>r_2\). Show that

\begin{eqnarray*}
y_1={x^{r_1}\over\alpha_0+\alpha_1x+\alpha_2x^2}\quad\mbox{ and }\quad y_2={x^{r_2}\over\alpha_0+\alpha_1x+\alpha_2x^2}
\end{eqnarray*}

form a fundamental set of Frobenius solutions of \eqref{eq:3.5E.4} on any interval \((0,\rho)\) on which \(\alpha_0+\alpha_1x+\alpha_2x^2\) has no zeros.

Answer

\(\displaystyle (\alpha_0+\alpha_1x+\alpha_2x^2)\sum_{n=0}^\infty a_nx^n=
\alpha_0a_0+ (\alpha_0a_1+\alpha_1a_0)x+
\sum_{n=2}^\infty(\alpha_0a_n+\alpha_1a_{n-1}+\alpha_2a_{n-2})x^n=1 \),
so \(\displaystyle \sum_{n=0}^\infty
a_nx^n={\alpha_0a_0\over\alpha_0+\alpha_1x+\alpha_2x^2} \).

\part{b}  If
\(\displaystyle {p_1(r-1)\over p_0(r)}=
{\alpha_1\over\alpha_0} \) and
\(\displaystyle {p_2(r-2)\over p_0(r)}=
{\alpha_2\over\alpha_0} \), then Eqn.~(3.5.12) is equivalent to
\(a_0(r)=1 \), \(\alpha_0a_1(r)+\alpha_1a_0(r)=0 \), \(\alpha_0a_n(r)
+\alpha_1a_{n-1}(r)+\alpha_2a_{n-2}(r)=0 ,\quad   n\ge2 \).
Therefore, Theorem~3.5.2 implies the conclusion.

In Exercises \((3.5E.61)\) to \((3.5E.68)\), use the method suggested by Exercise \((3.5E.60)\) to find the general solution on some interval \((0,\rho)\).

Exercise \(\PageIndex{61}\)

\(2x^2(1+x)y''-x(1-3x)y'+y=0\)

Exercise \(\PageIndex{62}\)

\(6x^2(1+2x^2)y''+x(1+50x^2)y'+(1+30x^2)y=0\)

Answer

\(p_0(r)=(2r-1)(3r-1) \);
\(p_1(r)=0 \);
\(p_2(r)=2(2r+3)(3r+5) \);
\(\displaystyle {p_1(r-1)\over p_0(r)}=0={\alpha_1\over\alpha_0} \);
\(\displaystyle {p_2(r-2)\over p_0(r)}=2={\alpha_2\over\alpha_0} \);
\(y_1=\displaystyle{x^{1/3}\over1+2x^2} \); \(y_2=\displaystyle{x^{1/2}\over1+2x^2} \).

Exercise \(\PageIndex{63}\)

\(28x^2(1-3x)y''-7x(5+9x)y'+7(2+9x)y=0\)

Exercise \(\PageIndex{64}\)

\(9x^2(5+x)y''+9x(5+3x)y'-(5-8x)y=0\)

Answer

\(p_0(r)=5(3r-1)(3r+1) \);
\(p_1(r)=(3r+2)(3r+4) \);
\(p_2(r)=0 \);
\(\displaystyle {p_1(r-1)\over p_0(r)}={1\over5}={\alpha_1\over\alpha_0} \);
\(\displaystyle {p_2(r-2)\over p_0(r)}=0={\alpha_2\over\alpha_0} \);
\(y_1=\displaystyle{x^{1/3}\over5+x} \); \(y_2=\displaystyle{x^{-1/3}\over5+x} \).

Exercise \(\PageIndex{65}\)

\(8x^2(2-x^2)y''+2x(10-21x^2)y'-(2+35x^2)y=0\)

Exercise \(\PageIndex{66}\)

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

Answer

\(p_0(r)=(2r-3)(2r-1) \);
\(p_1(r)=3(2r-1)(2r+1) \);
\(p_2(r)=(2r+1)(2r+3) \);
\(\displaystyle {p_1(r-1)\over p_0(r)}=3={\alpha_1\over\alpha_0} \);
\(\displaystyle {p_2(r-2)\over p_0(r)}=1={\alpha_2\over\alpha_0} \);
\(y_1=\displaystyle{x^{1/2}\over1+3x+x^2} \);  \(y_2=\displaystyle{x^{3/2}\over1+3x+x^2} \).

Exercise \(\PageIndex{67}\)

\(3x^2(1+x)^2y''-x(1-10x-11x^2)y'+(1+5x^2)y=0\)

Exercise \(\PageIndex{68}\)

\(4x^2(3+2x+x^2)y''-x(3-14x-15x^2)y'+(3+7x^2)y=0\)

Answer

\(p_0(r)=3(r-1)(4r-1) \);
\(p_1(r)=2r(4r+3) \);
\(p_2(r)=(r+1)(4r+7) \);
\(\displaystyle {p_1(r-1)\over p_0(r)}={2\over3}={\alpha_1\over\alpha_0} \);
\(\displaystyle {p_2(r-2)\over p_0(r)}={1\over3}={\alpha_2\over\alpha_0} \);
\(y_1=\displaystyle{x\over3+2x+x^2} \); \(y_2=\displaystyle{x^{1/4}\over3+2x+x^2} \).