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Mathematics LibreTexts

7.5E: The Method of Frobenius I (Exercises)

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    18308
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    This set contains exercises specifically identified by 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 [exer:7.5.1} –[exer:7.5.10} , [exer:7.5.14} -[exer:7.5.25} , and [exer:7.5.28} –[exer:7.5.51}.

    [exer:7.5.1] \(2x^2(1+x+x^2)y''+x(3+3x+5x^2)y'-y=0\)

    [exer:7.5.2] \(3x^2y''+2x(1+x-2x^2)y'+(2x-8x^2)y=0\)

    [exer:7.5.3] \(x^2(3+3x+x^2)y''+x(5+8x+7x^2)y'-(1-2x-9x^2)y=0\)

    [exer:7.5.4] \(4x^2y''+x(7+2x+4x^2)y'-(1-4x-7x^2)y=0\)

    [exer:7.5.5] \(12x^2(1+x)y''+x(11+35x+3x^2)y'-(1-10x-5x^2)y=0\)

    [exer:7.5.6] \(x^2(5+x+10x^2)y''+x(4+3x+48x^2)y'+(x+36x^2)y=0\)

    [exer:7.5.7] \(8x^2y''-2x(3-4x-x^2)y'+(3+6x+x^2)y=0\)

    [exer:7.5.8] \(18x^2(1+x)y''+3x(5+11x+x^2)y'-(1-2x-5x^2)y=0\)

    [exer:7.5.9] \(x(3+x+x^2)y''+(4+x-x^2)y'+xy=0\)

    [exer:7.5.10] \(10x^2(1+x+2x^2)y''+x(13+13x+66x^2)y'-(1+4x+10x^2)y=0\)

    [exer:7.5.11] The Frobenius solutions of

    \[2x^2(1+x+x^2)y''+x(9+11x+11x^2)y'+(6+10x+7x^2)y=0\]

    obtained in Example

    Example \(\PageIndex{1}\):

    Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at7.5.1} are defined on \((0,\rho)\), where \(\rho\) is defined in Theorem [thmtype:7.5.2}. 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.

    Compute \(\sigma_N(\delta)\) (see Eqn. Equation \ref{eq:7.5.28} ) for \(N=5\), \(10\), \(15\),…, \(50\).

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

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

    [exer:7.5.12] By Theorem [thmtype:7.5.2} the Frobenius solutions of the equation in Exercise [exer:7.5.4} are defined on \((0,\infty)\). Do experiments

    a

    ,

    b

    , and

    c

    of Exercise [exer:7.5.11} for each Frobenius solution, with \(M=20\) and \(\delta=1\), \(2\), and \(3\) in the verification procedure described at the end of this section.

    [exer:7.5.13] The Frobenius solutions of the equation in Exercise [exer:7.5.6} are defined on \((0,\rho)\), where \(\rho\) is defined in Theorem [thmtype:7.5.2}. Find \(\rho\) and do experiments

    a

    ,

    b

    , and

    c

    of Exercise [exer:7.5.11} 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.

    [exer:7.5.14] \(2x^2y''+x(3+2x)y'-(1-x)y=0\)

    [exer:7.5.15] \(x^2(3+x)y''+x(5+4x)y'-(1-2x)y=0\)

    [exer:7.5.16] \(2x^2y''+x(5+x)y'-(2-3x)y=0\)

    [exer:7.5.17] \(3x^2y''+x(1+x)y'-y=0\)

    [exer:7.5.18] \(2x^2y''-xy'+(1-2x)y=0\)

    [exer:7.5.19] \(9x^2y''+9xy'-(1+3x)y=0\)

    [exer:7.5.20] \(3x^2y''+x(1+x)y'-(1+3x)y=0\)

    [exer:7.5.21] \(2x^2(3+x)y''+x(1+5x)y'+(1+x)y=0\)

    [exer:7.5.22] \(x^2(4+x)y''-x(1-3x)y'+y=0\)

    [exer:7.5.23] \(2x^2y''+5xy'+(1+x)y=0\)

    [exer:7.5.24] \(x^2(3+4x)y''+x(5+18x)y'-(1-12x)y=0\)

    [exer:7.5.25] \(6x^2y''+x(10-x)y'-(2+x)y=0\)

    [exer:7.5.26] By Theorem [thmtype:7.5.2} the Frobenius solutions of the equation in Exercise [exer:7.5.17} are defined on \((0,\infty)\). Do experiments

    a

    ,

    b

    , and

    c

    of Exercise [exer:7.5.11} 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.

    [exer:7.5.27] The Frobenius solutions of the equation in Exercise [exer:7.5.22} are defined on \((0,\rho)\), where \(\rho\) is defined in Theorem [thmtype:7.5.2}. Find \(\rho\) and do experiments

    a

    ,

    b

    , and

    c

    of Exercise [exer:7.5.11} 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.

    [exer:7.5.28] \(x^2(8+x)y''+x(2+3x)y'+(1+x)y=0\)

    [exer:7.5.29] \(x^2(3+4x)y''+x(11+4x)y'-(3+4x)y=0\)

    [exer:7.5.30] \(2x^2(2+3x)y''+x(4+11x)y'-(1-x)y=0\)

    [exer:7.5.31] \(x^2(2+x)y''+5x(1-x)y'-(2-8x)y\)

    [exer:7.5.32] \(x^2(6+x)y''+x(11+4x)y'+(1+2x)y=0\)

    [exer:7.5.33] \(8x^2y''+x(2+x^2)y'+y=0\)

    [exer:7.5.34] \(8x^2(1-x^2)y''+2x(1-13x^2)y'+(1-9x^2)y=0\)

    [exer:7.5.35] \(x^2(1+x^2)y''-2x(2-x^2)y'+4y=0\)

    [exer:7.5.36] \(x(3+x^2)y''+(2-x^2)y'-8xy=0\)

    [exer:7.5.37] \(4x^2(1-x^2)y''+x(7-19x^2)y'-(1+14x^2)y=0\)

    [exer:7.5.38] \(3x^2(2-x^2)y''+x(1-11x^2)y'+(1-5x^2)y=0\)

    [exer:7.5.39] \(2x^2(2+x^2)y''-x(12-7x^2)y'+(7+3x^2)y=0\)

    [exer:7.5.40] \(2x^2(2+x^2)y''+x(4+7x^2)y'-(1-3x^2)y=0\)

    [exer:7.5.41] \(2x^2(1+2x^2)y''+5x(1+6x^2)y'-(2-40x^2)y=0\)

    [exer:7.5.42] \(3x^2(1+x^2)y''+5x(1+x^2)y'-(1+5x^2)y=0\)

    [exer:7.5.43] \(x(1+x^2)y''+(4+7x^2)y'+8xy=0\)

    [exer:7.5.44] \(x^2(2+x^2)y''+x(3+x^2)y'-y=0\)

    [exer:7.5.45] \(2x^2(1+x^2)y''+x(3+8x^2)y'-(3-4x^2)y=0\)

    [exer:7.5.46] \(9x^2y''+3x(3+x^2)y'-(1-5x^2)y=0\)

    [exer:7.5.47] \(6x^2y''+x(1+6x^2)y'+(1+9x^2)y=0\)

    [exer:7.5.48] \(x^2(8+x^2)y''+7x(2+x^2)y'-(2-9x^2)y=0\)

    [exer:7.5.49] \(9x^2(1+x^2)y''+3x(3+13x^2)y'-(1-25x^2)y=0\)

    [exer:7.5.50] \(4x^2(1+x^2)y''+4x(1+6x^2)y'-(1-25x^2)y=0\)

    [exer:7.5.51] \(8x^2(1+2x^2)y''+2x(5+34x^2)y'-(1-30x^2)y=0\)

    [exer:7.5.52] Suppose \(r_1>r_2\), \(a_0=b_0=1\), and the Frobenius series

    \[y_1=x^{r_1}\sum_{n=0}^\infty a_nx^n\quad\mbox{ and } \quad y_2=x^{r_2}\sum_{n=0}^\infty b_nx^n\]

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

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

    Use the result of

    b

    to complete the proof of Theorem [thmtype:7.5.3}.

    [exer:7.5.53] The equation

    \[\label{eq:7.5.{exer:7.5.53}A} x^2y''+xy'+(x^2-\nu^2)y=0\]

    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 Equation \ref{eq:7.5.{exer:7.5.53}A} are Bessel functions of order \(\nu\).

    Assuming that \(\nu\) isn’t an integer, find a fundamental set of Frobenius solutions of Equation \ref{eq:7.5.{exer:7.5.53}A}.

    If \(\nu=1/2\), the solutions of Equation \ref{eq:7.5.{exer:7.5.53}A} reduce to familiar elementary functions. Identify these functions.

    [exer:7.5.54]

    Verify that

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

    if \(x\ne0\).

    Let

    \[Ly= x^2(\alpha_0+\alpha_1x+\alpha_2x^2)y''+x(\beta_0+\beta_1x+\beta_2x^2)y' +(\gamma_0+\gamma_1x+\gamma_2x^2)y=0.\]

    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 \((0,\rho)\).

    [exer:7.5.55]

    Deduce from Eqn. Equation \ref{eq:7.5.20} that

    \[a_n(r)=(-1)^n\prod_{j=1}^n{p_1(j+r-1)\over p_0(j+r)}.\]

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

    \[y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n\quad\mbox{ and }\quad y_2=x^{r_2}\sum_{n=0}^\infty a_n(r_2)x^n\]

    form a fundamental set of Frobenius solutions of

    \[x^2(\alpha_0+\alpha_1x)y''+x(\beta_0+\beta_1x)y'+(\gamma_0+\gamma_1x)y=0.\]

    Show that if \(p_0\) satisfies the hypotheses of

    b

    then

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

    and

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

    form a fundamental set of Frobenius solutions of

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

    [exer:7.5.56] Let

    \[Ly=x^2(\alpha_0+\alpha_2x^2)y''+x(\beta_0+\beta_2x^2)y'+ (\gamma_0+\gamma_2x^2)y=0\]

    and define

    \[p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0\quad\mbox{ and }\quad p_2(r)=\alpha_2r(r-1)+\beta_2r+\gamma_2.\]

    Use Theorem [thmtype:7.5.2} to show that if

    \[\label{eq:7.5.{exer:7.5.56}A} \begin{array}{rcl} a_0(r)&=&1,\\ p_0(2m+r)a_{2m}(r)+p_2(2m+r-2)a_{2m-2}(r)&=&0,\quad m\ge1, \end{array}\]

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

    Deduce from Equation \ref{eq:7.5.{exer:7.5.56}A} that if \(p_0(2m+r)\) is nonzero for every positive integer \(m\) then

    \[a_{2m}(r)=(-1)^m\prod_{j=1}^m{p_2(2j+r-2)\over p_0(2j+r)}.\]

    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

    \[y_1=x^{r_1}\sum_{m=0}^\infty a_{2m}(r_1)x^{2m}\quad\mbox{ and }\quad y_2=x^{r_2}\sum_{m=0}^\infty a_{2m}(r_2)x^{2m}\]

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

    Show that if \(p_0\) satisfies the hypotheses of

    c

    then

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

    and

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

    form a fundamental set of Frobenius solutions of

    \[\alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_2x^2)y=0.\]

    [exer:7.5.57] Let

    \[Ly=x^2q_0(x)y''+xq_1(x)y'+q_2(x)y,\]

    where

    \[q_0(x)=\sum_{j=0}^\infty \alpha_jx^j,\quad q_1(x)=\sum_{j=0}^\infty \beta_jx^j,\quad q_2(x)=\sum_{j=0}^\infty \gamma_jx^j,\]

    and define

    \[p_j(r)=\alpha_jr(r-1)+\beta_jr+\gamma_j,\quad j=0,1,\dots.\]

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

    \[Ly=x^r\sum_{n=0}^\infty b_nx^n,\]

    where

    \[b_n=\sum_{j=0}^np_j(n+r-j)a_{n-j}.\]

    [exer:7.5.58]

    Let \(L\) be as in Exercise [exer:7.5.57}. Show that if

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

    where

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

    then

    \[Ly(x,r)=p_0(r)x^r.\]

    Conclude that if

    \[p_0(r)=\alpha_0(r-r_1)(r-r_2)\]

    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 \(Ly=0\).

    [exer:7.5.59] 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

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

    Show that if

    \[y(x,r)=x^{r}\sum_{m=0}^\infty a_{qm}(r)x^{qm}\]

    where

    \[\label{eq:7.5.{exer:7.5.59}A} \begin{array}{rcl} a_0(r)&=&1,\\ a_{qm}(r)&=&-{p_q\left(q(m-1)+r\right)\over p_0(qm+r)}a_{q(m-1)}(r),\quad m\ge1, \end{array}\]

    then

    \[Ly(x,r)=p_0(r)x^r.\]

    Deduce from Equation \ref{eq:7.5.{exer:7.5.59}A} that

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

    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

    \[y_1=x^{r_1}\sum_{m=0}^\infty a_{qm}(r_1)x^{qm}\quad\mbox{ and }\quad y_2=x^{r_2}\sum_{m=0}^\infty a_{qm}(r_2)x^{qm}\]

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

    Show that if \(p_0\) satisfies the hypotheses of

    c

    then

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

    and

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

    form a fundamental set of Frobenius solutions of

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

    [exer:7.5.60]

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

    \[\alpha_0a_1+\alpha_1a_0=0\]

    and

    \[\alpha_0a_n+\alpha_1a_{n-1}+\alpha_2a_{n-2}=0,\quad n\ge2.\]

    Show that

    \[(\alpha_0+\alpha_1x+\alpha_2x^2)\sum_{n=0}^\infty a_nx^n=\alpha_0a_0,\]

    and infer that

    \[\sum_{n=0}^\infty a_nx^n={\alpha_0a_0\over\alpha_0+\alpha_1x+\alpha_2x^2}.\]

    With \(\alpha_0,\alpha_1\), and \(\alpha_2\) as in

    a

    , consider the equation

    \[\label{eq:7.5.{exer:7.5.60}A} x^2(\alpha_0+\alpha_1x+\alpha_2 x^2)y''+x(\beta_0+\beta_1x+\beta_2x^2)y'+ (\gamma_0+\gamma_1x+\gamma_2x^2)y=0,\]

    and define

    \[p_j(r)=\alpha_jr(r-1)+\beta_jr+\gamma_j,\quad j=0,1,2.\]

    Suppose

    \[{p_1(r-1)\over p_0(r)}= {\alpha_1\over\alpha_0},\qquad {p_2(r-2)\over p_0(r)}= {\alpha_2\over\alpha_0},\]

    and

    \[p_0(r)=\alpha_0(r-r_1)(r-r_2),\]

    where \(r_1>r_2\). Show that

    \[y_1={x^{r_1}\over\alpha_0+\alpha_1x+\alpha_2x^2}\quad\mbox{ and }\quad y_2={x^{r_2}\over\alpha_0+\alpha_1x+\alpha_2x^2}\]

    form a fundamental set of Frobenius solutions of Equation \ref{eq:7.5.{exer:7.5.60}A} on any interval \((0,\rho)\) on which \(\alpha_0+\alpha_1x+\alpha_2x^2\) has no zeros.

    [exer:7.5.61] \(2x^2(1+x)y''-x(1-3x)y'+y=0\)

    [exer:7.5.62] \(6x^2(1+2x^2)y''+x(1+50x^2)y'+(1+30x^2)y=0\)

    [exer:7.5.63] \(28x^2(1-3x)y''-7x(5+9x)y'+7(2+9x)y=0\)

    [exer:7.5.64] \(9x^2(5+x)y''+9x(5+3x)y'-(5-8x)y=0\)

    [exer:7.5.65] \(8x^2(2-x^2)y''+2x(10-21x^2)y'-(2+35x^2)y=0\)

    [exer:7.5.66] \(4x^2(1+3x+x^2)y''-4x(1-3x-3x^2)y'+3(1-x+x^2)y=0\)

    [exer:7.5.67] \(3x^2(1+x)^2y''-x(1-10x-11x^2)y'+(1+5x^2)y=0\)

    [exer:7.5.68] \(4x^2(3+2x+x^2)y''-x(3-14x-15x^2)y'+(3+7x^2)y=0\)