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Mathematics LibreTexts

7.6E: The Method of Frobenius II (Exercises)

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    18312
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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

    a

    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

    a

    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 \(Ly=0\).