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Mathematics LibreTexts

7.7E: The Method of Frobenius III (Exercises)

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    18313
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    In Exercises [exer:7.7.1} –[exer:7.7.40} find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients.

    [exer:7.7.1] \(x^2y''-3xy'+(3+4x)y=0\)

    [exer:7.7.2] \(xy''+y=0\)

    [exer:7.7.3] \(4x^2(1+x)y''+4x(1+2x)y'-(1+3x)y=0\)

    [exer:7.7.4] \(xy''+xy'+y=0\)

    [exer:7.7.5] \(2x^2(2+3x)y''+x(4+21x)y'-(1-9x)y=0\)

    [exer:7.7.6] \(x^2y''+x(2+x)y'-(2-3x)y=0\)

    [exer:7.7.7] \(4x^2y''+4xy'-(9-x)y=0\)

    [exer:7.7.8] \(x^2y''+10xy'+(14+x)y=0\)

    [exer:7.7.9] \(4x^2(1+x)y''+4x(3+8x)y'-(5-49x)y=0\)

    [exer:7.7.10] \(x^2(1+x)y''-x(3+10x)y'+30xy=0\)

    [exer:7.7.11] \(x^2y''+x(1+x)y'-3(3+x)y=0\)

    [exer:7.7.12] \(x^2y''+x(1-2x)y'-(4+x)y=0\)

    [exer:7.7.13] \(x(1+x)y''-4y'-2y=0\)

    [exer:7.7.14] \(x^2(1+2x)y''+x(9+13x)y'+(7+5x)y=0\)

    [exer:7.7.15] \(4x^2y''-2x(4-x)y'-(7+5x)y=0\)

    [exer:7.7.16] \(3x^2(3+x)y''-x(15+x)y'-20y=0\)

    [exer:7.7.17] \(x^2(1+x)y''+x(1-10x)y'-(9-10x)y=0\)

    [exer:7.7.18] \(x^2(1+x)y''+3x^2y'-(6-x)y=0\)

    [exer:7.7.19] \(x^2(1+2x)y''-2x(3+14x)y'+(6+100x)y=0\)

    [exer:7.7.20] \(x^2(1+x)y''-x(6+11x)y'+(6+32x)y=0\)

    [exer:7.7.21] \(4x^2(1+x)y''+4x(1+4x)y'-(49+27x)y=0\)

    [exer:7.7.22] \(x^2(1+2x)y''-x(9+8x)y'-12xy=0\)

    [exer:7.7.23] \(x^2(1+x^2)y''-x(7-2x^2)y'+12y=0\)

    [exer:7.7.24] \(x^2y''-x(7-x^2)y'+12y=0\)

    [exer:7.7.25] \(xy''-5y'+xy=0\)

    [exer:7.7.26] \(x^2y''+x(1+2x^2)y'-(1-10x^2)y=0\)

    [exer:7.7.27] \(x^2y''-xy'-(3-x^2)y=0\)

    [exer:7.7.28] \(4x^2y''+2x(8+x^2)y'+(5+3x^2)y=0\)

    [exer:7.7.29] \(x^2y''+x(1+x^2)y'-(1-3x^2)y=0\)

    [exer:7.7.30] \(x^2y''+x(1-2x^2)y'-4(1+2x^2)y=0\)

    [exer:7.7.31] \(4x^2y''+8xy'-(35-x^2)y=0\)

    [exer:7.7.32] \(9x^2y''-3x(11+2x^2)y'+(13+10x^2)y=0\)

    [exer:7.7.33] \(x^2y''+x(1-2x^2)y'-4(1-x^2)y=0\)

    [exer:7.7.34] \(x^2y''+x(1-3x^2)y'-4(1-3x^2)y=0\)

    [exer:7.7.35] \(x^2(1+x^2)y''+x(5+11x^2)y'+24x^2y=0\)

    [exer:7.7.36] \(4x^2(1+x^2)y''+8xy'-(35-x^2)y=0\)

    [exer:7.7.37] \(x^2(1+x^2)y''-x(5-x^2)y'-(7+25x^2)y=0\)

    [exer:7.7.38] \(x^2(1+x^2)y''+x(5+2x^2)y'-21y=0\)

    [exer:7.7.39] \(x^2(1+2x^2)y''-x(3+x^2)y'-2x^2y=0\)

    [exer:7.7.40] \(4x^2(1+x^2)y''+4x(2+x^2)y'-(15+x^2)y=0\)

    [exer:7.7.41] Under the assumptions of Theorem [thmtype: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.

    Use the result of (a) to complete the proof of Theorem [thmtype:7.7.1} .

    [exer:7.7.42] 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.

    [exer:7.7.43] Prove Theorem [thmtype:7.7.2}.

    [exer:7.7.44] Under the assumptions of Theorem [thmtype:7.7.1} , show that \(C=0\) if and only if \(p_1(r_2+\l)=0\) for some integer \(\l\) in \(\{0,1,\dots,k-1\}\).

    [exer:7.7.45] Under the assumptions of Theorem [thmtype:7.7.2} , show that \(C=0\) if and only if \(p_2(r_2+2\l)=0\) for some integer \(\ell\) in \(\{0,1,\dots,k-1\}\).

    [exer:7.7.46] 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

    \[\begin{aligned} y_2&=&x^{r_2}\sum_{n=0}^{k-1} {(-1)^n\over n!\prod_{j=1}^n(j-k)} \left(\gamma_1\over\alpha_0\right)^n x^n\\[10pt]&&-{1\over k!(k-1)!}\left(\gamma_1\over\alpha_0\right)^k\left(y_1\ln x- x^{r_1}\sum_{n=1}^\infty {(-1)^n\over n!\prod_{j=1}^n(j+k)}\left(\gamma_1\over\alpha_0\right)^n \left(\sum_{j=1}^n{2j+k\over j(j+k)}\right)x^n\right).\end{aligned}\nonumber\]

    [exer:7.7.47] 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

    \[\begin{aligned} y_2&=&x^{r_2}\sum_{m=0}^{k-1} {(-1)^m\over4^mm!\prod_{j=1}^m(j-k)} \left(\gamma_2\over\alpha_0\right)^m x^{2m}\\[10pt]&&-{2\over 4^kk!(k-1)!}\left(\gamma_2\over\alpha_0\right)^k\left(y_1\ln x- {x^{r_1}\over2}\sum_{m=1}^\infty {(-1)^m\over 4^mm!\prod_{j=1}^m(j+k)}\left(\gamma_2\over\alpha_0\right)^m \left(\sum_{j=1}^m{2j+k\over j(j+k)}\right)x^{2m}\right).\end{aligned}\nonumber\]

    [exer:7.7.48] Let \(L\) be as in Exercises 7.5. [exer:7.5.57} and 7.5. [exer: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 \[y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n.\nonumber\]

    1. 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.
    2. Conclude from Exercise 7.6. .[exer:7.6.66} that \[L\left(y_1\ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n\right)=k\alpha_0x^{r_1}.\nonumber\]
    3. Conclude from Exercise 7.5. .[exer: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 \[A=\sum_{j=1}^k p_j(r_1-j)a_{k-j}(r_2).\nonumber\]
    4. Show that \(y_1\) and \[y_2=x^{r_2}\sum_{n=0}^\infty a_n(r_2)x^n -{A\over k\alpha_0} \left(y_1 \ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n\right)\nonumber\]
    5. form a fundamental set of Frobenius solutions of \(Ly=0\).
    6. 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 \(a_k(r_2)~=~0\).