Internal problem ID [1448]
Internal file name [OUTPUT/1449_Sunday_June_05_2022_02_18_09_AM_38044428/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 7 Series Solutions of Linear Second Equations. 7.6 THE METHOD OF
FROBENIUS III. Exercises 7.7. Page 389
Problem number: 32.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference is integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {9 x^{2} y^{\prime \prime }-3 x \left (2 x^{2}+11\right ) y^{\prime }+\left (10 x^{2}+13\right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ 9 x^{2} y^{\prime \prime }+\left (-6 x^{3}-33 x \right ) y^{\prime }+\left (10 x^{2}+13\right ) y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= -\frac {2 x^{2}+11}{3 x}\\ q(x) &= \frac {10 x^{2}+13}{9 x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty , -\infty ]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 9 x^{2} y^{\prime \prime }+\left (-6 x^{3}-33 x \right ) y^{\prime }+\left (10 x^{2}+13\right ) y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} 9 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (-6 x^{3}-33 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (10 x^{2}+13\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{n +r +2} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-33 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}13 a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{n +r +2} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-6 a_{n -2} \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}10 a_{n -2} x^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-6 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-33 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}10 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}13 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-33 x^{n +r} a_{n} \left (n +r \right )+13 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 9 x^{r} a_{0} r \left (-1+r \right )-33 x^{r} a_{0} r +13 a_{0} x^{r} = 0 \] Or \[ \left (9 x^{r} r \left (-1+r \right )-33 x^{r} r +13 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (9 r^{2}-42 r +13\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 9 r^{2}-42 r +13 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {13}{3}}\\ r_2 &= {\frac {1}{3}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (9 r^{2}-42 r +13\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {13}{3}}, {\frac {1}{3}}\right ]\).
Since \(r_1 - r_2 = 4\) is an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= x^{\frac {13}{3}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+x^{\frac {1}{3}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {13}{3}}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{3}}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 9 a_{n} \left (n +r \right ) \left (n +r -1\right )-6 a_{n -2} \left (n +r -2\right )-33 a_{n} \left (n +r \right )+10 a_{n -2}+13 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {2 a_{n -2} \left (3 n +3 r -11\right )}{9 n^{2}+18 n r +9 r^{2}-42 n -42 r +13}\tag {4} \] Which for the root \(r = {\frac {13}{3}}\) becomes \[ a_{n} = \frac {2 a_{n -2} \left (3 n +2\right )}{9 n \left (n +4\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {13}{3}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-10+6 r}{9 r^{2}-6 r -35} \] Which for the root \(r = {\frac {13}{3}}\) becomes \[ a_{2}={\frac {4}{27}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-10+6 r}{9 r^{2}-6 r -35}\) | \(\frac {4}{27}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-10+6 r}{9 r^{2}-6 r -35}\) | \(\frac {4}{27}\) |
| \(a_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {36 r^{2}-48 r -20}{\left (9 r^{2}-6 r -35\right ) \left (9 r^{2}+30 r -11\right )} \] Which for the root \(r = {\frac {13}{3}}\) becomes \[ a_{4}={\frac {7}{486}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-10+6 r}{9 r^{2}-6 r -35}\) | \(\frac {4}{27}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {36 r^{2}-48 r -20}{\left (9 r^{2}-6 r -35\right ) \left (9 r^{2}+30 r -11\right )}\) | \(\frac {7}{486}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-10+6 r}{9 r^{2}-6 r -35}\) | \(\frac {4}{27}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {36 r^{2}-48 r -20}{\left (9 r^{2}-6 r -35\right ) \left (9 r^{2}+30 r -11\right )}\) | \(\frac {7}{486}\) |
| \(a_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {13}{3}} \left (a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}\dots \right ) \\ &= x^{\frac {13}{3}} \left (1+\frac {4 x^{2}}{27}+\frac {7 x^{4}}{486}+O\left (x^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=4\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{4}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{4} \\ &= \frac {36 r^{2}-48 r -20}{\left (9 r^{2}-6 r -35\right ) \left (9 r^{2}+30 r -11\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {36 r^{2}-48 r -20}{\left (9 r^{2}-6 r -35\right ) \left (9 r^{2}+30 r -11\right )}&= \lim _{r\rightarrow {\frac {1}{3}}}\frac {36 r^{2}-48 r -20}{\left (9 r^{2}-6 r -35\right ) \left (9 r^{2}+30 r -11\right )}\\ &= \textit {undefined} \end {align*}
Since the limit does not exist then the log term is needed. Therefore the second solution has the form \[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \] Therefore \begin{align*} \frac {d}{d x}y_{2}\left (x \right ) &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) \\ &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) \\ \frac {d^{2}}{d x^{2}}y_{2}\left (x \right ) &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right ) \\ &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) \\ \end{align*} Substituting these back into the given ode \(9 x^{2} y^{\prime \prime }+\left (-6 x^{3}-33 x \right ) y^{\prime }+\left (10 x^{2}+13\right ) y = 0\) gives \[ 9 x^{2} \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (-6 x^{3}-33 x \right ) \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+\left (10 x^{2}+13\right ) \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (9 x^{2} y_{1}^{\prime \prime }\left (x \right )+\left (-6 x^{3}-33 x \right ) y_{1}^{\prime }\left (x \right )+\left (10 x^{2}+13\right ) y_{1}\left (x \right )\right ) \ln \left (x \right )+9 x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (-6 x^{3}-33 x \right ) y_{1}\left (x \right )}{x}\right ) C +9 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (-6 x^{3}-33 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (10 x^{2}+13\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} But since \(y_{1}\left (x \right )\) is a solution to the ode, then \[ 9 x^{2} y_{1}^{\prime \prime }\left (x \right )+\left (-6 x^{3}-33 x \right ) y_{1}^{\prime }\left (x \right )+\left (10 x^{2}+13\right ) y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (9 x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (-6 x^{3}-33 x \right ) y_{1}\left (x \right )}{x}\right ) C +9 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (-6 x^{3}-33 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (10 x^{2}+13\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Substituting \(y_{1} = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\) into the above gives \begin{equation} \tag{9} \left (18 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right ) x +6 \left (-x^{2}-7\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C +9 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) x^{2}+3 \left (-2 x^{3}-11 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )+\left (10 x^{2}+13\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Since \(r_{1} = {\frac {13}{3}}\) and \(r_{2} = {\frac {1}{3}}\) then the above becomes \begin{equation} \tag{10} \left (18 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{\frac {10}{3}+n} a_{n} \left (n +\frac {13}{3}\right )\right ) x +6 \left (-x^{2}-7\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {13}{3}}\right )\right ) C +9 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-\frac {5}{3}+n} b_{n} \left (n +\frac {1}{3}\right ) \left (-\frac {2}{3}+n \right )\right ) x^{2}+3 \left (-2 x^{3}-11 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-\frac {2}{3}+n} b_{n} \left (n +\frac {1}{3}\right )\right )+\left (10 x^{2}+13\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{3}}\right ) = 0 \end{equation} Expanding \(-6 C \,x^{\frac {19}{3}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -6 C \,x^{\frac {19}{3}} &= -6 C \,x^{\frac {19}{3}} + \dots \\ &= -6 C \,x^{\frac {19}{3}} \end {align*}
Expanding \(-42 C \,x^{\frac {13}{3}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -42 C \,x^{\frac {13}{3}} &= -42 C \,x^{\frac {13}{3}} + \dots \\ &= -42 C \,x^{\frac {13}{3}} \end {align*}
Expanding \(-2 x^{\frac {7}{3}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -2 x^{\frac {7}{3}} &= -2 x^{\frac {7}{3}} + \dots \\ &= -2 x^{\frac {7}{3}} \end {align*}
Expanding \(-11 x^{\frac {1}{3}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -11 x^{\frac {1}{3}} &= -11 x^{\frac {1}{3}} + \dots \\ &= -11 x^{\frac {1}{3}} \end {align*}
Expanding \(10 x^{\frac {7}{3}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} 10 x^{\frac {7}{3}} &= 10 x^{\frac {7}{3}} + \dots \\ &= 10 x^{\frac {7}{3}} \end {align*}
Expanding \(13 x^{\frac {1}{3}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} 13 x^{\frac {1}{3}} &= 13 x^{\frac {1}{3}} + \dots \\ &= 13 x^{\frac {1}{3}} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (18 n +78\right ) C a_{n} x^{n +\frac {13}{3}}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 C \,x^{n +\frac {19}{3}} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-42 C \,x^{n +\frac {13}{3}} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +\frac {1}{3}} b_{n} \left (9 n^{2}-3 n -2\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 n -2\right ) b_{n} x^{n +\frac {7}{3}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (-33 n -11\right ) b_{n} x^{n +\frac {1}{3}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n +\frac {7}{3}} b_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}13 b_{n} x^{n +\frac {1}{3}}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +\frac {1}{3}\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +\frac {1}{3}}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (18 n +78\right ) C a_{n} x^{n +\frac {13}{3}} &= \moverset {\infty }{\munderset {n =4}{\sum }}C a_{n -4} \left (18 n +6\right ) x^{n +\frac {1}{3}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 C \,x^{n +\frac {19}{3}} a_{n}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-6 C a_{n -6} x^{n +\frac {1}{3}}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-42 C \,x^{n +\frac {13}{3}} a_{n}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-42 C a_{n -4} x^{n +\frac {1}{3}}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 n -2\right ) b_{n} x^{n +\frac {7}{3}} &= \moverset {\infty }{\munderset {n =2}{\sum }}b_{n -2} \left (-6 n +10\right ) x^{n +\frac {1}{3}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n +\frac {7}{3}} b_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}10 b_{n -2} x^{n +\frac {1}{3}} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +\frac {1}{3}\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =4}{\sum }}C a_{n -4} \left (18 n +6\right ) x^{n +\frac {1}{3}}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-6 C a_{n -6} x^{n +\frac {1}{3}}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-42 C a_{n -4} x^{n +\frac {1}{3}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +\frac {1}{3}} b_{n} \left (9 n^{2}-3 n -2\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}b_{n -2} \left (-6 n +10\right ) x^{n +\frac {1}{3}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (-33 n -11\right ) b_{n} x^{n +\frac {1}{3}}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}10 b_{n -2} x^{n +\frac {1}{3}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}13 b_{n} x^{n +\frac {1}{3}}\right ) = 0 \end{equation} For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=1\), Eq (2B) gives \[ -27 b_{1} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -27 b_{1} = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}=0 \] For \(n=2\), Eq (2B) gives \[ -36 b_{2}+8 b_{0} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -36 b_{2}+8 = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}={\frac {2}{9}} \] For \(n=3\), Eq (2B) gives \[ -27 b_{3}+2 b_{1} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -27 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=0 \] For \(n=N\), where \(N=4\) which is the difference between the two roots, we are free to choose \(b_{4} = 0\). Hence for \(n=4\), Eq (2B) gives \[ 36 C -\frac {8}{9} = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C={\frac {2}{81}} \] For \(n=5\), Eq (2B) gives \[ 54 C a_{1}-10 b_{3}+45 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 45 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=0 \] Now that we found all \(b_{n}\) and \(C\), we can calculate the second solution from \[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \] Using the above value found for \(C={\frac {2}{81}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= \frac {2}{81}\eslowast \left (x^{\frac {13}{3}} \left (1+\frac {4 x^{2}}{27}+\frac {7 x^{4}}{486}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+x^{\frac {1}{3}} \left (1+\frac {2 x^{2}}{9}+O\left (x^{6}\right )\right ) \] Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x^{\frac {13}{3}} \left (1+\frac {4 x^{2}}{27}+\frac {7 x^{4}}{486}+O\left (x^{6}\right )\right ) + c_{2} \left (\frac {2}{81}\eslowast \left (x^{\frac {13}{3}} \left (1+\frac {4 x^{2}}{27}+\frac {7 x^{4}}{486}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+x^{\frac {1}{3}} \left (1+\frac {2 x^{2}}{9}+O\left (x^{6}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {13}{3}} \left (1+\frac {4 x^{2}}{27}+\frac {7 x^{4}}{486}+O\left (x^{6}\right )\right )+c_{2} \left (\frac {2 x^{\frac {13}{3}} \left (1+\frac {4 x^{2}}{27}+\frac {7 x^{4}}{486}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{81}+x^{\frac {1}{3}} \left (1+\frac {2 x^{2}}{9}+O\left (x^{6}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {13}{3}} \left (1+\frac {4 x^{2}}{27}+\frac {7 x^{4}}{486}+O\left (x^{6}\right )\right )+c_{2} \left (\frac {2 x^{\frac {13}{3}} \left (1+\frac {4 x^{2}}{27}+\frac {7 x^{4}}{486}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{81}+x^{\frac {1}{3}} \left (1+\frac {2 x^{2}}{9}+O\left (x^{6}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {13}{3}} \left (1+\frac {4 x^{2}}{27}+\frac {7 x^{4}}{486}+O\left (x^{6}\right )\right )+c_{2} \left (\frac {2 x^{\frac {13}{3}} \left (1+\frac {4 x^{2}}{27}+\frac {7 x^{4}}{486}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{81}+x^{\frac {1}{3}} \left (1+\frac {2 x^{2}}{9}+O\left (x^{6}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 9 x^{2} y^{\prime \prime }+\left (-6 x^{3}-33 x \right ) y^{\prime }+\left (10 x^{2}+13\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {\left (10 x^{2}+13\right ) y}{9 x^{2}}+\frac {\left (2 x^{2}+11\right ) y^{\prime }}{3 x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-\frac {\left (2 x^{2}+11\right ) y^{\prime }}{3 x}+\frac {\left (10 x^{2}+13\right ) y}{9 x^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {2 x^{2}+11}{3 x}, P_{3}\left (x \right )=\frac {10 x^{2}+13}{9 x^{2}}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-\frac {11}{3} \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=\frac {13}{9} \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 9 x^{2} y^{\prime \prime }-3 x \left (2 x^{2}+11\right ) y^{\prime }+\left (10 x^{2}+13\right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (-1+3 r \right ) \left (-13+3 r \right ) x^{r}+a_{1} \left (2+3 r \right ) \left (-10+3 r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (3 k +3 r -1\right ) \left (3 k +3 r -13\right )-2 a_{k -2} \left (3 k -11+3 r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (-1+3 r \right ) \left (-13+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {1}{3}, \frac {13}{3}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (2+3 r \right ) \left (-10+3 r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 9 \left (k +r -\frac {13}{3}\right ) \left (k +r -\frac {1}{3}\right ) a_{k}-6 a_{k -2} \left (k -\frac {11}{3}+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & 9 \left (k -\frac {7}{3}+r \right ) \left (k +\frac {5}{3}+r \right ) a_{k +2}-6 a_{k} \left (k -\frac {5}{3}+r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {2 a_{k} \left (3 k +3 r -5\right )}{\left (3 k -7+3 r \right ) \left (3 k +5+3 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{3} \\ {} & {} & a_{k +2}=\frac {2 a_{k} \left (3 k -4\right )}{\left (3 k -6\right ) \left (3 k +6\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =\frac {1}{3}\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =2 \\ {} & {} & a_{k +2}=\frac {2 a_{k} \left (3 k -4\right )}{\left (3 k -6\right ) \left (3 k +6\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {13}{3} \\ {} & {} & a_{k +2}=\frac {2 a_{k} \left (3 k +8\right )}{\left (3 k +6\right ) \left (3 k +18\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {13}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {13}{3}}, a_{k +2}=\frac {2 a_{k} \left (3 k +8\right )}{\left (3 k +6\right ) \left (3 k +18\right )}, a_{1}=0\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Whittaker successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 49
Order:=6; dsolve(9*x^2*diff(y(x),x$2)-3*x*(11+2*x^2)*diff(y(x),x)+(13+10*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = x^{\frac {1}{3}} \left (\left (1+\frac {4}{27} x^{2}+\frac {7}{486} x^{4}+\operatorname {O}\left (x^{6}\right )\right ) x^{4} c_{1} +c_{2} \left (\ln \left (x \right ) \left (-\frac {32}{9} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+\left (-144-32 x^{2}-\frac {8}{3} x^{4}+\operatorname {O}\left (x^{6}\right )\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.012 (sec). Leaf size: 62
AsymptoticDSolveValue[9*x^2*y''[x]-3*x*(11+2*x^2)*y'[x]+(13+10*x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {7 x^{25/3}}{486}+\frac {4 x^{19/3}}{27}+x^{13/3}\right )+c_1 \left (\frac {2}{81} x^{13/3} \log (x)+\frac {1}{81} \left (x^2+9\right )^2 \sqrt [3]{x}\right ) \]