Internal problem ID [6993]
Internal file name [OUTPUT/6236_Friday_August_12_2022_11_06_33_PM_16214539/index.tex]
Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th
edition. 1997.
Section: CHAPTER 18. Power series solutions. 18.9 Indicial Equation with Difference of Roots a
Positive Integer: Logarithmic Case. Exercises page 384
Problem number: 13.
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]]
\[ \boxed {9 x^{2} y^{\prime \prime }-15 y^{\prime } x +7 \left (x +1\right ) y=0} \] 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. \[ 9 x^{2} y^{\prime \prime }-15 y^{\prime } x +\left (7 x +7\right ) y = 0 \] 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 {5}{3 x}\\ q(x) &= \frac {\frac {7 x}{9}+\frac {7}{9}}{x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 9 x^{2} y^{\prime \prime }-15 y^{\prime } x +\left (7 x +7\right ) y = 0 \] 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 )-15 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x +\left (7 x +7\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 (-15 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}7 x^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}7 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 }}7 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}7 a_{n -1} 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 =0}{\sum }}\left (-15 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}7 a_{n -1} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}7 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-15 x^{n +r} a_{n} \left (n +r \right )+7 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 9 x^{r} a_{0} r \left (-1+r \right )-15 x^{r} a_{0} r +7 a_{0} x^{r} = 0 \] Or \[ \left (9 x^{r} r \left (-1+r \right )-15 x^{r} r +7 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (9 r^{2}-24 r +7\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 9 r^{2}-24 r +7 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {7}{3}}\\ r_2 &= {\frac {1}{3}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (9 r^{2}-24 r +7\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {7}{3}}, {\frac {1}{3}}\right ]\).
Since \(r_1 - r_2 = 2\) 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 {7}{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 {7}{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 \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 9 a_{n} \left (n +r \right ) \left (n +r -1\right )-15 a_{n} \left (n +r \right )+7 a_{n -1}+7 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {7 a_{n -1}}{9 n^{2}+18 n r +9 r^{2}-24 n -24 r +7}\tag {4} \] Which for the root \(r = {\frac {7}{3}}\) becomes \[ a_{n} = -\frac {7 a_{n -1}}{9 n \left (n +2\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {7}{3}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=-\frac {7}{9 r^{2}-6 r -8} \] Which for the root \(r = {\frac {7}{3}}\) becomes \[ a_{1}=-{\frac {7}{27}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {7}{9 r^{2}-6 r -8}\) | \(-{\frac {7}{27}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {49}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right )} \] Which for the root \(r = {\frac {7}{3}}\) becomes \[ a_{2}={\frac {49}{1944}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {7}{9 r^{2}-6 r -8}\) | \(-{\frac {7}{27}}\) |
| \(a_{2}\) | \(\frac {49}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right )}\) | \(\frac {49}{1944}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {343}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right ) \left (9 r^{2}+30 r +16\right )} \] Which for the root \(r = {\frac {7}{3}}\) becomes \[ a_{3}=-{\frac {343}{262440}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {7}{9 r^{2}-6 r -8}\) | \(-{\frac {7}{27}}\) |
| \(a_{2}\) | \(\frac {49}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right )}\) | \(\frac {49}{1944}\) |
| \(a_{3}\) | \(-\frac {343}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right ) \left (9 r^{2}+30 r +16\right )}\) | \(-{\frac {343}{262440}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {2401}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right ) \left (9 r^{2}+30 r +16\right ) \left (9 r^{2}+48 r +55\right )} \] Which for the root \(r = {\frac {7}{3}}\) becomes \[ a_{4}={\frac {2401}{56687040}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {7}{9 r^{2}-6 r -8}\) | \(-{\frac {7}{27}}\) |
| \(a_{2}\) | \(\frac {49}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right )}\) | \(\frac {49}{1944}\) |
| \(a_{3}\) | \(-\frac {343}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right ) \left (9 r^{2}+30 r +16\right )}\) | \(-{\frac {343}{262440}}\) |
| \(a_{4}\) | \(\frac {2401}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right ) \left (9 r^{2}+30 r +16\right ) \left (9 r^{2}+48 r +55\right )}\) | \(\frac {2401}{56687040}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {16807}{59049 \left (r +\frac {11}{3}\right ) \left (r +\frac {2}{3}\right )^{2} \left (r +\frac {5}{3}\right )^{2} \left (r +\frac {8}{3}\right )^{2} \left (r -\frac {4}{3}\right ) \left (r -\frac {1}{3}\right ) \left (r +\frac {14}{3}\right )} \] Which for the root \(r = {\frac {7}{3}}\) becomes \[ a_{5}=-{\frac {2401}{2550916800}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {7}{9 r^{2}-6 r -8}\) | \(-{\frac {7}{27}}\) |
| \(a_{2}\) | \(\frac {49}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right )}\) | \(\frac {49}{1944}\) |
| \(a_{3}\) | \(-\frac {343}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right ) \left (9 r^{2}+30 r +16\right )}\) | \(-{\frac {343}{262440}}\) |
| \(a_{4}\) | \(\frac {2401}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right ) \left (9 r^{2}+30 r +16\right ) \left (9 r^{2}+48 r +55\right )}\) | \(\frac {2401}{56687040}\) |
| \(a_{5}\) | \(-\frac {16807}{59049 \left (r +\frac {11}{3}\right ) \left (r +\frac {2}{3}\right )^{2} \left (r +\frac {5}{3}\right )^{2} \left (r +\frac {8}{3}\right )^{2} \left (r -\frac {4}{3}\right ) \left (r -\frac {1}{3}\right ) \left (r +\frac {14}{3}\right )}\) | \(-{\frac {2401}{2550916800}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {117649}{\left (3 r +11\right ) \left (3 r +2\right )^{2} \left (3 r +5\right )^{2} \left (3 r +8\right )^{2} \left (3 r -4\right ) \left (3 r -1\right ) \left (3 r +14\right ) \left (9 r^{2}+84 r +187\right )} \] Which for the root \(r = {\frac {7}{3}}\) becomes \[ a_{6}={\frac {16807}{1101996057600}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {7}{9 r^{2}-6 r -8}\) | \(-{\frac {7}{27}}\) |
| \(a_{2}\) | \(\frac {49}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right )}\) | \(\frac {49}{1944}\) |
| \(a_{3}\) | \(-\frac {343}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right ) \left (9 r^{2}+30 r +16\right )}\) | \(-{\frac {343}{262440}}\) |
| \(a_{4}\) | \(\frac {2401}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right ) \left (9 r^{2}+30 r +16\right ) \left (9 r^{2}+48 r +55\right )}\) | \(\frac {2401}{56687040}\) |
| \(a_{5}\) | \(-\frac {16807}{59049 \left (r +\frac {11}{3}\right ) \left (r +\frac {2}{3}\right )^{2} \left (r +\frac {5}{3}\right )^{2} \left (r +\frac {8}{3}\right )^{2} \left (r -\frac {4}{3}\right ) \left (r -\frac {1}{3}\right ) \left (r +\frac {14}{3}\right )}\) | \(-{\frac {2401}{2550916800}}\) |
| \(a_{6}\) | \(\frac {117649}{\left (3 r +11\right ) \left (3 r +2\right )^{2} \left (3 r +5\right )^{2} \left (3 r +8\right )^{2} \left (3 r -4\right ) \left (3 r -1\right ) \left (3 r +14\right ) \left (9 r^{2}+84 r +187\right )}\) | \(\frac {16807}{1101996057600}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=-\frac {823543}{\left (3 r +11\right ) \left (3 r +2\right )^{2} \left (3 r +5\right )^{2} \left (3 r +8\right )^{2} \left (3 r -4\right ) \left (3 r -1\right ) \left (3 r +14\right ) \left (9 r^{2}+84 r +187\right ) \left (9 r^{2}+102 r +280\right )} \] Which for the root \(r = {\frac {7}{3}}\) becomes \[ a_{7}=-{\frac {16807}{89261680665600}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {7}{9 r^{2}-6 r -8}\) | \(-{\frac {7}{27}}\) |
| \(a_{2}\) | \(\frac {49}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right )}\) | \(\frac {49}{1944}\) |
| \(a_{3}\) | \(-\frac {343}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right ) \left (9 r^{2}+30 r +16\right )}\) | \(-{\frac {343}{262440}}\) |
| \(a_{4}\) | \(\frac {2401}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right ) \left (9 r^{2}+30 r +16\right ) \left (9 r^{2}+48 r +55\right )}\) | \(\frac {2401}{56687040}\) |
| \(a_{5}\) | \(-\frac {16807}{59049 \left (r +\frac {11}{3}\right ) \left (r +\frac {2}{3}\right )^{2} \left (r +\frac {5}{3}\right )^{2} \left (r +\frac {8}{3}\right )^{2} \left (r -\frac {4}{3}\right ) \left (r -\frac {1}{3}\right ) \left (r +\frac {14}{3}\right )}\) | \(-{\frac {2401}{2550916800}}\) |
| \(a_{6}\) | \(\frac {117649}{\left (3 r +11\right ) \left (3 r +2\right )^{2} \left (3 r +5\right )^{2} \left (3 r +8\right )^{2} \left (3 r -4\right ) \left (3 r -1\right ) \left (3 r +14\right ) \left (9 r^{2}+84 r +187\right )}\) | \(\frac {16807}{1101996057600}\) |
| \(a_{7}\) | \(-\frac {823543}{\left (3 r +11\right ) \left (3 r +2\right )^{2} \left (3 r +5\right )^{2} \left (3 r +8\right )^{2} \left (3 r -4\right ) \left (3 r -1\right ) \left (3 r +14\right ) \left (9 r^{2}+84 r +187\right ) \left (9 r^{2}+102 r +280\right )}\) | \(-{\frac {16807}{89261680665600}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {7}{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}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= x^{\frac {7}{3}} \left (1-\frac {7 x}{27}+\frac {49 x^{2}}{1944}-\frac {343 x^{3}}{262440}+\frac {2401 x^{4}}{56687040}-\frac {2401 x^{5}}{2550916800}+\frac {16807 x^{6}}{1101996057600}-\frac {16807 x^{7}}{89261680665600}+O\left (x^{8}\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 \(N=2\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{2}\left (r \right )\). 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_{2} \\ &= \frac {49}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {49}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\right )}&= \lim _{r\rightarrow {\frac {1}{3}}}\frac {49}{\left (9 r^{2}-6 r -8\right ) \left (9 r^{2}+12 r -5\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 \(9 x^{2} y^{\prime \prime }-15 y^{\prime } x +\left (7 x +7\right ) y = 0\) gives \[ 9 x^{2} \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )-15 \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right ) x +\left (7 x +7\right ) \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (9 x^{2} y_{1}^{\prime \prime }\left (x \right )-15 y_{1}^{\prime }\left (x \right ) x +\left (7 x +7\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 )-15 y_{1}\left (x \right )\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 )-15 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) x +\left (7 x +7\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 \[ 9 x^{2} y_{1}^{\prime \prime }\left (x \right )-15 y_{1}^{\prime }\left (x \right ) x +\left (7 x +7\right ) y_{1}\left (x \right ) = 0 \] 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 )-15 y_{1}\left (x \right )\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 )-15 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) x +\left (7 x +7\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 -24 \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}-15 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) x +\left (7 x +7\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Since \(r_{1} = {\frac {7}{3}}\) 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 {4}{3}+n} a_{n} \left (n +\frac {7}{3}\right )\right ) x -24 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {7}{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}-15 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-\frac {2}{3}+n} b_{n} \left (n +\frac {1}{3}\right )\right ) x +\left (7 x +7\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{3}}\right ) = 0 \end{equation} Expanding \(7 x^{\frac {4}{3}}\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} 7 x^{\frac {4}{3}} &= 7 x^{\frac {4}{3}} + \dots \\ &= 7 x^{\frac {4}{3}} \end {align*}
Expanding \(7 x^{\frac {1}{3}}\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} 7 x^{\frac {1}{3}} &= 7 x^{\frac {1}{3}} + \dots \\ &= 7 x^{\frac {1}{3}} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (18 n +42\right ) C a_{n} x^{n +\frac {7}{3}}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-24 C a_{n} x^{n +\frac {7}{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 =0}{\sum }}\left (-15 n -5\right ) b_{n} x^{n +\frac {1}{3}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}7 x^{\frac {4}{3}+n} b_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}7 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 +42\right ) C a_{n} x^{n +\frac {7}{3}} &= \moverset {\infty }{\munderset {n =2}{\sum }}C a_{n -2} \left (18 n +6\right ) x^{n +\frac {1}{3}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-24 C a_{n} x^{n +\frac {7}{3}}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-24 C a_{n -2} x^{n +\frac {1}{3}}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}7 x^{\frac {4}{3}+n} b_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}7 b_{n -1} 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 =2}{\sum }}C a_{n -2} \left (18 n +6\right ) x^{n +\frac {1}{3}}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-24 C a_{n -2} 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 =0}{\sum }}\left (-15 n -5\right ) b_{n} x^{n +\frac {1}{3}}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}7 b_{n -1} x^{n +\frac {1}{3}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}7 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 \[ -9 b_{1}+7 b_{0} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -9 b_{1}+7 = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}={\frac {7}{9}} \] For \(n=N\), where \(N=2\) which is the difference between the two roots, we are free to choose \(b_{2} = 0\). Hence for \(n=2\), Eq (2B) gives \[ 18 C +\frac {49}{9} = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-{\frac {49}{162}} \] For \(n=3\), Eq (2B) gives \[ 36 C a_{1}+7 b_{2}+27 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 27 b_{3}+\frac {686}{243} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=-{\frac {686}{6561}} \] For \(n=4\), Eq (2B) gives \[ 54 C a_{2}+7 b_{3}+72 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 72 b_{4}-\frac {60025}{52488} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}={\frac {60025}{3779136}} \] For \(n=5\), Eq (2B) gives \[ 72 C a_{3}+7 b_{4}+135 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 135 b_{5}+\frac {2638699}{18895680} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=-{\frac {2638699}{2550916800}} \] For \(n=6\), Eq (2B) gives \[ 90 C a_{4}+7 b_{5}+216 b_{6} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 216 b_{6}-\frac {10706059}{1275458400} = 0 \] Solving the above for \(b_{6}\) gives \[ b_{6}={\frac {10706059}{275499014400}} \] For \(n=7\), Eq (2B) gives \[ 108 C a_{5}+7 b_{6}+315 b_{7} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 315 b_{7}+\frac {83413141}{275499014400} = 0 \] Solving the above for \(b_{7}\) gives \[ b_{7}=-{\frac {11916163}{12397455648000}} \] 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 \(C=-{\frac {49}{162}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= -\frac {49}{162}\eslowast \left (x^{\frac {7}{3}} \left (1-\frac {7 x}{27}+\frac {49 x^{2}}{1944}-\frac {343 x^{3}}{262440}+\frac {2401 x^{4}}{56687040}-\frac {2401 x^{5}}{2550916800}+\frac {16807 x^{6}}{1101996057600}-\frac {16807 x^{7}}{89261680665600}+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+x^{\frac {1}{3}} \left (1+\frac {7 x}{9}-\frac {686 x^{3}}{6561}+\frac {60025 x^{4}}{3779136}-\frac {2638699 x^{5}}{2550916800}+\frac {10706059 x^{6}}{275499014400}-\frac {11916163 x^{7}}{12397455648000}+O\left (x^{8}\right )\right ) \] 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 {7}{3}} \left (1-\frac {7 x}{27}+\frac {49 x^{2}}{1944}-\frac {343 x^{3}}{262440}+\frac {2401 x^{4}}{56687040}-\frac {2401 x^{5}}{2550916800}+\frac {16807 x^{6}}{1101996057600}-\frac {16807 x^{7}}{89261680665600}+O\left (x^{8}\right )\right ) + c_{2} \left (-\frac {49}{162}\eslowast \left (x^{\frac {7}{3}} \left (1-\frac {7 x}{27}+\frac {49 x^{2}}{1944}-\frac {343 x^{3}}{262440}+\frac {2401 x^{4}}{56687040}-\frac {2401 x^{5}}{2550916800}+\frac {16807 x^{6}}{1101996057600}-\frac {16807 x^{7}}{89261680665600}+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+x^{\frac {1}{3}} \left (1+\frac {7 x}{9}-\frac {686 x^{3}}{6561}+\frac {60025 x^{4}}{3779136}-\frac {2638699 x^{5}}{2550916800}+\frac {10706059 x^{6}}{275499014400}-\frac {11916163 x^{7}}{12397455648000}+O\left (x^{8}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {7}{3}} \left (1-\frac {7 x}{27}+\frac {49 x^{2}}{1944}-\frac {343 x^{3}}{262440}+\frac {2401 x^{4}}{56687040}-\frac {2401 x^{5}}{2550916800}+\frac {16807 x^{6}}{1101996057600}-\frac {16807 x^{7}}{89261680665600}+O\left (x^{8}\right )\right )+c_{2} \left (-\frac {49 x^{\frac {7}{3}} \left (1-\frac {7 x}{27}+\frac {49 x^{2}}{1944}-\frac {343 x^{3}}{262440}+\frac {2401 x^{4}}{56687040}-\frac {2401 x^{5}}{2550916800}+\frac {16807 x^{6}}{1101996057600}-\frac {16807 x^{7}}{89261680665600}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{162}+x^{\frac {1}{3}} \left (1+\frac {7 x}{9}-\frac {686 x^{3}}{6561}+\frac {60025 x^{4}}{3779136}-\frac {2638699 x^{5}}{2550916800}+\frac {10706059 x^{6}}{275499014400}-\frac {11916163 x^{7}}{12397455648000}+O\left (x^{8}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {7}{3}} \left (1-\frac {7 x}{27}+\frac {49 x^{2}}{1944}-\frac {343 x^{3}}{262440}+\frac {2401 x^{4}}{56687040}-\frac {2401 x^{5}}{2550916800}+\frac {16807 x^{6}}{1101996057600}-\frac {16807 x^{7}}{89261680665600}+O\left (x^{8}\right )\right )+c_{2} \left (-\frac {49 x^{\frac {7}{3}} \left (1-\frac {7 x}{27}+\frac {49 x^{2}}{1944}-\frac {343 x^{3}}{262440}+\frac {2401 x^{4}}{56687040}-\frac {2401 x^{5}}{2550916800}+\frac {16807 x^{6}}{1101996057600}-\frac {16807 x^{7}}{89261680665600}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{162}+x^{\frac {1}{3}} \left (1+\frac {7 x}{9}-\frac {686 x^{3}}{6561}+\frac {60025 x^{4}}{3779136}-\frac {2638699 x^{5}}{2550916800}+\frac {10706059 x^{6}}{275499014400}-\frac {11916163 x^{7}}{12397455648000}+O\left (x^{8}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {7}{3}} \left (1-\frac {7 x}{27}+\frac {49 x^{2}}{1944}-\frac {343 x^{3}}{262440}+\frac {2401 x^{4}}{56687040}-\frac {2401 x^{5}}{2550916800}+\frac {16807 x^{6}}{1101996057600}-\frac {16807 x^{7}}{89261680665600}+O\left (x^{8}\right )\right )+c_{2} \left (-\frac {49 x^{\frac {7}{3}} \left (1-\frac {7 x}{27}+\frac {49 x^{2}}{1944}-\frac {343 x^{3}}{262440}+\frac {2401 x^{4}}{56687040}-\frac {2401 x^{5}}{2550916800}+\frac {16807 x^{6}}{1101996057600}-\frac {16807 x^{7}}{89261680665600}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{162}+x^{\frac {1}{3}} \left (1+\frac {7 x}{9}-\frac {686 x^{3}}{6561}+\frac {60025 x^{4}}{3779136}-\frac {2638699 x^{5}}{2550916800}+\frac {10706059 x^{6}}{275499014400}-\frac {11916163 x^{7}}{12397455648000}+O\left (x^{8}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 9 x^{2} \left (\frac {d}{d x}y^{\prime }\right )-15 y^{\prime } x +\left (7 x +7\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {5 y^{\prime }}{3 x}-\frac {7 \left (x +1\right ) y}{9 x^{2}} \\ \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} \\ {} & {} & \frac {d}{d x}y^{\prime }-\frac {5 y^{\prime }}{3 x}+\frac {7 \left (x +1\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 {5}{3 x}, P_{3}\left (x \right )=\frac {7 \left (x +1\right )}{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 {5}{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 {7}{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} \left (\frac {d}{d x}y^{\prime }\right )-15 y^{\prime } x +\left (7 x +7\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..1 \\ {} & {} & 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 \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\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 (-7+3 r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (3 k +3 r -1\right ) \left (3 k +3 r -7\right )+7 a_{k -1}\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 (-7+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {1}{3}, \frac {7}{3}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 9 \left (k +r -\frac {1}{3}\right ) \left (k +r -\frac {7}{3}\right ) a_{k}+7 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 9 \left (k +\frac {2}{3}+r \right ) \left (k -\frac {4}{3}+r \right ) a_{k +1}+7 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {7 a_{k}}{\left (3 k +2+3 r \right ) \left (3 k -4+3 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{3} \\ {} & {} & a_{k +1}=-\frac {7 a_{k}}{\left (3 k +3\right ) \left (3 k -3\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 =1 \\ {} & {} & a_{k +1}=-\frac {7 a_{k}}{\left (3 k +3\right ) \left (3 k -3\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {7}{3} \\ {} & {} & a_{k +1}=-\frac {7 a_{k}}{\left (3 k +9\right ) \left (3 k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {7}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {7}{3}}, a_{k +1}=-\frac {7 a_{k}}{\left (3 k +9\right ) \left (3 k +3\right )}\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 <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.11 (sec). Leaf size: 77
Order:=8; dsolve(9*x^2*diff(y(x),x$2)-15*x*diff(y(x),x)+7*(x+1)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = x^{\frac {1}{3}} \left (x^{2} \left (1-\frac {7}{27} x +\frac {49}{1944} x^{2}-\frac {343}{262440} x^{3}+\frac {2401}{56687040} x^{4}-\frac {2401}{2550916800} x^{5}+\frac {16807}{1101996057600} x^{6}-\frac {16807}{89261680665600} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{1} +c_{2} \left (\ln \left (x \right ) \left (\frac {49}{81} x^{2}-\frac {343}{2187} x^{3}+\frac {2401}{157464} x^{4}-\frac {16807}{21257640} x^{5}+\frac {117649}{4591650240} x^{6}-\frac {117649}{206624260800} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (-2-\frac {14}{9} x +\frac {1372}{6561} x^{3}-\frac {60025}{1889568} x^{4}+\frac {2638699}{1275458400} x^{5}-\frac {10706059}{137749507200} x^{6}+\frac {11916163}{6198727824000} x^{7}+\operatorname {O}\left (x^{8}\right )\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.064 (sec). Leaf size: 141
AsymptoticDSolveValue[9*x^2*y''[x]-15*x*y'[x]+7*(x+1)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_2 \left (\frac {16807 x^{25/3}}{1101996057600}-\frac {2401 x^{22/3}}{2550916800}+\frac {2401 x^{19/3}}{56687040}-\frac {343 x^{16/3}}{262440}+\frac {49 x^{13/3}}{1944}-\frac {7 x^{10/3}}{27}+x^{7/3}\right )+c_1 \left (\frac {\sqrt [3]{x} \left (6235397 x^6-169717086 x^5+2713009950 x^4-19803722400 x^3+20832487200 x^2+107138505600 x+137749507200\right )}{137749507200}-\frac {49 x^{7/3} \left (2401 x^4-74088 x^3+1428840 x^2-14696640 x+56687040\right ) \log (x)}{9183300480}\right ) \]