Internal problem ID [5559]
Internal file name [OUTPUT/4807_Sunday_June_05_2022_03_06_18_PM_7339101/index.tex]
Book: A FIRST COURSE IN DIFFERENTIAL EQUATIONS with Modeling Applications.
Dennis G. Zill. 9th edition. Brooks/Cole. CA, USA.
Section: Chapter 6. SERIES SOLUTIONS OF LINEAR EQUATIONS. Exercises. 6.2 page
239
Problem number: 4.
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]]
\[ \boxed {y^{\prime \prime }-\frac {y^{\prime }}{x}+\frac {y}{\left (x -1\right )^{3}}=0} \] 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. \[ y^{\prime \prime }-\frac {y^{\prime }}{x}+\frac {y}{\left (x -1\right )^{3}} = 0 \] 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 {1}{x}\\ q(x) &= \frac {1}{\left (x -1\right )^{3}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty ]\)
Irregular singular points : \([1]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ y^{\prime \prime } x \left (x -1\right )^{3}-y^{\prime } \left (x -1\right )^{3}+y x = 0 \] 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} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x \left (x -1\right )^{3}-\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) \left (x -1\right )^{3}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) x = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}a_{n -3} \left (n +r -3\right ) \left (n -4+r \right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-3 a_{n -2} \left (n +r -2\right ) \left (n +r -3\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-a_{n -3} \left (n +r -3\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}3 a_{n -2} \left (n +r -2\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r -1} \\ \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 \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =3}{\sum }}a_{n -3} \left (n +r -3\right ) \left (n -4+r \right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-3 a_{n -2} \left (n +r -2\right ) \left (n +r -3\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-a_{n -3} \left (n +r -3\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}3 a_{n -2} \left (n +r -2\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ -x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ -x^{-1+r} a_{0} r \left (-1+r \right )+r a_{0} x^{-1+r} = 0 \] Or \[ \left (-x^{-1+r} r \left (-1+r \right )+r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (2-r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ -r \left (-2+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 2\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (2-r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([2, 0]\).
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^{2} \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 )+\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 +2}\\ 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}\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 \[ a_{1} = \frac {3 r \left (-2+r \right )}{r^{2}-1} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {6 r^{2}-12 r +1}{r \left (2+r \right )} \] For \(3\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -3} \left (n +r -3\right ) \left (n -4+r \right )-3 a_{n -2} \left (n +r -2\right ) \left (n +r -3\right )+3 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )-a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -3} \left (n +r -3\right )+3 a_{n -2} \left (n +r -2\right )-3 a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )+a_{n -2} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {n^{2} a_{n -3}-3 n^{2} a_{n -2}+3 n^{2} a_{n -1}+2 n r a_{n -3}-6 n r a_{n -2}+6 n r a_{n -1}+r^{2} a_{n -3}-3 r^{2} a_{n -2}+3 r^{2} a_{n -1}-8 n a_{n -3}+18 n a_{n -2}-12 n a_{n -1}-8 r a_{n -3}+18 r a_{n -2}-12 r a_{n -1}+15 a_{n -3}-23 a_{n -2}+9 a_{n -1}}{n^{2}+2 n r +r^{2}-2 n -2 r}\tag {4} \] Which for the root \(r = 2\) becomes \[ a_{n} = \frac {\left (a_{n -3}-3 a_{n -2}+3 a_{n -1}\right ) n^{2}+\left (-4 a_{n -3}+6 a_{n -2}\right ) n +3 a_{n -3}+a_{n -2}-3 a_{n -1}}{n \left (n +2\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 2\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3 r \left (-2+r \right )}{r^{2}-1}\) | \(0\) |
| \(a_{2}\) | \(\frac {6 r^{2}-12 r +1}{r \left (2+r \right )}\) | \(\frac {1}{8}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {10 r^{4}-20 r^{3}-4 r^{2}+14 r -3}{\left (r +3\right ) \left (1+r \right )^{2} \left (-1+r \right )} \] Which for the root \(r = 2\) becomes \[ a_{3}={\frac {1}{5}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3 r \left (-2+r \right )}{r^{2}-1}\) | \(0\) |
| \(a_{2}\) | \(\frac {6 r^{2}-12 r +1}{r \left (2+r \right )}\) | \(\frac {1}{8}\) |
| \(a_{3}\) | \(\frac {10 r^{4}-20 r^{3}-4 r^{2}+14 r -3}{\left (r +3\right ) \left (1+r \right )^{2} \left (-1+r \right )}\) | \(\frac {1}{5}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {15 r^{6}-54 r^{4}+13 r^{2}-1}{\left (4+r \right ) \left (2+r \right )^{2} r \left (-1+r \right ) \left (1+r \right )} \] Which for the root \(r = 2\) becomes \[ a_{4}={\frac {49}{192}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3 r \left (-2+r \right )}{r^{2}-1}\) | \(0\) |
| \(a_{2}\) | \(\frac {6 r^{2}-12 r +1}{r \left (2+r \right )}\) | \(\frac {1}{8}\) |
| \(a_{3}\) | \(\frac {10 r^{4}-20 r^{3}-4 r^{2}+14 r -3}{\left (r +3\right ) \left (1+r \right )^{2} \left (-1+r \right )}\) | \(\frac {1}{5}\) |
| \(a_{4}\) | \(\frac {15 r^{6}-54 r^{4}+13 r^{2}-1}{\left (4+r \right ) \left (2+r \right )^{2} r \left (-1+r \right ) \left (1+r \right )}\) | \(\frac {49}{192}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {21 r^{8}+84 r^{7}+14 r^{6}-252 r^{5}-298 r^{4}-78 r^{3}+47 r^{2}+30 r -9}{\left (5+r \right ) \left (r +3\right )^{2} r \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )} \] Which for the root \(r = 2\) becomes \[ a_{5}={\frac {423}{1400}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3 r \left (-2+r \right )}{r^{2}-1}\) | \(0\) |
| \(a_{2}\) | \(\frac {6 r^{2}-12 r +1}{r \left (2+r \right )}\) | \(\frac {1}{8}\) |
| \(a_{3}\) | \(\frac {10 r^{4}-20 r^{3}-4 r^{2}+14 r -3}{\left (r +3\right ) \left (1+r \right )^{2} \left (-1+r \right )}\) | \(\frac {1}{5}\) |
| \(a_{4}\) | \(\frac {15 r^{6}-54 r^{4}+13 r^{2}-1}{\left (4+r \right ) \left (2+r \right )^{2} r \left (-1+r \right ) \left (1+r \right )}\) | \(\frac {49}{192}\) |
| \(a_{5}\) | \(\frac {21 r^{8}+84 r^{7}+14 r^{6}-252 r^{5}-298 r^{4}-78 r^{3}+47 r^{2}+30 r -9}{\left (5+r \right ) \left (r +3\right )^{2} r \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )}\) | \(\frac {423}{1400}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{2} \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^{2} \left (1+\frac {x^{2}}{8}+\frac {x^{3}}{5}+\frac {49 x^{4}}{192}+\frac {423 x^{5}}{1400}+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 \(N=2\). 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 {6 r^{2}-12 r +1}{r \left (2+r \right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {6 r^{2}-12 r +1}{r \left (2+r \right )}&= \lim _{r\rightarrow 0}\frac {6 r^{2}-12 r +1}{r \left (2+r \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 \(y^{\prime \prime } x \left (x -1\right )^{3}-y^{\prime } \left (x -1\right )^{3}+y x = 0\) gives \[ \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right ) x \left (x -1\right )^{3}-\left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right ) \left (x -1\right )^{3}+\left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) x = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (y_{1}^{\prime \prime }\left (x \right ) x \left (x -1\right )^{3}-y_{1}^{\prime }\left (x \right ) \left (x -1\right )^{3}+y_{1}\left (x \right ) x \right ) \ln \left (x \right )+\left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) x \left (x -1\right )^{3}-\frac {y_{1}\left (x \right ) \left (x -1\right )^{3}}{x}\right ) C +\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 ) x \left (x -1\right )^{3}-\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) \left (x -1\right )^{3}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x = 0 \end{equation} But since \(y_{1}\left (x \right )\) is a solution to the ode, then \[ y_{1}^{\prime \prime }\left (x \right ) x \left (x -1\right )^{3}-y_{1}^{\prime }\left (x \right ) \left (x -1\right )^{3}+y_{1}\left (x \right ) x = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (\left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) x \left (x -1\right )^{3}-\frac {y_{1}\left (x \right ) \left (x -1\right )^{3}}{x}\right ) C +\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 ) x \left (x -1\right )^{3}-\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) \left (x -1\right )^{3}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x = 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} \frac {\left (2 x \left (x -1\right )^{3} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )-2 \left (x -1\right )^{3} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C}{x}+\frac {\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} \left (x -1\right )^{3}-\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) \left (x -1\right )^{3} x +\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x^{2}}{x} = 0 \end{equation} Since \(r_{1} = 2\) and \(r_{2} = 0\) then the above becomes \begin{equation} \tag{10} \frac {\left (2 x \left (x -1\right )^{3} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} a_{n} \left (n +2\right )\right )-2 \left (x -1\right )^{3} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +2}\right )\right ) C}{x}+\frac {\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} n \left (n -1\right )\right ) x^{2} \left (x -1\right )^{3}-\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} n \right ) \left (x -1\right )^{3} x +\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) x^{2}}{x} = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +4} a_{n} \left (n +2\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 C \,x^{n +3} a_{n} \left (n +2\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 C \,x^{n +2} a_{n} \left (n +2\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{1+n} a_{n} \left (n +2\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{n +4} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 C \,x^{n +3} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 C \,x^{n +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{1+n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n +2} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 n \,x^{1+n} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n} b_{n} n \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-n \,x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-n \,x^{n +2} b_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 n \,x^{1+n} b_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n} b_{n} n \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} n \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} b_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +4} a_{n} \left (n +2\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}2 C a_{n -5} \left (n -3\right ) x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 C \,x^{n +3} a_{n} \left (n +2\right )\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-6 C a_{n -4} \left (n -2\right ) x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}6 C \,x^{n +2} a_{n} \left (n +2\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}6 C a_{n -3} \left (n -1\right ) x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{1+n} a_{n} \left (n +2\right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 C a_{n -2} n \,x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{n +4} a_{n}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-2 C a_{n -5} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}6 C \,x^{n +3} a_{n} &= \moverset {\infty }{\munderset {n =4}{\sum }}6 C a_{n -4} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 C \,x^{n +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-6 C a_{n -3} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{1+n} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{n -2} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n +2} b_{n} \left (n -1\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (n -3\right ) b_{n -3} \left (n -4\right ) x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 n \,x^{1+n} b_{n} \left (n -1\right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-3 \left (n -2\right ) b_{n -2} \left (n -3\right ) x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n} b_{n} n \left (n -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 \left (n -1\right ) b_{n -1} \left (n -2\right ) x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-n \,x^{n +2} b_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-\left (n -3\right ) b_{n -3} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 n \,x^{1+n} b_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}3 \left (n -2\right ) b_{n -2} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n} b_{n} n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 \left (n -1\right ) b_{n -1} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} b_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}b_{n -2} x^{n -1} \\ \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 \(n -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =5}{\sum }}2 C a_{n -5} \left (n -3\right ) x^{n -1}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-6 C a_{n -4} \left (n -2\right ) x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}6 C a_{n -3} \left (n -1\right ) x^{n -1}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 C a_{n -2} n \,x^{n -1}\right )+\moverset {\infty }{\munderset {n =5}{\sum }}\left (-2 C a_{n -5} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}6 C a_{n -4} x^{n -1}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-6 C a_{n -3} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{n -2} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}\left (n -3\right ) b_{n -3} \left (n -4\right ) x^{n -1}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-3 \left (n -2\right ) b_{n -2} \left (n -3\right ) x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 \left (n -1\right ) b_{n -1} \left (n -2\right ) x^{n -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-n \,x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-\left (n -3\right ) b_{n -3} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}3 \left (n -2\right ) b_{n -2} x^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 \left (n -1\right ) b_{n -1} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} n \right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}b_{n -2} x^{n -1}\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 \[ b_{1} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ b_{1} = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}=0 \] 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 \[ -2 C +1 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C={\frac {1}{2}} \] For \(n=3\), Eq (2B) gives \[ \left (6 a_{0}-4 a_{1}\right ) C +4 b_{1}-3 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 3-3 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=1 \] For \(n=4\), Eq (2B) gives \[ \left (-6 a_{0}+12 a_{1}-6 a_{2}\right ) C -b_{1}+b_{2}+9 b_{3}-8 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {45}{8}-8 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}={\frac {45}{64}} \] For \(n=5\), Eq (2B) gives \[ 2 \left (a_{0}-6 a_{1}+9 a_{2}-4 a_{3}\right ) C -8 b_{3}+24 b_{4}-15 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {51}{5}-15 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}={\frac {17}{25}} \] 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 \(C={\frac {1}{2}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= \frac {1}{2}\eslowast \left (x^{2} \left (1+\frac {x^{2}}{8}+\frac {x^{3}}{5}+\frac {49 x^{4}}{192}+\frac {423 x^{5}}{1400}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1+x^{3}+\frac {45 x^{4}}{64}+\frac {17 x^{5}}{25}+O\left (x^{6}\right ) \] 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^{2} \left (1+\frac {x^{2}}{8}+\frac {x^{3}}{5}+\frac {49 x^{4}}{192}+\frac {423 x^{5}}{1400}+O\left (x^{6}\right )\right ) + c_{2} \left (\frac {1}{2}\eslowast \left (x^{2} \left (1+\frac {x^{2}}{8}+\frac {x^{3}}{5}+\frac {49 x^{4}}{192}+\frac {423 x^{5}}{1400}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1+x^{3}+\frac {45 x^{4}}{64}+\frac {17 x^{5}}{25}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{2} \left (1+\frac {x^{2}}{8}+\frac {x^{3}}{5}+\frac {49 x^{4}}{192}+\frac {423 x^{5}}{1400}+O\left (x^{6}\right )\right )+c_{2} \left (\frac {x^{2} \left (1+\frac {x^{2}}{8}+\frac {x^{3}}{5}+\frac {49 x^{4}}{192}+\frac {423 x^{5}}{1400}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{2}+1+x^{3}+\frac {45 x^{4}}{64}+\frac {17 x^{5}}{25}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{2} \left (1+\frac {x^{2}}{8}+\frac {x^{3}}{5}+\frac {49 x^{4}}{192}+\frac {423 x^{5}}{1400}+O\left (x^{6}\right )\right )+c_{2} \left (\frac {x^{2} \left (1+\frac {x^{2}}{8}+\frac {x^{3}}{5}+\frac {49 x^{4}}{192}+\frac {423 x^{5}}{1400}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{2}+1+x^{3}+\frac {45 x^{4}}{64}+\frac {17 x^{5}}{25}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{2} \left (1+\frac {x^{2}}{8}+\frac {x^{3}}{5}+\frac {49 x^{4}}{192}+\frac {423 x^{5}}{1400}+O\left (x^{6}\right )\right )+c_{2} \left (\frac {x^{2} \left (1+\frac {x^{2}}{8}+\frac {x^{3}}{5}+\frac {49 x^{4}}{192}+\frac {423 x^{5}}{1400}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{2}+1+x^{3}+\frac {45 x^{4}}{64}+\frac {17 x^{5}}{25}+O\left (x^{6}\right )\right ) \] Verified OK.
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 -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunC ODE, case a = 0, e <> 0, c <> 0 `
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 52
Order:=6; dsolve(diff(y(x),x$2)-1/x*diff(y(x),x)+1/(x-1)^3*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{2} \left (1+\frac {1}{8} x^{2}+\frac {1}{5} x^{3}+\frac {49}{192} x^{4}+\frac {423}{1400} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (-x^{2}-\frac {1}{8} x^{4}-\frac {1}{5} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (-2-2 x^{3}-\frac {45}{32} x^{4}-\frac {34}{25} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.047 (sec). Leaf size: 71
AsymptoticDSolveValue[y''[x]-1/x*y'[x]+1/(x-1)^3*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {1}{16} \left (x^2+8\right ) x^2 \log (x)+\frac {1}{64} \left (-5 x^4+64 x^3-400 x^2+64\right )\right )+c_2 \left (\frac {49 x^6}{192}+\frac {x^5}{5}+\frac {x^4}{8}+x^2\right ) \]