Internal problem ID [2919]
Internal file name [OUTPUT/2411_Sunday_June_05_2022_03_06_51_AM_76483647/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 11, Series Solutions to Linear Differential Equations. Exercises for 11.4. page
758
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]]
\[ \boxed {\left (-2+x \right )^{2} y^{\prime \prime }+\left (-2+x \right ) {\mathrm e}^{x} y^{\prime }+\frac {4 y}{x}=0} \] 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. \[ \left (x^{2}-4 x +4\right ) y^{\prime \prime }+\left (x \,{\mathrm e}^{x}-2 \,{\mathrm e}^{x}\right ) y^{\prime }+\frac {4 y}{x} = 0 \] 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 {{\mathrm e}^{x}}{-2+x}\\ q(x) &= \frac {4}{x \left (-2+x \right )^{2}}\\ \end {align*}
| \(p(x)=\frac {{\mathrm e}^{x}}{-2+x}\) | |
| singularity | type |
| \(x = 2\) | \(\text {``regular''}\) |
| \(x = \infty \) | \(\text {``regular''}\) |
| \(q(x)=\frac {4}{x \left (-2+x \right )^{2}}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 2\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([2, \infty , 0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x \left (x^{2}-4 x +4\right ) y^{\prime \prime }+{\mathrm e}^{x} x \left (-2+x \right ) y^{\prime }+4 y = 0 \] 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} x \left (x^{2}-4 x +4\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+{\mathrm e}^{x} x \left (-2+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Expanding \(\left (-2+x \right ) {\mathrm e}^{x} x\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \left (-2+x \right ) {\mathrm e}^{x} x &= -2 x -x^{2}+\frac {1}{6} x^{4}+\frac {1}{12} x^{5}+\frac {1}{40} x^{6} + \dots \\ &= -2 x -x^{2}+\frac {1}{6} x^{4}+\frac {1}{12} x^{5}+\frac {1}{40} x^{6} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n} \left (n +r \right )}{40}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n} \left (n +r \right )}{12}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +3} a_{n} \left (n +r \right )}{6}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 a_{n} x^{n +r}\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^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n} \left (n +r \right )}{40} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {a_{n -6} \left (n +r -6\right ) x^{n +r -1}}{40} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n} \left (n +r \right )}{12} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} \left (n +r -5\right ) x^{n +r -1}}{12} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +3} a_{n} \left (n +r \right )}{6} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n +r -4\right ) x^{n +r -1}}{6} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} \left (n +r -2\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} 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 =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =6}{\sum }}\frac {a_{n -6} \left (n +r -6\right ) x^{n +r -1}}{40}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} \left (n +r -5\right ) x^{n +r -1}}{12}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n +r -4\right ) x^{n +r -1}}{6}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} \left (n +r -2\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right ) = 0 \] When \(n = 0\) the above becomes \[ 4 x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Or \[ 4 x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ 4 x^{-1+r} r \left (-1+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r \left (-1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ 4 x^{-1+r} r \left (-1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([1, 0]\).
Since \(r_1 - r_2 = 1\) 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 \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^{1+n}\\ 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 \[ a_{1} = \frac {2 r^{2}-r -2}{2 r \left (1+r \right )} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {3 r^{4}+5 r^{3}-7 r^{2}-5 r +2}{4 r \left (1+r \right )^{2} \left (2+r \right )} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {2 r^{6}+13 r^{5}+18 r^{4}-13 r^{3}-28 r^{2}-7 r +2}{4 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {15 r^{8}+208 r^{7}+984 r^{6}+1798 r^{5}+363 r^{4}-2576 r^{3}-2562 r^{2}-486 r +156}{48 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = \frac {18 r^{10}+425 r^{9}+3902 r^{8}+17872 r^{7}+41868 r^{6}+38587 r^{5}-27540 r^{4}-88804 r^{3}-61280 r^{2}-7068 r +4200}{96 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )} \] For \(6\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )-4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+4 a_{n} \left (n +r \right ) \left (n +r -1\right )+\frac {a_{n -6} \left (n +r -6\right )}{40}+\frac {a_{n -5} \left (n +r -5\right )}{12}+\frac {a_{n -4} \left (n +r -4\right )}{6}-a_{n -2} \left (n +r -2\right )-2 a_{n -1} \left (n +r -1\right )+4 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {120 n^{2} a_{n -2}-480 n^{2} a_{n -1}+240 n r a_{n -2}-960 n r a_{n -1}+120 r^{2} a_{n -2}-480 r^{2} a_{n -1}+3 n a_{n -6}+10 n a_{n -5}+20 n a_{n -4}-720 n a_{n -2}+1200 n a_{n -1}+3 r a_{n -6}+10 r a_{n -5}+20 r a_{n -4}-720 r a_{n -2}+1200 r a_{n -1}-18 a_{n -6}-50 a_{n -5}-80 a_{n -4}+960 a_{n -2}-240 a_{n -1}}{480 \left (n +r \right ) \left (n +r -1\right )}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {\left (-120 a_{n -2}+480 a_{n -1}\right ) n^{2}+\left (-3 a_{n -6}-10 a_{n -5}-20 a_{n -4}+480 a_{n -2}-240 a_{n -1}\right ) n +15 a_{n -6}+40 a_{n -5}+60 a_{n -4}-360 a_{n -2}-480 a_{n -1}}{480 n \left (1+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}-r -2}{2 r \left (1+r \right )}\) | \(-{\frac {1}{4}}\) |
| \(a_{2}\) | \(\frac {3 r^{4}+5 r^{3}-7 r^{2}-5 r +2}{4 r \left (1+r \right )^{2} \left (2+r \right )}\) | \(-{\frac {1}{24}}\) |
| \(a_{3}\) | \(\frac {2 r^{6}+13 r^{5}+18 r^{4}-13 r^{3}-28 r^{2}-7 r +2}{4 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\) | \(-{\frac {13}{576}}\) |
| \(a_{4}\) | \(\frac {15 r^{8}+208 r^{7}+984 r^{6}+1798 r^{5}+363 r^{4}-2576 r^{3}-2562 r^{2}-486 r +156}{48 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\) | \(-{\frac {35}{2304}}\) |
| \(a_{5}\) | \(\frac {18 r^{10}+425 r^{9}+3902 r^{8}+17872 r^{7}+41868 r^{6}+38587 r^{5}-27540 r^{4}-88804 r^{3}-61280 r^{2}-7068 r +4200}{96 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )}\) | \(-{\frac {1297}{138240}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x \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 \left (1-\frac {x}{4}-\frac {x^{2}}{24}-\frac {13 x^{3}}{576}-\frac {35 x^{4}}{2304}-\frac {1297 x^{5}}{138240}+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 \(N=1\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{1}\left (r \right )\). 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_{1} \\ &= \frac {2 r^{2}-r -2}{2 r \left (1+r \right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {2 r^{2}-r -2}{2 r \left (1+r \right )}&= \lim _{r\rightarrow 0}\frac {2 r^{2}-r -2}{2 r \left (1+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 \(x \left (x^{2}-4 x +4\right ) y^{\prime \prime }+{\mathrm e}^{x} x \left (-2+x \right ) y^{\prime }+4 y = 0\) gives \[ x \left (x^{2}-4 x +4\right ) \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+{\mathrm e}^{x} x \left (-2+x \right ) \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+4 C y_{1}\left (x \right ) \ln \left (x \right )+4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (x \left (x^{2}-4 x +4\right ) y_{1}^{\prime \prime }\left (x \right )+{\mathrm e}^{x} \left (-2+x \right ) x y_{1}^{\prime }\left (x \right )+4 y_{1}\left (x \right )\right ) \ln \left (x \right )+x \left (x^{2}-4 x +4\right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+{\mathrm e}^{x} \left (-2+x \right ) y_{1}\left (x \right )\right ) C +x \left (x^{2}-4 x +4\right ) \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 )+{\mathrm e}^{x} \left (-2+x \right ) x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+4 \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 \[ x \left (x^{2}-4 x +4\right ) y_{1}^{\prime \prime }\left (x \right )+{\mathrm e}^{x} \left (-2+x \right ) x y_{1}^{\prime }\left (x \right )+4 y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (x \left (x^{2}-4 x +4\right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+{\mathrm e}^{x} \left (-2+x \right ) y_{1}\left (x \right )\right ) C +x \left (x^{2}-4 x +4\right ) \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 )+{\mathrm e}^{x} \left (-2+x \right ) x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+4 \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} \frac {\left (2 x \left (-2+x \right )^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )+\left (-2+x \right ) \left (x \,{\mathrm e}^{x}-x +2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C}{x}+\frac {x^{2} \left (-2+x \right )^{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 )+{\mathrm e}^{x} x^{2} \left (-2+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )+4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x}{x} = 0 \end{equation} Since \(r_{1} = 1\) and \(r_{2} = 0\) then the above becomes \begin{equation} \tag{10} \frac {\left (2 x \left (-2+x \right )^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} a_{n} \left (1+n \right )\right )+\left (-2+x \right ) \left (x \,{\mathrm e}^{x}-x +2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{1+n}\right )\right ) C}{x}+\frac {x^{2} \left (-2+x \right )^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} n \left (n -1\right )\right )+{\mathrm e}^{x} x^{2} \left (-2+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} n \right )+4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) x}{x} = 0 \end{equation} Expanding \(C \,{\mathrm e}^{x} x^{2}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} C \,{\mathrm e}^{x} x^{2} &= C \,x^{2}+C \,x^{3}+\frac {1}{2} C \,x^{4}+\frac {1}{6} C \,x^{5}+\frac {1}{24} C \,x^{6} + \dots \\ &= C \,x^{2}+C \,x^{3}+\frac {1}{2} C \,x^{4}+\frac {1}{6} C \,x^{5}+\frac {1}{24} C \,x^{6} \end {align*}
Expanding \(-2 C \,{\mathrm e}^{x} x\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -2 C \,{\mathrm e}^{x} x &= -2 C x -2 C \,x^{2}-C \,x^{3}-\frac {1}{3} C \,x^{4}-\frac {1}{12} C \,x^{5}-\frac {1}{60} C \,x^{6} + \dots \\ &= -2 C x -2 C \,x^{2}-C \,x^{3}-\frac {1}{3} C \,x^{4}-\frac {1}{12} C \,x^{5}-\frac {1}{60} C \,x^{6} \end {align*}
Expanding \(-2 \,{\mathrm e}^{x}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -2 \,{\mathrm e}^{x} &= -2-2 x -x^{2}-\frac {1}{3} x^{3}-\frac {1}{12} x^{4}-\frac {1}{60} x^{5}-\frac {1}{360} x^{6} + \dots \\ &= -2-2 x -x^{2}-\frac {1}{3} x^{3}-\frac {1}{12} x^{4}-\frac {1}{60} x^{5}-\frac {1}{360} x^{6} \end {align*}
Expanding \(x \,{\mathrm e}^{x}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} x \,{\mathrm e}^{x} &= x +x^{2}+\frac {1}{2} x^{3}+\frac {1}{6} x^{4}+\frac {1}{24} x^{5}+\frac {1}{120} x^{6} + \dots \\ &= x +x^{2}+\frac {1}{2} x^{3}+\frac {1}{6} x^{4}+\frac {1}{24} x^{5}+\frac {1}{120} x^{6} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}4 n \,x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {n \,x^{n +3} b_{n}}{3}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +5} a_{n}}{6}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {n \,x^{n +5} b_{n}}{60}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +6} a_{n}}{24}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +6} b_{n}}{120}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n +2} b_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {n \,x^{n +4} b_{n}}{12}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +4} a_{n}}{2}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n +3} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +5} b_{n}}{24}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{1+n} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +4} a_{n}}{3}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +4} b_{n}}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 b_{n} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 C \,x^{1+n} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 C \,x^{1+n} a_{n} \left (1+n \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 n \,x^{1+n} b_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{1+n} b_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{n +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}C \,x^{n +3} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +5} a_{n}}{12}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +3} b_{n}}{2}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-n \,x^{n +2} b_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {n \,x^{n +6} b_{n}}{360}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n} b_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{1+n} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 a_{n} x^{n} C \right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n} b_{n} n \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +2} a_{n} \left (1+n \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 C \,x^{n} a_{n} \left (1+n \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}C \,x^{n +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +6} a_{n}}{60}\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 +2} a_{n} \left (1+n \right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}2 C a_{n -3} \left (n -2\right ) x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 C \,x^{1+n} a_{n} \left (1+n \right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-8 C a_{n -2} \left (n -1\right ) x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}8 C \,x^{n} a_{n} \left (1+n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}8 C a_{n -1} n \,x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}C \,x^{n +2} a_{n} &= \moverset {\infty }{\munderset {n =3}{\sum }}C a_{n -3} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}C \,x^{n +3} a_{n} &= \moverset {\infty }{\munderset {n =4}{\sum }}C a_{n -4} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +4} a_{n}}{2} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {C a_{n -5} x^{n -1}}{2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +5} a_{n}}{6} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {C a_{n -6} x^{n -1}}{6} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +6} a_{n}}{24} &= \moverset {\infty }{\munderset {n =7}{\sum }}\frac {C a_{n -7} x^{n -1}}{24} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-C a_{n -3} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{1+n} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 C a_{n -2} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{n +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-2 C a_{n -3} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n +3} a_{n}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-C a_{n -4} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +4} a_{n}}{3}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {C a_{n -5} x^{n -1}}{3}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +5} a_{n}}{12}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {C a_{n -6} x^{n -1}}{12}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +6} a_{n}}{60}\right ) &= \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {C a_{n -7} x^{n -1}}{60}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 C \,x^{1+n} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}4 C a_{n -2} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 a_{n} x^{n} C \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 C a_{n -1} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{1+n} b_{n} \left (n -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (n -2\right ) b_{n -2} \left (n -3\right ) x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n} b_{n} n \left (n -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 \left (n -1\right ) b_{n -1} \left (n -2\right ) x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n} b_{n} n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 \left (n -1\right ) b_{n -1} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 n \,x^{1+n} b_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 \left (n -2\right ) b_{n -2} x^{n -1}\right ) \\ \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 }}\left (-\frac {n \,x^{n +3} b_{n}}{3}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {\left (n -4\right ) b_{n -4} x^{n -1}}{3}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {n \,x^{n +4} b_{n}}{12}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {\left (n -5\right ) b_{n -5} x^{n -1}}{12}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {n \,x^{n +5} b_{n}}{60}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {\left (n -6\right ) b_{n -6} x^{n -1}}{60}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {n \,x^{n +6} b_{n}}{360}\right ) &= \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {\left (n -7\right ) b_{n -7} x^{n -1}}{360}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{1+n} b_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (n -2\right ) b_{n -2} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n +2} b_{n} &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (n -3\right ) b_{n -3} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +3} b_{n}}{2} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {\left (n -4\right ) b_{n -4} x^{n -1}}{2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +4} b_{n}}{6} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {\left (n -5\right ) b_{n -5} x^{n -1}}{6} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +5} b_{n}}{24} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {\left (n -6\right ) b_{n -6} x^{n -1}}{24} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +6} b_{n}}{120} &= \moverset {\infty }{\munderset {n =7}{\sum }}\frac {\left (n -7\right ) b_{n -7} x^{n -1}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 b_{n} x^{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}4 b_{n -1} 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} \text {Expression too large to display} \end{equation} For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=N\), where \(N=1\) which is the difference between the two roots, we are free to choose \(b_{1} = 0\). Hence for \(n=1\), Eq (2B) gives \[ 4 C +4 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-1 \] For \(n=2\), Eq (2B) gives \[ \left (-6 a_{0}+12 a_{1}\right ) C +2 b_{1}+8 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 9+8 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=-{\frac {9}{8}} \] For \(n=3\), Eq (2B) gives \[ \left (-14 a_{1}+20 a_{2}\right ) C -b_{1}-8 b_{2}+24 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {19}{3}+24 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=-{\frac {19}{72}} \] For \(n=4\), Eq (2B) gives \[ \left (2 a_{1}-22 a_{2}+28 a_{3}\right ) C -26 b_{3}+48 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {1019}{144}+48 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-{\frac {1019}{6912}} \] For \(n=5\), Eq (2B) gives \[ \frac {\left (a_{0}+24 a_{2}-180 a_{3}+216 a_{4}\right ) C}{6}+\frac {b_{1}}{6}+3 b_{3}-52 b_{4}+80 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {5827}{864}+80 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=-{\frac {5827}{69120}} \] 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 \(C=-1\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= \left (-1\right )\eslowast \left (x \left (1-\frac {x}{4}-\frac {x^{2}}{24}-\frac {13 x^{3}}{576}-\frac {35 x^{4}}{2304}-\frac {1297 x^{5}}{138240}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-\frac {9 x^{2}}{8}-\frac {19 x^{3}}{72}-\frac {1019 x^{4}}{6912}-\frac {5827 x^{5}}{69120}+O\left (x^{6}\right ) \] 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 \left (1-\frac {x}{4}-\frac {x^{2}}{24}-\frac {13 x^{3}}{576}-\frac {35 x^{4}}{2304}-\frac {1297 x^{5}}{138240}+O\left (x^{6}\right )\right ) + c_{2} \left (\left (-1\right )\eslowast \left (x \left (1-\frac {x}{4}-\frac {x^{2}}{24}-\frac {13 x^{3}}{576}-\frac {35 x^{4}}{2304}-\frac {1297 x^{5}}{138240}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-\frac {9 x^{2}}{8}-\frac {19 x^{3}}{72}-\frac {1019 x^{4}}{6912}-\frac {5827 x^{5}}{69120}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1-\frac {x}{4}-\frac {x^{2}}{24}-\frac {13 x^{3}}{576}-\frac {35 x^{4}}{2304}-\frac {1297 x^{5}}{138240}+O\left (x^{6}\right )\right )+c_{2} \left (-x \left (1-\frac {x}{4}-\frac {x^{2}}{24}-\frac {13 x^{3}}{576}-\frac {35 x^{4}}{2304}-\frac {1297 x^{5}}{138240}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {9 x^{2}}{8}-\frac {19 x^{3}}{72}-\frac {1019 x^{4}}{6912}-\frac {5827 x^{5}}{69120}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1-\frac {x}{4}-\frac {x^{2}}{24}-\frac {13 x^{3}}{576}-\frac {35 x^{4}}{2304}-\frac {1297 x^{5}}{138240}+O\left (x^{6}\right )\right )+c_{2} \left (-x \left (1-\frac {x}{4}-\frac {x^{2}}{24}-\frac {13 x^{3}}{576}-\frac {35 x^{4}}{2304}-\frac {1297 x^{5}}{138240}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {9 x^{2}}{8}-\frac {19 x^{3}}{72}-\frac {1019 x^{4}}{6912}-\frac {5827 x^{5}}{69120}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-\frac {x}{4}-\frac {x^{2}}{24}-\frac {13 x^{3}}{576}-\frac {35 x^{4}}{2304}-\frac {1297 x^{5}}{138240}+O\left (x^{6}\right )\right )+c_{2} \left (-x \left (1-\frac {x}{4}-\frac {x^{2}}{24}-\frac {13 x^{3}}{576}-\frac {35 x^{4}}{2304}-\frac {1297 x^{5}}{138240}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {9 x^{2}}{8}-\frac {19 x^{3}}{72}-\frac {1019 x^{4}}{6912}-\frac {5827 x^{5}}{69120}+O\left (x^{6}\right )\right ) \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying differential order: 2; exact nonlinear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying to convert to an ODE of Bessel type -> trying reduction of order to Riccati trying Riccati sub-methods: trying Riccati_symmetries -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3`[0, y]
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 60
Order:=6; dsolve((x-2)^2*diff(y(x),x$2)+(x-2)*exp(x)*diff(y(x),x)+4/x*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x \left (1-\frac {1}{4} x -\frac {1}{24} x^{2}-\frac {13}{576} x^{3}-\frac {35}{2304} x^{4}-\frac {1297}{138240} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (-x +\frac {1}{4} x^{2}+\frac {1}{24} x^{3}+\frac {13}{576} x^{4}+\frac {35}{2304} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (1+\frac {1}{2} x -\frac {5}{4} x^{2}-\frac {41}{144} x^{3}-\frac {1097}{6912} x^{4}-\frac {397}{4320} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.063 (sec). Leaf size: 87
AsymptoticDSolveValue[(x-2)^2*y''[x]+(x-2)*Exp[x]*y'[x]+4/x*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {1}{576} x \left (13 x^3+24 x^2+144 x-576\right ) \log (x)+\frac {-1097 x^4-1968 x^3-8640 x^2+3456 x+6912}{6912}\right )+c_2 \left (-\frac {35 x^5}{2304}-\frac {13 x^4}{576}-\frac {x^3}{24}-\frac {x^2}{4}+x\right ) \]