4.12 problem 20

Internal problem ID [6654]
Internal file name [OUTPUT/5902_Sunday_June_05_2022_04_00_51_PM_15720401/index.tex]

Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section: CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS. CHAPTER 6 IN REVIEW. Page 271
Problem number: 20.
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]]

\[ \boxed {\left ({\mathrm e}^{x}-1-x \right ) y^{\prime \prime }+y x=0} \] 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. \[ \left ({\mathrm e}^{x}-1-x \right ) y^{\prime \prime }+y x = 0 \] 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) &= 0\\ q(x) &= \frac {x}{{\mathrm e}^{x}-1-x}\\ \end {align*}

Table 48: Table \(p(x),q(x)\) singularites.

\(p(x)=0\)
singularity	type

\(q(x)=\frac {x}{{\mathrm e}^{x}-1-x}\)
singularity	type
\(x = -\operatorname {LambertW}\left (Z , -{\mathrm e}^{-1}\right )-1\)	\(\text {``regular''}\)

Combining everything together gives the following summary of singularities for the ode as

Regular singular points : \([-\operatorname {LambertW}\left (Z , -{\mathrm e}^{-1}\right )-1]\)

Irregular singular points : \([\infty ]\)

Since \(x = 0\) is regular singular point, then Frobenius power series is used. 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 ({\mathrm e}^{x}-1-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) x = 0 \end{equation} Expanding \({\mathrm e}^{x}-1-x\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} {\mathrm e}^{x}-1-x &= \frac {1}{2} x^{2}+\frac {1}{6} x^{3}+\frac {1}{24} x^{4}+\frac {1}{120} x^{5}+\frac {1}{720} x^{6}+\frac {1}{5040} x^{7}+\frac {1}{40320} x^{8} + \dots \\ &= \frac {1}{2} x^{2}+\frac {1}{6} x^{3}+\frac {1}{24} x^{4}+\frac {1}{120} x^{5}+\frac {1}{720} x^{6}+\frac {1}{5040} x^{7}+\frac {1}{40320} x^{8} \end {align*}

Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +6} a_{n} \left (n +r \right ) \left (n +r -1\right )}{40320}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n} \left (n +r \right ) \left (n +r -1\right )}{5040}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n} \left (n +r \right ) \left (n +r -1\right )}{720}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +3} a_{n} \left (n +r \right ) \left (n +r -1\right )}{120}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )}{24}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{2}\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 \(n +r\) 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 }}\frac {x^{n +r +6} a_{n} \left (n +r \right ) \left (n +r -1\right )}{40320} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {a_{n -6} \left (n -6+r \right ) \left (n -7+r \right ) x^{n +r}}{40320} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n} \left (n +r \right ) \left (n +r -1\right )}{5040} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} \left (-5+n +r \right ) \left (n -6+r \right ) x^{n +r}}{5040} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n} \left (n +r \right ) \left (n +r -1\right )}{720} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n +r -4\right ) \left (-5+n +r \right ) x^{n +r}}{720} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +3} a_{n} \left (n +r \right ) \left (n +r -1\right )}{120} &= \moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} \left (n +r -3\right ) \left (n +r -4\right ) x^{n +r}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )}{24} &= \moverset {\infty }{\munderset {n =2}{\sum }}\frac {a_{n -2} \left (n +r -2\right ) \left (n +r -3\right ) x^{n +r}}{24} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{6} &= \moverset {\infty }{\munderset {n =1}{\sum }}\frac {a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}}{6} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}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 =6}{\sum }}\frac {a_{n -6} \left (n -6+r \right ) \left (n -7+r \right ) x^{n +r}}{40320}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} \left (-5+n +r \right ) \left (n -6+r \right ) x^{n +r}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n +r -4\right ) \left (-5+n +r \right ) x^{n +r}}{720}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} \left (n +r -3\right ) \left (n +r -4\right ) x^{n +r}}{120}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\frac {a_{n -2} \left (n +r -2\right ) \left (n +r -3\right ) x^{n +r}}{24}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\frac {a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ \frac {x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{2} = 0 \] When \(n = 0\) the above becomes \[ \frac {x^{r} a_{0} r \left (-1+r \right )}{2} = 0 \] Or \[ \frac {x^{r} a_{0} r \left (-1+r \right )}{2} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \frac {x^{r} r \left (-1+r \right )}{2} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \frac {r \left (-1+r \right )}{2} = 0 \] 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 \[ \frac {x^{r} r \left (-1+r \right )}{2} = 0 \] 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 \[ a_{1} = \frac {-r^{2}+r -6}{3 r \left (1+r \right )} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {r^{4}+47 r^{2}+144}{36 r \left (1+r \right )^{2} \left (2+r \right )} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {r^{6}+3 r^{5}-179 r^{4}-363 r^{3}-2162 r^{2}-1980 r -5760}{540 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {-r^{8}-8 r^{7}+170 r^{6}+1132 r^{5}+12443 r^{4}+42076 r^{3}+116700 r^{2}+152352 r +250560}{6480 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = \frac {-5 r^{10}-75 r^{9}-240 r^{8}+1170 r^{7}-36405 r^{6}-390015 r^{5}-2082886 r^{4}-6427176 r^{3}-14403600 r^{2}-19843488 r -24167808}{136080 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )} \] For \(6\le n\) the recursive equation is \begin{equation} \tag{3} \frac {a_{n -6} \left (n -6+r \right ) \left (n -7+r \right )}{40320}+\frac {a_{n -5} \left (-5+n +r \right ) \left (n -6+r \right )}{5040}+\frac {a_{n -4} \left (n +r -4\right ) \left (-5+n +r \right )}{720}+\frac {a_{n -3} \left (n +r -3\right ) \left (n +r -4\right )}{120}+\frac {a_{n -2} \left (n +r -2\right ) \left (n +r -3\right )}{24}+\frac {a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )}{6}+\frac {a_{n} \left (n +r \right ) \left (n +r -1\right )}{2}+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n^{2} a_{n -6}+8 n^{2} a_{n -5}+56 n^{2} a_{n -4}+336 n^{2} a_{n -3}+1680 n^{2} a_{n -2}+6720 n^{2} a_{n -1}+2 n r a_{n -6}+16 n r a_{n -5}+112 n r a_{n -4}+672 n r a_{n -3}+3360 n r a_{n -2}+13440 n r a_{n -1}+r^{2} a_{n -6}+8 r^{2} a_{n -5}+56 r^{2} a_{n -4}+336 r^{2} a_{n -3}+1680 r^{2} a_{n -2}+6720 r^{2} a_{n -1}-13 n a_{n -6}-88 n a_{n -5}-504 n a_{n -4}-2352 n a_{n -3}-8400 n a_{n -2}-20160 n a_{n -1}-13 r a_{n -6}-88 r a_{n -5}-504 r a_{n -4}-2352 r a_{n -3}-8400 r a_{n -2}-20160 r a_{n -1}+42 a_{n -6}+240 a_{n -5}+1120 a_{n -4}+4032 a_{n -3}+10080 a_{n -2}+53760 a_{n -1}}{20160 \left (n +r \right ) \left (n +r -1\right )}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {\left (-a_{n -6}-8 a_{n -5}-56 a_{n -4}-336 a_{n -3}-1680 a_{n -2}-6720 a_{n -1}\right ) n^{2}+\left (11 a_{n -6}+72 a_{n -5}+392 a_{n -4}+1680 a_{n -3}+5040 a_{n -2}+6720 a_{n -1}\right ) n -30 a_{n -6}-160 a_{n -5}-672 a_{n -4}-2016 a_{n -3}-3360 a_{n -2}-40320 a_{n -1}}{20160 n \left (1+n \right )}\tag {5} \] 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.

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {-r^{2}+r -6}{3 r \left (1+r \right )}\)	\(-1\)
\(a_{2}\)	\(\frac {r^{4}+47 r^{2}+144}{36 r \left (1+r \right )^{2} \left (2+r \right )}\)	\(\frac {4}{9}\)
\(a_{3}\)	\(\frac {r^{6}+3 r^{5}-179 r^{4}-363 r^{3}-2162 r^{2}-1980 r -5760}{540 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\)	\(-{\frac {29}{216}}\)
\(a_{4}\)	\(\frac {-r^{8}-8 r^{7}+170 r^{6}+1132 r^{5}+12443 r^{4}+42076 r^{3}+116700 r^{2}+152352 r +250560}{6480 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\)	\(\frac {37}{1200}\)
\(a_{5}\)	\(\frac {-5 r^{10}-75 r^{9}-240 r^{8}+1170 r^{7}-36405 r^{6}-390015 r^{5}-2082886 r^{4}-6427176 r^{3}-14403600 r^{2}-19843488 r -24167808}{136080 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )}\)	\(-{\frac {58}{10125}}\)

For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {-r^{12}-24 r^{11}-1325 r^{10}-22980 r^{9}-52863 r^{8}+1292688 r^{7}[phone] r^{6}[phone] r^{5}[phone] r^{4}[phone] r^{3}[phone] r^{2}[phone] r [phone]}{4082400 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{6}={\frac {14209}{15876000}} \] And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {-r^{2}+r -6}{3 r \left (1+r \right )}\)	\(-1\)
\(a_{2}\)	\(\frac {r^{4}+47 r^{2}+144}{36 r \left (1+r \right )^{2} \left (2+r \right )}\)	\(\frac {4}{9}\)
\(a_{3}\)	\(\frac {r^{6}+3 r^{5}-179 r^{4}-363 r^{3}-2162 r^{2}-1980 r -5760}{540 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\)	\(-{\frac {29}{216}}\)
\(a_{4}\)	\(\frac {-r^{8}-8 r^{7}+170 r^{6}+1132 r^{5}+12443 r^{4}+42076 r^{3}+116700 r^{2}+152352 r +250560}{6480 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\)	\(\frac {37}{1200}\)
\(a_{5}\)	\(\frac {-5 r^{10}-75 r^{9}-240 r^{8}+1170 r^{7}-36405 r^{6}-390015 r^{5}-2082886 r^{4}-6427176 r^{3}-14403600 r^{2}-19843488 r -24167808}{136080 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )}\)	\(-{\frac {58}{10125}}\)
\(a_{6}\)	\(\frac {-r^{12}-24 r^{11}-1325 r^{10}-22980 r^{9}-52863 r^{8}+1292688 r^{7}[phone] r^{6}[phone] r^{5}[phone] r^{4}[phone] r^{3}[phone] r^{2}[phone] r [phone]}{4082400 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )}\)	\(\frac {14209}{15876000}\)

For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {149 r^{14}+5215 r^{13}+79387 r^{12}+686735 r^{11}+3713973 r^{10}[phone] r^{9}[phone] r^{8}-104697475 r^{7}-1414855462 r^{6}-8172285680 r^{5}-30218766312 r^{4}-77117343120 r^{3}-142070328000 r^{2}-182743776000 r -159104424960}{24494400 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (7+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{7}=-{\frac {107329}{889056000}} \] And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {-r^{2}+r -6}{3 r \left (1+r \right )}\)	\(-1\)
\(a_{2}\)	\(\frac {r^{4}+47 r^{2}+144}{36 r \left (1+r \right )^{2} \left (2+r \right )}\)	\(\frac {4}{9}\)
\(a_{3}\)	\(\frac {r^{6}+3 r^{5}-179 r^{4}-363 r^{3}-2162 r^{2}-1980 r -5760}{540 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\)	\(-{\frac {29}{216}}\)
\(a_{4}\)	\(\frac {-r^{8}-8 r^{7}+170 r^{6}+1132 r^{5}+12443 r^{4}+42076 r^{3}+116700 r^{2}+152352 r +250560}{6480 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\)	\(\frac {37}{1200}\)
\(a_{5}\)	\(\frac {-5 r^{10}-75 r^{9}-240 r^{8}+1170 r^{7}-36405 r^{6}-390015 r^{5}-2082886 r^{4}-6427176 r^{3}-14403600 r^{2}-19843488 r -24167808}{136080 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )}\)	\(-{\frac {58}{10125}}\)
\(a_{6}\)	\(\frac {-r^{12}-24 r^{11}-1325 r^{10}-22980 r^{9}-52863 r^{8}+1292688 r^{7}[phone] r^{6}[phone] r^{5}[phone] r^{4}[phone] r^{3}[phone] r^{2}[phone] r [phone]}{4082400 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )}\)	\(\frac {14209}{15876000}\)
\(a_{7}\)	\(\frac {149 r^{14}+5215 r^{13}+79387 r^{12}+686735 r^{11}+3713973 r^{10}[phone] r^{9}[phone] r^{8}-104697475 r^{7}-1414855462 r^{6}-8172285680 r^{5}-30218766312 r^{4}-77117343120 r^{3}-142070328000 r^{2}-182743776000 r -159104424960}{24494400 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (7+r \right )}\)	\(-{\frac {107329}{889056000}}\)

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}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= x \left (1-x +\frac {4 x^{2}}{9}-\frac {29 x^{3}}{216}+\frac {37 x^{4}}{1200}-\frac {58 x^{5}}{10125}+\frac {14209 x^{6}}{15876000}-\frac {107329 x^{7}}{889056000}+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 \(N=1\). 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 {-r^{2}+r -6}{3 r \left (1+r \right )} \end {align*}

Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {-r^{2}+r -6}{3 r \left (1+r \right )}&= \lim _{r\rightarrow 0}\frac {-r^{2}+r -6}{3 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 \(\left ({\mathrm e}^{x}-1-x \right ) y^{\prime \prime }+y x = 0\) gives \[ \left ({\mathrm e}^{x}-1-x \right ) \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) x = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (\left ({\mathrm e}^{x}-1-x \right ) y_{1}^{\prime \prime }\left (x \right )+y_{1}\left (x \right ) x \right ) \ln \left (x \right )+\left ({\mathrm e}^{x}-1-x \right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )\right ) C +\left ({\mathrm e}^{x}-1-x \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 )+\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 \[ \left ({\mathrm e}^{x}-1-x \right ) y_{1}^{\prime \prime }\left (x \right )+y_{1}\left (x \right ) x = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left ({\mathrm e}^{x}-1-x \right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) C +\left ({\mathrm e}^{x}-1-x \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 )+\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 (-{\mathrm e}^{x}+1+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )+\left (-{\mathrm e}^{x}+1+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C}{x^{2}}+\frac {-x^{2} \left (-{\mathrm e}^{x}+1+x \right ) \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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x^{3}}{x^{2}} = 0 \end{equation} Since \(r_{1} = 1\) and \(r_{2} = 0\) then the above becomes \begin{equation} \tag{10} \frac {\left (-2 x \left (-{\mathrm e}^{x}+1+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} a_{n} \left (1+n \right )\right )+\left (-{\mathrm e}^{x}+1+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{1+n}\right )\right ) C}{x^{2}}+\frac {-x^{2} \left (-{\mathrm e}^{x}+1+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) x^{3}}{x^{2}} = 0 \end{equation} Expanding \(\frac {2 C \,{\mathrm e}^{x}}{x}\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} \frac {2 C \,{\mathrm e}^{x}}{x} &= \frac {2 C}{x}+2 C +C x +\frac {C \,x^{2}}{3}+\frac {C \,x^{3}}{12}+\frac {C \,x^{4}}{60}+\frac {C \,x^{5}}{360}+\frac {C \,x^{6}}{2520}+\frac {C \,x^{7}}{20160}+\frac {C \,x^{8}}{181440} + \dots \\ &= \frac {2 C}{x}+2 C +C x +\frac {C \,x^{2}}{3}+\frac {C \,x^{3}}{12}+\frac {C \,x^{4}}{60}+\frac {C \,x^{5}}{360}+\frac {C \,x^{6}}{2520}+\frac {C \,x^{7}}{20160}+\frac {C \,x^{8}}{181440} \end {align*}

Expanding \(-\frac {C \,{\mathrm e}^{x}}{x}\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} -\frac {C \,{\mathrm e}^{x}}{x} &= -\frac {C}{x}-C -\frac {C x}{2}-\frac {C \,x^{2}}{6}-\frac {C \,x^{3}}{24}-\frac {C \,x^{4}}{120}-\frac {C \,x^{5}}{720}-\frac {C \,x^{6}}{5040}-\frac {C \,x^{7}}{40320}-\frac {C \,x^{8}}{362880} + \dots \\ &= -\frac {C}{x}-C -\frac {C x}{2}-\frac {C \,x^{2}}{6}-\frac {C \,x^{3}}{24}-\frac {C \,x^{4}}{120}-\frac {C \,x^{5}}{720}-\frac {C \,x^{6}}{5040}-\frac {C \,x^{7}}{40320}-\frac {C \,x^{8}}{362880} \end {align*}

Expanding \(\frac {{\mathrm e}^{x}}{x^{2}}\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} \frac {{\mathrm e}^{x}}{x^{2}} &= \frac {1}{x^{2}}+\frac {1}{x}+\frac {1}{2}+\frac {x}{6}+\frac {x^{2}}{24}+\frac {x^{3}}{120}+\frac {x^{4}}{720}+\frac {x^{5}}{5040}+\frac {x^{6}}{40320}+\frac {x^{7}}{362880}+\frac {x^{8}}{3628800} + \dots \\ &= \frac {1}{x^{2}}+\frac {1}{x}+\frac {1}{2}+\frac {x}{6}+\frac {x^{2}}{24}+\frac {x^{3}}{120}+\frac {x^{4}}{720}+\frac {x^{5}}{5040}+\frac {x^{6}}{40320}+\frac {x^{7}}{362880}+\frac {x^{8}}{3628800} \end {align*}

Which simplifies to \begin{equation} \tag{2A} \text {Expression too large to display} \end{equation} The next step is to make all powers of \(x\) be \(n -2\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -2}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +8} a_{n} \left (1+n \right )}{181440} &= \moverset {\infty }{\munderset {n =10}{\sum }}\frac {C a_{n -10} \left (n -9\right ) x^{n -2}}{181440} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +7} a_{n} \left (1+n \right )}{20160} &= \moverset {\infty }{\munderset {n =9}{\sum }}\frac {C a_{n -9} \left (-8+n \right ) x^{n -2}}{20160} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +6} a_{n} \left (1+n \right )}{2520} &= \moverset {\infty }{\munderset {n =8}{\sum }}\frac {C a_{-8+n} \left (-7+n \right ) x^{n -2}}{2520} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +5} a_{n} \left (1+n \right )}{360} &= \moverset {\infty }{\munderset {n =7}{\sum }}\frac {C a_{-7+n} \left (n -6\right ) x^{n -2}}{360} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +4} a_{n} \left (1+n \right )}{60} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {C a_{n -6} \left (n -5\right ) x^{n -2}}{60} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +3} a_{n} \left (1+n \right )}{12} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {C a_{n -5} \left (n -4\right ) x^{n -2}}{12} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +2} a_{n} \left (1+n \right )}{3} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {C a_{n -4} \left (n -3\right ) x^{n -2}}{3} \\ \moverset {\infty }{\munderset {n =0}{\sum }}C \,x^{1+n} a_{n} \left (1+n \right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}C a_{n -3} \left (n -2\right ) x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n} C \left (1+n \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{n -2} \left (n -1\right ) x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n -1} a_{n} \left (1+n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} n \,x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{n -1} a_{n} \left (1+n \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 C a_{n -1} n \,x^{n -2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n} C \left (1+n \right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 C a_{n -2} \left (n -1\right ) x^{n -2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n -1} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} x^{n -2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n} C \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-C a_{n -2} x^{n -2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{1+n} a_{n}}{2}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {C a_{n -3} x^{n -2}}{2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +2} a_{n}}{6}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {C a_{n -4} x^{n -2}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +3} a_{n}}{24}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {C a_{n -5} x^{n -2}}{24}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +4} a_{n}}{120}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {C a_{n -6} x^{n -2}}{120}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +5} a_{n}}{720}\right ) &= \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {C a_{-7+n} x^{n -2}}{720}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +6} a_{n}}{5040}\right ) &= \moverset {\infty }{\munderset {n =8}{\sum }}\left (-\frac {C a_{-8+n} x^{n -2}}{5040}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +7} a_{n}}{40320}\right ) &= \moverset {\infty }{\munderset {n =9}{\sum }}\left (-\frac {C a_{n -9} x^{n -2}}{40320}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +8} a_{n}}{362880}\right ) &= \moverset {\infty }{\munderset {n =10}{\sum }}\left (-\frac {C a_{n -10} x^{n -2}}{362880}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}C \,x^{n -1} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}C a_{n -1} x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n} C &= \moverset {\infty }{\munderset {n =2}{\sum }}C a_{n -2} x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +8} b_{n} \left (n -1\right )}{3628800} &= \moverset {\infty }{\munderset {n =10}{\sum }}\frac {\left (n -10\right ) b_{n -10} \left (n -11\right ) x^{n -2}}{3628800} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +7} b_{n} \left (n -1\right )}{362880} &= \moverset {\infty }{\munderset {n =9}{\sum }}\frac {\left (n -9\right ) b_{n -9} \left (n -10\right ) x^{n -2}}{362880} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +6} b_{n} \left (n -1\right )}{40320} &= \moverset {\infty }{\munderset {n =8}{\sum }}\frac {\left (-8+n \right ) b_{-8+n} \left (n -9\right ) x^{n -2}}{40320} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +5} b_{n} \left (n -1\right )}{5040} &= \moverset {\infty }{\munderset {n =7}{\sum }}\frac {\left (-7+n \right ) b_{-7+n} \left (-8+n \right ) x^{n -2}}{5040} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +4} b_{n} \left (n -1\right )}{720} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {\left (n -6\right ) b_{n -6} \left (-7+n \right ) x^{n -2}}{720} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +3} b_{n} \left (n -1\right )}{120} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {\left (n -5\right ) b_{n -5} \left (n -6\right ) x^{n -2}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +2} b_{n} \left (n -1\right )}{24} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {\left (n -4\right ) b_{n -4} \left (n -5\right ) x^{n -2}}{24} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{1+n} b_{n} \left (n -1\right )}{6} &= \moverset {\infty }{\munderset {n =3}{\sum }}\frac {\left (n -3\right ) b_{n -3} \left (n -4\right ) x^{n -2}}{6} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n} b_{n} n \left (n -1\right )}{2} &= \moverset {\infty }{\munderset {n =2}{\sum }}\frac {\left (n -2\right ) b_{n -2} \left (n -3\right ) x^{n -2}}{2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n -1} b_{n} \left (n -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} \left (n -2\right ) x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-n \,x^{n -1} b_{n} \left (n -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-\left (n -1\right ) b_{n -1} \left (n -2\right ) x^{n -2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} b_{n} &= \moverset {\infty }{\munderset {n =3}{\sum }}b_{n -3} x^{n -2} \\ \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 \(n -2\). \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 \[ 0 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=1 \] For \(n=3\), Eq (2B) gives \[ \frac {C a_{0}}{2}+b_{0} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ {\frac {3}{2}} = 0 \] For \(n=4\), Eq (2B) gives \[ \frac {\left (a_{0}+9 a_{1}\right ) C}{6}+b_{1}+b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {4}{3}+b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}={\frac {4}{3}} \] For \(n=5\), Eq (2B) gives \[ \frac {\left (a_{0}+12 a_{1}+60 a_{2}\right ) C}{24}+\frac {4 b_{2}}{3}+3 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {175}{72}+3 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=-{\frac {175}{216}} \] For \(n=6\), Eq (2B) gives \[ \frac {\left (a_{0}+15 a_{1}+100 a_{2}+420 a_{3}\right ) C}{120}+\frac {b_{2}}{12}+2 b_{3}+6 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {3727}{2160}+6 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}={\frac {3727}{12960}} \] For \(n=7\), Eq (2B) gives \[ \frac {\left (a_{0}+18 a_{1}+150 a_{2}+840 a_{3}+3240 a_{4}\right ) C}{720}+10 b_{5}+\frac {b_{2}}{60}+\frac {b_{3}}{4}+3 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {47531}{64800}+10 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=-{\frac {47531}{648000}} \] 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 \[ y_{2}\left (x \right )= 1\eslowast \left (x \left (1-x +\frac {4 x^{2}}{9}-\frac {29 x^{3}}{216}+\frac {37 x^{4}}{1200}-\frac {58 x^{5}}{10125}+\frac {14209 x^{6}}{15876000}-\frac {107329 x^{7}}{889056000}+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+1+\frac {4 x^{2}}{3}-\frac {175 x^{3}}{216}+\frac {3727 x^{4}}{12960}-\frac {47531 x^{5}}{648000}+O\left (x^{8}\right ) \] 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-x +\frac {4 x^{2}}{9}-\frac {29 x^{3}}{216}+\frac {37 x^{4}}{1200}-\frac {58 x^{5}}{10125}+\frac {14209 x^{6}}{15876000}-\frac {107329 x^{7}}{889056000}+O\left (x^{8}\right )\right ) + c_{2} \left (1\eslowast \left (x \left (1-x +\frac {4 x^{2}}{9}-\frac {29 x^{3}}{216}+\frac {37 x^{4}}{1200}-\frac {58 x^{5}}{10125}+\frac {14209 x^{6}}{15876000}-\frac {107329 x^{7}}{889056000}+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+1+\frac {4 x^{2}}{3}-\frac {175 x^{3}}{216}+\frac {3727 x^{4}}{12960}-\frac {47531 x^{5}}{648000}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1-x +\frac {4 x^{2}}{9}-\frac {29 x^{3}}{216}+\frac {37 x^{4}}{1200}-\frac {58 x^{5}}{10125}+\frac {14209 x^{6}}{15876000}-\frac {107329 x^{7}}{889056000}+O\left (x^{8}\right )\right )+c_{2} \left (x \left (1-x +\frac {4 x^{2}}{9}-\frac {29 x^{3}}{216}+\frac {37 x^{4}}{1200}-\frac {58 x^{5}}{10125}+\frac {14209 x^{6}}{15876000}-\frac {107329 x^{7}}{889056000}+O\left (x^{8}\right )\right ) \ln \left (x \right )+1+\frac {4 x^{2}}{3}-\frac {175 x^{3}}{216}+\frac {3727 x^{4}}{12960}-\frac {47531 x^{5}}{648000}+O\left (x^{8}\right )\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1-x +\frac {4 x^{2}}{9}-\frac {29 x^{3}}{216}+\frac {37 x^{4}}{1200}-\frac {58 x^{5}}{10125}+\frac {14209 x^{6}}{15876000}-\frac {107329 x^{7}}{889056000}+O\left (x^{8}\right )\right )+c_{2} \left (x \left (1-x +\frac {4 x^{2}}{9}-\frac {29 x^{3}}{216}+\frac {37 x^{4}}{1200}-\frac {58 x^{5}}{10125}+\frac {14209 x^{6}}{15876000}-\frac {107329 x^{7}}{889056000}+O\left (x^{8}\right )\right ) \ln \left (x \right )+1+\frac {4 x^{2}}{3}-\frac {175 x^{3}}{216}+\frac {3727 x^{4}}{12960}-\frac {47531 x^{5}}{648000}+O\left (x^{8}\right )\right ) \\ \end{align*}

Verification of solutions

\[ y = c_{1} x \left (1-x +\frac {4 x^{2}}{9}-\frac {29 x^{3}}{216}+\frac {37 x^{4}}{1200}-\frac {58 x^{5}}{10125}+\frac {14209 x^{6}}{15876000}-\frac {107329 x^{7}}{889056000}+O\left (x^{8}\right )\right )+c_{2} \left (x \left (1-x +\frac {4 x^{2}}{9}-\frac {29 x^{3}}{216}+\frac {37 x^{4}}{1200}-\frac {58 x^{5}}{10125}+\frac {14209 x^{6}}{15876000}-\frac {107329 x^{7}}{889056000}+O\left (x^{8}\right )\right ) \ln \left (x \right )+1+\frac {4 x^{2}}{3}-\frac {175 x^{3}}{216}+\frac {3727 x^{4}}{12960}-\frac {47531 x^{5}}{648000}+O\left (x^{8}\right )\right ) \] 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 differential order: 2; exact nonlinear 
   trying symmetries linear in x and y(x) 
   trying to convert to a linear ODE with constant coefficients 
   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: 0.312 (sec). Leaf size: 70

Order:=8; 
dsolve((exp(x)-1-x)*diff(y(x),x$2)+x*y(x)=0,y(x),type='series',x=0);
 

\[ y \left (x \right ) = c_{1} x \left (1-x +\frac {4}{9} x^{2}-\frac {29}{216} x^{3}+\frac {37}{1200} x^{4}-\frac {58}{10125} x^{5}+\frac {14209}{15876000} x^{6}-\frac {107329}{889056000} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (\left (-2\right ) x +2 x^{2}-\frac {8}{9} x^{3}+\frac {29}{108} x^{4}-\frac {37}{600} x^{5}+\frac {116}{10125} x^{6}-\frac {14209}{7938000} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (1-\frac {8}{3} x^{2}+\frac {175}{108} x^{3}-\frac {3727}{6480} x^{4}+\frac {47531}{324000} x^{5}-\frac {3003737}{102060000} x^{6}+\frac {48833381}{10001880000} x^{7}+\operatorname {O}\left (x^{8}\right )\right )\right ) \]

✓ Solution by Mathematica

Time used: 0.366 (sec). Leaf size: 133

AsymptoticDSolveValue[(Exp[x]-1-x)*y''[x]+x*y[x]==0,y[x],{x,0,7}]
 

\[ y(x)\to c_1 \left (x^6 \left (\frac {116 \log (x)}{10125}-\frac {3003737}{102060000}\right )+x^5 \left (\frac {47531}{324000}-\frac {37 \log (x)}{600}\right )+x^4 \left (\frac {29 \log (x)}{108}-\frac {3727}{6480}\right )+x^3 \left (\frac {175}{108}-\frac {8 \log (x)}{9}\right )+x^2 \left (2 \log (x)-\frac {8}{3}\right )-2 x \log (x)+1\right )+c_2 x \left (-\frac {107329 x^7}{889056000}+\frac {14209 x^6}{15876000}-\frac {58 x^5}{10125}+\frac {37 x^4}{1200}-\frac {29 x^3}{216}+\frac {4 x^2}{9}-x+1\right ) \]