Internal problem ID [5523]
Internal file name [OUTPUT/4771_Sunday_June_05_2022_03_05_27_PM_18805796/index.tex]
Book: Notes on Diffy Qs. Differential Equations for Engineers. By by Jiri Lebl, 2013.
Section: Chapter 7. POWER SERIES METHODS. 7.3.2 The method of Frobenius. Exercises.
page 300
Problem number: 7.3.8 (d).
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]]
\[ \boxed {x y^{\prime \prime }+x y^{\prime }-{\mathrm e}^{x} y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ x y^{\prime \prime }+x y^{\prime }-{\mathrm e}^{x} y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= 1\\ q(x) &= -\frac {{\mathrm e}^{x}}{x}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty ]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x y^{\prime \prime }+x y^{\prime }-{\mathrm e}^{x} y = 0 \] 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 (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-{\mathrm e}^{x} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Expanding \(-{\mathrm e}^{x}\) as Taylor series around \(x=0\) and keeping only the first \(6\) 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} + \dots \\ &= -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} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +2} a_{n}}{2}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +3} a_{n}}{6}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +4} a_{n}}{24}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +5} a_{n}}{120}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n}}{720}\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} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +2} a_{n}}{2}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {a_{n -3} x^{n +r -1}}{2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +3} a_{n}}{6}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {a_{n -4} x^{n +r -1}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +4} a_{n}}{24}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {a_{n -5} x^{n +r -1}}{24}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +5} a_{n}}{120}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} x^{n +r -1}}{120}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n}}{720}\right ) &= \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {a_{n -7} x^{n +r -1}}{720}\right ) \\ \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 =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {a_{n -3} x^{n +r -1}}{2}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {a_{n -4} x^{n +r -1}}{6}\right )+\moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {a_{n -5} x^{n +r -1}}{24}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} x^{n +r -1}}{120}\right )+\moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {a_{n -7} x^{n +r -1}}{720}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right ) = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Or \[ x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{-1+r} r \left (-1+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r \left (-1+r \right ) = 0 \] 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 \[ x^{-1+r} r \left (-1+r \right ) = 0 \] 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 \[ a_{1} = \frac {1-r}{r \left (1+r \right )} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {2 r}{\left (1+r \right )^{2} \left (2+r \right )} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {\left (r -2\right )^{2}}{2 r \left (2+r \right )^{2} \left (3+r \right )} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {r^{4}-r^{3}+22 r^{2}+3 r +3}{6 \left (1+r \right )^{2} r \left (3+r \right )^{2} \left (4+r \right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = \frac {r^{6}+2 r^{5}+41 r^{4}+32 r^{3}+32 r^{2}+128 r +112}{24 \left (2+r \right )^{2} r \left (4+r \right )^{2} \left (1+r \right )^{2} \left (5+r \right )} \] Substituting \(n = 6\) in Eq. (2B) gives \[ a_{6} = \frac {r^{8}+7 r^{7}+113 r^{6}+554 r^{5}+1943 r^{4}+4807 r^{3}+5931 r^{2}+3240 r +1740}{120 \left (3+r \right )^{2} r \left (5+r \right )^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (6+r \right )} \] For \(7\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )-a_{n -1}-a_{n -2}-\frac {a_{n -3}}{2}-\frac {a_{n -4}}{6}-\frac {a_{n -5}}{24}-\frac {a_{n -6}}{120}-\frac {a_{n -7}}{720} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {720 n a_{n -1}+720 r a_{n -1}-a_{n -7}-6 a_{n -6}-30 a_{n -5}-120 a_{n -4}-360 a_{n -3}-720 a_{n -2}-1440 a_{n -1}}{720 \left (n +r \right ) \left (n +r -1\right )}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {-720 n a_{n -1}+a_{n -7}+6 a_{n -6}+30 a_{n -5}+120 a_{n -4}+360 a_{n -3}+720 a_{n -2}+720 a_{n -1}}{720 \left (1+n \right ) n}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1-r}{r \left (1+r \right )}\) | \(0\) |
| \(a_{2}\) | \(\frac {2 r}{\left (1+r \right )^{2} \left (2+r \right )}\) | \(\frac {1}{6}\) |
| \(a_{3}\) | \(\frac {\left (r -2\right )^{2}}{2 r \left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {1}{72}\) |
| \(a_{4}\) | \(\frac {r^{4}-r^{3}+22 r^{2}+3 r +3}{6 \left (1+r \right )^{2} r \left (3+r \right )^{2} \left (4+r \right )}\) | \(\frac {7}{480}\) |
| \(a_{5}\) | \(\frac {r^{6}+2 r^{5}+41 r^{4}+32 r^{3}+32 r^{2}+128 r +112}{24 \left (2+r \right )^{2} r \left (4+r \right )^{2} \left (1+r \right )^{2} \left (5+r \right )}\) | \(\frac {29}{10800}\) |
| \(a_{6}\) | \(\frac {r^{8}+7 r^{7}+113 r^{6}+554 r^{5}+1943 r^{4}+4807 r^{3}+5931 r^{2}+3240 r +1740}{120 \left (3+r \right )^{2} r \left (5+r \right )^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (6+r \right )}\) | \(\frac {191}{181440}\) |
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^{2}}{6}+\frac {x^{3}}{72}+\frac {7 x^{4}}{480}+\frac {29 x^{5}}{10800}+\frac {191 x^{6}}{181440}+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 {1-r}{r \left (1+r \right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {1-r}{r \left (1+r \right )}&= \lim _{r\rightarrow 0}\frac {1-r}{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 \(x y^{\prime \prime }+x y^{\prime }-{\mathrm e}^{x} y = 0\) gives \[ x \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+x \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )-{\mathrm e}^{x} \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (y_{1}^{\prime \prime }\left (x \right ) x -{\mathrm e}^{x} y_{1}\left (x \right )+y_{1}^{\prime }\left (x \right ) x \right ) \ln \left (x \right )+x \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+y_{1}\left (x \right )\right ) C -{\mathrm e}^{x} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )+x \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 (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) = 0 \end{equation} But since \(y_{1}\left (x \right )\) is a solution to the ode, then \[ y_{1}^{\prime \prime }\left (x \right ) x -{\mathrm e}^{x} y_{1}\left (x \right )+y_{1}^{\prime }\left (x \right ) x = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (x \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+y_{1}\left (x \right )\right ) C -{\mathrm e}^{x} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )+x \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 (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\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 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right ) x +\left (x -1\right ) \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}-{\mathrm e}^{x} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x +\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) x^{2}}{x} = 0 \end{equation} Since \(r_{1} = 1\) and \(r_{2} = 0\) then the above becomes \begin{equation} \tag{10} \frac {\left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} a_{n} \left (1+n \right )\right ) x +\left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{1+n}\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}-{\mathrm e}^{x} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) x +\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} n \right ) x^{2}}{x} = 0 \end{equation} Expanding \(-{\mathrm e}^{x}\) as Taylor series around \(x=0\) and keeping only the first \(6\) 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} + \dots \\ &= -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} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n} a_{n} \left (1+n \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}C \,x^{1+n} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-C a_{n} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-b_{n} x^{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n} b_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +2} b_{n}}{2}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +3} b_{n}}{6}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +4} b_{n}}{24}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +5} b_{n}}{120}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +6} b_{n}}{720}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} b_{n} 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} a_{n} \left (1+n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} n \,x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}C \,x^{1+n} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}C a_{n -2} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-C a_{n} x^{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-b_{n} x^{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n} b_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-b_{n -2} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +2} b_{n}}{2}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {b_{n -3} x^{n -1}}{2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +3} b_{n}}{6}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {b_{n -4} x^{n -1}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +4} b_{n}}{24}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {b_{n -5} x^{n -1}}{24}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +5} b_{n}}{120}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {b_{n -6} x^{n -1}}{120}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +6} b_{n}}{720}\right ) &= \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {b_{n -7} x^{n -1}}{720}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n} b_{n} n &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) 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} \left (\moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} n \,x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}C a_{n -2} x^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} x^{n -1}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-b_{n -2} x^{n -1}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {b_{n -3} x^{n -1}}{2}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {b_{n -4} x^{n -1}}{6}\right )+\moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {b_{n -5} x^{n -1}}{24}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {b_{n -6} x^{n -1}}{120}\right )+\moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {b_{n -7} x^{n -1}}{720}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} 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 \(n=N\), 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 \[ C -1 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=1 \] For \(n=2\), Eq (2B) gives \[ \left (a_{0}+3 a_{1}\right ) C -b_{0}+2 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 2 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=0 \] For \(n=3\), Eq (2B) gives \[ \left (a_{1}+5 a_{2}\right ) C -\frac {b_{0}}{2}-b_{1}+b_{2}+6 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {1}{3}+6 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=-{\frac {1}{18}} \] For \(n=4\), Eq (2B) gives \[ \left (a_{2}+7 a_{3}\right ) C -\frac {b_{0}}{6}-\frac {b_{1}}{2}-b_{2}+2 b_{3}+12 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {1}{72}+12 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}={\frac {1}{864}} \] For \(n=5\), Eq (2B) gives \[ \left (a_{3}+9 a_{4}\right ) C -\frac {b_{0}}{24}-\frac {b_{1}}{6}-\frac {b_{2}}{2}-b_{3}+3 b_{4}+20 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {13}{80}+20 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=-{\frac {13}{1600}} \] 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 \[ y_{2}\left (x \right )= 1\eslowast \left (x \left (1+\frac {x^{2}}{6}+\frac {x^{3}}{72}+\frac {7 x^{4}}{480}+\frac {29 x^{5}}{10800}+\frac {191 x^{6}}{181440}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-\frac {x^{3}}{18}+\frac {x^{4}}{864}-\frac {13 x^{5}}{1600}+O\left (x^{6}\right ) \] 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^{2}}{6}+\frac {x^{3}}{72}+\frac {7 x^{4}}{480}+\frac {29 x^{5}}{10800}+\frac {191 x^{6}}{181440}+O\left (x^{6}\right )\right ) + c_{2} \left (1\eslowast \left (x \left (1+\frac {x^{2}}{6}+\frac {x^{3}}{72}+\frac {7 x^{4}}{480}+\frac {29 x^{5}}{10800}+\frac {191 x^{6}}{181440}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-\frac {x^{3}}{18}+\frac {x^{4}}{864}-\frac {13 x^{5}}{1600}+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^{2}}{6}+\frac {x^{3}}{72}+\frac {7 x^{4}}{480}+\frac {29 x^{5}}{10800}+\frac {191 x^{6}}{181440}+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1+\frac {x^{2}}{6}+\frac {x^{3}}{72}+\frac {7 x^{4}}{480}+\frac {29 x^{5}}{10800}+\frac {191 x^{6}}{181440}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {x^{3}}{18}+\frac {x^{4}}{864}-\frac {13 x^{5}}{1600}+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^{2}}{6}+\frac {x^{3}}{72}+\frac {7 x^{4}}{480}+\frac {29 x^{5}}{10800}+\frac {191 x^{6}}{181440}+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1+\frac {x^{2}}{6}+\frac {x^{3}}{72}+\frac {7 x^{4}}{480}+\frac {29 x^{5}}{10800}+\frac {191 x^{6}}{181440}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {x^{3}}{18}+\frac {x^{4}}{864}-\frac {13 x^{5}}{1600}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1+\frac {x^{2}}{6}+\frac {x^{3}}{72}+\frac {7 x^{4}}{480}+\frac {29 x^{5}}{10800}+\frac {191 x^{6}}{181440}+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1+\frac {x^{2}}{6}+\frac {x^{3}}{72}+\frac {7 x^{4}}{480}+\frac {29 x^{5}}{10800}+\frac {191 x^{6}}{181440}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {x^{3}}{18}+\frac {x^{4}}{864}-\frac {13 x^{5}}{1600}+O\left (x^{6}\right )\right ) \] 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 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: 0.016 (sec). Leaf size: 54
Order:=6; dsolve(x*diff(y(x),x$2)+x*diff(y(x),x)-exp(x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x \left (1+\frac {1}{6} x^{2}+\frac {1}{72} x^{3}+\frac {7}{480} x^{4}+\frac {29}{10800} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (x +\frac {1}{6} x^{3}+\frac {1}{72} x^{4}+\frac {7}{480} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (1-x -\frac {2}{9} x^{3}-\frac {11}{864} x^{4}-\frac {109}{4800} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.031 (sec). Leaf size: 70
AsymptoticDSolveValue[x*y''[x]+x*y'[x]-Exp[x]*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {7 x^5}{480}+\frac {x^4}{72}+\frac {x^3}{6}+x\right )+c_1 \left (\frac {1}{864} \left (-23 x^4-336 x^3-1728 x+864\right )+\frac {1}{72} x \left (x^3+12 x^2+72\right ) \log (x)\right ) \]