Internal problem ID [6049]
Internal file name [OUTPUT/5297_Sunday_June_05_2022_03_31_14_PM_49103670/index.tex]
Book: An introduction to Ordinary Differential Equations. Earl A. Coddington. Dover. NY
1961
Section: Chapter 4. Linear equations with Regular Singular Points. Page 154
Problem number: 2(d).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Complex roots"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} y^{\prime \prime }+\left (-3 x^{2}+x \right ) y^{\prime }+{\mathrm e}^{x} y=0} \] 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. \[ x^{2} y^{\prime \prime }+\left (-3 x^{2}+x \right ) y^{\prime }+{\mathrm e}^{x} y = 0 \] 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) &= -\frac {3 x -1}{x}\\ q(x) &= \frac {{\mathrm e}^{x}}{x^{2}}\\ \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 \[ x^{2} y^{\prime \prime }+\left (-3 x^{2}+x \right ) y^{\prime }+{\mathrm e}^{x} y = 0 \] 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^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (-3 x^{2}+x \right ) \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 \(8\) 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 \\ &= 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} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +2} a_{n}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +3} a_{n}}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n}}{24}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n}}{120}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +6} a_{n}}{720}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +7} a_{n}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +8} a_{n}}{40320}\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 }}\left (-3 x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 a_{n -1} \left (n +r -1\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +2} a_{n}}{2} &= \moverset {\infty }{\munderset {n =2}{\sum }}\frac {a_{n -2} x^{n +r}}{2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +3} a_{n}}{6} &= \moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} x^{n +r}}{6} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n}}{24} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} x^{n +r}}{24} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n}}{120} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} x^{n +r}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +6} a_{n}}{720} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {a_{n -6} x^{n +r}}{720} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +7} a_{n}}{5040} &= \moverset {\infty }{\munderset {n =7}{\sum }}\frac {a_{n -7} x^{n +r}}{5040} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +8} a_{n}}{40320} &= \moverset {\infty }{\munderset {n =8}{\sum }}\frac {a_{n -8} x^{n +r}}{40320} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\frac {a_{n -2} x^{n +r}}{2}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} x^{n +r}}{6}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} x^{n +r}}{24}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} x^{n +r}}{120}\right )+\left (\moverset {\infty }{\munderset {n =6}{\sum }}\frac {a_{n -6} x^{n +r}}{720}\right )+\left (\moverset {\infty }{\munderset {n =7}{\sum }}\frac {a_{n -7} x^{n +r}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =8}{\sum }}\frac {a_{n -8} x^{n +r}}{40320}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}+1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}+1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= i\\ r_2 &= -i \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}+1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([i, -i]\).
Since the roots are complex conjugates, then two linearly independent solutions can be constructed using \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 ) &= 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 ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +i}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -i} \end {align*}
\(y_{1}\left (x \right )\) is found first. Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {3 r -1}{r^{2}+2 r +2} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {17 r^{2}+4 r -6}{2 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {-r^{4}+138 r^{3}+243 r^{2}-45 r -85}{6 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {-r^{6}-36 r^{5}+1345 r^{4}+6100 r^{3}+5156 r^{2}-3044 r -2260}{24 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = \frac {-r^{8}-50 r^{7}-1194 r^{6}+11665 r^{5}+128676 r^{4}+315715 r^{3}+145319 r^{2}-192430 r -97100}{120 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right )} \] Substituting \(n = 6\) in Eq. (2B) gives \[ a_{6} = \frac {-r^{10}-66 r^{9}-2102 r^{8}-42690 r^{7}-76157 r^{6}+2029806 r^{5}+11717702 r^{4}+20421330 r^{3}+3645968 r^{2}-15274320 r -6037400}{720 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right ) \left (r^{2}+12 r +37\right )} \] Substituting \(n = 7\) in Eq. (2B) gives \[ a_{7} = \frac {-r^{12}-84 r^{11}-3401 r^{10}-89145 r^{9}-1687260 r^{8}-12804498 r^{7}-15286821 r^{6}+243174141 r^{5}+1126587431 r^{4}+1540608062 r^{3}-207286888 r^{2}-1516188226 r -493562010}{5040 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right ) \left (r^{2}+12 r +37\right ) \left (r^{2}+14 r +50\right )} \] For \(8\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )-3 a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )+a_{n}+a_{n -1}+\frac {a_{n -2}}{2}+\frac {a_{n -3}}{6}+\frac {a_{n -4}}{24}+\frac {a_{n -5}}{120}+\frac {a_{n -6}}{720}+\frac {a_{n -7}}{5040}+\frac {a_{n -8}}{40320} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {120960 n a_{n -1}+120960 r a_{n -1}-a_{n -8}-8 a_{n -7}-56 a_{n -6}-336 a_{n -5}-1680 a_{n -4}-6720 a_{n -3}-20160 a_{n -2}-161280 a_{n -1}}{40320 n^{2}+80640 n r +40320 r^{2}+40320}\tag {4} \] Which for the root \(r = i\) becomes \[ a_{n} = -\frac {40320 \left (4-3 i-3 n \right ) a_{n -1}+a_{n -8}+8 a_{n -7}+56 a_{n -6}+336 a_{n -5}+1680 a_{n -4}+6720 a_{n -3}+20160 a_{n -2}}{40320 n \left (2 i+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = i\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3 r -1}{r^{2}+2 r +2}\) | \(1+i\) |
| \(a_{2}\) | \(\frac {17 r^{2}+4 r -6}{2 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}\) | \(\frac {7}{16}+\frac {13 i}{16}\) |
| \(a_{3}\) | \(\frac {-r^{4}+138 r^{3}+243 r^{2}-45 r -85}{6 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}\) | \(\frac {7}{39}+\frac {395 i}{936}\) |
| \(a_{4}\) | \(\frac {-r^{6}-36 r^{5}+1345 r^{4}+6100 r^{3}+5156 r^{2}-3044 r -2260}{24 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )}\) | \(\frac {2117}{29952}+\frac {5197 i}{29952}\) |
| \(a_{5}\) | \(\frac {-r^{8}-50 r^{7}-1194 r^{6}+11665 r^{5}+128676 r^{4}+315715 r^{3}+145319 r^{2}-192430 r -97100}{120 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right )}\) | \(\frac {5521}{217152}+\frac {642043 i}{10857600}\) |
| \(a_{6}\) | \(\frac {-r^{10}-66 r^{9}-2102 r^{8}-42690 r^{7}-76157 r^{6}+2029806 r^{5}+11717702 r^{4}+20421330 r^{3}+3645968 r^{2}-15274320 r -6037400}{720 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right ) \left (r^{2}+12 r +37\right )}\) | \(\frac {782461}{97718400}+\frac {8813057 i}{521164800}\) |
| \(a_{7}\) | \(\frac {-r^{12}-84 r^{11}-3401 r^{10}-89145 r^{9}-1687260 r^{8}-12804498 r^{7}-15286821 r^{6}+243174141 r^{5}+1126587431 r^{4}+1540608062 r^{3}-207286888 r^{2}-1516188226 r -493562010}{5040 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right ) \left (r^{2}+12 r +37\right ) \left (r^{2}+14 r +50\right )}\) | \(\frac {1238071931}{580056422400}+\frac {3271304833 i}{812078991360}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= x^{i} \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^{i} \left (1+\left (1+i\right ) x +\left (\frac {7}{16}+\frac {13 i}{16}\right ) x^{2}+\left (\frac {7}{39}+\frac {395 i}{936}\right ) x^{3}+\left (\frac {2117}{29952}+\frac {5197 i}{29952}\right ) x^{4}+\left (\frac {5521}{217152}+\frac {642043 i}{10857600}\right ) x^{5}+\left (\frac {782461}{97718400}+\frac {8813057 i}{521164800}\right ) x^{6}+\left (\frac {1238071931}{580056422400}+\frac {3271304833 i}{812078991360}\right ) x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*} The second solution \(y_{2}\left (x \right )\) is found by taking the complex conjugate of \(y_{1}\left (x \right )\) which gives \[ y_{2}\left (x \right )= x^{-i} \left (1+\left (1-i\right ) x +\left (\frac {7}{16}-\frac {13 i}{16}\right ) x^{2}+\left (\frac {7}{39}-\frac {395 i}{936}\right ) x^{3}+\left (\frac {2117}{29952}-\frac {5197 i}{29952}\right ) x^{4}+\left (\frac {5521}{217152}-\frac {642043 i}{10857600}\right ) x^{5}+\left (\frac {782461}{97718400}-\frac {8813057 i}{521164800}\right ) x^{6}+\left (\frac {1238071931}{580056422400}-\frac {3271304833 i}{812078991360}\right ) x^{7}+O\left (x^{8}\right )\right ) \] Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x^{i} \left (1+\left (1+i\right ) x +\left (\frac {7}{16}+\frac {13 i}{16}\right ) x^{2}+\left (\frac {7}{39}+\frac {395 i}{936}\right ) x^{3}+\left (\frac {2117}{29952}+\frac {5197 i}{29952}\right ) x^{4}+\left (\frac {5521}{217152}+\frac {642043 i}{10857600}\right ) x^{5}+\left (\frac {782461}{97718400}+\frac {8813057 i}{521164800}\right ) x^{6}+\left (\frac {1238071931}{580056422400}+\frac {3271304833 i}{812078991360}\right ) x^{7}+O\left (x^{8}\right )\right ) + c_{2} x^{-i} \left (1+\left (1-i\right ) x +\left (\frac {7}{16}-\frac {13 i}{16}\right ) x^{2}+\left (\frac {7}{39}-\frac {395 i}{936}\right ) x^{3}+\left (\frac {2117}{29952}-\frac {5197 i}{29952}\right ) x^{4}+\left (\frac {5521}{217152}-\frac {642043 i}{10857600}\right ) x^{5}+\left (\frac {782461}{97718400}-\frac {8813057 i}{521164800}\right ) x^{6}+\left (\frac {1238071931}{580056422400}-\frac {3271304833 i}{812078991360}\right ) x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{i} \left (1+\left (1+i\right ) x +\left (\frac {7}{16}+\frac {13 i}{16}\right ) x^{2}+\left (\frac {7}{39}+\frac {395 i}{936}\right ) x^{3}+\left (\frac {2117}{29952}+\frac {5197 i}{29952}\right ) x^{4}+\left (\frac {5521}{217152}+\frac {642043 i}{10857600}\right ) x^{5}+\left (\frac {782461}{97718400}+\frac {8813057 i}{521164800}\right ) x^{6}+\left (\frac {1238071931}{580056422400}+\frac {3271304833 i}{812078991360}\right ) x^{7}+O\left (x^{8}\right )\right )+c_{2} x^{-i} \left (1+\left (1-i\right ) x +\left (\frac {7}{16}-\frac {13 i}{16}\right ) x^{2}+\left (\frac {7}{39}-\frac {395 i}{936}\right ) x^{3}+\left (\frac {2117}{29952}-\frac {5197 i}{29952}\right ) x^{4}+\left (\frac {5521}{217152}-\frac {642043 i}{10857600}\right ) x^{5}+\left (\frac {782461}{97718400}-\frac {8813057 i}{521164800}\right ) x^{6}+\left (\frac {1238071931}{580056422400}-\frac {3271304833 i}{812078991360}\right ) x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{i} \left (1+\left (1+i\right ) x +\left (\frac {7}{16}+\frac {13 i}{16}\right ) x^{2}+\left (\frac {7}{39}+\frac {395 i}{936}\right ) x^{3}+\left (\frac {2117}{29952}+\frac {5197 i}{29952}\right ) x^{4}+\left (\frac {5521}{217152}+\frac {642043 i}{10857600}\right ) x^{5}+\left (\frac {782461}{97718400}+\frac {8813057 i}{521164800}\right ) x^{6}+\left (\frac {1238071931}{580056422400}+\frac {3271304833 i}{812078991360}\right ) x^{7}+O\left (x^{8}\right )\right )+c_{2} x^{-i} \left (1+\left (1-i\right ) x +\left (\frac {7}{16}-\frac {13 i}{16}\right ) x^{2}+\left (\frac {7}{39}-\frac {395 i}{936}\right ) x^{3}+\left (\frac {2117}{29952}-\frac {5197 i}{29952}\right ) x^{4}+\left (\frac {5521}{217152}-\frac {642043 i}{10857600}\right ) x^{5}+\left (\frac {782461}{97718400}-\frac {8813057 i}{521164800}\right ) x^{6}+\left (\frac {1238071931}{580056422400}-\frac {3271304833 i}{812078991360}\right ) x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{i} \left (1+\left (1+i\right ) x +\left (\frac {7}{16}+\frac {13 i}{16}\right ) x^{2}+\left (\frac {7}{39}+\frac {395 i}{936}\right ) x^{3}+\left (\frac {2117}{29952}+\frac {5197 i}{29952}\right ) x^{4}+\left (\frac {5521}{217152}+\frac {642043 i}{10857600}\right ) x^{5}+\left (\frac {782461}{97718400}+\frac {8813057 i}{521164800}\right ) x^{6}+\left (\frac {1238071931}{580056422400}+\frac {3271304833 i}{812078991360}\right ) x^{7}+O\left (x^{8}\right )\right )+c_{2} x^{-i} \left (1+\left (1-i\right ) x +\left (\frac {7}{16}-\frac {13 i}{16}\right ) x^{2}+\left (\frac {7}{39}-\frac {395 i}{936}\right ) x^{3}+\left (\frac {2117}{29952}-\frac {5197 i}{29952}\right ) x^{4}+\left (\frac {5521}{217152}-\frac {642043 i}{10857600}\right ) x^{5}+\left (\frac {782461}{97718400}-\frac {8813057 i}{521164800}\right ) x^{6}+\left (\frac {1238071931}{580056422400}-\frac {3271304833 i}{812078991360}\right ) x^{7}+O\left (x^{8}\right )\right ) \] 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 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: 0.047 (sec). Leaf size: 85
Order:=8; dsolve(x^2*diff(y(x),x$2)+(x-3*x^2)*diff(y(x),x)+exp(x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{-i} \left (1+\left (1-i\right ) x +\left (\frac {7}{16}-\frac {13 i}{16}\right ) x^{2}+\left (\frac {7}{39}-\frac {395 i}{936}\right ) x^{3}+\left (\frac {2117}{29952}-\frac {5197 i}{29952}\right ) x^{4}+\left (\frac {5521}{217152}-\frac {642043 i}{10857600}\right ) x^{5}+\left (\frac {782461}{97718400}-\frac {8813057 i}{521164800}\right ) x^{6}+\left (\frac {1238071931}{580056422400}-\frac {3271304833 i}{812078991360}\right ) x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} x^{i} \left (1+\left (1+i\right ) x +\left (\frac {7}{16}+\frac {13 i}{16}\right ) x^{2}+\left (\frac {7}{39}+\frac {395 i}{936}\right ) x^{3}+\left (\frac {2117}{29952}+\frac {5197 i}{29952}\right ) x^{4}+\left (\frac {5521}{217152}+\frac {642043 i}{10857600}\right ) x^{5}+\left (\frac {782461}{97718400}+\frac {8813057 i}{521164800}\right ) x^{6}+\left (\frac {1238071931}{580056422400}+\frac {3271304833 i}{812078991360}\right ) x^{7}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.043 (sec). Leaf size: 122
AsymptoticDSolveValue[x^2*y''[x]+(x-3*x^2)*y'[x]+Exp[x]*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to \left (\frac {1}{97718400}+\frac {11 i}{1563494400}\right ) c_1 x^i \left ((1302761+756800 i) x^6+(4384656+2763936 i) x^5+(12605400+8289000 i) x^4+(31161600+19814400 i) x^3+(66096000+33955200 i) x^2+(111974400+20736000 i) x+(66355200-45619200 i)\right )-\left (\frac {11}{1563494400}+\frac {i}{97718400}\right ) c_2 x^{-i} \left ((756800+1302761 i) x^6+(2763936+4384656 i) x^5+(8289000+12605400 i) x^4+(19814400+31161600 i) x^3+(33955200+66096000 i) x^2+(20736000+111974400 i) x-(45619200-66355200 i)\right ) \]