Internal problem ID [6052]
Internal file name [OUTPUT/5300_Sunday_June_05_2022_03_32_50_PM_52482633/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 159
Problem number: 2.
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
[[_Emden, _Fowler]]
\[ \boxed {x^{2} y^{\prime \prime }+y^{\prime } {\mathrm e}^{x} x +y=0} \] 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. \[ x^{2} y^{\prime \prime }+y^{\prime } {\mathrm e}^{x} x +y = 0 \] 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 {{\mathrm e}^{x}}{x}\\ q(x) &= \frac {1}{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 \[ x^{2} y^{\prime \prime }+y^{\prime } {\mathrm e}^{x} x +y = 0 \] 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 (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) {\mathrm e}^{x} x +\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Expanding \(x \,{\mathrm e}^{x}\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} x \,{\mathrm e}^{x} &= x +x^{2}+\frac {1}{2} x^{3}+\frac {1}{6} x^{4}+\frac {1}{24} x^{5}+\frac {1}{120} x^{6}+\frac {1}{720} x^{7}+\frac {1}{5040} x^{8} + \dots \\ &= x +x^{2}+\frac {1}{2} x^{3}+\frac {1}{6} x^{4}+\frac {1}{24} x^{5}+\frac {1}{120} x^{6}+\frac {1}{720} x^{7}+\frac {1}{5040} 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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +7} a_{n} \left (n +r \right )}{5040}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +6} a_{n} \left (n +r \right )}{720}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n} \left (n +r \right )}{120}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n} \left (n +r \right )}{24}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +3} a_{n} \left (n +r \right )}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +2} a_{n} \left (n +r \right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 ) = 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 +7} a_{n} \left (n +r \right )}{5040} &= \moverset {\infty }{\munderset {n =7}{\sum }}\frac {a_{n -7} \left (n -7+r \right ) x^{n +r}}{5040} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +6} a_{n} \left (n +r \right )}{720} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {a_{n -6} \left (n -6+r \right ) x^{n +r}}{720} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n} \left (n +r \right )}{120} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} \left (n -5+r \right ) x^{n +r}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n} \left (n +r \right )}{24} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n -4+r \right ) x^{n +r}}{24} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +3} a_{n} \left (n +r \right )}{6} &= \moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} \left (n -3+r \right ) x^{n +r}}{6} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +2} a_{n} \left (n +r \right )}{2} &= \moverset {\infty }{\munderset {n =2}{\sum }}\frac {a_{n -2} \left (n +r -2\right ) x^{n +r}}{2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+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} \\ \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 )+\left (\moverset {\infty }{\munderset {n =7}{\sum }}\frac {a_{n -7} \left (n -7+r \right ) x^{n +r}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =6}{\sum }}\frac {a_{n -6} \left (n -6+r \right ) x^{n +r}}{720}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} \left (n -5+r \right ) x^{n +r}}{120}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n -4+r \right ) x^{n +r}}{24}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} \left (n -3+r \right ) x^{n +r}}{6}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\frac {a_{n -2} \left (n +r -2\right ) x^{n +r}}{2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}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 ) = 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 \[ a_{1} = -\frac {r}{r^{2}+2 r +2} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = -\frac {r^{3}}{2 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = -\frac {r \left (r^{4}-6 r^{2}-9 r -5\right )}{6 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = -\frac {r \left (r^{6}-2 r^{5}-43 r^{4}-136 r^{3}-180 r^{2}-112 r -40\right )}{24 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = -\frac {r \left (r^{8}-10 r^{7}-201 r^{6}-1035 r^{5}-2331 r^{4}-2105 r^{3}+321 r^{2}+1760 r +800\right )}{120 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right )} \] Substituting \(n = 6\) in Eq. (2B) gives \[ a_{6} = -\frac {r \left (2+r \right ) \left (r^{9}-34 r^{8}-666 r^{7}-3942 r^{6}-7855 r^{5}+11152 r^{4}+77700 r^{3}+138084 r^{2}+111970 r +41300\right )}{720 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right ) \left (r^{2}+12 r +37\right )} \] For \(7\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+\frac {a_{n -7} \left (n -7+r \right )}{5040}+\frac {a_{n -6} \left (n -6+r \right )}{720}+\frac {a_{n -5} \left (n -5+r \right )}{120}+\frac {a_{n -4} \left (n -4+r \right )}{24}+\frac {a_{n -3} \left (n -3+r \right )}{6}+\frac {a_{n -2} \left (n +r -2\right )}{2}+a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n a_{n -7}+7 n a_{n -6}+42 n a_{n -5}+210 n a_{n -4}+840 n a_{n -3}+2520 n a_{n -2}+5040 n a_{n -1}+r a_{n -7}+7 r a_{n -6}+42 r a_{n -5}+210 r a_{n -4}+840 r a_{n -3}+2520 r a_{n -2}+5040 r a_{n -1}-7 a_{n -7}-42 a_{n -6}-210 a_{n -5}-840 a_{n -4}-2520 a_{n -3}-5040 a_{n -2}-5040 a_{n -1}}{5040 \left (n^{2}+2 n r +r^{2}+1\right )}\tag {4} \] Which for the root \(r = i\) becomes \[ a_{n} = \frac {\left (-a_{n -7}-7 a_{n -6}-42 a_{n -5}-210 a_{n -4}-840 a_{n -3}-2520 a_{n -2}-5040 a_{n -1}\right ) n +\left (7-i\right ) a_{n -7}+\left (42-7 i\right ) a_{n -6}+\left (210-42 i\right ) a_{n -5}+\left (840-210 i\right ) a_{n -4}+\left (2520-840 i\right ) a_{n -3}+\left (5040-2520 i\right ) a_{n -2}+\left (5040-5040 i\right ) a_{n -1}}{5040 n \left (2 i+n \right )}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{r^{2}+2 r +2}\) | \(-\frac {2}{5}-\frac {i}{5}\) |
| \(a_{2}\) | \(-\frac {r^{3}}{2 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}\) | \(\frac {3}{80}-\frac {i}{80}\) |
| \(a_{3}\) | \(-\frac {r \left (r^{4}-6 r^{2}-9 r -5\right )}{6 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}\) | \(\frac {67}{9360}+\frac {9 i}{1040}\) |
| \(a_{4}\) | \(-\frac {r \left (r^{6}-2 r^{5}-43 r^{4}-136 r^{3}-180 r^{2}-112 r -40\right )}{24 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )}\) | \(-\frac {103}{149760}+\frac {229 i}{149760}\) |
| \(a_{5}\) | \(-\frac {r \left (r^{8}-10 r^{7}-201 r^{6}-1035 r^{5}-2331 r^{4}-2105 r^{3}+321 r^{2}+1760 r +800\right )}{120 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right )}\) | \(-\frac {2831}{7238400}-\frac {607 i}{4343040}\) |
| \(a_{6}\) | \(-\frac {r \left (2+r \right ) \left (r^{9}-34 r^{8}-666 r^{7}-3942 r^{6}-7855 r^{5}+11152 r^{4}+77700 r^{3}+138084 r^{2}+111970 r +41300\right )}{720 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right ) \left (r^{2}+12 r +37\right )}\) | \(-\frac {59077}{1563494400}-\frac {26063 i}{260582400}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=-\frac {r \left (r^{12}-84 r^{11}-2248 r^{10}-19677 r^{9}-49342 r^{8}+352058 r^{7}+3397664 r^{6}+13171067 r^{5}+29036801 r^{4}+38565016 r^{3}+30552414 r^{2}+13873510 r +3307350\right )}{5040 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right ) \left (r^{2}+12 r +37\right ) \left (r^{2}+14 r +50\right )} \] Which for the root \(r = i\) becomes \[ a_{7}=\frac {22952047}{2030197478400}-\frac {8634893 i}{580056422400} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{r^{2}+2 r +2}\) | \(-\frac {2}{5}-\frac {i}{5}\) |
| \(a_{2}\) | \(-\frac {r^{3}}{2 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}\) | \(\frac {3}{80}-\frac {i}{80}\) |
| \(a_{3}\) | \(-\frac {r \left (r^{4}-6 r^{2}-9 r -5\right )}{6 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}\) | \(\frac {67}{9360}+\frac {9 i}{1040}\) |
| \(a_{4}\) | \(-\frac {r \left (r^{6}-2 r^{5}-43 r^{4}-136 r^{3}-180 r^{2}-112 r -40\right )}{24 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )}\) | \(-\frac {103}{149760}+\frac {229 i}{149760}\) |
| \(a_{5}\) | \(-\frac {r \left (r^{8}-10 r^{7}-201 r^{6}-1035 r^{5}-2331 r^{4}-2105 r^{3}+321 r^{2}+1760 r +800\right )}{120 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right )}\) | \(-\frac {2831}{7238400}-\frac {607 i}{4343040}\) |
| \(a_{6}\) | \(-\frac {r \left (2+r \right ) \left (r^{9}-34 r^{8}-666 r^{7}-3942 r^{6}-7855 r^{5}+11152 r^{4}+77700 r^{3}+138084 r^{2}+111970 r +41300\right )}{720 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right ) \left (r^{2}+12 r +37\right )}\) | \(-\frac {59077}{1563494400}-\frac {26063 i}{260582400}\) |
| \(a_{7}\) | \(-\frac {r \left (r^{12}-84 r^{11}-2248 r^{10}-19677 r^{9}-49342 r^{8}+352058 r^{7}+3397664 r^{6}+13171067 r^{5}+29036801 r^{4}+38565016 r^{3}+30552414 r^{2}+13873510 r +3307350\right )}{5040 \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right ) \left (r^{2}+12 r +37\right ) \left (r^{2}+14 r +50\right )}\) | \(\frac {22952047}{2030197478400}-\frac {8634893 i}{580056422400}\) |
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 (-\frac {2}{5}-\frac {i}{5}\right ) x +\left (\frac {3}{80}-\frac {i}{80}\right ) x^{2}+\left (\frac {67}{9360}+\frac {9 i}{1040}\right ) x^{3}+\left (-\frac {103}{149760}+\frac {229 i}{149760}\right ) x^{4}+\left (-\frac {2831}{7238400}-\frac {607 i}{4343040}\right ) x^{5}+\left (-\frac {59077}{1563494400}-\frac {26063 i}{260582400}\right ) x^{6}+\left (\frac {22952047}{2030197478400}-\frac {8634893 i}{580056422400}\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 \[ y_{2}\left (x \right )= x^{-i} \left (1+\left (-\frac {2}{5}+\frac {i}{5}\right ) x +\left (\frac {3}{80}+\frac {i}{80}\right ) x^{2}+\left (\frac {67}{9360}-\frac {9 i}{1040}\right ) x^{3}+\left (-\frac {103}{149760}-\frac {229 i}{149760}\right ) x^{4}+\left (-\frac {2831}{7238400}+\frac {607 i}{4343040}\right ) x^{5}+\left (-\frac {59077}{1563494400}+\frac {26063 i}{260582400}\right ) x^{6}+\left (\frac {22952047}{2030197478400}+\frac {8634893 i}{580056422400}\right ) x^{7}+O\left (x^{8}\right )\right ) \] 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 (-\frac {2}{5}-\frac {i}{5}\right ) x +\left (\frac {3}{80}-\frac {i}{80}\right ) x^{2}+\left (\frac {67}{9360}+\frac {9 i}{1040}\right ) x^{3}+\left (-\frac {103}{149760}+\frac {229 i}{149760}\right ) x^{4}+\left (-\frac {2831}{7238400}-\frac {607 i}{4343040}\right ) x^{5}+\left (-\frac {59077}{1563494400}-\frac {26063 i}{260582400}\right ) x^{6}+\left (\frac {22952047}{2030197478400}-\frac {8634893 i}{580056422400}\right ) x^{7}+O\left (x^{8}\right )\right ) + c_{2} x^{-i} \left (1+\left (-\frac {2}{5}+\frac {i}{5}\right ) x +\left (\frac {3}{80}+\frac {i}{80}\right ) x^{2}+\left (\frac {67}{9360}-\frac {9 i}{1040}\right ) x^{3}+\left (-\frac {103}{149760}-\frac {229 i}{149760}\right ) x^{4}+\left (-\frac {2831}{7238400}+\frac {607 i}{4343040}\right ) x^{5}+\left (-\frac {59077}{1563494400}+\frac {26063 i}{260582400}\right ) x^{6}+\left (\frac {22952047}{2030197478400}+\frac {8634893 i}{580056422400}\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 (-\frac {2}{5}-\frac {i}{5}\right ) x +\left (\frac {3}{80}-\frac {i}{80}\right ) x^{2}+\left (\frac {67}{9360}+\frac {9 i}{1040}\right ) x^{3}+\left (-\frac {103}{149760}+\frac {229 i}{149760}\right ) x^{4}+\left (-\frac {2831}{7238400}-\frac {607 i}{4343040}\right ) x^{5}+\left (-\frac {59077}{1563494400}-\frac {26063 i}{260582400}\right ) x^{6}+\left (\frac {22952047}{2030197478400}-\frac {8634893 i}{580056422400}\right ) x^{7}+O\left (x^{8}\right )\right )+c_{2} x^{-i} \left (1+\left (-\frac {2}{5}+\frac {i}{5}\right ) x +\left (\frac {3}{80}+\frac {i}{80}\right ) x^{2}+\left (\frac {67}{9360}-\frac {9 i}{1040}\right ) x^{3}+\left (-\frac {103}{149760}-\frac {229 i}{149760}\right ) x^{4}+\left (-\frac {2831}{7238400}+\frac {607 i}{4343040}\right ) x^{5}+\left (-\frac {59077}{1563494400}+\frac {26063 i}{260582400}\right ) x^{6}+\left (\frac {22952047}{2030197478400}+\frac {8634893 i}{580056422400}\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 (-\frac {2}{5}-\frac {i}{5}\right ) x +\left (\frac {3}{80}-\frac {i}{80}\right ) x^{2}+\left (\frac {67}{9360}+\frac {9 i}{1040}\right ) x^{3}+\left (-\frac {103}{149760}+\frac {229 i}{149760}\right ) x^{4}+\left (-\frac {2831}{7238400}-\frac {607 i}{4343040}\right ) x^{5}+\left (-\frac {59077}{1563494400}-\frac {26063 i}{260582400}\right ) x^{6}+\left (\frac {22952047}{2030197478400}-\frac {8634893 i}{580056422400}\right ) x^{7}+O\left (x^{8}\right )\right )+c_{2} x^{-i} \left (1+\left (-\frac {2}{5}+\frac {i}{5}\right ) x +\left (\frac {3}{80}+\frac {i}{80}\right ) x^{2}+\left (\frac {67}{9360}-\frac {9 i}{1040}\right ) x^{3}+\left (-\frac {103}{149760}-\frac {229 i}{149760}\right ) x^{4}+\left (-\frac {2831}{7238400}+\frac {607 i}{4343040}\right ) x^{5}+\left (-\frac {59077}{1563494400}+\frac {26063 i}{260582400}\right ) x^{6}+\left (\frac {22952047}{2030197478400}+\frac {8634893 i}{580056422400}\right ) x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{i} \left (1+\left (-\frac {2}{5}-\frac {i}{5}\right ) x +\left (\frac {3}{80}-\frac {i}{80}\right ) x^{2}+\left (\frac {67}{9360}+\frac {9 i}{1040}\right ) x^{3}+\left (-\frac {103}{149760}+\frac {229 i}{149760}\right ) x^{4}+\left (-\frac {2831}{7238400}-\frac {607 i}{4343040}\right ) x^{5}+\left (-\frac {59077}{1563494400}-\frac {26063 i}{260582400}\right ) x^{6}+\left (\frac {22952047}{2030197478400}-\frac {8634893 i}{580056422400}\right ) x^{7}+O\left (x^{8}\right )\right )+c_{2} x^{-i} \left (1+\left (-\frac {2}{5}+\frac {i}{5}\right ) x +\left (\frac {3}{80}+\frac {i}{80}\right ) x^{2}+\left (\frac {67}{9360}-\frac {9 i}{1040}\right ) x^{3}+\left (-\frac {103}{149760}-\frac {229 i}{149760}\right ) x^{4}+\left (-\frac {2831}{7238400}+\frac {607 i}{4343040}\right ) x^{5}+\left (-\frac {59077}{1563494400}+\frac {26063 i}{260582400}\right ) x^{6}+\left (\frac {22952047}{2030197478400}+\frac {8634893 i}{580056422400}\right ) x^{7}+O\left (x^{8}\right )\right ) \] 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*exp(x)*diff(y(x),x)+y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{-i} \left (1+\left (-\frac {2}{5}+\frac {i}{5}\right ) x +\left (\frac {3}{80}+\frac {i}{80}\right ) x^{2}+\left (\frac {67}{9360}-\frac {9 i}{1040}\right ) x^{3}+\left (-\frac {103}{149760}-\frac {229 i}{149760}\right ) x^{4}+\left (-\frac {2831}{7238400}+\frac {607 i}{4343040}\right ) x^{5}+\left (-\frac {59077}{1563494400}+\frac {26063 i}{260582400}\right ) x^{6}+\left (\frac {22952047}{2030197478400}+\frac {8634893 i}{580056422400}\right ) x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} x^{i} \left (1+\left (-\frac {2}{5}-\frac {i}{5}\right ) x +\left (\frac {3}{80}-\frac {i}{80}\right ) x^{2}+\left (\frac {67}{9360}+\frac {9 i}{1040}\right ) x^{3}+\left (-\frac {103}{149760}+\frac {229 i}{149760}\right ) x^{4}+\left (-\frac {2831}{7238400}-\frac {607 i}{4343040}\right ) x^{5}+\left (-\frac {59077}{1563494400}-\frac {26063 i}{260582400}\right ) x^{6}+\left (\frac {22952047}{2030197478400}-\frac {8634893 i}{580056422400}\right ) x^{7}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.028 (sec). Leaf size: 122
AsymptoticDSolveValue[x^2*y''[x]+x*Exp[x]*y'[x]+y[x]==0,y[x],{x,0,7}]
\[ y(x)\to \left (\frac {11}{1563494400}+\frac {i}{97718400}\right ) c_2 x^{-i} \left ((4913+7070 i) x^6-(8568-32328 i) x^5-(132840+24120 i) x^4-(247680+869760 i) x^3+(2540160-1918080 i) x^2-(4976640-35665920 i) x+(45619200-66355200 i)\right )-\left (\frac {1}{97718400}+\frac {11 i}{1563494400}\right ) c_1 x^i \left ((7070+4913 i) x^6+(32328-8568 i) x^5-(24120+132840 i) x^4-(869760+247680 i) x^3-(1918080-2540160 i) x^2+(35665920-4976640 i) x-(66355200-45619200 i)\right ) \]