Internal problem ID [2922]
Internal file name [OUTPUT/2414_Sunday_June_05_2022_03_07_14_AM_72399094/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 11, Series Solutions to Linear Differential Equations. Exercises for 11.4. page
758
Problem number: 7.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[[_Emden, _Fowler]]
\[ \boxed {4 x^{2} y^{\prime \prime }+y^{\prime } x \,{\mathrm e}^{x}-y=0} \] 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. \[ 4 x^{2} y^{\prime \prime }+y^{\prime } x \,{\mathrm e}^{x}-y = 0 \] 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 {{\mathrm e}^{x}}{4 x}\\ q(x) &= -\frac {1}{4 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 \[ 4 x^{2} y^{\prime \prime }+y^{\prime } x \,{\mathrm e}^{x}-y = 0 \] 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} 4 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 ) x \,{\mathrm e}^{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 \(6\) terms gives \begin {align*} x \,{\mathrm e}^{x} &= x +x^{2}+\frac {1}{2} x^{3}+\frac {1}{6} x^{4}+\frac {1}{24} x^{5}+\frac {1}{120} x^{6} + \dots \\ &= x +x^{2}+\frac {1}{2} x^{3}+\frac {1}{6} x^{4}+\frac {1}{24} x^{5}+\frac {1}{120} x^{6} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}4 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 +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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-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 +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 }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )-a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r -a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )+x^{r} r -x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}-3 r -1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2}-3 r -1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= -{\frac {1}{4}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (4 r^{2}-3 r -1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [1, -{\frac {1}{4}}\right ]\).
Since \(r_1 - r_2 = {\frac {5}{4}}\) is not 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 ) &= 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^{1+n}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {1}{4}} \end {align*}
We start by finding \(y_{1}\left (x \right )\). 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 \[ a_{1} = -\frac {1}{4 r +5} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {-4 r^{2}-3 r +2}{32 r^{3}+144 r^{2}+202 r +90} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {-16 r^{4}-48 r^{3}-17 r^{2}+33 r +15}{384 r^{5}+3744 r^{4}+13992 r^{3}+25038 r^{2}+21426 r +7020} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {-64 r^{6}-400 r^{5}-572 r^{4}+773 r^{3}+2385 r^{2}+1538 r +132}{24 \left (64 r^{5}+624 r^{4}+2332 r^{3}+4173 r^{2}+3571 r +1170\right ) \left (4 r^{2}+29 r +51\right )} \] For \(5\le n\) the recursive equation is \begin{equation} \tag{3} 4 a_{n} \left (n +r \right ) \left (n +r -1\right )+\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 \[ a_{n} = -\frac {n a_{n -5}+5 n a_{n -4}+20 n a_{n -3}+60 n a_{n -2}+120 n a_{n -1}+r a_{n -5}+5 r a_{n -4}+20 r a_{n -3}+60 r a_{n -2}+120 r a_{n -1}-5 a_{n -5}-20 a_{n -4}-60 a_{n -3}-120 a_{n -2}-120 a_{n -1}}{120 \left (4 n^{2}+8 n r +4 r^{2}-3 n -3 r -1\right )}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {\left (-a_{n -5}-5 a_{n -4}-20 a_{n -3}-60 a_{n -2}-120 a_{n -1}\right ) n +4 a_{n -5}+15 a_{n -4}+40 a_{n -3}+60 a_{n -2}}{480 n^{2}+600 n}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{4 r +5}\) | \(-{\frac {1}{9}}\) |
| \(a_{2}\) | \(\frac {-4 r^{2}-3 r +2}{32 r^{3}+144 r^{2}+202 r +90}\) | \(-{\frac {5}{468}}\) |
| \(a_{3}\) | \(\frac {-16 r^{4}-48 r^{3}-17 r^{2}+33 r +15}{384 r^{5}+3744 r^{4}+13992 r^{3}+25038 r^{2}+21426 r +7020}\) | \(-{\frac {11}{23868}}\) |
| \(a_{4}\) | \(\frac {-64 r^{6}-400 r^{5}-572 r^{4}+773 r^{3}+2385 r^{2}+1538 r +132}{24 \left (64 r^{5}+624 r^{4}+2332 r^{3}+4173 r^{2}+3571 r +1170\right ) \left (4 r^{2}+29 r +51\right )}\) | \(\frac {79}{501228}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-256 r^{8}-2432 r^{7}-3808 r^{6}+33576 r^{5}+168107 r^{4}+308308 r^{3}+239784 r^{2}+45745 r -18960}{120 \left (4 r^{2}+37 r +84\right ) \left (64 r^{5}+624 r^{4}+2332 r^{3}+4173 r^{2}+3571 r +1170\right ) \left (4 r^{2}+29 r +51\right )} \] Which for the root \(r = 1\) becomes \[ a_{5}={\frac {16043}{313267500}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{4 r +5}\) | \(-{\frac {1}{9}}\) |
| \(a_{2}\) | \(\frac {-4 r^{2}-3 r +2}{32 r^{3}+144 r^{2}+202 r +90}\) | \(-{\frac {5}{468}}\) |
| \(a_{3}\) | \(\frac {-16 r^{4}-48 r^{3}-17 r^{2}+33 r +15}{384 r^{5}+3744 r^{4}+13992 r^{3}+25038 r^{2}+21426 r +7020}\) | \(-{\frac {11}{23868}}\) |
| \(a_{4}\) | \(\frac {-64 r^{6}-400 r^{5}-572 r^{4}+773 r^{3}+2385 r^{2}+1538 r +132}{24 \left (64 r^{5}+624 r^{4}+2332 r^{3}+4173 r^{2}+3571 r +1170\right ) \left (4 r^{2}+29 r +51\right )}\) | \(\frac {79}{501228}\) |
| \(a_{5}\) | \(\frac {-256 r^{8}-2432 r^{7}-3808 r^{6}+33576 r^{5}+168107 r^{4}+308308 r^{3}+239784 r^{2}+45745 r -18960}{120 \left (4 r^{2}+37 r +84\right ) \left (64 r^{5}+624 r^{4}+2332 r^{3}+4173 r^{2}+3571 r +1170\right ) \left (4 r^{2}+29 r +51\right )}\) | \(\frac {16043}{313267500}\) |
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}{9}-\frac {5 x^{2}}{468}-\frac {11 x^{3}}{23868}+\frac {79 x^{4}}{501228}+\frac {16043 x^{5}}{313267500}+O\left (x^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Eq (2B) derived above is now used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = -\frac {1}{4 r +5} \] Substituting \(n = 2\) in Eq. (2B) gives \[ b_{2} = \frac {-4 r^{2}-3 r +2}{32 r^{3}+144 r^{2}+202 r +90} \] Substituting \(n = 3\) in Eq. (2B) gives \[ b_{3} = \frac {-16 r^{4}-48 r^{3}-17 r^{2}+33 r +15}{384 r^{5}+3744 r^{4}+13992 r^{3}+25038 r^{2}+21426 r +7020} \] Substituting \(n = 4\) in Eq. (2B) gives \[ b_{4} = \frac {-64 r^{6}-400 r^{5}-572 r^{4}+773 r^{3}+2385 r^{2}+1538 r +132}{24 \left (64 r^{5}+624 r^{4}+2332 r^{3}+4173 r^{2}+3571 r +1170\right ) \left (4 r^{2}+29 r +51\right )} \] For \(5\le n\) the recursive equation is \begin{equation} \tag{3} 4 b_{n} \left (n +r \right ) \left (n +r -1\right )+\frac {b_{n -5} \left (n -5+r \right )}{120}+\frac {b_{n -4} \left (n -4+r \right )}{24}+\frac {b_{n -3} \left (n -3+r \right )}{6}+\frac {b_{n -2} \left (n +r -2\right )}{2}+b_{n -1} \left (n +r -1\right )+b_{n} \left (n +r \right )-b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {n b_{n -5}+5 n b_{n -4}+20 n b_{n -3}+60 n b_{n -2}+120 n b_{n -1}+r b_{n -5}+5 r b_{n -4}+20 r b_{n -3}+60 r b_{n -2}+120 r b_{n -1}-5 b_{n -5}-20 b_{n -4}-60 b_{n -3}-120 b_{n -2}-120 b_{n -1}}{120 \left (4 n^{2}+8 n r +4 r^{2}-3 n -3 r -1\right )}\tag {4} \] Which for the root \(r = -{\frac {1}{4}}\) becomes \[ b_{n} = \frac {\left (-4 b_{n -5}-20 b_{n -4}-80 b_{n -3}-240 b_{n -2}-480 b_{n -1}\right ) n +21 b_{n -5}+85 b_{n -4}+260 b_{n -3}+540 b_{n -2}+600 b_{n -1}}{1920 n^{2}-2400 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {1}{4}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{4 r +5}\) | \(-{\frac {1}{4}}\) |
| \(b_{2}\) | \(\frac {-4 r^{2}-3 r +2}{32 r^{3}+144 r^{2}+202 r +90}\) | \(\frac {5}{96}\) |
| \(b_{3}\) | \(\frac {-16 r^{4}-48 r^{3}-17 r^{2}+33 r +15}{384 r^{5}+3744 r^{4}+13992 r^{3}+25038 r^{2}+21426 r +7020}\) | \(\frac {17}{8064}\) |
| \(b_{4}\) | \(\frac {-64 r^{6}-400 r^{5}-572 r^{4}+773 r^{3}+2385 r^{2}+1538 r +132}{24 \left (64 r^{5}+624 r^{4}+2332 r^{3}+4173 r^{2}+3571 r +1170\right ) \left (4 r^{2}+29 r +51\right )}\) | \(-{\frac {313}{1419264}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {-256 r^{8}-2432 r^{7}-3808 r^{6}+33576 r^{5}+168107 r^{4}+308308 r^{3}+239784 r^{2}+45745 r -18960}{120 \left (4 r^{2}+37 r +84\right ) \left (64 r^{5}+624 r^{4}+2332 r^{3}+4173 r^{2}+3571 r +1170\right ) \left (4 r^{2}+29 r +51\right )} \] Which for the root \(r = -{\frac {1}{4}}\) becomes \[ b_{5}=-{\frac {69703}{709632000}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{4 r +5}\) | \(-{\frac {1}{4}}\) |
| \(b_{2}\) | \(\frac {-4 r^{2}-3 r +2}{32 r^{3}+144 r^{2}+202 r +90}\) | \(\frac {5}{96}\) |
| \(b_{3}\) | \(\frac {-16 r^{4}-48 r^{3}-17 r^{2}+33 r +15}{384 r^{5}+3744 r^{4}+13992 r^{3}+25038 r^{2}+21426 r +7020}\) | \(\frac {17}{8064}\) |
| \(b_{4}\) | \(\frac {-64 r^{6}-400 r^{5}-572 r^{4}+773 r^{3}+2385 r^{2}+1538 r +132}{24 \left (64 r^{5}+624 r^{4}+2332 r^{3}+4173 r^{2}+3571 r +1170\right ) \left (4 r^{2}+29 r +51\right )}\) | \(-{\frac {313}{1419264}}\) |
| \(b_{5}\) | \(\frac {-256 r^{8}-2432 r^{7}-3808 r^{6}+33576 r^{5}+168107 r^{4}+308308 r^{3}+239784 r^{2}+45745 r -18960}{120 \left (4 r^{2}+37 r +84\right ) \left (64 r^{5}+624 r^{4}+2332 r^{3}+4173 r^{2}+3571 r +1170\right ) \left (4 r^{2}+29 r +51\right )}\) | \(-{\frac {69703}{709632000}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x \left (b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \right ) \\ &= \frac {1-\frac {x}{4}+\frac {5 x^{2}}{96}+\frac {17 x^{3}}{8064}-\frac {313 x^{4}}{1419264}-\frac {69703 x^{5}}{709632000}+O\left (x^{6}\right )}{x^{\frac {1}{4}}} \end {align*}
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}{9}-\frac {5 x^{2}}{468}-\frac {11 x^{3}}{23868}+\frac {79 x^{4}}{501228}+\frac {16043 x^{5}}{313267500}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1-\frac {x}{4}+\frac {5 x^{2}}{96}+\frac {17 x^{3}}{8064}-\frac {313 x^{4}}{1419264}-\frac {69703 x^{5}}{709632000}+O\left (x^{6}\right )\right )}{x^{\frac {1}{4}}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1-\frac {x}{9}-\frac {5 x^{2}}{468}-\frac {11 x^{3}}{23868}+\frac {79 x^{4}}{501228}+\frac {16043 x^{5}}{313267500}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {x}{4}+\frac {5 x^{2}}{96}+\frac {17 x^{3}}{8064}-\frac {313 x^{4}}{1419264}-\frac {69703 x^{5}}{709632000}+O\left (x^{6}\right )\right )}{x^{\frac {1}{4}}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1-\frac {x}{9}-\frac {5 x^{2}}{468}-\frac {11 x^{3}}{23868}+\frac {79 x^{4}}{501228}+\frac {16043 x^{5}}{313267500}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {x}{4}+\frac {5 x^{2}}{96}+\frac {17 x^{3}}{8064}-\frac {313 x^{4}}{1419264}-\frac {69703 x^{5}}{709632000}+O\left (x^{6}\right )\right )}{x^{\frac {1}{4}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-\frac {x}{9}-\frac {5 x^{2}}{468}-\frac {11 x^{3}}{23868}+\frac {79 x^{4}}{501228}+\frac {16043 x^{5}}{313267500}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {x}{4}+\frac {5 x^{2}}{96}+\frac {17 x^{3}}{8064}-\frac {313 x^{4}}{1419264}-\frac {69703 x^{5}}{709632000}+O\left (x^{6}\right )\right )}{x^{\frac {1}{4}}} \] 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: 0.031 (sec). Leaf size: 45
Order:=6; dsolve(4*x^2*diff(y(x),x$2)+x*exp(x)*diff(y(x),x)-y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} \left (1-\frac {1}{4} x +\frac {5}{96} x^{2}+\frac {17}{8064} x^{3}-\frac {313}{1419264} x^{4}-\frac {69703}{709632000} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x^{\frac {1}{4}}}+c_{2} x \left (1-\frac {1}{9} x -\frac {5}{468} x^{2}-\frac {11}{23868} x^{3}+\frac {79}{501228} x^{4}+\frac {16043}{313267500} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 86
AsymptoticDSolveValue[4*x^2*y''[x]+x*Exp[x]*y'[x]-y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 x \left (\frac {16043 x^5}{313267500}+\frac {79 x^4}{501228}-\frac {11 x^3}{23868}-\frac {5 x^2}{468}-\frac {x}{9}+1\right )+\frac {c_2 \left (-\frac {69703 x^5}{709632000}-\frac {313 x^4}{1419264}+\frac {17 x^3}{8064}+\frac {5 x^2}{96}-\frac {x}{4}+1\right )}{\sqrt [4]{x}} \]