Internal problem ID [6054]
Internal file name [OUTPUT/5302_Sunday_June_05_2022_03_33_16_PM_20004479/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 166
Problem number: 1(ii).
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]]
\[ \boxed {4 x^{2} y^{\prime \prime }-4 x \,{\mathrm e}^{x} y^{\prime }+3 \cos \left (x \right ) y=0} \] 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. \[ 4 x^{2} y^{\prime \prime }-4 x \,{\mathrm e}^{x} y^{\prime }+3 \cos \left (x \right ) y = 0 \] 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 {3 \cos \left (x \right )}{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 \[ 4 x^{2} y^{\prime \prime }-4 x \,{\mathrm e}^{x} y^{\prime }+3 \cos \left (x \right ) y = 0 \] 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 )-4 x \,{\mathrm e}^{x} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+3 \cos \left (x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Expanding \(-4 x \,{\mathrm e}^{x}\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} -4 x \,{\mathrm e}^{x} &= -4 x -4 x^{2}-2 x^{3}-\frac {2}{3} x^{4}-\frac {1}{6} x^{5}-\frac {1}{30} x^{6}-\frac {1}{180} x^{7}-\frac {1}{1260} x^{8} + \dots \\ &= -4 x -4 x^{2}-2 x^{3}-\frac {2}{3} x^{4}-\frac {1}{6} x^{5}-\frac {1}{30} x^{6}-\frac {1}{180} x^{7}-\frac {1}{1260} x^{8} \end {align*}
Expanding \(3 \cos \left (x \right )\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} 3 \cos \left (x \right ) &= 3-\frac {3}{2} x^{2}+\frac {1}{8} x^{4}-\frac {1}{240} x^{6}+\frac {1}{13440} x^{8} + \dots \\ &= 3-\frac {3}{2} x^{2}+\frac {1}{8} x^{4}-\frac {1}{240} x^{6}+\frac {1}{13440} x^{8} \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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +7} a_{n} \left (n +r \right )}{1260}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n} \left (n +r \right )}{180}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +5} a_{n} \left (n +r \right )}{30}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +4} a_{n} \left (n +r \right )}{6}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {2 x^{n +r +3} a_{n} \left (n +r \right )}{3}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r +2} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 a_{n} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {3 x^{n +r +2} a_{n}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n}}{8}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n}}{240}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +8} a_{n}}{13440}\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 (-\frac {x^{n +r +7} a_{n} \left (n +r \right )}{1260}\right ) &= \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {a_{n -7} \left (n -7+r \right ) x^{n +r}}{1260}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n} \left (n +r \right )}{180}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} \left (n -6+r \right ) x^{n +r}}{180}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +5} a_{n} \left (n +r \right )}{30}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {a_{n -5} \left (n -5+r \right ) x^{n +r}}{30}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +4} a_{n} \left (n +r \right )}{6}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {a_{n -4} \left (n -4+r \right ) x^{n +r}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {2 x^{n +r +3} a_{n} \left (n +r \right )}{3}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {2 a_{n -3} \left (n -3+r \right ) x^{n +r}}{3}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r +2} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 a_{n -2} \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} \left (n +r -1\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {3 x^{n +r +2} a_{n}}{2}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {3 a_{n -2} x^{n +r}}{2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n}}{8} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} x^{n +r}}{8} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n}}{240}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} x^{n +r}}{240}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +8} a_{n}}{13440} &= \moverset {\infty }{\munderset {n =8}{\sum }}\frac {a_{n -8} x^{n +r}}{13440} \\ \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 )+\moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {a_{n -7} \left (n -7+r \right ) x^{n +r}}{1260}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} \left (n -6+r \right ) x^{n +r}}{180}\right )+\moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {a_{n -5} \left (n -5+r \right ) x^{n +r}}{30}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {a_{n -4} \left (n -4+r \right ) x^{n +r}}{6}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {2 a_{n -3} \left (n -3+r \right ) x^{n +r}}{3}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 a_{n} x^{n +r}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {3 a_{n -2} x^{n +r}}{2}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} x^{n +r}}{8}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} x^{n +r}}{240}\right )+\left (\moverset {\infty }{\munderset {n =8}{\sum }}\frac {a_{n -8} x^{n +r}}{13440}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-4 x^{n +r} a_{n} \left (n +r \right )+3 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )-4 x^{r} a_{0} r +3 a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )-4 x^{r} r +3 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}-8 r +3\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2}-8 r +3 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {3}{2}}\\ r_2 &= {\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (4 r^{2}-8 r +3\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {3}{2}}, {\frac {1}{2}}\right ]\).
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^{\frac {3}{2}} \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 )+\sqrt {x}\, \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 +\frac {3}{2}}\\ 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 +\frac {1}{2}}\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 \[ a_{1} = \frac {4 r}{4 r^{2}-1} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {8 r^{2}+10 r -1}{16 r^{3}+8 r^{2}-4 r -2} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {16 r^{4}+104 r^{3}+260 r^{2}+154 r -12}{96 r^{5}+432 r^{4}+528 r^{3}+72 r^{2}-138 r -45} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {128 r^{6}+2272 r^{5}+14096 r^{4}+40832 r^{3}+54968 r^{2}+23922 r -2097}{3072 \left (r -\frac {1}{2}\right ) \left (r +\frac {3}{2}\right ) \left (r +\frac {7}{2}\right ) \left (r +\frac {5}{2}\right )^{2} \left (r +\frac {1}{2}\right )^{2}} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = \frac {128 r^{8}+5184 r^{7}+63840 r^{6}+386800 r^{5}+1320392 r^{4}+2573796 r^{3}+2595110 r^{2}+925655 r -83940}{30 \left (2 r -1\right ) \left (3+2 r \right ) \left (2 r +7\right ) \left (2 r +5\right )^{2} \left (2 r +1\right )^{2} \left (4 r^{2}+32 r +63\right )} \] Substituting \(n = 6\) in Eq. (2B) gives \[ a_{6} = \frac {2048 r^{10}+166400 r^{9}+3645184 r^{8}+39318272 r^{7}+248963968 r^{6}+989572160 r^{5}+2499293216 r^{4}+3862419888 r^{3}+3238012464 r^{2}+1002158970 r -93223485}{1474560 \left (r +\frac {11}{2}\right ) \left (r -\frac {1}{2}\right ) \left (r +\frac {3}{2}\right ) \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {9}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2} \left (r +\frac {1}{2}\right )^{2}} \] Substituting \(n = 7\) in Eq. (2B) gives \[ a_{7} = \frac {2048 r^{12}+316416 r^{11}+11362816 r^{10}+198127872 r^{9}+2053881088 r^{8}+13811070336 r^{7}+62663041472 r^{6}+193775095008 r^{5}+401584712904 r^{4}+528516962388 r^{3}+388228716162 r^{2}+107690411745 r -10310307210}{1260 \left (2 r +11\right ) \left (2 r -1\right ) \left (3+2 r \right ) \left (2 r +7\right )^{2} \left (2 r +9\right )^{2} \left (2 r +5\right )^{2} \left (2 r +1\right )^{2} \left (4 r^{2}+48 r +143\right )} \] For \(8\le n\) the recursive equation is \begin{equation} \tag{3} 4 a_{n} \left (n +r \right ) \left (n +r -1\right )-\frac {a_{n -7} \left (n -7+r \right )}{1260}-\frac {a_{n -6} \left (n -6+r \right )}{180}-\frac {a_{n -5} \left (n -5+r \right )}{30}-\frac {a_{n -4} \left (n -4+r \right )}{6}-\frac {2 a_{n -3} \left (n -3+r \right )}{3}-2 a_{n -2} \left (n +r -2\right )-4 a_{n -1} \left (n +r -1\right )-4 a_{n} \left (n +r \right )+3 a_{n}-\frac {3 a_{n -2}}{2}+\frac {a_{n -4}}{8}-\frac {a_{n -6}}{240}+\frac {a_{n -8}}{13440} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {32 n a_{n -7}+224 n a_{n -6}+1344 n a_{n -5}+6720 n a_{n -4}+26880 n a_{n -3}+80640 n a_{n -2}+161280 n a_{n -1}+32 r a_{n -7}+224 r a_{n -6}+1344 r a_{n -5}+6720 r a_{n -4}+26880 r a_{n -3}+80640 r a_{n -2}+161280 r a_{n -1}-3 a_{n -8}-224 a_{n -7}-1176 a_{n -6}-6720 a_{n -5}-31920 a_{n -4}-80640 a_{n -3}-100800 a_{n -2}-161280 a_{n -1}}{161280 n^{2}+322560 n r +161280 r^{2}-322560 n -322560 r +120960}\tag {4} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{n} = \frac {32 \left (a_{n -7}+7 a_{n -6}+42 a_{n -5}+210 a_{n -4}+840 a_{n -3}+2520 a_{n -2}+5040 a_{n -1}\right ) n -3 a_{n -8}-176 a_{n -7}-840 a_{n -6}-4704 a_{n -5}-21840 a_{n -4}-40320 a_{n -3}+20160 a_{n -2}+80640 a_{n -1}}{161280 n \left (n +1\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {3}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r}{4 r^{2}-1}\) | \(\frac {3}{4}\) |
| \(a_{2}\) | \(\frac {8 r^{2}+10 r -1}{16 r^{3}+8 r^{2}-4 r -2}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {16 r^{4}+104 r^{3}+260 r^{2}+154 r -12}{96 r^{5}+432 r^{4}+528 r^{3}+72 r^{2}-138 r -45}\) | \(\frac {103}{384}\) |
| \(a_{4}\) | \(\frac {128 r^{6}+2272 r^{5}+14096 r^{4}+40832 r^{3}+54968 r^{2}+23922 r -2097}{3072 \left (r -\frac {1}{2}\right ) \left (r +\frac {3}{2}\right ) \left (r +\frac {7}{2}\right ) \left (r +\frac {5}{2}\right )^{2} \left (r +\frac {1}{2}\right )^{2}}\) | \(\frac {669}{5120}\) |
| \(a_{5}\) | \(\frac {128 r^{8}+5184 r^{7}+63840 r^{6}+386800 r^{5}+1320392 r^{4}+2573796 r^{3}+2595110 r^{2}+925655 r -83940}{30 \left (2 r -1\right ) \left (3+2 r \right ) \left (2 r +7\right ) \left (2 r +5\right )^{2} \left (2 r +1\right )^{2} \left (4 r^{2}+32 r +63\right )}\) | \(\frac {54731}{921600}\) |
| \(a_{6}\) | \(\frac {2048 r^{10}+166400 r^{9}+3645184 r^{8}+39318272 r^{7}+248963968 r^{6}+989572160 r^{5}+2499293216 r^{4}+3862419888 r^{3}+3238012464 r^{2}+1002158970 r -93223485}{1474560 \left (r +\frac {11}{2}\right ) \left (r -\frac {1}{2}\right ) \left (r +\frac {3}{2}\right ) \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {9}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2} \left (r +\frac {1}{2}\right )^{2}}\) | \(\frac {123443}{4838400}\) |
| \(a_{7}\) | \(\frac {2048 r^{12}+316416 r^{11}+11362816 r^{10}+198127872 r^{9}+2053881088 r^{8}+13811070336 r^{7}+62663041472 r^{6}+193775095008 r^{5}+401584712904 r^{4}+528516962388 r^{3}+388228716162 r^{2}+107690411745 r -10310307210}{1260 \left (2 r +11\right ) \left (2 r -1\right ) \left (3+2 r \right ) \left (2 r +7\right )^{2} \left (2 r +9\right )^{2} \left (2 r +5\right )^{2} \left (2 r +1\right )^{2} \left (4 r^{2}+48 r +143\right )}\) | \(\frac {30273113}{2890137600}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {3}{2}} \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^{\frac {3}{2}} \left (1+\frac {3 x}{4}+\frac {x^{2}}{2}+\frac {103 x^{3}}{384}+\frac {669 x^{4}}{5120}+\frac {54731 x^{5}}{921600}+\frac {123443 x^{6}}{4838400}+\frac {30273113 x^{7}}{2890137600}+O\left (x^{8}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=1\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{1}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{1} \\ &= \frac {4 r}{4 r^{2}-1} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {4 r}{4 r^{2}-1}&= \lim _{r\rightarrow {\frac {1}{2}}}\frac {4 r}{4 r^{2}-1}\\ &= \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 \(4 x^{2} y^{\prime \prime }-4 x \,{\mathrm e}^{x} y^{\prime }+3 \cos \left (x \right ) y = 0\) gives \[ 4 x^{2} \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )-4 x \,{\mathrm e}^{x} \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+3 \cos \left (x \right ) \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (4 y_{1}^{\prime \prime }\left (x \right ) x^{2}-4 y_{1}^{\prime }\left (x \right ) {\mathrm e}^{x} x +3 y_{1}\left (x \right ) \cos \left (x \right )\right ) \ln \left (x \right )+4 x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )-4 \,{\mathrm e}^{x} y_{1}\left (x \right )\right ) C +4 x^{2} \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 )-4 x \,{\mathrm e}^{x} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+3 \cos \left (x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} But since \(y_{1}\left (x \right )\) is a solution to the ode, then \[ 4 y_{1}^{\prime \prime }\left (x \right ) x^{2}-4 y_{1}^{\prime }\left (x \right ) {\mathrm e}^{x} x +3 y_{1}\left (x \right ) \cos \left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (4 x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )-4 \,{\mathrm e}^{x} y_{1}\left (x \right )\right ) C +4 x^{2} \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 )-4 x \,{\mathrm e}^{x} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+3 \cos \left (x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\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} \left (8 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right ) x -4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right ) \left (1+{\mathrm e}^{x}\right )\right ) C -4 x \,{\mathrm e}^{x} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )+4 \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}+3 \cos \left (x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Since \(r_{1} = {\frac {3}{2}}\) and \(r_{2} = {\frac {1}{2}}\) then the above becomes \begin{equation} \tag{10} \left (8 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +\frac {1}{2}} a_{n} \left (n +\frac {3}{2}\right )\right ) x -4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {3}{2}}\right ) \left (1+{\mathrm e}^{x}\right )\right ) C -4 x \,{\mathrm e}^{x} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-\frac {1}{2}+n} b_{n} \left (n +\frac {1}{2}\right )\right )+4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-\frac {3}{2}+n} b_{n} \left (n +\frac {1}{2}\right ) \left (-\frac {1}{2}+n \right )\right ) x^{2}+3 \cos \left (x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{2}}\right ) = 0 \end{equation} Expanding \(-4 C \,x^{\frac {3}{2}}\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} -4 C \,x^{\frac {3}{2}} &= -4 C \,x^{\frac {3}{2}} + \dots \\ &= -4 C \,x^{\frac {3}{2}} \end {align*}
Expanding \(-4 C \,x^{\frac {3}{2}} {\mathrm e}^{x}\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} -4 C \,x^{\frac {3}{2}} {\mathrm e}^{x} &= -4 C \,x^{\frac {3}{2}}-4 C \,x^{\frac {5}{2}}-2 C \,x^{\frac {7}{2}}-\frac {2 C \,x^{\frac {9}{2}}}{3}-\frac {C \,x^{\frac {11}{2}}}{6}-\frac {C \,x^{\frac {13}{2}}}{30}-\frac {C \,x^{\frac {15}{2}}}{180}-\frac {C \,x^{\frac {17}{2}}}{1260} + \dots \\ &= -4 C \,x^{\frac {3}{2}}-4 C \,x^{\frac {5}{2}}-2 C \,x^{\frac {7}{2}}-\frac {2 C \,x^{\frac {9}{2}}}{3}-\frac {C \,x^{\frac {11}{2}}}{6}-\frac {C \,x^{\frac {13}{2}}}{30}-\frac {C \,x^{\frac {15}{2}}}{180}-\frac {C \,x^{\frac {17}{2}}}{1260} \end {align*}
Expanding \(-2 \sqrt {x}\, {\mathrm e}^{x}\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} -2 \sqrt {x}\, {\mathrm e}^{x} &= -2 \sqrt {x}-2 x^{\frac {3}{2}}-x^{\frac {5}{2}}-\frac {x^{\frac {7}{2}}}{3}-\frac {x^{\frac {9}{2}}}{12}-\frac {x^{\frac {11}{2}}}{60}-\frac {x^{\frac {13}{2}}}{360}-\frac {x^{\frac {15}{2}}}{2520} + \dots \\ &= -2 \sqrt {x}-2 x^{\frac {3}{2}}-x^{\frac {5}{2}}-\frac {x^{\frac {7}{2}}}{3}-\frac {x^{\frac {9}{2}}}{12}-\frac {x^{\frac {11}{2}}}{60}-\frac {x^{\frac {13}{2}}}{360}-\frac {x^{\frac {15}{2}}}{2520} \end {align*}
Expanding \(3 \cos \left (x \right )\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} 3 \cos \left (x \right ) &= 3-\frac {3}{2} x^{2}+\frac {1}{8} x^{4}-\frac {1}{240} x^{6}+\frac {1}{13440} x^{8} + \dots \\ &= 3-\frac {3}{2} x^{2}+\frac {1}{8} x^{4}-\frac {1}{240} x^{6}+\frac {1}{13440} x^{8} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (8 n +12\right ) C a_{n} x^{n +\frac {3}{2}}\right )+2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C \,x^{n +\frac {3}{2}} a_{n}\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C \,x^{n +\frac {5}{2}} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{n +\frac {7}{2}} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {2 C \,x^{n +\frac {9}{2}} a_{n}}{3}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +\frac {11}{2}} a_{n}}{6}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +\frac {13}{2}} a_{n}}{30}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +\frac {15}{2}} a_{n}}{180}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +\frac {17}{2}} a_{n}}{1260}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +\frac {15}{2}} b_{n} \left (2 n +1\right )}{2520}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +\frac {13}{2}} b_{n} \left (2 n +1\right )}{360}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +\frac {11}{2}} b_{n} \left (2 n +1\right )}{60}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +\frac {9}{2}} b_{n} \left (2 n +1\right )}{12}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +\frac {7}{2}} b_{n} \left (2 n +1\right )}{3}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +\frac {5}{2}} b_{n} \left (-2 n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 n -2\right ) b_{n} x^{n +\frac {3}{2}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 n -2\right ) b_{n} x^{n +\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +\frac {1}{2}} b_{n} \left (4 n^{2}-1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 b_{n} x^{n +\frac {1}{2}}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {3 x^{n +\frac {5}{2}} b_{n}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +\frac {9}{2}} b_{n}}{8}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +\frac {13}{2}} b_{n}}{240}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +\frac {17}{2}} b_{n}}{13440}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +\frac {1}{2}\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +\frac {1}{2}}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (8 n +12\right ) C a_{n} x^{n +\frac {3}{2}} &= \moverset {\infty }{\munderset {n =1}{\sum }}C a_{n -1} \left (8 n +4\right ) x^{n +\frac {1}{2}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C \,x^{n +\frac {3}{2}} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 C a_{n -1} x^{n +\frac {1}{2}}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C \,x^{n +\frac {3}{2}} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 C a_{n -1} x^{n +\frac {1}{2}}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C \,x^{n +\frac {5}{2}} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-4 C a_{n -2} x^{n +\frac {1}{2}}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{n +\frac {7}{2}} a_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-2 C a_{n -3} x^{n +\frac {1}{2}}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {2 C \,x^{n +\frac {9}{2}} a_{n}}{3}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {2 C a_{n -4} x^{n +\frac {1}{2}}}{3}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +\frac {11}{2}} a_{n}}{6}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {C a_{n -5} x^{n +\frac {1}{2}}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +\frac {13}{2}} a_{n}}{30}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {C a_{n -6} x^{n +\frac {1}{2}}}{30}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +\frac {15}{2}} a_{n}}{180}\right ) &= \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {C a_{n -7} x^{n +\frac {1}{2}}}{180}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +\frac {17}{2}} a_{n}}{1260}\right ) &= \moverset {\infty }{\munderset {n =8}{\sum }}\left (-\frac {C a_{n -8} x^{n +\frac {1}{2}}}{1260}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +\frac {15}{2}} b_{n} \left (2 n +1\right )}{2520}\right ) &= \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {b_{n -7} \left (2 n -13\right ) x^{n +\frac {1}{2}}}{2520}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +\frac {13}{2}} b_{n} \left (2 n +1\right )}{360}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {b_{n -6} \left (2 n -11\right ) x^{n +\frac {1}{2}}}{360}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +\frac {11}{2}} b_{n} \left (2 n +1\right )}{60}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {b_{n -5} \left (2 n -9\right ) x^{n +\frac {1}{2}}}{60}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +\frac {9}{2}} b_{n} \left (2 n +1\right )}{12}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {b_{n -4} \left (2 n -7\right ) x^{n +\frac {1}{2}}}{12}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +\frac {7}{2}} b_{n} \left (2 n +1\right )}{3}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {b_{n -3} \left (2 n -5\right ) x^{n +\frac {1}{2}}}{3}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +\frac {5}{2}} b_{n} \left (-2 n -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}b_{n -2} \left (-2 n +3\right ) x^{n +\frac {1}{2}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 n -2\right ) b_{n} x^{n +\frac {3}{2}} &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} \left (-4 n +2\right ) x^{n +\frac {1}{2}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {3 x^{n +\frac {5}{2}} b_{n}}{2}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {3 b_{n -2} x^{n +\frac {1}{2}}}{2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +\frac {9}{2}} b_{n}}{8} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {b_{n -4} x^{n +\frac {1}{2}}}{8} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +\frac {13}{2}} b_{n}}{240}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {b_{n -6} x^{n +\frac {1}{2}}}{240}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +\frac {17}{2}} b_{n}}{13440} &= \moverset {\infty }{\munderset {n =8}{\sum }}\frac {b_{n -8} x^{n +\frac {1}{2}}}{13440} \\ \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 \(n +\frac {1}{2}\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =1}{\sum }}C a_{n -1} \left (8 n +4\right ) x^{n +\frac {1}{2}}\right )+2 \left (\moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 C a_{n -1} x^{n +\frac {1}{2}}\right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-4 C a_{n -2} x^{n +\frac {1}{2}}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-2 C a_{n -3} x^{n +\frac {1}{2}}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {2 C a_{n -4} x^{n +\frac {1}{2}}}{3}\right )+\moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {C a_{n -5} x^{n +\frac {1}{2}}}{6}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {C a_{n -6} x^{n +\frac {1}{2}}}{30}\right )+\moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {C a_{n -7} x^{n +\frac {1}{2}}}{180}\right )+\moverset {\infty }{\munderset {n =8}{\sum }}\left (-\frac {C a_{n -8} x^{n +\frac {1}{2}}}{1260}\right )+\moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {b_{n -7} \left (2 n -13\right ) x^{n +\frac {1}{2}}}{2520}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {b_{n -6} \left (2 n -11\right ) x^{n +\frac {1}{2}}}{360}\right )+\moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {b_{n -5} \left (2 n -9\right ) x^{n +\frac {1}{2}}}{60}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {b_{n -4} \left (2 n -7\right ) x^{n +\frac {1}{2}}}{12}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {b_{n -3} \left (2 n -5\right ) x^{n +\frac {1}{2}}}{3}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}b_{n -2} \left (-2 n +3\right ) x^{n +\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} \left (-4 n +2\right ) x^{n +\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 n -2\right ) b_{n} x^{n +\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +\frac {1}{2}} b_{n} \left (4 n^{2}-1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 b_{n} x^{n +\frac {1}{2}}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {3 b_{n -2} x^{n +\frac {1}{2}}}{2}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {b_{n -4} x^{n +\frac {1}{2}}}{8}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {b_{n -6} x^{n +\frac {1}{2}}}{240}\right )+\left (\moverset {\infty }{\munderset {n =8}{\sum }}\frac {b_{n -8} x^{n +\frac {1}{2}}}{13440}\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 \[ 4 C -2 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C={\frac {1}{2}} \] For \(n=2\), Eq (2B) gives \[ 4 \left (-a_{0}+3 a_{1}\right ) C -\frac {5 b_{0}}{2}-6 b_{1}+8 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 8 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=0 \] For \(n=3\), Eq (2B) gives \[ 2 \left (-a_{0}-2 a_{1}+10 a_{2}\right ) C -\frac {b_{0}}{3}-\frac {9 b_{1}}{2}-10 b_{2}+24 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {13}{6}+24 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=-{\frac {13}{144}} \] For \(n=4\), Eq (2B) gives \[ \frac {2 \left (-a_{0}-3 a_{1}-6 a_{2}+42 a_{3}\right ) C}{3}+48 b_{4}+\frac {b_{0}}{24}-b_{1}-\frac {13 b_{2}}{2}-14 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {1715}{576}+48 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-{\frac {1715}{27648}} \] For \(n=5\), Eq (2B) gives \[ \frac {\left (-a_{0}-4 a_{1}-12 a_{2}-24 a_{3}+216 a_{4}\right ) C}{6}-\frac {17 b_{3}}{2}-18 b_{4}+80 b_{5}-\frac {b_{0}}{60}-\frac {b_{1}}{8}-\frac {5 b_{2}}{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {6565}{2304}+80 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=-{\frac {1313}{36864}} \] For \(n=6\), Eq (2B) gives \[ \frac {\left (-a_{0}-5 a_{1}-20 a_{2}-60 a_{3}-120 a_{4}+1320 a_{5}\right ) C}{30}-\frac {7 b_{2}}{24}-\frac {7 b_{3}}{3}-\frac {21 b_{4}}{2}-22 b_{5}+120 b_{6}-\frac {b_{0}}{144}-\frac {b_{1}}{20} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {2999423}{1382400}+120 b_{6} = 0 \] Solving the above for \(b_{6}\) gives \[ b_{6}=-{\frac {2999423}{165888000}} \] For \(n=7\), Eq (2B) gives \[ \frac {\left (-a_{0}-6 a_{1}-30 a_{2}-120 a_{3}-360 a_{4}-720 a_{5}+9360 a_{6}\right ) C}{180}-\frac {b_{1}}{80}-\frac {b_{2}}{12}-\frac {11 b_{3}}{24}-3 b_{4}-\frac {25 b_{5}}{2}-26 b_{6}+168 b_{7}-\frac {b_{0}}{2520} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {204656267}{145152000}+168 b_{7} = 0 \] Solving the above for \(b_{7}\) gives \[ b_{7}=-{\frac {204656267}{24385536000}} \] 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 \(C={\frac {1}{2}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= \frac {1}{2}\eslowast \left (x^{\frac {3}{2}} \left (1+\frac {3 x}{4}+\frac {x^{2}}{2}+\frac {103 x^{3}}{384}+\frac {669 x^{4}}{5120}+\frac {54731 x^{5}}{921600}+\frac {123443 x^{6}}{4838400}+\frac {30273113 x^{7}}{2890137600}+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (1-\frac {13 x^{3}}{144}-\frac {1715 x^{4}}{27648}-\frac {1313 x^{5}}{36864}-\frac {2999423 x^{6}}{165888000}-\frac {204656267 x^{7}}{24385536000}+O\left (x^{8}\right )\right ) \] 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^{\frac {3}{2}} \left (1+\frac {3 x}{4}+\frac {x^{2}}{2}+\frac {103 x^{3}}{384}+\frac {669 x^{4}}{5120}+\frac {54731 x^{5}}{921600}+\frac {123443 x^{6}}{4838400}+\frac {30273113 x^{7}}{2890137600}+O\left (x^{8}\right )\right ) + c_{2} \left (\frac {1}{2}\eslowast \left (x^{\frac {3}{2}} \left (1+\frac {3 x}{4}+\frac {x^{2}}{2}+\frac {103 x^{3}}{384}+\frac {669 x^{4}}{5120}+\frac {54731 x^{5}}{921600}+\frac {123443 x^{6}}{4838400}+\frac {30273113 x^{7}}{2890137600}+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (1-\frac {13 x^{3}}{144}-\frac {1715 x^{4}}{27648}-\frac {1313 x^{5}}{36864}-\frac {2999423 x^{6}}{165888000}-\frac {204656267 x^{7}}{24385536000}+O\left (x^{8}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {3}{2}} \left (1+\frac {3 x}{4}+\frac {x^{2}}{2}+\frac {103 x^{3}}{384}+\frac {669 x^{4}}{5120}+\frac {54731 x^{5}}{921600}+\frac {123443 x^{6}}{4838400}+\frac {30273113 x^{7}}{2890137600}+O\left (x^{8}\right )\right )+c_{2} \left (\frac {x^{\frac {3}{2}} \left (1+\frac {3 x}{4}+\frac {x^{2}}{2}+\frac {103 x^{3}}{384}+\frac {669 x^{4}}{5120}+\frac {54731 x^{5}}{921600}+\frac {123443 x^{6}}{4838400}+\frac {30273113 x^{7}}{2890137600}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{2}+\sqrt {x}\, \left (1-\frac {13 x^{3}}{144}-\frac {1715 x^{4}}{27648}-\frac {1313 x^{5}}{36864}-\frac {2999423 x^{6}}{165888000}-\frac {204656267 x^{7}}{24385536000}+O\left (x^{8}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {3}{2}} \left (1+\frac {3 x}{4}+\frac {x^{2}}{2}+\frac {103 x^{3}}{384}+\frac {669 x^{4}}{5120}+\frac {54731 x^{5}}{921600}+\frac {123443 x^{6}}{4838400}+\frac {30273113 x^{7}}{2890137600}+O\left (x^{8}\right )\right )+c_{2} \left (\frac {x^{\frac {3}{2}} \left (1+\frac {3 x}{4}+\frac {x^{2}}{2}+\frac {103 x^{3}}{384}+\frac {669 x^{4}}{5120}+\frac {54731 x^{5}}{921600}+\frac {123443 x^{6}}{4838400}+\frac {30273113 x^{7}}{2890137600}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{2}+\sqrt {x}\, \left (1-\frac {13 x^{3}}{144}-\frac {1715 x^{4}}{27648}-\frac {1313 x^{5}}{36864}-\frac {2999423 x^{6}}{165888000}-\frac {204656267 x^{7}}{24385536000}+O\left (x^{8}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {3}{2}} \left (1+\frac {3 x}{4}+\frac {x^{2}}{2}+\frac {103 x^{3}}{384}+\frac {669 x^{4}}{5120}+\frac {54731 x^{5}}{921600}+\frac {123443 x^{6}}{4838400}+\frac {30273113 x^{7}}{2890137600}+O\left (x^{8}\right )\right )+c_{2} \left (\frac {x^{\frac {3}{2}} \left (1+\frac {3 x}{4}+\frac {x^{2}}{2}+\frac {103 x^{3}}{384}+\frac {669 x^{4}}{5120}+\frac {54731 x^{5}}{921600}+\frac {123443 x^{6}}{4838400}+\frac {30273113 x^{7}}{2890137600}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{2}+\sqrt {x}\, \left (1-\frac {13 x^{3}}{144}-\frac {1715 x^{4}}{27648}-\frac {1313 x^{5}}{36864}-\frac {2999423 x^{6}}{165888000}-\frac {204656267 x^{7}}{24385536000}+O\left (x^{8}\right )\right )\right ) \] 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 -> trying with_periodic_functions in the coefficients --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 5 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 -> trying with_periodic_functions in the coefficients --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 5 <- 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 -> trying with_periodic_functions in the coefficients 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 -> trying with_periodic_functions in the 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 with_periodic_functions in the coefficients --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 5 --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3`[0, y]
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 81
Order:=8; dsolve(4*x^2*diff(y(x),x$2)-4*x*exp(x)*diff(y(x),x)+3*cos(x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (x \left (1+\frac {3}{4} x +\frac {1}{2} x^{2}+\frac {103}{384} x^{3}+\frac {669}{5120} x^{4}+\frac {54731}{921600} x^{5}+\frac {123443}{4838400} x^{6}+\frac {30273113}{2890137600} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{1} +c_{2} \left (\ln \left (x \right ) \left (\frac {1}{2} x +\frac {3}{8} x^{2}+\frac {1}{4} x^{3}+\frac {103}{768} x^{4}+\frac {669}{10240} x^{5}+\frac {54731}{1843200} x^{6}+\frac {123443}{9676800} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (1+x +\frac {3}{4} x^{2}+\frac {59}{144} x^{3}+\frac {5701}{27648} x^{4}+\frac {17519}{184320} x^{5}+\frac {6852157}{165888000} x^{6}+\frac {417496453}{24385536000} x^{7}+\operatorname {O}\left (x^{8}\right )\right )\right )\right ) \sqrt {x} \]
✓ Solution by Mathematica
Time used: 0.146 (sec). Leaf size: 146
AsymptoticDSolveValue[4*x^2*y''[x]-4*x*Exp[x]*y'[x]+3*Cos[x]*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_2 \left (\frac {123443 x^{15/2}}{4838400}+\frac {54731 x^{13/2}}{921600}+\frac {669 x^{11/2}}{5120}+\frac {103 x^{9/2}}{384}+\frac {x^{7/2}}{2}+\frac {3 x^{5/2}}{4}+x^{3/2}\right )+c_1 \left (\frac {\left (54731 x^5+120420 x^4+247200 x^3+460800 x^2+691200 x+921600\right ) x^{3/2} \log (x)}{1843200}+\frac {\left (1926367 x^6+4929300 x^5+11958000 x^4+26496000 x^3+62208000 x^2+82944000 x+165888000\right ) \sqrt {x}}{165888000}\right ) \]