Internal problem ID [1413]
Internal file name [OUTPUT/1414_Sunday_June_05_2022_02_15_47_AM_30723664/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 7 Series Solutions of Linear Second Equations. 7.6 THE METHOD OF
FROBENIUS III. Exercises 7.7. Page 389
Problem number: Example 7.7.1 page 381.
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]]
\[ \boxed {2 x^{2} \left (x +2\right ) y^{\prime \prime }-x \left (4-7 x \right ) y^{\prime }-\left (5-3 x \right ) y=0} \] 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. \[ \left (2 x^{3}+4 x^{2}\right ) y^{\prime \prime }+\left (7 x^{2}-4 x \right ) y^{\prime }+\left (3 x -5\right ) y = 0 \] 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 {7 x -4}{2 x \left (x +2\right )}\\ q(x) &= \frac {3 x -5}{2 x^{2} \left (x +2\right )}\\ \end {align*}
| \(p(x)=\frac {7 x -4}{2 x \left (x +2\right )}\) | |
| singularity | type |
| \(x = -2\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=\frac {3 x -5}{2 x^{2} \left (x +2\right )}\) | |
| singularity | type |
| \(x = -2\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-2, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 2 x^{2} \left (x +2\right ) y^{\prime \prime }+\left (7 x^{2}-4 x \right ) y^{\prime }+\left (3 x -5\right ) y = 0 \] 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} 2 x^{2} \left (x +2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (7 x^{2}-4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (3 x -5\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\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 }}7 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 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 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 }}2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}7 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}7 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} 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 =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\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 =1}{\sum }}7 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 =1}{\sum }}3 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-4 x^{n +r} a_{n} \left (n +r \right )-5 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )-4 x^{r} a_{0} r -5 a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )-4 x^{r} r -5 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}-8 r -5\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2}-8 r -5 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {5}{2}}\\ r_2 &= -{\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (4 r^{2}-8 r -5\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {5}{2}}, -{\frac {1}{2}}\right ]\).
Since \(r_1 - r_2 = 3\) 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 {5}{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 )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{\sqrt {x}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {5}{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 \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+4 a_{n} \left (n +r \right ) \left (n +r -1\right )+7 a_{n -1} \left (n +r -1\right )-4 a_{n} \left (n +r \right )+3 a_{n -1}-5 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {\left (n +r \right ) a_{n -1}}{2 n +2 r -5}\tag {4} \] Which for the root \(r = {\frac {5}{2}}\) becomes \[ a_{n} = -\frac {\left (2 n +5\right ) a_{n -1}}{4 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {5}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=\frac {-1-r}{-3+2 r} \] Which for the root \(r = {\frac {5}{2}}\) becomes \[ a_{1}=-{\frac {7}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{-3+2 r}\) | \(-{\frac {7}{4}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r^{2}+3 r +2}{4 r^{2}-8 r +3} \] Which for the root \(r = {\frac {5}{2}}\) becomes \[ a_{2}={\frac {63}{32}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{-3+2 r}\) | \(-{\frac {7}{4}}\) |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{4 r^{2}-8 r +3}\) | \(\frac {63}{32}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}-12 r^{2}-2 r +3} \] Which for the root \(r = {\frac {5}{2}}\) becomes \[ a_{3}=-{\frac {231}{128}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{-3+2 r}\) | \(-{\frac {7}{4}}\) |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{4 r^{2}-8 r +3}\) | \(\frac {63}{32}\) |
| \(a_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}-12 r^{2}-2 r +3}\) | \(-{\frac {231}{128}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{16 r^{4}-40 r^{2}+9} \] Which for the root \(r = {\frac {5}{2}}\) becomes \[ a_{4}={\frac {3003}{2048}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{-3+2 r}\) | \(-{\frac {7}{4}}\) |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{4 r^{2}-8 r +3}\) | \(\frac {63}{32}\) |
| \(a_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}-12 r^{2}-2 r +3}\) | \(-{\frac {231}{128}}\) |
| \(a_{4}\) | \(\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{16 r^{4}-40 r^{2}+9}\) | \(\frac {3003}{2048}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-r^{5}-15 r^{4}-85 r^{3}-225 r^{2}-274 r -120}{\left (16 r^{4}-40 r^{2}+9\right ) \left (5+2 r \right )} \] Which for the root \(r = {\frac {5}{2}}\) becomes \[ a_{5}=-{\frac {9009}{8192}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{-3+2 r}\) | \(-{\frac {7}{4}}\) |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{4 r^{2}-8 r +3}\) | \(\frac {63}{32}\) |
| \(a_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}-12 r^{2}-2 r +3}\) | \(-{\frac {231}{128}}\) |
| \(a_{4}\) | \(\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{16 r^{4}-40 r^{2}+9}\) | \(\frac {3003}{2048}\) |
| \(a_{5}\) | \(\frac {-r^{5}-15 r^{4}-85 r^{3}-225 r^{2}-274 r -120}{\left (16 r^{4}-40 r^{2}+9\right ) \left (5+2 r \right )}\) | \(-{\frac {9009}{8192}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {5}{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}\dots \right ) \\ &= x^{\frac {5}{2}} \left (1-\frac {7 x}{4}+\frac {63 x^{2}}{32}-\frac {231 x^{3}}{128}+\frac {3003 x^{4}}{2048}-\frac {9009 x^{5}}{8192}+O\left (x^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=3\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{3}\left (r \right )\). 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_{3} \\ &= \frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}-12 r^{2}-2 r +3} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}-12 r^{2}-2 r +3}&= \lim _{r\rightarrow -{\frac {1}{2}}}\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}-12 r^{2}-2 r +3}\\ &= \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 \(2 x^{2} \left (x +2\right ) y^{\prime \prime }+\left (7 x^{2}-4 x \right ) y^{\prime }+\left (3 x -5\right ) y = 0\) gives \[ 2 x^{2} \left (x +2\right ) \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (7 x^{2}-4 x \right ) \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+\left (3 x -5\right ) \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (2 x^{2} \left (x +2\right ) y_{1}^{\prime \prime }\left (x \right )+\left (7 x^{2}-4 x \right ) y_{1}^{\prime }\left (x \right )+\left (3 x -5\right ) y_{1}\left (x \right )\right ) \ln \left (x \right )+2 x^{2} \left (x +2\right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (7 x^{2}-4 x \right ) y_{1}\left (x \right )}{x}\right ) C +2 x^{2} \left (x +2\right ) \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 )+\left (7 x^{2}-4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (3 x -5\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 \[ 2 x^{2} \left (x +2\right ) y_{1}^{\prime \prime }\left (x \right )+\left (7 x^{2}-4 x \right ) y_{1}^{\prime }\left (x \right )+\left (3 x -5\right ) y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (2 x^{2} \left (x +2\right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (7 x^{2}-4 x \right ) y_{1}\left (x \right )}{x}\right ) C +2 x^{2} \left (x +2\right ) \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 )+\left (7 x^{2}-4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (3 x -5\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 (4 x \left (x +2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )+\left (5 x -8\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C +2 \left (x^{3}+2 x^{2}\right ) \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 )+\left (7 x^{2}-4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )+\left (3 x -5\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Since \(r_{1} = {\frac {5}{2}}\) and \(r_{2} = -{\frac {1}{2}}\) then the above becomes \begin{equation} \tag{10} \left (4 x \left (x +2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{\frac {3}{2}+n} a_{n} \left (n +\frac {5}{2}\right )\right )+\left (5 x -8\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {5}{2}}\right )\right ) C +2 \left (x^{3}+2 x^{2}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-\frac {5}{2}+n} b_{n} \left (n -\frac {1}{2}\right ) \left (-\frac {3}{2}+n \right )\right )+\left (7 x^{2}-4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-\frac {3}{2}+n} b_{n} \left (n -\frac {1}{2}\right )\right )+\left (3 x -5\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {1}{2}}\right ) = 0 \end{equation} Expanding \(4 C \,x^{\frac {5}{2}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} 4 C \,x^{\frac {5}{2}} &= 4 C \,x^{\frac {5}{2}} + \dots \\ &= 4 C \,x^{\frac {5}{2}} \end {align*}
Expanding \(5 C \,x^{\frac {7}{2}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} 5 C \,x^{\frac {7}{2}} &= 5 C \,x^{\frac {7}{2}} + \dots \\ &= 5 C \,x^{\frac {7}{2}} \end {align*}
Expanding \(-8 C \,x^{\frac {5}{2}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -8 C \,x^{\frac {5}{2}} &= -8 C \,x^{\frac {5}{2}} + \dots \\ &= -8 C \,x^{\frac {5}{2}} \end {align*}
Expanding \(\frac {\sqrt {x}}{2}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \frac {\sqrt {x}}{2} &= \frac {\sqrt {x}}{2} + \dots \\ &= \frac {\sqrt {x}}{2} \end {align*}
Expanding \(\frac {1}{\sqrt {x}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \frac {1}{\sqrt {x}} &= \frac {1}{\sqrt {x}} + \dots \\ &= \frac {1}{\sqrt {x}} \end {align*}
Expanding \(\frac {7 \sqrt {x}}{2}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \frac {7 \sqrt {x}}{2} &= \frac {7 \sqrt {x}}{2} + \dots \\ &= \frac {7 \sqrt {x}}{2} \end {align*}
Expanding \(-\frac {2}{\sqrt {x}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -\frac {2}{\sqrt {x}} &= -\frac {2}{\sqrt {x}} + \dots \\ &= -\frac {2}{\sqrt {x}} \end {align*}
Expanding \(3 \sqrt {x}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} 3 \sqrt {x} &= 3 \sqrt {x} + \dots \\ &= 3 \sqrt {x} \end {align*}
Expanding \(-\frac {5}{\sqrt {x}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -\frac {5}{\sqrt {x}} &= -\frac {5}{\sqrt {x}} + \dots \\ &= -\frac {5}{\sqrt {x}} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (4 n +10\right ) C a_{n} x^{\frac {7}{2}+n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (8 n +20\right ) C a_{n} x^{n +\frac {5}{2}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 C \,x^{\frac {7}{2}+n} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 C \,x^{n +\frac {5}{2}} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +\frac {1}{2}} b_{n} \left (4 n^{2}-8 n +3\right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -\frac {1}{2}} b_{n} \left (4 n^{2}-8 n +3\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {7 x^{n +\frac {1}{2}} b_{n} \left (2 n -1\right )}{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 }}3 x^{n +\frac {1}{2}} b_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 b_{n} x^{n -\frac {1}{2}}\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 (4 n +10\right ) C a_{n} x^{\frac {7}{2}+n} &= \moverset {\infty }{\munderset {n =4}{\sum }}C a_{n -4} \left (4 n -6\right ) x^{n -\frac {1}{2}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (8 n +20\right ) C a_{n} x^{n +\frac {5}{2}} &= \moverset {\infty }{\munderset {n =3}{\sum }}C a_{n -3} \left (8 n -4\right ) x^{n -\frac {1}{2}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}5 C \,x^{\frac {7}{2}+n} a_{n} &= \moverset {\infty }{\munderset {n =4}{\sum }}5 C a_{n -4} x^{n -\frac {1}{2}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 C \,x^{n +\frac {5}{2}} a_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-8 C a_{n -3} x^{n -\frac {1}{2}}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +\frac {1}{2}} b_{n} \left (4 n^{2}-8 n +3\right )}{2} &= \moverset {\infty }{\munderset {n =1}{\sum }}\frac {b_{n -1} \left (4 \left (n -1\right )^{2}-8 n +11\right ) x^{n -\frac {1}{2}}}{2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {7 x^{n +\frac {1}{2}} b_{n} \left (2 n -1\right )}{2} &= \moverset {\infty }{\munderset {n =1}{\sum }}\frac {7 b_{n -1} \left (-3+2 n \right ) x^{n -\frac {1}{2}}}{2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +\frac {1}{2}} b_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}3 b_{n -1} x^{n -\frac {1}{2}} \\ \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 =4}{\sum }}C a_{n -4} \left (4 n -6\right ) x^{n -\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}C a_{n -3} \left (8 n -4\right ) x^{n -\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}5 C a_{n -4} x^{n -\frac {1}{2}}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-8 C a_{n -3} x^{n -\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\frac {b_{n -1} \left (4 \left (n -1\right )^{2}-8 n +11\right ) x^{n -\frac {1}{2}}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -\frac {1}{2}} b_{n} \left (4 n^{2}-8 n +3\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\frac {7 b_{n -1} \left (-3+2 n \right ) x^{n -\frac {1}{2}}}{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 =1}{\sum }}3 b_{n -1} x^{n -\frac {1}{2}}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 b_{n} x^{n -\frac {1}{2}}\right ) = 0 \end{equation} For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=1\), Eq (2B) gives \[ b_{0}-8 b_{1} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 1-8 b_{1} = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}={\frac {1}{8}} \] For \(n=2\), Eq (2B) gives \[ 6 b_{1}-8 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {3}{4}-8 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}={\frac {3}{32}} \] For \(n=N\), where \(N=3\) which is the difference between the two roots, we are free to choose \(b_{3} = 0\). Hence for \(n=3\), Eq (2B) gives \[ 12 C +\frac {45}{32} = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-{\frac {15}{128}} \] For \(n=4\), Eq (2B) gives \[ \left (15 a_{0}+20 a_{1}\right ) C +28 b_{3}+16 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {75}{32}+16 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-{\frac {75}{512}} \] For \(n=5\), Eq (2B) gives \[ \left (19 a_{1}+28 a_{2}\right ) C +45 b_{4}+40 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {9375}{1024}+40 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}={\frac {1875}{8192}} \] 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 \(C=-{\frac {15}{128}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= -\frac {15}{128}\eslowast \left (x^{\frac {5}{2}} \left (1-\frac {7 x}{4}+\frac {63 x^{2}}{32}-\frac {231 x^{3}}{128}+\frac {3003 x^{4}}{2048}-\frac {9009 x^{5}}{8192}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+\frac {1+\frac {x}{8}+\frac {3 x^{2}}{32}-\frac {75 x^{4}}{512}+\frac {1875 x^{5}}{8192}+O\left (x^{6}\right )}{\sqrt {x}} \] 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 {5}{2}} \left (1-\frac {7 x}{4}+\frac {63 x^{2}}{32}-\frac {231 x^{3}}{128}+\frac {3003 x^{4}}{2048}-\frac {9009 x^{5}}{8192}+O\left (x^{6}\right )\right ) + c_{2} \left (-\frac {15}{128}\eslowast \left (x^{\frac {5}{2}} \left (1-\frac {7 x}{4}+\frac {63 x^{2}}{32}-\frac {231 x^{3}}{128}+\frac {3003 x^{4}}{2048}-\frac {9009 x^{5}}{8192}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+\frac {1+\frac {x}{8}+\frac {3 x^{2}}{32}-\frac {75 x^{4}}{512}+\frac {1875 x^{5}}{8192}+O\left (x^{6}\right )}{\sqrt {x}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {5}{2}} \left (1-\frac {7 x}{4}+\frac {63 x^{2}}{32}-\frac {231 x^{3}}{128}+\frac {3003 x^{4}}{2048}-\frac {9009 x^{5}}{8192}+O\left (x^{6}\right )\right )+c_{2} \left (-\frac {15 x^{\frac {5}{2}} \left (1-\frac {7 x}{4}+\frac {63 x^{2}}{32}-\frac {231 x^{3}}{128}+\frac {3003 x^{4}}{2048}-\frac {9009 x^{5}}{8192}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{128}+\frac {1+\frac {x}{8}+\frac {3 x^{2}}{32}-\frac {75 x^{4}}{512}+\frac {1875 x^{5}}{8192}+O\left (x^{6}\right )}{\sqrt {x}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {5}{2}} \left (1-\frac {7 x}{4}+\frac {63 x^{2}}{32}-\frac {231 x^{3}}{128}+\frac {3003 x^{4}}{2048}-\frac {9009 x^{5}}{8192}+O\left (x^{6}\right )\right )+c_{2} \left (-\frac {15 x^{\frac {5}{2}} \left (1-\frac {7 x}{4}+\frac {63 x^{2}}{32}-\frac {231 x^{3}}{128}+\frac {3003 x^{4}}{2048}-\frac {9009 x^{5}}{8192}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{128}+\frac {1+\frac {x}{8}+\frac {3 x^{2}}{32}-\frac {75 x^{4}}{512}+\frac {1875 x^{5}}{8192}+O\left (x^{6}\right )}{\sqrt {x}}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {5}{2}} \left (1-\frac {7 x}{4}+\frac {63 x^{2}}{32}-\frac {231 x^{3}}{128}+\frac {3003 x^{4}}{2048}-\frac {9009 x^{5}}{8192}+O\left (x^{6}\right )\right )+c_{2} \left (-\frac {15 x^{\frac {5}{2}} \left (1-\frac {7 x}{4}+\frac {63 x^{2}}{32}-\frac {231 x^{3}}{128}+\frac {3003 x^{4}}{2048}-\frac {9009 x^{5}}{8192}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{128}+\frac {1+\frac {x}{8}+\frac {3 x^{2}}{32}-\frac {75 x^{4}}{512}+\frac {1875 x^{5}}{8192}+O\left (x^{6}\right )}{\sqrt {x}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x^{2} \left (x +2\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (7 x^{2}-4 x \right ) y^{\prime }+\left (3 x -5\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {\left (3 x -5\right ) y}{2 x^{2} \left (x +2\right )}-\frac {\left (7 x -4\right ) y^{\prime }}{2 x \left (x +2\right )} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {\left (7 x -4\right ) y^{\prime }}{2 x \left (x +2\right )}+\frac {\left (3 x -5\right ) y}{2 x^{2} \left (x +2\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {7 x -4}{2 x \left (x +2\right )}, P_{3}\left (x \right )=\frac {3 x -5}{2 x^{2} \left (x +2\right )}\right ] \\ {} & \circ & \left (x +2\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-2 \\ {} & {} & \left (\left (x +2\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-2}}}=\frac {9}{2} \\ {} & \circ & \left (x +2\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-2 \\ {} & {} & \left (\left (x +2\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-2}}}=0 \\ {} & \circ & x =-2\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-2 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 2 x^{2} \left (x +2\right ) \left (\frac {d}{d x}y^{\prime }\right )+x \left (7 x -4\right ) y^{\prime }+\left (3 x -5\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -2\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (2 u^{3}-8 u^{2}+8 u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (7 u^{2}-32 u +36\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (3 u -11\right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 4 a_{0} r \left (7+2 r \right ) u^{-1+r}+\left (4 a_{1} \left (1+r \right ) \left (9+2 r \right )-a_{0} \left (8 r^{2}+24 r +11\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (4 a_{k +1} \left (k +r +1\right ) \left (2 k +9+2 r \right )-a_{k} \left (8 k^{2}+16 k r +8 r^{2}+24 k +24 r +11\right )+a_{k -1} \left (2 k +1+2 r \right ) \left (k +r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 4 r \left (7+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {7}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 4 a_{1} \left (1+r \right ) \left (9+2 r \right )-a_{0} \left (8 r^{2}+24 r +11\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (-4 a_{k}+a_{k -1}+4 a_{k +1}\right ) k^{2}+\left (4 \left (-4 a_{k}+a_{k -1}+4 a_{k +1}\right ) r -24 a_{k}+a_{k -1}+44 a_{k +1}\right ) k +2 \left (-4 a_{k}+a_{k -1}+4 a_{k +1}\right ) r^{2}+\left (-24 a_{k}+a_{k -1}+44 a_{k +1}\right ) r -11 a_{k}+36 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 2 \left (-4 a_{k +1}+a_{k}+4 a_{k +2}\right ) \left (k +1\right )^{2}+\left (4 \left (-4 a_{k +1}+a_{k}+4 a_{k +2}\right ) r -24 a_{k +1}+a_{k}+44 a_{k +2}\right ) \left (k +1\right )+2 \left (-4 a_{k +1}+a_{k}+4 a_{k +2}\right ) r^{2}+\left (-24 a_{k +1}+a_{k}+44 a_{k +2}\right ) r -11 a_{k +1}+36 a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}+4 k r a_{k}-16 k r a_{k +1}+2 r^{2} a_{k}-8 r^{2} a_{k +1}+5 k a_{k}-40 k a_{k +1}+5 r a_{k}-40 r a_{k +1}+3 a_{k}-43 a_{k +1}}{4 \left (2 k^{2}+4 k r +2 r^{2}+15 k +15 r +22\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}+5 k a_{k}-40 k a_{k +1}+3 a_{k}-43 a_{k +1}}{4 \left (2 k^{2}+15 k +22\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}+5 k a_{k}-40 k a_{k +1}+3 a_{k}-43 a_{k +1}}{4 \left (2 k^{2}+15 k +22\right )}, 36 a_{1}-11 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k}, a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}+5 k a_{k}-40 k a_{k +1}+3 a_{k}-43 a_{k +1}}{4 \left (2 k^{2}+15 k +22\right )}, 36 a_{1}-11 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {7}{2} \\ {} & {} & a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}-9 k a_{k}+16 k a_{k +1}+10 a_{k}-a_{k +1}}{4 \left (2 k^{2}+k -6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {7}{2} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {7}{2}}, a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}-9 k a_{k}+16 k a_{k +1}+10 a_{k}-a_{k +1}}{4 \left (2 k^{2}+k -6\right )}, -20 a_{1}-25 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k -\frac {7}{2}}, a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}-9 k a_{k}+16 k a_{k +1}+10 a_{k}-a_{k +1}}{4 \left (2 k^{2}+k -6\right )}, -20 a_{1}-25 a_{0}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +2\right )^{k -\frac {7}{2}}\right ), a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}+5 k a_{k}-40 k a_{k +1}+3 a_{k}-43 a_{k +1}}{4 \left (2 k^{2}+15 k +22\right )}, 36 a_{1}-11 a_{0}=0, b_{k +2}=-\frac {2 k^{2} b_{k}-8 k^{2} b_{k +1}-9 k b_{k}+16 k b_{k +1}+10 b_{k}-b_{k +1}}{4 \left (2 k^{2}+k -6\right )}, -20 b_{1}-25 b_{0}=0\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 65
Order:=6; dsolve(2*x^2*(2+x)*diff(y(x),x$2)-x*(4-7*x)*diff(y(x),x)-(5-3*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} x^{3} \left (1-\frac {7}{4} x +\frac {63}{32} x^{2}-\frac {231}{128} x^{3}+\frac {3003}{2048} x^{4}-\frac {9009}{8192} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (-\frac {45}{32} x^{3}+\frac {315}{128} x^{4}-\frac {2835}{1024} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (12+\frac {3}{2} x +\frac {9}{8} x^{2}-\frac {981}{64} x^{3}+\frac {6417}{256} x^{4}-\frac {28089}{1024} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right )}{\sqrt {x}} \]
✓ Solution by Mathematica
Time used: 0.055 (sec). Leaf size: 98
AsymptoticDSolveValue[2*x^2*(2+x)*y''[x]-x*(4-7*x)*y'[x]-(5-3*x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {3003 x^{13/2}}{2048}-\frac {231 x^{11/2}}{128}+\frac {63 x^{9/2}}{32}-\frac {7 x^{7/2}}{4}+x^{5/2}\right )+c_1 \left (\frac {15}{512} (7 x-4) x^{5/2} \log (x)+\frac {809 x^4-548 x^3+96 x^2+128 x+1024}{1024 \sqrt {x}}\right ) \]