Internal problem ID [2943]
Internal file name [OUTPUT/2435_Sunday_June_05_2022_03_10_44_AM_75711485/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 11.5. page
771
Problem number: (d).
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]]
\[ \boxed {x^{2} y^{\prime \prime }+\left (4 x +\frac {1}{2} x^{2}-\frac {1}{3} x^{3}\right ) y^{\prime }-\frac {7 y}{4}=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ x^{2} y^{\prime \prime }+\left (4 x +\frac {1}{2} x^{2}-\frac {1}{3} x^{3}\right ) y^{\prime }-\frac {7 y}{4} = 0 \] 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 {2 x^{2}-3 x -24}{6 x}\\ q(x) &= -\frac {7}{4 x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty , -\infty ]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} y^{\prime \prime }+\left (4 x +\frac {1}{2} x^{2}-\frac {1}{3} x^{3}\right ) y^{\prime }-\frac {7 y}{4} = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (4 x +\frac {1}{2} x^{2}-\frac {1}{3} x^{3}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-\frac {7 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )}{4} = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +2} a_{n} \left (n +r \right )}{3}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{1+n +r} a_{n} \left (n +r \right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {7 a_{n} x^{n +r}}{4}\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 +2} a_{n} \left (n +r \right )}{3}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) x^{n +r}}{3}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{1+n +r} a_{n} \left (n +r \right )}{2} &= \moverset {\infty }{\munderset {n =1}{\sum }}\frac {a_{n -1} \left (n +r -1\right ) x^{n +r}}{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 \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) x^{n +r}}{3}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\frac {a_{n -1} \left (n +r -1\right ) x^{n +r}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {7 a_{n} x^{n +r}}{4}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+4 x^{n +r} a_{n} \left (n +r \right )-\frac {7 a_{n} x^{n +r}}{4} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+4 x^{r} a_{0} r -\frac {7 a_{0} x^{r}}{4} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+4 x^{r} r -\frac {7 x^{r}}{4}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \frac {\left (4 r^{2}+12 r -7\right ) x^{r}}{4} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}+3 r -\frac {7}{4} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= -{\frac {7}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \frac {\left (4 r^{2}+12 r -7\right ) x^{r}}{4} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, -{\frac {7}{2}}\right ]\).
Since \(r_1 - r_2 = 4\) 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 ) &= \sqrt {x}\, \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}}{x^{\frac {7}{2}}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {1}{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 {7}{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 \[ a_{1} = -\frac {2 r}{4 r^{2}+20 r +9} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )-\frac {a_{n -2} \left (n +r -2\right )}{3}+\frac {a_{n -1} \left (n +r -1\right )}{2}+4 a_{n} \left (n +r \right )-\frac {7 a_{n}}{4} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {\frac {4}{3} n a_{n -2}-2 n a_{n -1}+\frac {4}{3} r a_{n -2}-2 r a_{n -1}-\frac {8}{3} a_{n -2}+2 a_{n -1}}{4 n^{2}+8 n r +4 r^{2}+12 n +12 r -7}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = \frac {\left (4 a_{n -2}-6 a_{n -1}\right ) n -6 a_{n -2}+3 a_{n -1}}{12 n \left (n +4\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{4 r^{2}+20 r +9}\) | \(-{\frac {1}{20}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {16 r^{3}+92 r^{2}+48 r}{48 r^{4}+576 r^{3}+2184 r^{2}+2736 r +891} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}={\frac {49}{2880}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{4 r^{2}+20 r +9}\) | \(-{\frac {1}{20}}\) |
| \(a_{2}\) | \(\frac {16 r^{3}+92 r^{2}+48 r}{48 r^{4}+576 r^{3}+2184 r^{2}+2736 r +891}\) | \(\frac {49}{2880}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {8 r \left (8 r^{3}+63 r^{2}+119 r +57\right )}{3 \left (16 r^{4}+192 r^{3}+728 r^{2}+912 r +297\right ) \left (4 r^{2}+36 r +65\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}=-{\frac {533}{241920}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{4 r^{2}+20 r +9}\) | \(-{\frac {1}{20}}\) |
| \(a_{2}\) | \(\frac {16 r^{3}+92 r^{2}+48 r}{48 r^{4}+576 r^{3}+2184 r^{2}+2736 r +891}\) | \(\frac {49}{2880}\) |
| \(a_{3}\) | \(-\frac {8 r \left (8 r^{3}+63 r^{2}+119 r +57\right )}{3 \left (16 r^{4}+192 r^{3}+728 r^{2}+912 r +297\right ) \left (4 r^{2}+36 r +65\right )}\) | \(-{\frac {533}{241920}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {16 r \left (16 r^{5}+292 r^{4}+1869 r^{3}+5123 r^{2}+5876 r +2073\right )}{9 \left (16 r^{4}+192 r^{3}+728 r^{2}+912 r +297\right ) \left (4 r^{2}+36 r +65\right ) \left (4 r^{2}+44 r +105\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}={\frac {277}{491520}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{4 r^{2}+20 r +9}\) | \(-{\frac {1}{20}}\) |
| \(a_{2}\) | \(\frac {16 r^{3}+92 r^{2}+48 r}{48 r^{4}+576 r^{3}+2184 r^{2}+2736 r +891}\) | \(\frac {49}{2880}\) |
| \(a_{3}\) | \(-\frac {8 r \left (8 r^{3}+63 r^{2}+119 r +57\right )}{3 \left (16 r^{4}+192 r^{3}+728 r^{2}+912 r +297\right ) \left (4 r^{2}+36 r +65\right )}\) | \(-{\frac {533}{241920}}\) |
| \(a_{4}\) | \(\frac {16 r \left (16 r^{5}+292 r^{4}+1869 r^{3}+5123 r^{2}+5876 r +2073\right )}{9 \left (16 r^{4}+192 r^{3}+728 r^{2}+912 r +297\right ) \left (4 r^{2}+36 r +65\right ) \left (4 r^{2}+44 r +105\right )}\) | \(\frac {277}{491520}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {32 r \left (48 r^{6}+1056 r^{5}+8937 r^{4}+36942 r^{3}+77608 r^{2}+76571 r +26247\right )}{9 \left (16 r^{4}+192 r^{3}+728 r^{2}+912 r +297\right ) \left (4 r^{2}+36 r +65\right ) \left (4 r^{2}+44 r +105\right ) \left (4 r^{2}+52 r +153\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}=-{\frac {203759}{2388787200}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{4 r^{2}+20 r +9}\) | \(-{\frac {1}{20}}\) |
| \(a_{2}\) | \(\frac {16 r^{3}+92 r^{2}+48 r}{48 r^{4}+576 r^{3}+2184 r^{2}+2736 r +891}\) | \(\frac {49}{2880}\) |
| \(a_{3}\) | \(-\frac {8 r \left (8 r^{3}+63 r^{2}+119 r +57\right )}{3 \left (16 r^{4}+192 r^{3}+728 r^{2}+912 r +297\right ) \left (4 r^{2}+36 r +65\right )}\) | \(-{\frac {533}{241920}}\) |
| \(a_{4}\) | \(\frac {16 r \left (16 r^{5}+292 r^{4}+1869 r^{3}+5123 r^{2}+5876 r +2073\right )}{9 \left (16 r^{4}+192 r^{3}+728 r^{2}+912 r +297\right ) \left (4 r^{2}+36 r +65\right ) \left (4 r^{2}+44 r +105\right )}\) | \(\frac {277}{491520}\) |
| \(a_{5}\) | \(-\frac {32 r \left (48 r^{6}+1056 r^{5}+8937 r^{4}+36942 r^{3}+77608 r^{2}+76571 r +26247\right )}{9 \left (16 r^{4}+192 r^{3}+728 r^{2}+912 r +297\right ) \left (4 r^{2}+36 r +65\right ) \left (4 r^{2}+44 r +105\right ) \left (4 r^{2}+52 r +153\right )}\) | \(-{\frac {203759}{2388787200}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= \sqrt {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 ) \\ &= \sqrt {x}\, \left (1-\frac {x}{20}+\frac {49 x^{2}}{2880}-\frac {533 x^{3}}{241920}+\frac {277 x^{4}}{491520}-\frac {203759 x^{5}}{2388787200}+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 \(N=4\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{4}\left (r \right )\). 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_{4} \\ &= \frac {16 r \left (16 r^{5}+292 r^{4}+1869 r^{3}+5123 r^{2}+5876 r +2073\right )}{9 \left (16 r^{4}+192 r^{3}+728 r^{2}+912 r +297\right ) \left (4 r^{2}+36 r +65\right ) \left (4 r^{2}+44 r +105\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {16 r \left (16 r^{5}+292 r^{4}+1869 r^{3}+5123 r^{2}+5876 r +2073\right )}{9 \left (16 r^{4}+192 r^{3}+728 r^{2}+912 r +297\right ) \left (4 r^{2}+36 r +65\right ) \left (4 r^{2}+44 r +105\right )}&= \lim _{r\rightarrow -{\frac {7}{2}}}\frac {16 r \left (16 r^{5}+292 r^{4}+1869 r^{3}+5123 r^{2}+5876 r +2073\right )}{9 \left (16 r^{4}+192 r^{3}+728 r^{2}+912 r +297\right ) \left (4 r^{2}+36 r +65\right ) \left (4 r^{2}+44 r +105\right )}\\ &= \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 \(x^{2} y^{\prime \prime }+\left (4 x +\frac {1}{2} x^{2}-\frac {1}{3} x^{3}\right ) y^{\prime }-\frac {7 y}{4} = 0\) gives \[ x^{2} \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (4 x +\frac {1}{2} x^{2}-\frac {1}{3} x^{3}\right ) \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )-\frac {7 C y_{1}\left (x \right ) \ln \left (x \right )}{4}-\frac {7 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )}{4} = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (x^{2} y_{1}^{\prime \prime }\left (x \right )+\left (4 x +\frac {1}{2} x^{2}-\frac {1}{3} x^{3}\right ) y_{1}^{\prime }\left (x \right )-\frac {7 y_{1}\left (x \right )}{4}\right ) \ln \left (x \right )+x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (4 x +\frac {1}{2} x^{2}-\frac {1}{3} x^{3}\right ) y_{1}\left (x \right )}{x}\right ) C +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 )+\left (4 x +\frac {1}{2} x^{2}-\frac {1}{3} x^{3}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )-\frac {7 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )}{4} = 0 \end{equation} But since \(y_{1}\left (x \right )\) is a solution to the ode, then \[ x^{2} y_{1}^{\prime \prime }\left (x \right )+\left (4 x +\frac {1}{2} x^{2}-\frac {1}{3} x^{3}\right ) y_{1}^{\prime }\left (x \right )-\frac {7 y_{1}\left (x \right )}{4} = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (4 x +\frac {1}{2} x^{2}-\frac {1}{3} x^{3}\right ) y_{1}\left (x \right )}{x}\right ) C +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 )+\left (4 x +\frac {1}{2} x^{2}-\frac {1}{3} x^{3}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )-\frac {7 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )}{4} = 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 (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right ) x -\frac {\left (x^{2}-\frac {3}{2} x -9\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )}{3}\right ) C +\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}+\frac {\left (-2 x^{3}+3 x^{2}+24 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )}{6}-\frac {7 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )}{4} = 0 \end{equation} Since \(r_{1} = {\frac {1}{2}}\) and \(r_{2} = -{\frac {7}{2}}\) then the above becomes \begin{equation} \tag{10} \left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-\frac {1}{2}+n} a_{n} \left (n +\frac {1}{2}\right )\right ) x -\frac {\left (x^{2}-\frac {3}{2} x -9\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {1}{2}}\right )}{3}\right ) C +\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-\frac {11}{2}+n} b_{n} \left (n -\frac {7}{2}\right ) \left (-\frac {9}{2}+n \right )\right ) x^{2}+\frac {\left (-2 x^{3}+3 x^{2}+24 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-\frac {9}{2}+n} b_{n} \left (n -\frac {7}{2}\right )\right )}{6}-\frac {7 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {7}{2}}\right )}{4} = 0 \end{equation} Expanding \(-\frac {C \,x^{\frac {5}{2}}}{3}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -\frac {C \,x^{\frac {5}{2}}}{3} &= -\frac {C \,x^{\frac {5}{2}}}{3} + \dots \\ &= -\frac {C \,x^{\frac {5}{2}}}{3} \end {align*}
Expanding \(\frac {C \,x^{\frac {3}{2}}}{2}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \frac {C \,x^{\frac {3}{2}}}{2} &= \frac {C \,x^{\frac {3}{2}}}{2} + \dots \\ &= \frac {C \,x^{\frac {3}{2}}}{2} \end {align*}
Expanding \(3 C \sqrt {x}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} 3 C \sqrt {x} &= 3 C \sqrt {x} + \dots \\ &= 3 C \sqrt {x} \end {align*}
Expanding \(-\frac {1}{6 x^{\frac {3}{2}}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -\frac {1}{6 x^{\frac {3}{2}}} &= -\frac {1}{6 x^{\frac {3}{2}}} + \dots \\ &= -\frac {1}{6 x^{\frac {3}{2}}} \end {align*}
Expanding \(\frac {1}{4 x^{\frac {5}{2}}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \frac {1}{4 x^{\frac {5}{2}}} &= \frac {1}{4 x^{\frac {5}{2}}} + \dots \\ &= \frac {1}{4 x^{\frac {5}{2}}} \end {align*}
Expanding \(\frac {2}{x^{\frac {7}{2}}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \frac {2}{x^{\frac {7}{2}}} &= \frac {2}{x^{\frac {7}{2}}} + \dots \\ &= \frac {2}{x^{\frac {7}{2}}} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}C \,x^{n +\frac {1}{2}} a_{n} \left (2 n +1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +\frac {5}{2}} a_{n}}{3}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +\frac {3}{2}} a_{n}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 C \,x^{n +\frac {1}{2}} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n -\frac {7}{2}} b_{n} \left (4 n^{2}-32 n +63\right )}{4}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n -\frac {3}{2}} b_{n} \left (2 n -7\right )}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n -\frac {5}{2}} b_{n} \left (2 n -7\right )}{4}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (4 n -14\right ) b_{n} x^{n -\frac {7}{2}}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {7 b_{n} x^{n -\frac {7}{2}}}{4}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n -\frac {7}{2}\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -\frac {7}{2}}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}C \,x^{n +\frac {1}{2}} a_{n} \left (2 n +1\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}C a_{n -4} \left (2 n -7\right ) x^{n -\frac {7}{2}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +\frac {5}{2}} a_{n}}{3}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {C a_{n -6} x^{n -\frac {7}{2}}}{3}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +\frac {3}{2}} a_{n}}{2} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {C a_{n -5} x^{n -\frac {7}{2}}}{2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 C \,x^{n +\frac {1}{2}} a_{n} &= \moverset {\infty }{\munderset {n =4}{\sum }}3 C a_{n -4} x^{n -\frac {7}{2}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n -\frac {3}{2}} b_{n} \left (2 n -7\right )}{6}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {b_{n -2} \left (-11+2 n \right ) x^{n -\frac {7}{2}}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n -\frac {5}{2}} b_{n} \left (2 n -7\right )}{4} &= \moverset {\infty }{\munderset {n =1}{\sum }}\frac {b_{n -1} \left (-9+2 n \right ) x^{n -\frac {7}{2}}}{4} \\ \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 \(n -\frac {7}{2}\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =4}{\sum }}C a_{n -4} \left (2 n -7\right ) x^{n -\frac {7}{2}}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {C a_{n -6} x^{n -\frac {7}{2}}}{3}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {C a_{n -5} x^{n -\frac {7}{2}}}{2}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}3 C a_{n -4} x^{n -\frac {7}{2}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n -\frac {7}{2}} b_{n} \left (4 n^{2}-32 n +63\right )}{4}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {b_{n -2} \left (-11+2 n \right ) x^{n -\frac {7}{2}}}{6}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\frac {b_{n -1} \left (-9+2 n \right ) x^{n -\frac {7}{2}}}{4}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (4 n -14\right ) b_{n} x^{n -\frac {7}{2}}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {7 b_{n} x^{n -\frac {7}{2}}}{4}\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 \[ -3 b_{1}-\frac {7 b_{0}}{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -3 b_{1}-\frac {7}{4} = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}=-{\frac {7}{12}} \] For \(n=2\), Eq (2B) gives \[ -4 b_{2}+\frac {7 b_{0}}{6}-\frac {5 b_{1}}{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -4 b_{2}+\frac {91}{48} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}={\frac {91}{192}} \] For \(n=3\), Eq (2B) gives \[ -3 b_{3}+\frac {5 b_{1}}{6}-\frac {3 b_{2}}{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -3 b_{3}-\frac {1939}{2304} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=-{\frac {1939}{6912}} \] For \(n=N\), where \(N=4\) which is the difference between the two roots, we are free to choose \(b_{4} = 0\). Hence for \(n=4\), Eq (2B) gives \[ 4 C +\frac {8491}{27648} = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-{\frac {8491}{110592}} \] For \(n=5\), Eq (2B) gives \[ \frac {\left (a_{0}+12 a_{1}\right ) C}{2}+\frac {b_{3}}{6}+\frac {b_{4}}{4}+5 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {103033}{1658880}+5 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}={\frac {103033}{8294400}} \] 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 \(C=-{\frac {8491}{110592}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= -\frac {8491}{110592}\eslowast \left (\sqrt {x}\, \left (1-\frac {x}{20}+\frac {49 x^{2}}{2880}-\frac {533 x^{3}}{241920}+\frac {277 x^{4}}{491520}-\frac {203759 x^{5}}{2388787200}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+\frac {1-\frac {7 x}{12}+\frac {91 x^{2}}{192}-\frac {1939 x^{3}}{6912}+\frac {103033 x^{5}}{8294400}+O\left (x^{6}\right )}{x^{\frac {7}{2}}} \] 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} \sqrt {x}\, \left (1-\frac {x}{20}+\frac {49 x^{2}}{2880}-\frac {533 x^{3}}{241920}+\frac {277 x^{4}}{491520}-\frac {203759 x^{5}}{2388787200}+O\left (x^{6}\right )\right ) + c_{2} \left (-\frac {8491}{110592}\eslowast \left (\sqrt {x}\, \left (1-\frac {x}{20}+\frac {49 x^{2}}{2880}-\frac {533 x^{3}}{241920}+\frac {277 x^{4}}{491520}-\frac {203759 x^{5}}{2388787200}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+\frac {1-\frac {7 x}{12}+\frac {91 x^{2}}{192}-\frac {1939 x^{3}}{6912}+\frac {103033 x^{5}}{8294400}+O\left (x^{6}\right )}{x^{\frac {7}{2}}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {x}\, \left (1-\frac {x}{20}+\frac {49 x^{2}}{2880}-\frac {533 x^{3}}{241920}+\frac {277 x^{4}}{491520}-\frac {203759 x^{5}}{2388787200}+O\left (x^{6}\right )\right )+c_{2} \left (-\frac {8491 \sqrt {x}\, \left (1-\frac {x}{20}+\frac {49 x^{2}}{2880}-\frac {533 x^{3}}{241920}+\frac {277 x^{4}}{491520}-\frac {203759 x^{5}}{2388787200}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{110592}+\frac {1-\frac {7 x}{12}+\frac {91 x^{2}}{192}-\frac {1939 x^{3}}{6912}+\frac {103033 x^{5}}{8294400}+O\left (x^{6}\right )}{x^{\frac {7}{2}}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {x}\, \left (1-\frac {x}{20}+\frac {49 x^{2}}{2880}-\frac {533 x^{3}}{241920}+\frac {277 x^{4}}{491520}-\frac {203759 x^{5}}{2388787200}+O\left (x^{6}\right )\right )+c_{2} \left (-\frac {8491 \sqrt {x}\, \left (1-\frac {x}{20}+\frac {49 x^{2}}{2880}-\frac {533 x^{3}}{241920}+\frac {277 x^{4}}{491520}-\frac {203759 x^{5}}{2388787200}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{110592}+\frac {1-\frac {7 x}{12}+\frac {91 x^{2}}{192}-\frac {1939 x^{3}}{6912}+\frac {103033 x^{5}}{8294400}+O\left (x^{6}\right )}{x^{\frac {7}{2}}}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1-\frac {x}{20}+\frac {49 x^{2}}{2880}-\frac {533 x^{3}}{241920}+\frac {277 x^{4}}{491520}-\frac {203759 x^{5}}{2388787200}+O\left (x^{6}\right )\right )+c_{2} \left (-\frac {8491 \sqrt {x}\, \left (1-\frac {x}{20}+\frac {49 x^{2}}{2880}-\frac {533 x^{3}}{241920}+\frac {277 x^{4}}{491520}-\frac {203759 x^{5}}{2388787200}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{110592}+\frac {1-\frac {7 x}{12}+\frac {91 x^{2}}{192}-\frac {1939 x^{3}}{6912}+\frac {103033 x^{5}}{8294400}+O\left (x^{6}\right )}{x^{\frac {7}{2}}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+\left (4 x +\frac {1}{2} x^{2}-\frac {1}{3} x^{3}\right ) y^{\prime }-\frac {7 y}{4}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {7 y}{4 x^{2}}+\frac {\left (2 x^{2}-3 x -24\right ) y^{\prime }}{6 x} \\ \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} \\ {} & {} & y^{\prime \prime }-\frac {\left (2 x^{2}-3 x -24\right ) y^{\prime }}{6 x}-\frac {7 y}{4 x^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {2 x^{2}-3 x -24}{6 x}, P_{3}\left (x \right )=-\frac {7}{4 x^{2}}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=4 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-\frac {7}{4} \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 12 x^{2} y^{\prime \prime }-2 x \left (2 x^{2}-3 x -24\right ) y^{\prime }-21 y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 3 a_{0} \left (7+2 r \right ) \left (-1+2 r \right ) x^{r}+\left (3 a_{1} \left (9+2 r \right ) \left (1+2 r \right )+6 a_{0} r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (3 a_{k} \left (2 k +2 r +7\right ) \left (2 k +2 r -1\right )+6 a_{k -1} \left (k +r -1\right )-4 a_{k -2} \left (k -2+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 3 \left (7+2 r \right ) \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {7}{2}, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 3 a_{1} \left (9+2 r \right ) \left (1+2 r \right )+6 a_{0} r =0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {2 a_{0} r}{4 r^{2}+20 r +9} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 3 a_{k} \left (2 k +2 r +7\right ) \left (2 k +2 r -1\right )+6 a_{k -1} \left (k +r -1\right )-4 a_{k -2} \left (k -2+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & 3 a_{k +2} \left (2 k +11+2 r \right ) \left (2 k +3+2 r \right )+6 a_{k +1} \left (k +1+r \right )-4 a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {2 \left (2 k a_{k}-3 k a_{k +1}+2 r a_{k}-3 r a_{k +1}-3 a_{k +1}\right )}{3 \left (2 k +11+2 r \right ) \left (2 k +3+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {7}{2} \\ {} & {} & a_{k +2}=\frac {2 \left (2 k a_{k}-3 k a_{k +1}-7 a_{k}+\frac {15}{2} a_{k +1}\right )}{3 \left (2 k +4\right ) \left (2 k -4\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-\frac {7}{2}\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =2 \\ {} & {} & a_{k +2}=\frac {2 \left (2 k a_{k}-3 k a_{k +1}-7 a_{k}+\frac {15}{2} a_{k +1}\right )}{3 \left (2 k +4\right ) \left (2 k -4\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +2}=\frac {2 \left (2 k a_{k}-3 k a_{k +1}+a_{k}-\frac {9}{2} a_{k +1}\right )}{3 \left (2 k +12\right ) \left (2 k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}, a_{k +2}=\frac {2 \left (2 k a_{k}-3 k a_{k +1}+a_{k}-\frac {9}{2} a_{k +1}\right )}{3 \left (2 k +12\right ) \left (2 k +4\right )}, a_{1}=-\frac {a_{0}}{20}\right ] \end {array} \]
Maple trace
`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 <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunB ODE, case c = 0 `
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 63
Order:=6; dsolve(x^2*diff(y(x),x$2)+(4*x+1/2*x^2-1/3*x^3)*diff(y(x),x)-7/4*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} x^{4} \left (1-\frac {1}{20} x +\frac {49}{2880} x^{2}-\frac {533}{241920} x^{3}+\frac {277}{491520} x^{4}-\frac {203759}{2388787200} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (\frac {8491}{768} x^{4}-\frac {8491}{15360} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (-144+84 x -\frac {273}{4} x^{2}+\frac {1939}{48} x^{3}-\frac {221}{12} x^{4}-\frac {49993}{57600} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right )}{x^{\frac {7}{2}}} \]
✓ Solution by Mathematica
Time used: 0.038 (sec). Leaf size: 93
AsymptoticDSolveValue[x^2*y''[x]+(4*x+1/2*x^2-1/3*x^3)*y'[x]-7/4*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {277 x^{9/2}}{491520}-\frac {533 x^{7/2}}{241920}+\frac {49 x^{5/2}}{2880}-\frac {x^{3/2}}{20}+\sqrt {x}\right )+c_1 \left (\frac {65067 x^4-124096 x^3+209664 x^2-258048 x+442368}{442368 x^{7/2}}-\frac {8491 \sqrt {x} \log (x)}{110592}\right ) \]