Internal problem ID [13984]
Internal file name [OUTPUT/13156_Friday_February_23_2024_06_58_00_AM_98569054/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 35. Modified Power series solutions and basic method of Frobenius. Additional
Exercises. page 715
Problem number: 35.3 (g).
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]]
\[ \boxed {\left (4 x^{2}-1\right ) y^{\prime \prime }+\left (4-\frac {2}{x}\right ) y^{\prime }+\frac {\left (-x^{2}+1\right ) y}{x^{2}+1}=0} \] 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. \[ \left (4 x^{2}-1\right ) y^{\prime \prime }+\left (4-\frac {2}{x}\right ) y^{\prime }+\left (-\frac {x^{2}}{x^{2}+1}+\frac {1}{x^{2}+1}\right ) y = 0 \] 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 {2}{x \left (2 x +1\right )}\\ q(x) &= -\frac {x^{2}-1}{\left (4 x^{2}-1\right ) \left (x^{2}+1\right )}\\ \end {align*}
| \(p(x)=\frac {2}{x \left (2 x +1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -{\frac {1}{2}}\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {x^{2}-1}{\left (4 x^{2}-1\right ) \left (x^{2}+1\right )}\) | |
| singularity | type |
| \(x = -{\frac {1}{2}}\) | \(\text {``regular''}\) |
| \(x = {\frac {1}{2}}\) | \(\text {``regular''}\) |
| \(x = -i\) | \(\text {``regular''}\) |
| \(x = i\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -{\frac {1}{2}}, {\frac {1}{2}}, -i, i, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ y^{\prime \prime } \left (4 x^{2}-1\right ) x \left (x^{2}+1\right )+\left (4 x^{3}-2 x^{2}+4 x -2\right ) y^{\prime }+\left (-x^{3}+x \right ) y = 0 \] 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} \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^{2}-1\right ) x \left (x^{2}+1\right )+\left (4 x^{3}-2 x^{2}+4 x -2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-x^{3}+x \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 }}4 x^{n +r +3} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r +2} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n} \left (n +r \right )\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 (-2 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +3} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r +3} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}4 a_{n -4} \left (n +r -4\right ) \left (n -5+r \right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}3 a_{n -2} \left (n +r -2\right ) \left (n +r -3\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}4 a_{n -3} \left (n +r -3\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} 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 -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +3} a_{n}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-a_{n -4} x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r -1} \\ \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 \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =4}{\sum }}4 a_{n -4} \left (n +r -4\right ) \left (n -5+r \right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}3 a_{n -2} \left (n +r -2\right ) \left (n +r -3\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}4 a_{n -3} \left (n +r -3\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 a_{n -2} \left (n +r -2\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-a_{n -4} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ -x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )-2 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ -x^{-1+r} a_{0} r \left (-1+r \right )-2 r a_{0} x^{-1+r} = 0 \] Or \[ \left (-x^{-1+r} r \left (-1+r \right )-2 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (-1-r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ -r \left (1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (-1-r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, -1]\).
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 ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ 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} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ 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 -1}\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 \[ a_{1} = \frac {4 r}{r^{2}+3 r +2} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {3 r^{3}+r^{2}+7 r +2}{\left (2+r \right )^{2} \left (3+r \right )} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {28 r^{4}+80 r^{3}+84 r^{2}+48 r +8}{\left (3+r \right )^{2} \left (2+r \right ) \left (1+r \right ) \left (r +4\right )} \] For \(4\le n\) the recursive equation is \begin{equation} \tag{3} 4 a_{n -4} \left (n +r -4\right ) \left (n -5+r \right )+3 a_{n -2} \left (n +r -2\right ) \left (n +r -3\right )-a_{n} \left (n +r \right ) \left (n +r -1\right )+4 a_{n -3} \left (n +r -3\right )-2 a_{n -2} \left (n +r -2\right )+4 a_{n -1} \left (n +r -1\right )-2 a_{n} \left (n +r \right )-a_{n -4}+a_{n -2} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {4 n^{2} a_{n -4}+3 n^{2} a_{n -2}+8 n r a_{n -4}+6 n r a_{n -2}+4 r^{2} a_{n -4}+3 r^{2} a_{n -2}-36 n a_{n -4}+4 n a_{n -3}-17 n a_{n -2}+4 n a_{n -1}-36 r a_{n -4}+4 r a_{n -3}-17 r a_{n -2}+4 r a_{n -1}+79 a_{n -4}-12 a_{n -3}+23 a_{n -2}-4 a_{n -1}}{n^{2}+2 n r +r^{2}+n +r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {\left (4 a_{n -4}+3 a_{n -2}\right ) n^{2}+\left (-36 a_{n -4}+4 a_{n -3}-17 a_{n -2}+4 a_{n -1}\right ) n +79 a_{n -4}-12 a_{n -3}+23 a_{n -2}-4 a_{n -1}}{n \left (1+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r}{r^{2}+3 r +2}\) | \(0\) |
| \(a_{2}\) | \(\frac {3 r^{3}+r^{2}+7 r +2}{\left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {1}{6}\) |
| \(a_{3}\) | \(\frac {28 r^{4}+80 r^{3}+84 r^{2}+48 r +8}{\left (3+r \right )^{2} \left (2+r \right ) \left (1+r \right ) \left (r +4\right )}\) | \(\frac {1}{9}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {13 r^{7}+113 r^{6}+492 r^{5}+1291 r^{4}+1970 r^{3}+1653 r^{2}+654 r +40}{\left (r +5\right ) \left (r +4\right )^{2} \left (1+r \right ) \left (2+r \right )^{2} \left (3+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {1}{24}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r}{r^{2}+3 r +2}\) | \(0\) |
| \(a_{2}\) | \(\frac {3 r^{3}+r^{2}+7 r +2}{\left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {1}{6}\) |
| \(a_{3}\) | \(\frac {28 r^{4}+80 r^{3}+84 r^{2}+48 r +8}{\left (3+r \right )^{2} \left (2+r \right ) \left (1+r \right ) \left (r +4\right )}\) | \(\frac {1}{9}\) |
| \(a_{4}\) | \(\frac {13 r^{7}+113 r^{6}+492 r^{5}+1291 r^{4}+1970 r^{3}+1653 r^{2}+654 r +40}{\left (r +5\right ) \left (r +4\right )^{2} \left (1+r \right ) \left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {1}{24}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {164 r^{8}+2272 r^{7}+13232 r^{6}+42056 r^{5}+79024 r^{4}+89040 r^{3}+58436 r^{2}+20128 r +2480}{\left (r +6\right ) \left (r +5\right )^{2} \left (r +4\right ) \left (1+r \right ) \left (2+r \right )^{2} \left (3+r \right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{5}={\frac {31}{270}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r}{r^{2}+3 r +2}\) | \(0\) |
| \(a_{2}\) | \(\frac {3 r^{3}+r^{2}+7 r +2}{\left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {1}{6}\) |
| \(a_{3}\) | \(\frac {28 r^{4}+80 r^{3}+84 r^{2}+48 r +8}{\left (3+r \right )^{2} \left (2+r \right ) \left (1+r \right ) \left (r +4\right )}\) | \(\frac {1}{9}\) |
| \(a_{4}\) | \(\frac {13 r^{7}+113 r^{6}+492 r^{5}+1291 r^{4}+1970 r^{3}+1653 r^{2}+654 r +40}{\left (r +5\right ) \left (r +4\right )^{2} \left (1+r \right ) \left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {1}{24}\) |
| \(a_{5}\) | \(\frac {164 r^{8}+2272 r^{7}+13232 r^{6}+42056 r^{5}+79024 r^{4}+89040 r^{3}+58436 r^{2}+20128 r +2480}{\left (r +6\right ) \left (r +5\right )^{2} \left (r +4\right ) \left (1+r \right ) \left (2+r \right )^{2} \left (3+r \right )^{2}}\) | \(\frac {31}{270}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= 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 \\ &= 1+\frac {x^{2}}{6}+\frac {x^{3}}{9}+\frac {x^{4}}{24}+\frac {31 x^{5}}{270}+O\left (x^{6}\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}{r^{2}+3 r +2} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {4 r}{r^{2}+3 r +2}&= \lim _{r\rightarrow -1}\frac {4 r}{r^{2}+3 r +2}\\ &= \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 \(y^{\prime \prime } \left (4 x^{2}-1\right ) x \left (x^{2}+1\right )+\left (4 x^{3}-2 x^{2}+4 x -2\right ) y^{\prime }+\left (-x^{3}+x \right ) y = 0\) gives \[ \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right ) \left (4 x^{2}-1\right ) x \left (x^{2}+1\right )+\left (4 x^{3}-2 x^{2}+4 x -2\right ) \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+\left (-x^{3}+x \right ) \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (y_{1}^{\prime \prime }\left (x \right ) \left (4 x^{2}-1\right ) x \left (x^{2}+1\right )+\left (4 x^{3}-2 x^{2}+4 x -2\right ) y_{1}^{\prime }\left (x \right )+\left (-x^{3}+x \right ) y_{1}\left (x \right )\right ) \ln \left (x \right )+\left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) \left (4 x^{2}-1\right ) x \left (x^{2}+1\right )+\frac {\left (4 x^{3}-2 x^{2}+4 x -2\right ) y_{1}\left (x \right )}{x}\right ) C +\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^{2}-1\right ) x \left (x^{2}+1\right )+\left (4 x^{3}-2 x^{2}+4 x -2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (-x^{3}+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 \[ y_{1}^{\prime \prime }\left (x \right ) \left (4 x^{2}-1\right ) x \left (x^{2}+1\right )+\left (4 x^{3}-2 x^{2}+4 x -2\right ) y_{1}^{\prime }\left (x \right )+\left (-x^{3}+x \right ) y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (\left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) \left (4 x^{2}-1\right ) x \left (x^{2}+1\right )+\frac {\left (4 x^{3}-2 x^{2}+4 x -2\right ) y_{1}\left (x \right )}{x}\right ) C +\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^{2}-1\right ) x \left (x^{2}+1\right )+\left (4 x^{3}-2 x^{2}+4 x -2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (-x^{3}+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} \frac {\left (2 \left (4 x^{5}+3 x^{3}-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )-4 \left (x^{2}+1\right ) \left (x -\frac {1}{2}\right )^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C}{x}+\frac {\left (4 x^{6}+3 x^{4}-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 )+2 \left (2 x^{4}-x^{3}+2 x^{2}-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )+\left (-x^{4}+x^{2}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )}{x} = 0 \end{equation} Since \(r_{1} = 0\) and \(r_{2} = -1\) then the above becomes \begin{equation} \tag{10} \frac {\left (2 \left (4 x^{5}+3 x^{3}-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} a_{n} n \right )-4 \left (x^{2}+1\right ) \left (x -\frac {1}{2}\right )^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\right ) C}{x}+\frac {\left (4 x^{6}+3 x^{4}-x^{2}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -3} b_{n} \left (n -1\right ) \left (n -2\right )\right )+2 \left (2 x^{4}-x^{3}+2 x^{2}-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} \left (n -1\right )\right )+\left (-x^{4}+x^{2}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -1}\right )}{x} = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}8 C n \,x^{n +3} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 C n \,x^{1+n} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C n \,x^{n -1} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C \,x^{n +3} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 C \,x^{n +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 C \,x^{1+n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 C a_{n} x^{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n -1} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 b_{n} x^{n +2} \left (n -1\right ) \left (n -2\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n} b_{n} \left (n^{2}-3 n +2\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-b_{n} x^{n -2} \left (n -1\right ) \left (n -2\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n -2} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +2} b_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n -2\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -2}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}8 C n \,x^{n +3} a_{n} &= \moverset {\infty }{\munderset {n =5}{\sum }}8 C \left (n -5\right ) a_{n -5} x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}6 C n \,x^{1+n} a_{n} &= \moverset {\infty }{\munderset {n =3}{\sum }}6 C \left (n -3\right ) a_{n -3} x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C n \,x^{n -1} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 C \left (n -1\right ) a_{n -1} x^{n -2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C \,x^{n +3} a_{n}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-4 C a_{n -5} x^{n -2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 C \,x^{n +2} a_{n} &= \moverset {\infty }{\munderset {n =4}{\sum }}4 C a_{n -4} x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 C \,x^{1+n} a_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-5 C a_{n -3} x^{n -2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 C a_{n} x^{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}4 C a_{n -2} x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n -1} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} x^{n -2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 b_{n} x^{n +2} \left (n -1\right ) \left (n -2\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}4 b_{n -4} \left (n -6\right ) \left (n -5\right ) x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n} b_{n} \left (n^{2}-3 n +2\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}3 b_{n -2} \left (\left (n -2\right )^{2}-3 n +8\right ) x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n} b_{n} \left (n -1\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}4 b_{n -3} \left (n -4\right ) x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n} b_{n} \left (n -1\right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 b_{n -2} \left (n -3\right ) x^{n -2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n -1} b_{n} \left (n -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}4 b_{n -1} \left (n -2\right ) x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +2} b_{n}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-b_{n -4} x^{n -2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}b_{n -2} x^{n -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 \(n -2\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =5}{\sum }}8 C \left (n -5\right ) a_{n -5} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}6 C \left (n -3\right ) a_{n -3} x^{n -2}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 C \left (n -1\right ) a_{n -1} x^{n -2}\right )+\moverset {\infty }{\munderset {n =5}{\sum }}\left (-4 C a_{n -5} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}4 C a_{n -4} x^{n -2}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-5 C a_{n -3} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}4 C a_{n -2} x^{n -2}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}4 b_{n -4} \left (n -6\right ) \left (n -5\right ) x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}3 b_{n -2} \left (\left (n -2\right )^{2}-3 n +8\right ) x^{n -2}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-b_{n} x^{n -2} \left (n -1\right ) \left (n -2\right )\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}4 b_{n -3} \left (n -4\right ) x^{n -2}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 b_{n -2} \left (n -3\right ) x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 b_{n -1} \left (n -2\right ) x^{n -2}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n -2} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-b_{n -4} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}b_{n -2} x^{n -2}\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 \[ -C -4 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-4 \] For \(n=2\), Eq (2B) gives \[ \left (4 a_{0}-3 a_{1}\right ) C +9 b_{0}-2 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -7-2 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=-{\frac {7}{2}} \] For \(n=3\), Eq (2B) gives \[ \left (-5 a_{0}+4 a_{1}-5 a_{2}\right ) C -4 b_{0}+b_{1}+4 b_{2}-6 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {16}{3}-6 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}={\frac {8}{9}} \] For \(n=4\), Eq (2B) gives \[ \left (4 a_{0}+a_{1}+4 a_{2}-7 a_{3}\right ) C +7 b_{0}-b_{2}+8 b_{3}-12 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {37}{18}-12 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}={\frac {37}{216}} \] For \(n=5\), Eq (2B) gives \[ \left (-4 a_{0}+4 a_{1}+7 a_{2}+4 a_{3}-9 a_{4}\right ) C +12 b_{4}-20 b_{5}-b_{1}+4 b_{2}+3 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {16}{9}-20 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}={\frac {4}{45}} \] 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 \(C=-4\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= \left (-4\right )\eslowast \left (1+\frac {x^{2}}{6}+\frac {x^{3}}{9}+\frac {x^{4}}{24}+\frac {31 x^{5}}{270}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\frac {1-\frac {7 x^{2}}{2}+\frac {8 x^{3}}{9}+\frac {37 x^{4}}{216}+\frac {4 x^{5}}{45}+O\left (x^{6}\right )}{x} \] 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} \left (1+\frac {x^{2}}{6}+\frac {x^{3}}{9}+\frac {x^{4}}{24}+\frac {31 x^{5}}{270}+O\left (x^{6}\right )\right ) + c_{2} \left (\left (-4\right )\eslowast \left (1+\frac {x^{2}}{6}+\frac {x^{3}}{9}+\frac {x^{4}}{24}+\frac {31 x^{5}}{270}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\frac {1-\frac {7 x^{2}}{2}+\frac {8 x^{3}}{9}+\frac {37 x^{4}}{216}+\frac {4 x^{5}}{45}+O\left (x^{6}\right )}{x}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1+\frac {x^{2}}{6}+\frac {x^{3}}{9}+\frac {x^{4}}{24}+\frac {31 x^{5}}{270}+O\left (x^{6}\right )\right )+c_{2} \left (\left (-4-\frac {2 x^{2}}{3}-\frac {4 x^{3}}{9}-\frac {x^{4}}{6}-\frac {62 x^{5}}{135}-4 O\left (x^{6}\right )\right ) \ln \left (x \right )+\frac {1-\frac {7 x^{2}}{2}+\frac {8 x^{3}}{9}+\frac {37 x^{4}}{216}+\frac {4 x^{5}}{45}+O\left (x^{6}\right )}{x}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1+\frac {x^{2}}{6}+\frac {x^{3}}{9}+\frac {x^{4}}{24}+\frac {31 x^{5}}{270}+O\left (x^{6}\right )\right )+c_{2} \left (\left (-4-\frac {2 x^{2}}{3}-\frac {4 x^{3}}{9}-\frac {x^{4}}{6}-\frac {62 x^{5}}{135}-4 O\left (x^{6}\right )\right ) \ln \left (x \right )+\frac {1-\frac {7 x^{2}}{2}+\frac {8 x^{3}}{9}+\frac {37 x^{4}}{216}+\frac {4 x^{5}}{45}+O\left (x^{6}\right )}{x}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1+\frac {x^{2}}{6}+\frac {x^{3}}{9}+\frac {x^{4}}{24}+\frac {31 x^{5}}{270}+O\left (x^{6}\right )\right )+c_{2} \left (\left (-4-\frac {2 x^{2}}{3}-\frac {4 x^{3}}{9}-\frac {x^{4}}{6}-\frac {62 x^{5}}{135}-4 O\left (x^{6}\right )\right ) \ln \left (x \right )+\frac {1-\frac {7 x^{2}}{2}+\frac {8 x^{3}}{9}+\frac {37 x^{4}}{216}+\frac {4 x^{5}}{45}+O\left (x^{6}\right )}{x}\right ) \] Verified OK.
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 -> 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 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 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 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying to convert to an ODE of Bessel type trying to convert to an ODE of Bessel type -> trying reduction of order to Riccati trying Riccati sub-methods: -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3`[0, y]
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 62
Order:=6; dsolve((4*x^2-1)*diff(y(x),x$2)+(4-2/x)*diff(y(x),x)+(1-x^2)/(1+x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {\ln \left (x \right ) \left (\left (-4\right ) x -\frac {2}{3} x^{3}-\frac {4}{9} x^{4}-\frac {1}{6} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} +c_{1} \left (1+\frac {1}{6} x^{2}+\frac {1}{9} x^{3}+\frac {1}{24} x^{4}+\frac {31}{270} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) x +\left (1+4 x -\frac {7}{2} x^{2}+\frac {14}{9} x^{3}+\frac {133}{216} x^{4}+\frac {23}{90} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2}}{x} \]
✓ Solution by Mathematica
Time used: 0.034 (sec). Leaf size: 79
AsymptoticDSolveValue[(4*x^2-1)*y''[x]+(4-2/x)*y'[x]+(1-x^2)/(1+x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {x^4}{24}+\frac {x^3}{9}+\frac {x^2}{6}+1\right )+c_1 \left (\frac {229 x^4+480 x^3-756 x^2+1728 x+216}{216 x}-\frac {2}{9} \left (2 x^3+3 x^2+18\right ) \log (x)\right ) \]