Internal problem ID [6996]
Internal file name [OUTPUT/6239_Thursday_August_18_2022_07_11_17_AM_85928680/index.tex]
Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th
edition. 1997.
Section: CHAPTER 18. Power series solutions. 18.11 Many-Term Recurrence Relations. Exercises
page 391
Problem number: 2.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[_Jacobi]
\[ \boxed {2 x \left (1-x \right ) y^{\prime \prime }+\left (1-2 x \right ) y^{\prime }+\left (x +2\right ) y=0} \] 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. \[ \left (-2 x^{2}+2 x \right ) y^{\prime \prime }+\left (1-2 x \right ) y^{\prime }+\left (x +2\right ) y = 0 \] 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 -1}{2 x \left (x -1\right )}\\ q(x) &= -\frac {x +2}{2 x \left (x -1\right )}\\ \end {align*}
| \(p(x)=\frac {2 x -1}{2 x \left (x -1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {x +2}{2 x \left (x -1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, 1]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ -2 y^{\prime \prime } x \left (x -1\right )+\left (1-2 x \right ) y^{\prime }+\left (x +2\right ) y = 0 \] 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 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x \left (x -1\right )+\left (1-2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (x +2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r}\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 }}\left (-2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) 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} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} 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} \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{-1+r} a_{0} r \left (-1+r \right )+r a_{0} x^{-1+r} = 0 \] Or \[ \left (2 x^{-1+r} r \left (-1+r \right )+r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (2 r -1\right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r^{2}-r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (2 r -1\right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {1}{2}}\) is not an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {1}{2}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} \end {align*}
We start by finding \(y_{1}\left (x \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {-2+2 r}{2 r +1} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} -2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+2 a_{n} \left (n +r \right ) \left (n +r -1\right )-2 a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )+a_{n -2}+2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {2 n^{2} a_{n -1}+4 n r a_{n -1}+2 r^{2} a_{n -1}-4 n a_{n -1}-4 r a_{n -1}-a_{n -2}}{2 n^{2}+4 n r +2 r^{2}-n -r}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = \frac {\left (4 n^{2}-4 n -3\right ) a_{n -1}-2 a_{n -2}}{4 n^{2}+2 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+2 r}{2 r +1}\) | \(-{\frac {1}{2}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {4 r^{3}+4 r^{2}-10 r -1}{4 r^{3}+16 r^{2}+19 r +6} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}=-{\frac {9}{40}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+2 r}{2 r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {4 r^{3}+4 r^{2}-10 r -1}{4 r^{3}+16 r^{2}+19 r +6}\) | \(-{\frac {9}{40}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {8 r^{5}+40 r^{4}+32 r^{3}-68 r^{2}-66 r +6}{8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}=-{\frac {149}{1680}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+2 r}{2 r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {4 r^{3}+4 r^{2}-10 r -1}{4 r^{3}+16 r^{2}+19 r +6}\) | \(-{\frac {9}{40}}\) |
| \(a_{3}\) | \(\frac {8 r^{5}+40 r^{4}+32 r^{3}-68 r^{2}-66 r +6}{8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90}\) | \(-{\frac {149}{1680}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {16 r^{7}+176 r^{6}+664 r^{5}+836 r^{4}-520 r^{3}-1816 r^{2}-823 r +111}{16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}=-{\frac {661}{13440}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+2 r}{2 r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {4 r^{3}+4 r^{2}-10 r -1}{4 r^{3}+16 r^{2}+19 r +6}\) | \(-{\frac {9}{40}}\) |
| \(a_{3}\) | \(\frac {8 r^{5}+40 r^{4}+32 r^{3}-68 r^{2}-66 r +6}{8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90}\) | \(-{\frac {149}{1680}}\) |
| \(a_{4}\) | \(\frac {16 r^{7}+176 r^{6}+664 r^{5}+836 r^{4}-520 r^{3}-1816 r^{2}-823 r +111}{16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520}\) | \(-{\frac {661}{13440}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {32 r^{9}+608 r^{8}+4608 r^{7}+17376 r^{6}+31368 r^{5}+11664 r^{4}-46046 r^{3}-64544 r^{2}-21156 r +3162}{\left (2 r^{2}+19 r +45\right ) \left (8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90\right ) \left (2 r^{2}+15 r +28\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}=-{\frac {16171}{492800}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+2 r}{2 r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {4 r^{3}+4 r^{2}-10 r -1}{4 r^{3}+16 r^{2}+19 r +6}\) | \(-{\frac {9}{40}}\) |
| \(a_{3}\) | \(\frac {8 r^{5}+40 r^{4}+32 r^{3}-68 r^{2}-66 r +6}{8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90}\) | \(-{\frac {149}{1680}}\) |
| \(a_{4}\) | \(\frac {16 r^{7}+176 r^{6}+664 r^{5}+836 r^{4}-520 r^{3}-1816 r^{2}-823 r +111}{16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520}\) | \(-{\frac {661}{13440}}\) |
| \(a_{5}\) | \(\frac {32 r^{9}+608 r^{8}+4608 r^{7}+17376 r^{6}+31368 r^{5}+11664 r^{4}-46046 r^{3}-64544 r^{2}-21156 r +3162}{\left (2 r^{2}+19 r +45\right ) \left (8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90\right ) \left (2 r^{2}+15 r +28\right )}\) | \(-{\frac {16171}{492800}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {64 r^{11}+1856 r^{10}+22880 r^{9}+155440 r^{8}+626048 r^{7}+1462528 r^{6}+1602128 r^{5}-514244 r^{4}-3483850 r^{3}-3417773 r^{2}-917322 r +146781}{\left (16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520\right ) \left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+23 r +66\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{6}=-{\frac {5530601}{230630400}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+2 r}{2 r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {4 r^{3}+4 r^{2}-10 r -1}{4 r^{3}+16 r^{2}+19 r +6}\) | \(-{\frac {9}{40}}\) |
| \(a_{3}\) | \(\frac {8 r^{5}+40 r^{4}+32 r^{3}-68 r^{2}-66 r +6}{8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90}\) | \(-{\frac {149}{1680}}\) |
| \(a_{4}\) | \(\frac {16 r^{7}+176 r^{6}+664 r^{5}+836 r^{4}-520 r^{3}-1816 r^{2}-823 r +111}{16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520}\) | \(-{\frac {661}{13440}}\) |
| \(a_{5}\) | \(\frac {32 r^{9}+608 r^{8}+4608 r^{7}+17376 r^{6}+31368 r^{5}+11664 r^{4}-46046 r^{3}-64544 r^{2}-21156 r +3162}{\left (2 r^{2}+19 r +45\right ) \left (8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90\right ) \left (2 r^{2}+15 r +28\right )}\) | \(-{\frac {16171}{492800}}\) |
| \(a_{6}\) | \(\frac {64 r^{11}+1856 r^{10}+22880 r^{9}+155440 r^{8}+626048 r^{7}+1462528 r^{6}+1602128 r^{5}-514244 r^{4}-3483850 r^{3}-3417773 r^{2}-917322 r +146781}{\left (16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520\right ) \left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+23 r +66\right )}\) | \(-{\frac {5530601}{230630400}}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {128 r^{13}+5248 r^{12}+94720 r^{11}+987968 r^{10}+6558944 r^{9}+28650144 r^{8}+81361776 r^{7}+137907936 r^{6}+90592936 r^{5}-126026704 r^{4}-323164836 r^{3}-256226108 r^{2}-59366226 r +10065978}{\left (2 r^{2}+15 r +28\right ) \left (8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90\right ) \left (2 r^{2}+23 r +66\right ) \left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+27 r +91\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{7}=-{\frac {299137703}{16144128000}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+2 r}{2 r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {4 r^{3}+4 r^{2}-10 r -1}{4 r^{3}+16 r^{2}+19 r +6}\) | \(-{\frac {9}{40}}\) |
| \(a_{3}\) | \(\frac {8 r^{5}+40 r^{4}+32 r^{3}-68 r^{2}-66 r +6}{8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90}\) | \(-{\frac {149}{1680}}\) |
| \(a_{4}\) | \(\frac {16 r^{7}+176 r^{6}+664 r^{5}+836 r^{4}-520 r^{3}-1816 r^{2}-823 r +111}{16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520}\) | \(-{\frac {661}{13440}}\) |
| \(a_{5}\) | \(\frac {32 r^{9}+608 r^{8}+4608 r^{7}+17376 r^{6}+31368 r^{5}+11664 r^{4}-46046 r^{3}-64544 r^{2}-21156 r +3162}{\left (2 r^{2}+19 r +45\right ) \left (8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90\right ) \left (2 r^{2}+15 r +28\right )}\) | \(-{\frac {16171}{492800}}\) |
| \(a_{6}\) | \(\frac {64 r^{11}+1856 r^{10}+22880 r^{9}+155440 r^{8}+626048 r^{7}+1462528 r^{6}+1602128 r^{5}-514244 r^{4}-3483850 r^{3}-3417773 r^{2}-917322 r +146781}{\left (16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520\right ) \left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+23 r +66\right )}\) | \(-{\frac {5530601}{230630400}}\) |
| \(a_{7}\) | \(\frac {128 r^{13}+5248 r^{12}+94720 r^{11}+987968 r^{10}+6558944 r^{9}+28650144 r^{8}+81361776 r^{7}+137907936 r^{6}+90592936 r^{5}-126026704 r^{4}-323164836 r^{3}-256226108 r^{2}-59366226 r +10065978}{\left (2 r^{2}+15 r +28\right ) \left (8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90\right ) \left (2 r^{2}+23 r +66\right ) \left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+27 r +91\right )}\) | \(-{\frac {299137703}{16144128000}}\) |
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}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= \sqrt {x}\, \left (1-\frac {x}{2}-\frac {9 x^{2}}{40}-\frac {149 x^{3}}{1680}-\frac {661 x^{4}}{13440}-\frac {16171 x^{5}}{492800}-\frac {5530601 x^{6}}{230630400}-\frac {299137703 x^{7}}{16144128000}+O\left (x^{8}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Eq (2B) derived above is now used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = \frac {-2+2 r}{2 r +1} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} -2 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+2 b_{n} \left (n +r \right ) \left (n +r -1\right )-2 b_{n -1} \left (n +r -1\right )+\left (n +r \right ) b_{n}+b_{n -2}+2 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {2 n^{2} b_{n -1}+4 n r b_{n -1}+2 r^{2} b_{n -1}-4 n b_{n -1}-4 r b_{n -1}-b_{n -2}}{2 n^{2}+4 n r +2 r^{2}-n -r}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {2 n^{2} b_{n -1}-4 n b_{n -1}-b_{n -2}}{n \left (2 n -1\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+2 r}{2 r +1}\) | \(-2\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {4 r^{3}+4 r^{2}-10 r -1}{4 r^{3}+16 r^{2}+19 r +6} \] Which for the root \(r = 0\) becomes \[ b_{2}=-{\frac {1}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+2 r}{2 r +1}\) | \(-2\) |
| \(b_{2}\) | \(\frac {4 r^{3}+4 r^{2}-10 r -1}{4 r^{3}+16 r^{2}+19 r +6}\) | \(-{\frac {1}{6}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {8 r^{5}+40 r^{4}+32 r^{3}-68 r^{2}-66 r +6}{8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90} \] Which for the root \(r = 0\) becomes \[ b_{3}={\frac {1}{15}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+2 r}{2 r +1}\) | \(-2\) |
| \(b_{2}\) | \(\frac {4 r^{3}+4 r^{2}-10 r -1}{4 r^{3}+16 r^{2}+19 r +6}\) | \(-{\frac {1}{6}}\) |
| \(b_{3}\) | \(\frac {8 r^{5}+40 r^{4}+32 r^{3}-68 r^{2}-66 r +6}{8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90}\) | \(\frac {1}{15}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {16 r^{7}+176 r^{6}+664 r^{5}+836 r^{4}-520 r^{3}-1816 r^{2}-823 r +111}{16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520} \] Which for the root \(r = 0\) becomes \[ b_{4}={\frac {37}{840}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+2 r}{2 r +1}\) | \(-2\) |
| \(b_{2}\) | \(\frac {4 r^{3}+4 r^{2}-10 r -1}{4 r^{3}+16 r^{2}+19 r +6}\) | \(-{\frac {1}{6}}\) |
| \(b_{3}\) | \(\frac {8 r^{5}+40 r^{4}+32 r^{3}-68 r^{2}-66 r +6}{8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90}\) | \(\frac {1}{15}\) |
| \(b_{4}\) | \(\frac {16 r^{7}+176 r^{6}+664 r^{5}+836 r^{4}-520 r^{3}-1816 r^{2}-823 r +111}{16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520}\) | \(\frac {37}{840}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {32 r^{9}+608 r^{8}+4608 r^{7}+17376 r^{6}+31368 r^{5}+11664 r^{4}-46046 r^{3}-64544 r^{2}-21156 r +3162}{\left (2 r^{2}+19 r +45\right ) \left (8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90\right ) \left (2 r^{2}+15 r +28\right )} \] Which for the root \(r = 0\) becomes \[ b_{5}={\frac {527}{18900}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+2 r}{2 r +1}\) | \(-2\) |
| \(b_{2}\) | \(\frac {4 r^{3}+4 r^{2}-10 r -1}{4 r^{3}+16 r^{2}+19 r +6}\) | \(-{\frac {1}{6}}\) |
| \(b_{3}\) | \(\frac {8 r^{5}+40 r^{4}+32 r^{3}-68 r^{2}-66 r +6}{8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90}\) | \(\frac {1}{15}\) |
| \(b_{4}\) | \(\frac {16 r^{7}+176 r^{6}+664 r^{5}+836 r^{4}-520 r^{3}-1816 r^{2}-823 r +111}{16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520}\) | \(\frac {37}{840}\) |
| \(b_{5}\) | \(\frac {32 r^{9}+608 r^{8}+4608 r^{7}+17376 r^{6}+31368 r^{5}+11664 r^{4}-46046 r^{3}-64544 r^{2}-21156 r +3162}{\left (2 r^{2}+19 r +45\right ) \left (8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90\right ) \left (2 r^{2}+15 r +28\right )}\) | \(\frac {527}{18900}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {64 r^{11}+1856 r^{10}+22880 r^{9}+155440 r^{8}+626048 r^{7}+1462528 r^{6}+1602128 r^{5}-514244 r^{4}-3483850 r^{3}-3417773 r^{2}-917322 r +146781}{\left (16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520\right ) \left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+23 r +66\right )} \] Which for the root \(r = 0\) becomes \[ b_{6}={\frac {16309}{831600}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+2 r}{2 r +1}\) | \(-2\) |
| \(b_{2}\) | \(\frac {4 r^{3}+4 r^{2}-10 r -1}{4 r^{3}+16 r^{2}+19 r +6}\) | \(-{\frac {1}{6}}\) |
| \(b_{3}\) | \(\frac {8 r^{5}+40 r^{4}+32 r^{3}-68 r^{2}-66 r +6}{8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90}\) | \(\frac {1}{15}\) |
| \(b_{4}\) | \(\frac {16 r^{7}+176 r^{6}+664 r^{5}+836 r^{4}-520 r^{3}-1816 r^{2}-823 r +111}{16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520}\) | \(\frac {37}{840}\) |
| \(b_{5}\) | \(\frac {32 r^{9}+608 r^{8}+4608 r^{7}+17376 r^{6}+31368 r^{5}+11664 r^{4}-46046 r^{3}-64544 r^{2}-21156 r +3162}{\left (2 r^{2}+19 r +45\right ) \left (8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90\right ) \left (2 r^{2}+15 r +28\right )}\) | \(\frac {527}{18900}\) |
| \(b_{6}\) | \(\frac {64 r^{11}+1856 r^{10}+22880 r^{9}+155440 r^{8}+626048 r^{7}+1462528 r^{6}+1602128 r^{5}-514244 r^{4}-3483850 r^{3}-3417773 r^{2}-917322 r +146781}{\left (16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520\right ) \left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+23 r +66\right )}\) | \(\frac {16309}{831600}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=\frac {128 r^{13}+5248 r^{12}+94720 r^{11}+987968 r^{10}+6558944 r^{9}+28650144 r^{8}+81361776 r^{7}+137907936 r^{6}+90592936 r^{5}-126026704 r^{4}-323164836 r^{3}-256226108 r^{2}-59366226 r +10065978}{\left (2 r^{2}+15 r +28\right ) \left (8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90\right ) \left (2 r^{2}+23 r +66\right ) \left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+27 r +91\right )} \] Which for the root \(r = 0\) becomes \[ b_{7}={\frac {14339}{970200}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+2 r}{2 r +1}\) | \(-2\) |
| \(b_{2}\) | \(\frac {4 r^{3}+4 r^{2}-10 r -1}{4 r^{3}+16 r^{2}+19 r +6}\) | \(-{\frac {1}{6}}\) |
| \(b_{3}\) | \(\frac {8 r^{5}+40 r^{4}+32 r^{3}-68 r^{2}-66 r +6}{8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90}\) | \(\frac {1}{15}\) |
| \(b_{4}\) | \(\frac {16 r^{7}+176 r^{6}+664 r^{5}+836 r^{4}-520 r^{3}-1816 r^{2}-823 r +111}{16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520}\) | \(\frac {37}{840}\) |
| \(b_{5}\) | \(\frac {32 r^{9}+608 r^{8}+4608 r^{7}+17376 r^{6}+31368 r^{5}+11664 r^{4}-46046 r^{3}-64544 r^{2}-21156 r +3162}{\left (2 r^{2}+19 r +45\right ) \left (8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90\right ) \left (2 r^{2}+15 r +28\right )}\) | \(\frac {527}{18900}\) |
| \(b_{6}\) | \(\frac {64 r^{11}+1856 r^{10}+22880 r^{9}+155440 r^{8}+626048 r^{7}+1462528 r^{6}+1602128 r^{5}-514244 r^{4}-3483850 r^{3}-3417773 r^{2}-917322 r +146781}{\left (16 r^{7}+272 r^{6}+1912 r^{5}+7160 r^{4}+15289 r^{3}+18353 r^{2}+11178 r +2520\right ) \left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+23 r +66\right )}\) | \(\frac {16309}{831600}\) |
| \(b_{7}\) | \(\frac {128 r^{13}+5248 r^{12}+94720 r^{11}+987968 r^{10}+6558944 r^{9}+28650144 r^{8}+81361776 r^{7}+137907936 r^{6}+90592936 r^{5}-126026704 r^{4}-323164836 r^{3}-256226108 r^{2}-59366226 r +10065978}{\left (2 r^{2}+15 r +28\right ) \left (8 r^{5}+76 r^{4}+274 r^{3}+461 r^{2}+351 r +90\right ) \left (2 r^{2}+23 r +66\right ) \left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+27 r +91\right )}\) | \(\frac {14339}{970200}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}+b_{7} x^{7}+b_{8} x^{8}\dots \\ &= 1-2 x -\frac {x^{2}}{6}+\frac {x^{3}}{15}+\frac {37 x^{4}}{840}+\frac {527 x^{5}}{18900}+\frac {16309 x^{6}}{831600}+\frac {14339 x^{7}}{970200}+O\left (x^{8}\right ) \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} \sqrt {x}\, \left (1-\frac {x}{2}-\frac {9 x^{2}}{40}-\frac {149 x^{3}}{1680}-\frac {661 x^{4}}{13440}-\frac {16171 x^{5}}{492800}-\frac {5530601 x^{6}}{230630400}-\frac {299137703 x^{7}}{16144128000}+O\left (x^{8}\right )\right ) + c_{2} \left (1-2 x -\frac {x^{2}}{6}+\frac {x^{3}}{15}+\frac {37 x^{4}}{840}+\frac {527 x^{5}}{18900}+\frac {16309 x^{6}}{831600}+\frac {14339 x^{7}}{970200}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {x}\, \left (1-\frac {x}{2}-\frac {9 x^{2}}{40}-\frac {149 x^{3}}{1680}-\frac {661 x^{4}}{13440}-\frac {16171 x^{5}}{492800}-\frac {5530601 x^{6}}{230630400}-\frac {299137703 x^{7}}{16144128000}+O\left (x^{8}\right )\right )+c_{2} \left (1-2 x -\frac {x^{2}}{6}+\frac {x^{3}}{15}+\frac {37 x^{4}}{840}+\frac {527 x^{5}}{18900}+\frac {16309 x^{6}}{831600}+\frac {14339 x^{7}}{970200}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {x}\, \left (1-\frac {x}{2}-\frac {9 x^{2}}{40}-\frac {149 x^{3}}{1680}-\frac {661 x^{4}}{13440}-\frac {16171 x^{5}}{492800}-\frac {5530601 x^{6}}{230630400}-\frac {299137703 x^{7}}{16144128000}+O\left (x^{8}\right )\right )+c_{2} \left (1-2 x -\frac {x^{2}}{6}+\frac {x^{3}}{15}+\frac {37 x^{4}}{840}+\frac {527 x^{5}}{18900}+\frac {16309 x^{6}}{831600}+\frac {14339 x^{7}}{970200}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1-\frac {x}{2}-\frac {9 x^{2}}{40}-\frac {149 x^{3}}{1680}-\frac {661 x^{4}}{13440}-\frac {16171 x^{5}}{492800}-\frac {5530601 x^{6}}{230630400}-\frac {299137703 x^{7}}{16144128000}+O\left (x^{8}\right )\right )+c_{2} \left (1-2 x -\frac {x^{2}}{6}+\frac {x^{3}}{15}+\frac {37 x^{4}}{840}+\frac {527 x^{5}}{18900}+\frac {16309 x^{6}}{831600}+\frac {14339 x^{7}}{970200}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -2 y^{\prime \prime } x \left (x -1\right )+\left (1-2 x \right ) y^{\prime }+\left (x +2\right ) y=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 {\left (x +2\right ) y}{2 x \left (x -1\right )}-\frac {\left (2 x -1\right ) y^{\prime }}{2 x \left (x -1\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} \\ {} & {} & y^{\prime \prime }+\frac {\left (2 x -1\right ) y^{\prime }}{2 x \left (x -1\right )}-\frac {\left (x +2\right ) y}{2 x \left (x -1\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 {2 x -1}{2 x \left (x -1\right )}, P_{3}\left (x \right )=-\frac {x +2}{2 x \left (x -1\right )}\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}}}=\frac {1}{2} \\ {} & \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}}}=0 \\ {} & \circ & x =0\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}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 2 y^{\prime \prime } x \left (x -1\right )+\left (2 x -1\right ) y^{\prime }+\left (-2-x \right ) 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\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & 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^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (-1+2 r \right ) x^{-1+r}+\left (-a_{1} \left (1+r \right ) \left (1+2 r \right )+2 a_{0} \left (1+r \right ) \left (-1+r \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k +1} \left (k +r +1\right ) \left (2 k +1+2 r \right )+2 a_{k} \left (k +r +1\right ) \left (k +r -1\right )-a_{k -1}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & -a_{1} \left (1+r \right ) \left (1+2 r \right )+2 a_{0} \left (1+r \right ) \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 \left (k +r +1\right ) \left (k +\frac {1}{2}+r \right ) a_{k +1}+2 k^{2} a_{k}+4 k r a_{k}+2 r^{2} a_{k}-2 a_{k}-a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & -2 \left (k +2+r \right ) \left (k +\frac {3}{2}+r \right ) a_{k +2}+2 \left (k +1\right )^{2} a_{k +1}+4 \left (k +1\right ) r a_{k +1}+2 r^{2} a_{k +1}-2 a_{k +1}-a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {2 k^{2} a_{k +1}+4 k r a_{k +1}+2 r^{2} a_{k +1}+4 k a_{k +1}+4 r a_{k +1}-a_{k}}{\left (k +2+r \right ) \left (2 k +3+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {2 k^{2} a_{k +1}+4 k a_{k +1}-a_{k}}{\left (k +2\right ) \left (2 k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=\frac {2 k^{2} a_{k +1}+4 k a_{k +1}-a_{k}}{\left (k +2\right ) \left (2 k +3\right )}, -a_{1}-2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +2}=\frac {2 k^{2} a_{k +1}+6 k a_{k +1}-a_{k}+\frac {5}{2} a_{k +1}}{\left (k +\frac {5}{2}\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 k^{2} a_{k +1}+6 k a_{k +1}-a_{k}+\frac {5}{2} a_{k +1}}{\left (k +\frac {5}{2}\right ) \left (2 k +4\right )}, -3 a_{1}-\frac {3 a_{0}}{2}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{2}}\right ), a_{k +2}=\frac {2 k^{2} a_{k +1}+4 k a_{k +1}-a_{k}}{\left (k +2\right ) \left (2 k +3\right )}, -a_{1}-2 a_{0}=0, b_{k +2}=\frac {2 k^{2} b_{k +1}+6 k b_{k +1}-b_{k}+\frac {5}{2} b_{k +1}}{\left (k +\frac {5}{2}\right ) \left (2 k +4\right )}, -3 b_{1}-\frac {3 b_{0}}{2}=0\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 -> Kummer -> 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 Equivalence transformation and function parameters: {x = t}, {kappa = 12, mu = -8} <- Equivalence to the rational form of Mathieu ODE successful <- Mathieu successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 52
Order:=8; dsolve(2*x*(1-x)*diff(y(x),x$2)+(1-2*x)*diff(y(x),x)+(2+x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} \sqrt {x}\, \left (1-\frac {1}{2} x -\frac {9}{40} x^{2}-\frac {149}{1680} x^{3}-\frac {661}{13440} x^{4}-\frac {16171}{492800} x^{5}-\frac {5530601}{230630400} x^{6}-\frac {299137703}{16144128000} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} \left (1-2 x -\frac {1}{6} x^{2}+\frac {1}{15} x^{3}+\frac {37}{840} x^{4}+\frac {527}{18900} x^{5}+\frac {16309}{831600} x^{6}+\frac {14339}{970200} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 111
AsymptoticDSolveValue[2*x*(1-x)*y''[x]+(1-2*x)*y'[x]+(2+x)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \sqrt {x} \left (-\frac {299137703 x^7}{16144128000}-\frac {5530601 x^6}{230630400}-\frac {16171 x^5}{492800}-\frac {661 x^4}{13440}-\frac {149 x^3}{1680}-\frac {9 x^2}{40}-\frac {x}{2}+1\right )+c_2 \left (\frac {14339 x^7}{970200}+\frac {16309 x^6}{831600}+\frac {527 x^5}{18900}+\frac {37 x^4}{840}+\frac {x^3}{15}-\frac {x^2}{6}-2 x+1\right ) \]