Internal problem ID [6055]
Internal file name [OUTPUT/5303_Sunday_June_05_2022_03_33_28_PM_47241310/index.tex]
Book: An introduction to Ordinary Differential Equations. Earl A. Coddington. Dover. NY
1961
Section: Chapter 4. Linear equations with Regular Singular Points. Page 166
Problem number: 1(iii).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Repeated root"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (-x^{2}+1\right ) x^{2} y^{\prime \prime }+3 \left (x^{2}+x \right ) y^{\prime }+y=0} \] 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. \[ \left (-x^{4}+x^{2}\right ) y^{\prime \prime }+\left (3 x^{2}+3 x \right ) y^{\prime }+y = 0 \] 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 {3}{x \left (-1+x \right )}\\ q(x) &= -\frac {1}{\left (x^{2}-1\right ) x^{2}}\\ \end {align*}
| \(p(x)=-\frac {3}{x \left (-1+x \right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {1}{\left (x^{2}-1\right ) x^{2}}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, 1, -1, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ -y^{\prime \prime } \left (x^{2}-1\right ) x^{2}+\left (3 x^{2}+3 x \right ) y^{\prime }+y = 0 \] 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 (x^{2}-1\right ) x^{2}+\left (3 x^{2}+3 x \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 }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} 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 )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+3 x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+3 x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+3 x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r +1\right )^{2} x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (r +1\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -1\\ r_2 &= -1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r +1\right )^{2} x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([-1, -1]\).
Since the root of the indicial equation is repeated, then we can construct two linearly independent solutions. The first solution has the form \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y = c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \] In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_{1}, c_{2}\}\) are two arbitray constants of integration which can be found from initial conditions. Using the value of the indicial root found earlier, \(r = -1\), Eqs (1A,1B) become \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n -1}\\ y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n -1}\right ) \end {align*}
We start by finding the first solution \(y_{1}\left (x \right )\). 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 \[ a_{1} = -\frac {3 r}{\left (r +2\right )^{2}} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} -a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+a_{n} \left (n +r \right ) \left (n +r -1\right )+3 a_{n -1} \left (n +r -1\right )+3 a_{n} \left (n +r \right )+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {n^{2} a_{n -2}+2 n r a_{n -2}+r^{2} a_{n -2}-5 n a_{n -2}-3 n a_{n -1}-5 r a_{n -2}-3 r a_{n -1}+6 a_{n -2}+3 a_{n -1}}{n^{2}+2 n r +r^{2}+2 n +2 r +1}\tag {4} \] Which for the root \(r = -1\) becomes \[ a_{n} = \frac {n^{2} a_{n -2}+\left (-7 a_{n -2}-3 a_{n -1}\right ) n +12 a_{n -2}+6 a_{n -1}}{n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = -1\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{\left (r +2\right )^{2}}\) | \(3\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r \left (r^{3}+3 r^{2}+9 r +5\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2}} \] Which for the root \(r = -1\) becomes \[ a_{2}={\frac {1}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{\left (r +2\right )^{2}}\) | \(3\) |
| \(a_{2}\) | \(\frac {r \left (r^{3}+3 r^{2}+9 r +5\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(\frac {1}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {6 r \left (r^{4}+6 r^{3}+15 r^{2}+16 r +5\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}} \] Which for the root \(r = -1\) becomes \[ a_{3}=-{\frac {1}{6}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{\left (r +2\right )^{2}}\) | \(3\) |
| \(a_{2}\) | \(\frac {r \left (r^{3}+3 r^{2}+9 r +5\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(-\frac {6 r \left (r^{4}+6 r^{3}+15 r^{2}+16 r +5\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(-{\frac {1}{6}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r \left (r^{7}+14 r^{6}+102 r^{5}+456 r^{4}+1251 r^{3}+1980 r^{2}+1562 r +430\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}} \] Which for the root \(r = -1\) becomes \[ a_{4}={\frac {1}{16}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{\left (r +2\right )^{2}}\) | \(3\) |
| \(a_{2}\) | \(\frac {r \left (r^{3}+3 r^{2}+9 r +5\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(-\frac {6 r \left (r^{4}+6 r^{3}+15 r^{2}+16 r +5\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(-{\frac {1}{6}}\) |
| \(a_{4}\) | \(\frac {r \left (r^{7}+14 r^{6}+102 r^{5}+456 r^{4}+1251 r^{3}+1980 r^{2}+1562 r +430\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(\frac {1}{16}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-9 r^{9}-180 r^{8}-1590 r^{7}-8064 r^{6}-25545 r^{5}-51228 r^{4}-62136 r^{3}-39984 r^{2}-9660 r}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}} \] Which for the root \(r = -1\) becomes \[ a_{5}=-{\frac {43}{1200}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{\left (r +2\right )^{2}}\) | \(3\) |
| \(a_{2}\) | \(\frac {r \left (r^{3}+3 r^{2}+9 r +5\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(-\frac {6 r \left (r^{4}+6 r^{3}+15 r^{2}+16 r +5\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(-{\frac {1}{6}}\) |
| \(a_{4}\) | \(\frac {r \left (r^{7}+14 r^{6}+102 r^{5}+456 r^{4}+1251 r^{3}+1980 r^{2}+1562 r +430\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(\frac {1}{16}\) |
| \(a_{5}\) | \(\frac {-9 r^{9}-180 r^{8}-1590 r^{7}-8064 r^{6}-25545 r^{5}-51228 r^{4}-62136 r^{3}-39984 r^{2}-9660 r}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}}\) | \(-{\frac {43}{1200}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {r \left (r^{11}+33 r^{10}+527 r^{9}+5313 r^{8}+36825 r^{7}+180423 r^{6}+626549 r^{5}+1520715 r^{4}+2493694 r^{3}+2582664 r^{2}+1473804 r +330660\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2} \left (r +7\right )^{2}} \] Which for the root \(r = -1\) becomes \[ a_{6}={\frac {161}{7200}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{\left (r +2\right )^{2}}\) | \(3\) |
| \(a_{2}\) | \(\frac {r \left (r^{3}+3 r^{2}+9 r +5\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(-\frac {6 r \left (r^{4}+6 r^{3}+15 r^{2}+16 r +5\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(-{\frac {1}{6}}\) |
| \(a_{4}\) | \(\frac {r \left (r^{7}+14 r^{6}+102 r^{5}+456 r^{4}+1251 r^{3}+1980 r^{2}+1562 r +430\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(\frac {1}{16}\) |
| \(a_{5}\) | \(\frac {-9 r^{9}-180 r^{8}-1590 r^{7}-8064 r^{6}-25545 r^{5}-51228 r^{4}-62136 r^{3}-39984 r^{2}-9660 r}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}}\) | \(-{\frac {43}{1200}}\) |
| \(a_{6}\) | \(\frac {r \left (r^{11}+33 r^{10}+527 r^{9}+5313 r^{8}+36825 r^{7}+180423 r^{6}+626549 r^{5}+1520715 r^{4}+2493694 r^{3}+2582664 r^{2}+1473804 r +330660\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2} \left (r +7\right )^{2}}\) | \(\frac {161}{7200}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=-\frac {12 r \left (r^{12}+42 r^{11}+805 r^{10}+9304 r^{9}+72148 r^{8}+394846 r^{7}+1560105 r^{6}+4468272 r^{5}+9155771 r^{4}+12971420 r^{3}+11876234 r^{2}+6139136 r +1284890\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2} \left (r +7\right )^{2} \left (r +8\right )^{2}} \] Which for the root \(r = -1\) becomes \[ a_{7}=-{\frac {1837}{117600}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{\left (r +2\right )^{2}}\) | \(3\) |
| \(a_{2}\) | \(\frac {r \left (r^{3}+3 r^{2}+9 r +5\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(-\frac {6 r \left (r^{4}+6 r^{3}+15 r^{2}+16 r +5\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(-{\frac {1}{6}}\) |
| \(a_{4}\) | \(\frac {r \left (r^{7}+14 r^{6}+102 r^{5}+456 r^{4}+1251 r^{3}+1980 r^{2}+1562 r +430\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(\frac {1}{16}\) |
| \(a_{5}\) | \(\frac {-9 r^{9}-180 r^{8}-1590 r^{7}-8064 r^{6}-25545 r^{5}-51228 r^{4}-62136 r^{3}-39984 r^{2}-9660 r}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}}\) | \(-{\frac {43}{1200}}\) |
| \(a_{6}\) | \(\frac {r \left (r^{11}+33 r^{10}+527 r^{9}+5313 r^{8}+36825 r^{7}+180423 r^{6}+626549 r^{5}+1520715 r^{4}+2493694 r^{3}+2582664 r^{2}+1473804 r +330660\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2} \left (r +7\right )^{2}}\) | \(\frac {161}{7200}\) |
| \(a_{7}\) | \(-\frac {12 r \left (r^{12}+42 r^{11}+805 r^{10}+9304 r^{9}+72148 r^{8}+394846 r^{7}+1560105 r^{6}+4468272 r^{5}+9155771 r^{4}+12971420 r^{3}+11876234 r^{2}+6139136 r +1284890\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2} \left (r +7\right )^{2} \left (r +8\right )^{2}}\) | \(-{\frac {1837}{117600}}\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= \frac {1}{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 ) \\ &= \frac {1+3 x +\frac {x^{2}}{2}-\frac {x^{3}}{6}+\frac {x^{4}}{16}-\frac {43 x^{5}}{1200}+\frac {161 x^{6}}{7200}-\frac {1837 x^{7}}{117600}+O\left (x^{8}\right )}{x} \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (x \right ) = y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = -1\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =-1\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(-\frac {3 r}{\left (r +2\right )^{2}}\) | \(3\) | \(\frac {3 r -6}{\left (r +2\right )^{3}}\) | \(-9\) |
| \(b_{2}\) | \(\frac {r \left (r^{3}+3 r^{2}+9 r +5\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(\frac {1}{2}\) | \(\frac {7 r^{4}+21 r^{3}+39 r^{2}+83 r +30}{\left (r +2\right )^{3} \left (r +3\right )^{3}}\) | \(-{\frac {7}{2}}\) |
| \(b_{3}\) | \(-\frac {6 r \left (r^{4}+6 r^{3}+15 r^{2}+16 r +5\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(-{\frac {1}{6}}\) | \(\frac {6 r^{7}+18 r^{6}-198 r^{5}-1398 r^{4}-3918 r^{3}-5670 r^{2}-3828 r -720}{\left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3}}\) | \(\frac {7}{9}\) |
| \(b_{4}\) | \(\frac {r \left (r^{7}+14 r^{6}+102 r^{5}+456 r^{4}+1251 r^{3}+1980 r^{2}+1562 r +430\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(\frac {1}{16}\) | \(\frac {14 r^{10}+276 r^{9}+2538 r^{8}+14836 r^{7}+62040 r^{6}+191580 r^{5}+427846 r^{4}+653496 r^{3}+621210 r^{2}+308660 r +51600}{\left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3}}\) | \(-{\frac {25}{96}}\) |
| \(b_{5}\) | \(\frac {-9 r^{9}-180 r^{8}-1590 r^{7}-8064 r^{6}-25545 r^{5}-51228 r^{4}-62136 r^{3}-39984 r^{2}-9660 r}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}}\) | \(-{\frac {43}{1200}}\) | \(\frac {9 r^{13}+180 r^{12}+585 r^{11}-17844 r^{10}-279537 r^{9}-2112372 r^{8}-10198173 r^{7}-34090812 r^{6}-81065160 r^{5}-136744704 r^{4}-158538684 r^{3}-117405360 r^{2}-47491920 r -6955200}{\left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3} \left (r +6\right )^{3}}\) | \(\frac {5141}{36000}\) |
| \(b_{6}\) | \(\frac {r \left (r^{11}+33 r^{10}+527 r^{9}+5313 r^{8}+36825 r^{7}+180423 r^{6}+626549 r^{5}+1520715 r^{4}+2493694 r^{3}+2582664 r^{2}+1473804 r +330660\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2} \left (r +7\right )^{2}}\) | \(\frac {161}{7200}\) | \(\frac {21 r^{16}+1017 r^{15}+23256 r^{14}+335736 r^{13}+3445794 r^{12}+26748522 r^{11}+162353496 r^{10}+782443656 r^{9}+3004366077 r^{8}+9145364481 r^{7}+21811384944 r^{6}+39982351680 r^{5}+54680504268 r^{4}+53209161900 r^{3}+33986813760 r^{2}+12201405840 r +1666526400}{\left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3} \left (r +6\right )^{3} \left (r +7\right )^{3}}\) | \(-{\frac {2083}{24000}}\) |
| \(b_{7}\) | \(-\frac {12 r \left (r^{12}+42 r^{11}+805 r^{10}+9304 r^{9}+72148 r^{8}+394846 r^{7}+1560105 r^{6}+4468272 r^{5}+9155771 r^{4}+12971420 r^{3}+11876234 r^{2}+6139136 r +1284890\right )}{\left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2} \left (r +7\right )^{2} \left (r +8\right )^{2}}\) | \(-{\frac {1837}{117600}}\) | \(\frac {12 r^{19}+588 r^{18}+10584 r^{17}+28104 r^{16}-2454036 r^{15}-58300164 r^{14}-757990332 r^{13}-6824487132 r^{12}-45901005984 r^{11}-238825017216 r^{10}-978228508620 r^{9}-3177743343804 r^{8}-8185489825296 r^{7}-16596780113568 r^{6}-26074413368568 r^{5}-30885647612568 r^{4}-26356147831872 r^{3}-14978521061280 r^{2}-4872759672960 r -621681177600}{\left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3} \left (r +6\right )^{3} \left (r +7\right )^{3} \left (r +8\right )^{3}}\) | \(\frac {489941}{8232000}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (x \right )&=y_{1}\left (x \right ) \ln \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 \\ &= \frac {\left (1+3 x +\frac {x^{2}}{2}-\frac {x^{3}}{6}+\frac {x^{4}}{16}-\frac {43 x^{5}}{1200}+\frac {161 x^{6}}{7200}-\frac {1837 x^{7}}{117600}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x}+\frac {-9 x -\frac {7 x^{2}}{2}+\frac {7 x^{3}}{9}-\frac {25 x^{4}}{96}+\frac {5141 x^{5}}{36000}-\frac {2083 x^{6}}{24000}+\frac {489941 x^{7}}{8232000}+O\left (x^{8}\right )}{x} \\ \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 ) \\ &= \frac {c_{1} \left (1+3 x +\frac {x^{2}}{2}-\frac {x^{3}}{6}+\frac {x^{4}}{16}-\frac {43 x^{5}}{1200}+\frac {161 x^{6}}{7200}-\frac {1837 x^{7}}{117600}+O\left (x^{8}\right )\right )}{x} + c_{2} \left (\frac {\left (1+3 x +\frac {x^{2}}{2}-\frac {x^{3}}{6}+\frac {x^{4}}{16}-\frac {43 x^{5}}{1200}+\frac {161 x^{6}}{7200}-\frac {1837 x^{7}}{117600}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x}+\frac {-9 x -\frac {7 x^{2}}{2}+\frac {7 x^{3}}{9}-\frac {25 x^{4}}{96}+\frac {5141 x^{5}}{36000}-\frac {2083 x^{6}}{24000}+\frac {489941 x^{7}}{8232000}+O\left (x^{8}\right )}{x}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= \frac {c_{1} \left (1+3 x +\frac {x^{2}}{2}-\frac {x^{3}}{6}+\frac {x^{4}}{16}-\frac {43 x^{5}}{1200}+\frac {161 x^{6}}{7200}-\frac {1837 x^{7}}{117600}+O\left (x^{8}\right )\right )}{x}+c_{2} \left (\frac {\left (1+3 x +\frac {x^{2}}{2}-\frac {x^{3}}{6}+\frac {x^{4}}{16}-\frac {43 x^{5}}{1200}+\frac {161 x^{6}}{7200}-\frac {1837 x^{7}}{117600}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x}+\frac {-9 x -\frac {7 x^{2}}{2}+\frac {7 x^{3}}{9}-\frac {25 x^{4}}{96}+\frac {5141 x^{5}}{36000}-\frac {2083 x^{6}}{24000}+\frac {489941 x^{7}}{8232000}+O\left (x^{8}\right )}{x}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (1+3 x +\frac {x^{2}}{2}-\frac {x^{3}}{6}+\frac {x^{4}}{16}-\frac {43 x^{5}}{1200}+\frac {161 x^{6}}{7200}-\frac {1837 x^{7}}{117600}+O\left (x^{8}\right )\right )}{x}+c_{2} \left (\frac {\left (1+3 x +\frac {x^{2}}{2}-\frac {x^{3}}{6}+\frac {x^{4}}{16}-\frac {43 x^{5}}{1200}+\frac {161 x^{6}}{7200}-\frac {1837 x^{7}}{117600}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x}+\frac {-9 x -\frac {7 x^{2}}{2}+\frac {7 x^{3}}{9}-\frac {25 x^{4}}{96}+\frac {5141 x^{5}}{36000}-\frac {2083 x^{6}}{24000}+\frac {489941 x^{7}}{8232000}+O\left (x^{8}\right )}{x}\right ) \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (1+3 x +\frac {x^{2}}{2}-\frac {x^{3}}{6}+\frac {x^{4}}{16}-\frac {43 x^{5}}{1200}+\frac {161 x^{6}}{7200}-\frac {1837 x^{7}}{117600}+O\left (x^{8}\right )\right )}{x}+c_{2} \left (\frac {\left (1+3 x +\frac {x^{2}}{2}-\frac {x^{3}}{6}+\frac {x^{4}}{16}-\frac {43 x^{5}}{1200}+\frac {161 x^{6}}{7200}-\frac {1837 x^{7}}{117600}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x}+\frac {-9 x -\frac {7 x^{2}}{2}+\frac {7 x^{3}}{9}-\frac {25 x^{4}}{96}+\frac {5141 x^{5}}{36000}-\frac {2083 x^{6}}{24000}+\frac {489941 x^{7}}{8232000}+O\left (x^{8}\right )}{x}\right ) \] 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: 0.031 (sec). Leaf size: 81
Order:=8; dsolve((1-x^2)*x^2*diff(y(x),x$2)+3*(x+x^2)*diff(y(x),x)+y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1+3 x +\frac {1}{2} x^{2}-\frac {1}{6} x^{3}+\frac {1}{16} x^{4}-\frac {43}{1200} x^{5}+\frac {161}{7200} x^{6}-\frac {1837}{117600} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (\left (-9\right ) x -\frac {7}{2} x^{2}+\frac {7}{9} x^{3}-\frac {25}{96} x^{4}+\frac {5141}{36000} x^{5}-\frac {2083}{24000} x^{6}+\frac {489941}{8232000} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2}}{x} \]
✓ Solution by Mathematica
Time used: 0.001 (sec). Leaf size: 84
AsymptoticDSolveValue[(1-x^2)*y''[x]+3*(x+x^2)*y'[x]+y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_2 \left (\frac {53 x^7}{630}+\frac {5 x^6}{24}+\frac {2 x^5}{15}-\frac {x^4}{4}-\frac {2 x^3}{3}+x\right )+c_1 \left (-\frac {19 x^7}{420}-\frac {x^6}{144}+\frac {3 x^5}{20}+\frac {5 x^4}{24}-\frac {x^2}{2}+1\right ) \]