Internal problem ID [7235]
Internal file name [OUTPUT/6221_Sunday_June_05_2022_04_32_52_PM_9038864/index.tex]
Book: Own collection of miscellaneous problems
Section: section 4.0
Problem number: 14.
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
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (x -2\right ) y^{\prime \prime }+\frac {y^{\prime }}{x}+\left (1+x \right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (x -2\right ) y^{\prime \prime }+\frac {y^{\prime }}{x}+\left (1+x \right ) y = 0 \] 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 {1}{x \left (x -2\right )}\\ q(x) &= \frac {1+x}{x -2}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, 2]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ y^{\prime \prime } x \left (x -2\right )+y^{\prime }+x \left (1+x \right ) y = 0 \] 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 ) x \left (x -2\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+x \left (1+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 }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\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^{n +r +2} 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 }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =3}{\sum }}a_{n -3} x^{n +r -1} \\ \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 =1}{\sum }}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 -1} a_{n} \left (n +r \right ) \left (n +r -1\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 =3}{\sum }}a_{n -3} 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 \[ -2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} x^{n +r -1} = 0 \] 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 \[ \left (-2 r^{2}+3 r \right ) x^{-1+r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ -2 r^{2}+3 r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {3}{2}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (-2 r^{2}+3 r \right ) x^{-1+r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {3}{2}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {3}{2}}\) 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 {3}{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 \[ a_{1} = \frac {r \left (-1+r \right )}{2 r^{2}+r -1} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {r^{3}-r^{2}+2 r -1}{4 r^{3}+8 r^{2}-r -2} \] For \(3\le n\) the recursive equation is \begin{equation} \tag{3} 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 )+a_{n} \left (n +r \right )+a_{n -3}+a_{n -2} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {n^{2} a_{n -1}+2 n r a_{n -1}+r^{2} a_{n -1}-3 n a_{n -1}-3 r a_{n -1}+a_{n -3}+a_{n -2}+2 a_{n -1}}{2 n^{2}+4 n r +2 r^{2}-3 n -3 r}\tag {4} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{n} = \frac {4 n^{2} a_{n -1}+4 a_{n -3}+4 a_{n -2}-a_{n -1}}{8 n^{2}+12 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {3}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r \left (-1+r \right )}{2 r^{2}+r -1}\) | \(\frac {3}{20}\) |
| \(a_{2}\) | \(\frac {r^{3}-r^{2}+2 r -1}{4 r^{3}+8 r^{2}-r -2}\) | \(\frac {25}{224}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {r^{5}+r^{4}+7 r^{3}+5 r^{2}-2 r -2}{8 r^{5}+44 r^{4}+70 r^{3}+25 r^{2}-18 r -9} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{3}={\frac {1361}{17280}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r \left (-1+r \right )}{2 r^{2}+r -1}\) | \(\frac {3}{20}\) |
| \(a_{2}\) | \(\frac {r^{3}-r^{2}+2 r -1}{4 r^{3}+8 r^{2}-r -2}\) | \(\frac {25}{224}\) |
| \(a_{3}\) | \(\frac {r^{5}+r^{4}+7 r^{3}+5 r^{2}-2 r -2}{8 r^{5}+44 r^{4}+70 r^{3}+25 r^{2}-18 r -9}\) | \(\frac {1361}{17280}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{7}+5 r^{6}+21 r^{5}+52 r^{4}+51 r^{3}+2 r^{2}-21 r -11}{\left (4 r^{3}+8 r^{2}-r -2\right ) \left (1+r \right ) \left (2 r +3\right ) \left (2 r^{2}+13 r +20\right )} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{4}={\frac {80753}{2365440}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r \left (-1+r \right )}{2 r^{2}+r -1}\) | \(\frac {3}{20}\) |
| \(a_{2}\) | \(\frac {r^{3}-r^{2}+2 r -1}{4 r^{3}+8 r^{2}-r -2}\) | \(\frac {25}{224}\) |
| \(a_{3}\) | \(\frac {r^{5}+r^{4}+7 r^{3}+5 r^{2}-2 r -2}{8 r^{5}+44 r^{4}+70 r^{3}+25 r^{2}-18 r -9}\) | \(\frac {1361}{17280}\) |
| \(a_{4}\) | \(\frac {r^{7}+5 r^{6}+21 r^{5}+52 r^{4}+51 r^{3}+2 r^{2}-21 r -11}{\left (4 r^{3}+8 r^{2}-r -2\right ) \left (1+r \right ) \left (2 r +3\right ) \left (2 r^{2}+13 r +20\right )}\) | \(\frac {80753}{2365440}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {r^{9}+11 r^{8}+66 r^{7}+262 r^{6}+652 r^{5}+936 r^{4}+648 r^{3}-11 r^{2}-311 r -164}{\left (8 r^{5}+44 r^{4}+70 r^{3}+25 r^{2}-18 r -9\right ) \left (2+r \right ) \left (2 r +5\right ) \left (2 r^{2}+17 r +35\right )} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{5}={\frac {616517}{38707200}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r \left (-1+r \right )}{2 r^{2}+r -1}\) | \(\frac {3}{20}\) |
| \(a_{2}\) | \(\frac {r^{3}-r^{2}+2 r -1}{4 r^{3}+8 r^{2}-r -2}\) | \(\frac {25}{224}\) |
| \(a_{3}\) | \(\frac {r^{5}+r^{4}+7 r^{3}+5 r^{2}-2 r -2}{8 r^{5}+44 r^{4}+70 r^{3}+25 r^{2}-18 r -9}\) | \(\frac {1361}{17280}\) |
| \(a_{4}\) | \(\frac {r^{7}+5 r^{6}+21 r^{5}+52 r^{4}+51 r^{3}+2 r^{2}-21 r -11}{\left (4 r^{3}+8 r^{2}-r -2\right ) \left (1+r \right ) \left (2 r +3\right ) \left (2 r^{2}+13 r +20\right )}\) | \(\frac {80753}{2365440}\) |
| \(a_{5}\) | \(\frac {r^{9}+11 r^{8}+66 r^{7}+262 r^{6}+652 r^{5}+936 r^{4}+648 r^{3}-11 r^{2}-311 r -164}{\left (8 r^{5}+44 r^{4}+70 r^{3}+25 r^{2}-18 r -9\right ) \left (2+r \right ) \left (2 r +5\right ) \left (2 r^{2}+17 r +35\right )}\) | \(\frac {616517}{38707200}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {3}{2}} \left (a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}\dots \right ) \\ &= x^{\frac {3}{2}} \left (1+\frac {3 x}{20}+\frac {25 x^{2}}{224}+\frac {1361 x^{3}}{17280}+\frac {80753 x^{4}}{2365440}+\frac {616517 x^{5}}{38707200}+O\left (x^{6}\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 \[ b_{1} = \frac {r \left (-1+r \right )}{2 r^{2}+r -1} \] Substituting \(n = 2\) in Eq. (2B) gives \[ b_{2} = \frac {r^{3}-r^{2}+2 r -1}{4 r^{3}+8 r^{2}-r -2} \] For \(3\le n\) the recursive equation is \begin{equation} \tag{3} 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 )+\left (n +r \right ) b_{n}+b_{n -3}+b_{n -2} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {n^{2} b_{n -1}+2 n r b_{n -1}+r^{2} b_{n -1}-3 n b_{n -1}-3 r b_{n -1}+b_{n -3}+b_{n -2}+2 b_{n -1}}{2 n^{2}+4 n r +2 r^{2}-3 n -3 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {\left (n^{2}-3 n +2\right ) b_{n -1}+b_{n -3}+b_{n -2}}{2 n^{2}-3 n}\tag {5} \] 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.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r \left (-1+r \right )}{2 r^{2}+r -1}\) | \(0\) |
| \(b_{2}\) | \(\frac {r^{3}-r^{2}+2 r -1}{4 r^{3}+8 r^{2}-r -2}\) | \(\frac {1}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {r^{5}+r^{4}+7 r^{3}+5 r^{2}-2 r -2}{8 r^{5}+44 r^{4}+70 r^{3}+25 r^{2}-18 r -9} \] Which for the root \(r = 0\) becomes \[ b_{3}={\frac {2}{9}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r \left (-1+r \right )}{2 r^{2}+r -1}\) | \(0\) |
| \(b_{2}\) | \(\frac {r^{3}-r^{2}+2 r -1}{4 r^{3}+8 r^{2}-r -2}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(\frac {r^{5}+r^{4}+7 r^{3}+5 r^{2}-2 r -2}{8 r^{5}+44 r^{4}+70 r^{3}+25 r^{2}-18 r -9}\) | \(\frac {2}{9}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {r^{7}+5 r^{6}+21 r^{5}+52 r^{4}+51 r^{3}+2 r^{2}-21 r -11}{\left (4 r^{3}+8 r^{2}-r -2\right ) \left (1+r \right ) \left (2 r +3\right ) \left (2 r^{2}+13 r +20\right )} \] Which for the root \(r = 0\) becomes \[ b_{4}={\frac {11}{120}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r \left (-1+r \right )}{2 r^{2}+r -1}\) | \(0\) |
| \(b_{2}\) | \(\frac {r^{3}-r^{2}+2 r -1}{4 r^{3}+8 r^{2}-r -2}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(\frac {r^{5}+r^{4}+7 r^{3}+5 r^{2}-2 r -2}{8 r^{5}+44 r^{4}+70 r^{3}+25 r^{2}-18 r -9}\) | \(\frac {2}{9}\) |
| \(b_{4}\) | \(\frac {r^{7}+5 r^{6}+21 r^{5}+52 r^{4}+51 r^{3}+2 r^{2}-21 r -11}{\left (4 r^{3}+8 r^{2}-r -2\right ) \left (1+r \right ) \left (2 r +3\right ) \left (2 r^{2}+13 r +20\right )}\) | \(\frac {11}{120}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {r^{9}+11 r^{8}+66 r^{7}+262 r^{6}+652 r^{5}+936 r^{4}+648 r^{3}-11 r^{2}-311 r -164}{\left (8 r^{5}+44 r^{4}+70 r^{3}+25 r^{2}-18 r -9\right ) \left (2+r \right ) \left (2 r +5\right ) \left (2 r^{2}+17 r +35\right )} \] Which for the root \(r = 0\) becomes \[ b_{5}={\frac {82}{1575}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r \left (-1+r \right )}{2 r^{2}+r -1}\) | \(0\) |
| \(b_{2}\) | \(\frac {r^{3}-r^{2}+2 r -1}{4 r^{3}+8 r^{2}-r -2}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(\frac {r^{5}+r^{4}+7 r^{3}+5 r^{2}-2 r -2}{8 r^{5}+44 r^{4}+70 r^{3}+25 r^{2}-18 r -9}\) | \(\frac {2}{9}\) |
| \(b_{4}\) | \(\frac {r^{7}+5 r^{6}+21 r^{5}+52 r^{4}+51 r^{3}+2 r^{2}-21 r -11}{\left (4 r^{3}+8 r^{2}-r -2\right ) \left (1+r \right ) \left (2 r +3\right ) \left (2 r^{2}+13 r +20\right )}\) | \(\frac {11}{120}\) |
| \(b_{5}\) | \(\frac {r^{9}+11 r^{8}+66 r^{7}+262 r^{6}+652 r^{5}+936 r^{4}+648 r^{3}-11 r^{2}-311 r -164}{\left (8 r^{5}+44 r^{4}+70 r^{3}+25 r^{2}-18 r -9\right ) \left (2+r \right ) \left (2 r +5\right ) \left (2 r^{2}+17 r +35\right )}\) | \(\frac {82}{1575}\) |
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}\dots \\ &= 1+\frac {x^{2}}{2}+\frac {2 x^{3}}{9}+\frac {11 x^{4}}{120}+\frac {82 x^{5}}{1575}+O\left (x^{6}\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} x^{\frac {3}{2}} \left (1+\frac {3 x}{20}+\frac {25 x^{2}}{224}+\frac {1361 x^{3}}{17280}+\frac {80753 x^{4}}{2365440}+\frac {616517 x^{5}}{38707200}+O\left (x^{6}\right )\right ) + c_{2} \left (1+\frac {x^{2}}{2}+\frac {2 x^{3}}{9}+\frac {11 x^{4}}{120}+\frac {82 x^{5}}{1575}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {3}{2}} \left (1+\frac {3 x}{20}+\frac {25 x^{2}}{224}+\frac {1361 x^{3}}{17280}+\frac {80753 x^{4}}{2365440}+\frac {616517 x^{5}}{38707200}+O\left (x^{6}\right )\right )+c_{2} \left (1+\frac {x^{2}}{2}+\frac {2 x^{3}}{9}+\frac {11 x^{4}}{120}+\frac {82 x^{5}}{1575}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {3}{2}} \left (1+\frac {3 x}{20}+\frac {25 x^{2}}{224}+\frac {1361 x^{3}}{17280}+\frac {80753 x^{4}}{2365440}+\frac {616517 x^{5}}{38707200}+O\left (x^{6}\right )\right )+c_{2} \left (1+\frac {x^{2}}{2}+\frac {2 x^{3}}{9}+\frac {11 x^{4}}{120}+\frac {82 x^{5}}{1575}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {3}{2}} \left (1+\frac {3 x}{20}+\frac {25 x^{2}}{224}+\frac {1361 x^{3}}{17280}+\frac {80753 x^{4}}{2365440}+\frac {616517 x^{5}}{38707200}+O\left (x^{6}\right )\right )+c_{2} \left (1+\frac {x^{2}}{2}+\frac {2 x^{3}}{9}+\frac {11 x^{4}}{120}+\frac {82 x^{5}}{1575}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime } x \left (x -2\right )+y^{\prime }+x \left (1+x \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 {y^{\prime }}{x \left (x -2\right )}-\frac {\left (1+x \right ) y}{x -2} \\ \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 {y^{\prime }}{x \left (x -2\right )}+\frac {\left (1+x \right ) y}{x -2}=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 {1}{x \left (x -2\right )}, P_{3}\left (x \right )=\frac {1+x}{x -2}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-\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} \\ {} & {} & y^{\prime \prime } x \left (x -2\right )+y^{\prime }+x \left (1+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 =1..2 \\ {} & {} & 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} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+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 (-3+2 r \right ) x^{-1+r}+\left (-a_{1} \left (1+r \right ) \left (-1+2 r \right )+a_{0} r \left (-1+r \right )\right ) x^{r}+\left (-a_{2} \left (2+r \right ) \left (1+2 r \right )+a_{1} \left (1+r \right ) r +a_{0}\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (2 k -1+2 r \right )+a_{k} \left (k +r \right ) \left (k +r -1\right )+a_{k -1}+a_{k -2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-3+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {3}{2}\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [-a_{1} \left (1+r \right ) \left (-1+2 r \right )+a_{0} r \left (-1+r \right )=0, -a_{2} \left (2+r \right ) \left (1+2 r \right )+a_{1} \left (1+r \right ) r +a_{0}=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {a_{0} r \left (-1+r \right )}{2 r^{2}+r -1}, a_{2}=\frac {a_{0} \left (r^{3}-r^{2}+2 r -1\right )}{4 r^{3}+8 r^{2}-r -2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 \left (k +r -\frac {1}{2}\right ) \left (k +1+r \right ) a_{k +1}+a_{k} \left (k +r \right ) \left (k +r -1\right )+a_{k -1}+a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & -2 \left (k +\frac {3}{2}+r \right ) \left (k +3+r \right ) a_{k +3}+a_{k +2} \left (k +2+r \right ) \left (k +1+r \right )+a_{k +1}+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k +2}+2 k r a_{k +2}+r^{2} a_{k +2}+3 k a_{k +2}+3 r a_{k +2}+a_{k}+a_{k +1}+2 a_{k +2}}{\left (2 k +3+2 r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k +2}+3 k a_{k +2}+a_{k}+a_{k +1}+2 a_{k +2}}{\left (2 k +3\right ) \left (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 +3}=\frac {k^{2} a_{k +2}+3 k a_{k +2}+a_{k}+a_{k +1}+2 a_{k +2}}{\left (2 k +3\right ) \left (k +3\right )}, a_{1}=0, a_{2}=\frac {a_{0}}{2}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k +2}+6 k a_{k +2}+a_{k}+a_{k +1}+\frac {35}{4} a_{k +2}}{\left (2 k +6\right ) \left (k +\frac {9}{2}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {3}{2}}, a_{k +3}=\frac {k^{2} a_{k +2}+6 k a_{k +2}+a_{k}+a_{k +1}+\frac {35}{4} a_{k +2}}{\left (2 k +6\right ) \left (k +\frac {9}{2}\right )}, a_{1}=\frac {3 a_{0}}{20}, a_{2}=\frac {25 a_{0}}{224}\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 {3}{2}}\right ), a_{k +3}=\frac {k^{2} a_{k +2}+3 k a_{k +2}+a_{k}+a_{k +1}+2 a_{k +2}}{\left (2 k +3\right ) \left (k +3\right )}, a_{1}=0, a_{2}=\frac {a_{0}}{2}, b_{k +3}=\frac {k^{2} b_{k +2}+6 k b_{k +2}+b_{k}+b_{k +1}+\frac {35}{4} b_{k +2}}{\left (2 k +6\right ) \left (k +\frac {9}{2}\right )}, b_{1}=\frac {3 b_{0}}{20}, b_{2}=\frac {25 b_{0}}{224}\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 trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunC ODE, case a <> 0, e <> 0, c = 0 `
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 42
Order:=6; dsolve((x-2)*diff(y(x), x$2) + 1/x*diff(y(x), x) + (x+1)*y(x) = 0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {3}{2}} \left (1+\frac {3}{20} x +\frac {25}{224} x^{2}+\frac {1361}{17280} x^{3}+\frac {80753}{2365440} x^{4}+\frac {616517}{38707200} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (1+\frac {1}{2} x^{2}+\frac {2}{9} x^{3}+\frac {11}{120} x^{4}+\frac {82}{1575} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 80
AsymptoticDSolveValue[(x-2)*y''[x] + 1/x*y'[x] + (x+1)*y[x] ==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {82 x^5}{1575}+\frac {11 x^4}{120}+\frac {2 x^3}{9}+\frac {x^2}{2}+1\right )+c_1 \left (\frac {616517 x^5}{38707200}+\frac {80753 x^4}{2365440}+\frac {1361 x^3}{17280}+\frac {25 x^2}{224}+\frac {3 x}{20}+1\right ) x^{3/2} \]