2.4.14 problem 14
Internal
problem
ID
[8329]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
14
Date
solved
:
Sunday, November 10, 2024 at 03:38:36 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} \left (x -2\right ) y^{\prime \prime }+\frac {y^{\prime }}{x}+\left (x +1\right ) y&=0 \end{align*}
Using series expansion around \(x=0\)
The type of the expansion point is first determined. This is done on the homogeneous part of
the ODE.
\[ \left (x -2\right ) y^{\prime \prime }+\frac {y^{\prime }}{x}+\left (x +1\right ) y = 0 \]
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 {1}{x \left (x -2\right )}\\ q(x) &= \frac {x +1}{x -2}\\ \end{align*}
Table 2.66: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=\frac {1}{x \left (x -2\right )}\) |
| |
|
singularity | type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| \(x = 2\) |
\(\text {``regular''}\) |
| |
| |
| \(q(x)=\frac {x +1}{x -2}\) |
| |
|
singularity | type |
| |
| \(x = 2\) | \(\text {``regular''}\) |
| |
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
\[ x \left (x -2\right ) y^{\prime \prime }+y^{\prime }+x \left (x +1\right ) y = 0 \]
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} x \left (x -2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+x \left (x +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} \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^{{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^{{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^{{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^{{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*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x -2\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\frac {\frac {d}{d x}y \left (x \right )}{x}+\left (x +1\right ) y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=-\frac {\left (x +1\right ) y \left (x \right )}{x -2}-\frac {\frac {d}{d x}y \left (x \right )}{\left (x -2\right ) x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )+\frac {\frac {d}{d x}y \left (x \right )}{\left (x -2\right ) x}+\frac {\left (x +1\right ) y \left (x \right )}{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 {x +1}{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} \\ {} & {} & x \left (x -2\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\frac {d}{d x}y \left (x \right )+x \left (x +1\right ) y \left (x \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\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 \left (x \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot y \left (x \right )=\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 \left (x \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d}{d x}y \left (x \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\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 \\ {} & {} & \frac {d}{d x}y \left (x \right )=\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 \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=\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 \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=\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 +1+r \right ) \left (k +r -\frac {1}{2}\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 +3+r \right ) \left (k +\frac {3}{2}+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 (k +3+r \right ) \left (2 k +3+2 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 (k +3\right ) \left (2 k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (x \right )=\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 (k +3\right ) \left (2 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 (k +\frac {9}{2}\right ) \left (2 k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & \left [y \left (x \right )=\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 (k +\frac {9}{2}\right ) \left (2 k +6\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 (x \right )=\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 (k +3\right ) \left (2 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 (k +\frac {9}{2}\right ) \left (2 k +6\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 `
Maple dsolve solution
Solving time : 0.020 (sec)
Leaf size : 42
dsolve((x-2)*diff(diff(y(x),x),x)+1/x*diff(y(x),x)+y(x)*(x+1) = 0,y(x),
series,x=0)
\[
y = c_{1} x^{{3}/{2}} \left (1+\frac {3}{20} x +\frac {25}{224} x^{2}+\frac {1361}{17280} x^{3}+\frac {80753}{2365440} x^{4}+\frac {616517}{38707200} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (1+\frac {1}{2} x^{2}+\frac {2}{9} x^{3}+\frac {11}{120} x^{4}+\frac {82}{1575} x^{5}+\operatorname {O}\left (x^{6}\right )\right )
\]
Mathematica DSolve solution
Solving time : 0.01 (sec)
Leaf size : 80
AsymptoticDSolveValue[{(x-2)*D[y[x],{x,2}] + 1/x*D[y[x],x] + (x+1)*y[x] ==0,{}},
y[x],{x,0,5}]
\[
y(x)\to c_2 \left (\frac {82 x^5}{1575}+\frac {11 x^4}{120}+\frac {2 x^3}{9}+\frac {x^2}{2}+1\right )+c_1 \left (\frac {616517 x^5}{38707200}+\frac {80753 x^4}{2365440}+\frac {1361 x^3}{17280}+\frac {25 x^2}{224}+\frac {3 x}{20}+1\right ) x^{3/2}
\]