2.4.29 problem 25
Internal
problem
ID
[8344]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
25
Date
solved
:
Sunday, November 10, 2024 at 03:38:56 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} 2 x^{2} \left (x^{2}+x +1\right ) y^{\prime \prime }+x \left (11 x^{2}+11 x +9\right ) y^{\prime }+\left (7 x^{2}+10 x +6\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.
\[ \left (2 x^{4}+2 x^{3}+2 x^{2}\right ) y^{\prime \prime }+\left (11 x^{3}+11 x^{2}+9 x \right ) y^{\prime }+\left (7 x^{2}+10 x +6\right ) y = 0 \]
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 {11 x^{2}+11 x +9}{2 x \left (x^{2}+x +1\right )}\\ q(x) &= \frac {7 x^{2}+10 x +6}{2 x^{2} \left (x^{2}+x +1\right )}\\ \end{align*}
Table 2.78: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=\frac {11 x^{2}+11 x +9}{2 x \left (x^{2}+x +1\right )}\) |
| |
|
singularity |
type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| \(x = -\frac {1}{2}-\frac {i \sqrt {3}}{2}\) | \(\text {``regular''}\) |
| |
| \(x = -\frac {1}{2}+\frac {i \sqrt {3}}{2}\) |
\(\text {``regular''}\) |
| |
| |
| \(q(x)=\frac {7 x^{2}+10 x +6}{2 x^{2} \left (x^{2}+x +1\right )}\) |
| |
|
singularity |
type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| \(x = -\frac {1}{2}-\frac {i \sqrt {3}}{2}\) | \(\text {``regular''}\) |
| |
| \(x = -\frac {1}{2}+\frac {i \sqrt {3}}{2}\) |
\(\text {``regular''}\) |
| |
Combining everything together gives the following summary of singularities for the ode
as
Regular singular points : \(\left [0, -\frac {1}{2}-\frac {i \sqrt {3}}{2}, -\frac {1}{2}+\frac {i \sqrt {3}}{2}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to
be
\[ 2 x^{2} \left (x^{2}+x +1\right ) y^{\prime \prime }+\left (11 x^{3}+11 x^{2}+9 x \right ) y^{\prime }+\left (7 x^{2}+10 x +6\right ) y = 0 \]
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 x^{2} \left (x^{2}+x +1\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 (11 x^{3}+11 x^{2}+9 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (7 x^{2}+10 x +6\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 }}2 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+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} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}11 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}11 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}7 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 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 }}2 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\
\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\
\moverset {\infty }{\munderset {n =0}{\sum }}11 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}11 a_{n -2} \left (n +r -2\right ) x^{n +r} \\
\moverset {\infty }{\munderset {n =0}{\sum }}11 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}11 a_{n -1} \left (n +r -1\right ) x^{n +r} \\
\moverset {\infty }{\munderset {n =0}{\sum }}7 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}7 a_{n -2} x^{n +r} \\
\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}10 a_{n -1} 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} \left (\moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}11 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}11 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}7 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}10 a_{n -1} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 a_{n} x^{n +r}\right ) = 0
\end{equation}
The indicial
equation is obtained from \(n = 0\). From Eq (2B) this gives
\[ 2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+9 x^{n +r} a_{n} \left (n +r \right )+6 a_{n} x^{n +r} = 0 \]
When \(n = 0\) the above becomes
\[ 2 x^{r} a_{0} r \left (-1+r \right )+9 x^{r} a_{0} r +6 a_{0} x^{r} = 0 \]
Or
\[ \left (2 x^{r} r \left (-1+r \right )+9 x^{r} r +6 x^{r}\right ) a_{0} = 0 \]
Since \(a_{0}\neq 0\) then the above simplifies to
\[ \left (2 r^{2}+7 r +6\right ) x^{r} = 0 \]
Since the above is true for all \(x\) then the
indicial equation becomes
\[ 2 r^{2}+7 r +6 = 0 \]
Solving for \(r\) gives the roots of the indicial equation as
\begin{align*} r_1 &= -{\frac {3}{2}}\\ r_2 &= -2 \end{align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes
\[ \left (2 r^{2}+7 r +6\right ) x^{r} = 0 \]
Solving for \(r\) gives the roots of the indicial equation
as \(\left [-{\frac {3}{2}}, -2\right ]\).
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 {3}{2}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -2} \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
\[ a_{1} = \frac {-r -2}{r +3} \]
For \(2\le n\) the recursive
equation is
\begin{equation}
\tag{3} 2 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+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 )+11 a_{n -2} \left (n +r -2\right )+11 a_{n -1} \left (n +r -1\right )+9 a_{n} \left (n +r \right )+7 a_{n -2}+10 a_{n -1}+6 a_{n} = 0
\end{equation}
Solving for \(a_{n}\) from recursive equation (4) gives
\[ a_{n} = -\frac {n a_{n -2}+n a_{n -1}+r a_{n -2}+r a_{n -1}-a_{n -2}+a_{n -1}}{n +r +2}\tag {4} \]
Which for the root \(r = -{\frac {3}{2}}\)
becomes
\[ a_{n} = \frac {\left (-2 a_{n -2}-2 a_{n -1}\right ) n +5 a_{n -2}+a_{n -1}}{2 n +1}\tag {5} \]
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.
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {-r -2}{r +3}\) | \(-{\frac {1}{3}}\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ a_{2}=\frac {1}{4+r} \]
Which for the root \(r = -{\frac {3}{2}}\) becomes
\[ a_{2}={\frac {2}{5}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {-r -2}{r +3}\) | \(-{\frac {1}{3}}\) |
| | |
| \(a_{2}\) | \(\frac {1}{4+r}\) | \(\frac {2}{5}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=\frac {r^{2}+3 r +1}{\left (r +3\right ) \left (5+r \right )} \]
Which for the root \(r = -{\frac {3}{2}}\) becomes
\[ a_{3}=-{\frac {5}{21}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {-r -2}{r +3}\) | \(-{\frac {1}{3}}\) |
| | |
| \(a_{2}\) | \(\frac {1}{4+r}\) | \(\frac {2}{5}\) |
| | |
| \(a_{3}\) |
\(\frac {r^{2}+3 r +1}{\left (r +3\right ) \left (5+r \right )}\) |
\(-{\frac {5}{21}}\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=\frac {-r^{3}-8 r^{2}-19 r -13}{\left (6+r \right ) \left (r +3\right ) \left (4+r \right )} \]
Which for the root \(r = -{\frac {3}{2}}\) becomes
\[ a_{4}={\frac {7}{135}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {-r -2}{r +3}\) | \(-{\frac {1}{3}}\) |
| | |
| \(a_{2}\) | \(\frac {1}{4+r}\) | \(\frac {2}{5}\) |
| | |
| \(a_{3}\) | \(\frac {r^{2}+3 r +1}{\left (r +3\right ) \left (5+r \right )}\) | \(-{\frac {5}{21}}\) |
| | |
| \(a_{4}\) |
\(\frac {-r^{3}-8 r^{2}-19 r -13}{\left (6+r \right ) \left (r +3\right ) \left (4+r \right )}\) |
\(\frac {7}{135}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=\frac {2 r^{3}+18 r^{2}+52 r +49}{\left (r +3\right ) \left (4+r \right ) \left (5+r \right ) \left (7+r \right )} \]
Which for the root \(r = -{\frac {3}{2}}\) becomes
\[ a_{5}={\frac {76}{1155}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {-r -2}{r +3}\) |
\(-{\frac {1}{3}}\) |
| | |
| \(a_{2}\) |
\(\frac {1}{4+r}\) | \(\frac {2}{5}\) |
| | |
| \(a_{3}\) | \(\frac {r^{2}+3 r +1}{\left (r +3\right ) \left (5+r \right )}\) | \(-{\frac {5}{21}}\) |
| | |
| \(a_{4}\) |
\(\frac {-r^{3}-8 r^{2}-19 r -13}{\left (6+r \right ) \left (r +3\right ) \left (4+r \right )}\) |
\(\frac {7}{135}\) |
| | |
| \(a_{5}\) |
\(\frac {2 r^{3}+18 r^{2}+52 r +49}{\left (r +3\right ) \left (4+r \right ) \left (5+r \right ) \left (7+r \right )}\) |
\(\frac {76}{1155}\) |
| | |
Using the above table, then the solution \(y_{1}\left (x \right )\) is
\begin{align*} y_{1}\left (x \right )&= \frac {1}{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 ) \\ &= \frac {1-\frac {x}{3}+\frac {2 x^{2}}{5}-\frac {5 x^{3}}{21}+\frac {7 x^{4}}{135}+\frac {76 x^{5}}{1155}+O\left (x^{6}\right )}{x^{{3}/{2}}} \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
\[ b_{1} = \frac {-r -2}{r +3} \]
For \(2\le n\) the recursive
equation is
\begin{equation}
\tag{3} 2 b_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+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 )+11 b_{n -2} \left (n +r -2\right )+11 b_{n -1} \left (n +r -1\right )+9 b_{n} \left (n +r \right )+7 b_{n -2}+10 b_{n -1}+6 b_{n} = 0
\end{equation}
Solving for \(b_{n}\) from recursive equation (4) gives
\[ b_{n} = -\frac {n b_{n -2}+n b_{n -1}+r b_{n -2}+r b_{n -1}-b_{n -2}+b_{n -1}}{n +r +2}\tag {4} \]
Which for the root \(r = -2\)
becomes
\[ b_{n} = \frac {\left (-b_{n -2}-b_{n -1}\right ) n +3 b_{n -2}+b_{n -1}}{n}\tag {5} \]
At this point, it is a good idea to keep track of \(b_{n}\) in a table both before
substituting \(r = -2\) and after as more terms are found using the above recursive equation.
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) | \(1\) |
| | |
| \(b_{1}\) | \(\frac {-r -2}{r +3}\) | \(0\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ b_{2}=\frac {1}{4+r} \]
Which for the root \(r = -2\) becomes
\[ b_{2}={\frac {1}{2}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) | \(1\) |
| | |
| \(b_{1}\) | \(\frac {-r -2}{r +3}\) | \(0\) |
| | |
| \(b_{2}\) | \(\frac {1}{4+r}\) | \(\frac {1}{2}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ b_{3}=\frac {r^{2}+3 r +1}{\left (r +3\right ) \left (5+r \right )} \]
Which for the root \(r = -2\) becomes
\[ b_{3}=-{\frac {1}{3}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(\frac {-r -2}{r +3}\) | \(0\) |
| | |
| \(b_{2}\) | \(\frac {1}{4+r}\) | \(\frac {1}{2}\) |
| | |
| \(b_{3}\) |
\(\frac {r^{2}+3 r +1}{\left (r +3\right ) \left (5+r \right )}\) |
\(-{\frac {1}{3}}\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ b_{4}=\frac {-r^{3}-8 r^{2}-19 r -13}{\left (6+r \right ) \left (r +3\right ) \left (4+r \right )} \]
Which for the root \(r = -2\) becomes
\[ b_{4}={\frac {1}{8}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(\frac {-r -2}{r +3}\) | \(0\) |
| | |
| \(b_{2}\) | \(\frac {1}{4+r}\) | \(\frac {1}{2}\) |
| | |
| \(b_{3}\) | \(\frac {r^{2}+3 r +1}{\left (r +3\right ) \left (5+r \right )}\) | \(-{\frac {1}{3}}\) |
| | |
| \(b_{4}\) |
\(\frac {-r^{3}-8 r^{2}-19 r -13}{\left (6+r \right ) \left (r +3\right ) \left (4+r \right )}\) |
\(\frac {1}{8}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ b_{5}=\frac {2 r^{3}+18 r^{2}+52 r +49}{\left (r +3\right ) \left (4+r \right ) \left (5+r \right ) \left (7+r \right )} \]
Which for the root \(r = -2\) becomes
\[ b_{5}={\frac {1}{30}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(\frac {-r -2}{r +3}\) |
\(0\) |
| | |
| \(b_{2}\) |
\(\frac {1}{4+r}\) | \(\frac {1}{2}\) |
| | |
| \(b_{3}\) | \(\frac {r^{2}+3 r +1}{\left (r +3\right ) \left (5+r \right )}\) | \(-{\frac {1}{3}}\) |
| | |
| \(b_{4}\) |
\(\frac {-r^{3}-8 r^{2}-19 r -13}{\left (6+r \right ) \left (r +3\right ) \left (4+r \right )}\) |
\(\frac {1}{8}\) |
| | |
| \(b_{5}\) |
\(\frac {2 r^{3}+18 r^{2}+52 r +49}{\left (r +3\right ) \left (4+r \right ) \left (5+r \right ) \left (7+r \right )}\) |
\(\frac {1}{30}\) |
| | |
Using the above table, then the solution \(y_{2}\left (x \right )\) is
\begin{align*} y_{2}\left (x \right )&= \frac {1}{x^{{3}/{2}}} \left (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 \right ) \\ &= \frac {1+\frac {x^{2}}{2}-\frac {x^{3}}{3}+\frac {x^{4}}{8}+\frac {x^{5}}{30}+O\left (x^{6}\right )}{x^{2}} \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-\frac {x}{3}+\frac {2 x^{2}}{5}-\frac {5 x^{3}}{21}+\frac {7 x^{4}}{135}+\frac {76 x^{5}}{1155}+O\left (x^{6}\right )\right )}{x^{{3}/{2}}} + \frac {c_2 \left (1+\frac {x^{2}}{2}-\frac {x^{3}}{3}+\frac {x^{4}}{8}+\frac {x^{5}}{30}+O\left (x^{6}\right )\right )}{x^{2}} \\
\end{align*}
Hence the final solution is
\begin{align*}
y &= y_h \\
&= \frac {c_1 \left (1-\frac {x}{3}+\frac {2 x^{2}}{5}-\frac {5 x^{3}}{21}+\frac {7 x^{4}}{135}+\frac {76 x^{5}}{1155}+O\left (x^{6}\right )\right )}{x^{{3}/{2}}}+\frac {c_2 \left (1+\frac {x^{2}}{2}-\frac {x^{3}}{3}+\frac {x^{4}}{8}+\frac {x^{5}}{30}+O\left (x^{6}\right )\right )}{x^{2}} \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x^{2} \left (x^{2}+x +1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+x \left (11 x^{2}+11 x +9\right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (7 x^{2}+10 x +6\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 (7 x^{2}+10 x +6\right ) y \left (x \right )}{2 x^{2} \left (x^{2}+x +1\right )}-\frac {\left (11 x^{2}+11 x +9\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{2 x \left (x^{2}+x +1\right )} \\ \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 {\left (11 x^{2}+11 x +9\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{2 x \left (x^{2}+x +1\right )}+\frac {\left (7 x^{2}+10 x +6\right ) y \left (x \right )}{2 x^{2} \left (x^{2}+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 {11 x^{2}+11 x +9}{2 x \left (x^{2}+x +1\right )}, P_{3}\left (x \right )=\frac {7 x^{2}+10 x +6}{2 x^{2} \left (x^{2}+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 {9}{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}}}=3 \\ {} & \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 x^{2} \left (x^{2}+x +1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+x \left (11 x^{2}+11 x +9\right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (7 x^{2}+10 x +6\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 =0..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} x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )=\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 \left (\frac {d}{d x}y \left (x \right )\right )=\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 \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & 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} \left (2+r \right ) \left (3+2 r \right ) x^{r}+\left (a_{1} \left (3+r \right ) \left (5+2 r \right )+a_{0} \left (5+2 r \right ) \left (2+r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +r +2\right ) \left (2 k +2 r +3\right )+a_{k -1} \left (2 k +2 r +3\right ) \left (k +r +1\right )+a_{k -2} \left (2 k +2 r +3\right ) \left (k +r -1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (2+r \right ) \left (3+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-2, -\frac {3}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (3+r \right ) \left (5+2 r \right )+a_{0} \left (5+2 r \right ) \left (2+r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {\left (2+r \right ) a_{0}}{3+r} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (k +r +\frac {3}{2}\right ) \left (\left (a_{k}+a_{k -2}+a_{k -1}\right ) k +\left (a_{k}+a_{k -2}+a_{k -1}\right ) r +2 a_{k}-a_{k -2}+a_{k -1}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & 2 \left (k +\frac {7}{2}+r \right ) \left (\left (a_{k +2}+a_{k}+a_{k +1}\right ) \left (k +2\right )+\left (a_{k +2}+a_{k}+a_{k +1}\right ) r +2 a_{k +2}-a_{k}+a_{k +1}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k a_{k}+k a_{k +1}+r a_{k}+r a_{k +1}+a_{k}+3 a_{k +1}}{k +4+r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2 \\ {} & {} & a_{k +2}=-\frac {k a_{k}+k a_{k +1}-a_{k}+a_{k +1}}{k +2} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-2 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -2}, a_{k +2}=-\frac {k a_{k}+k a_{k +1}-a_{k}+a_{k +1}}{k +2}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {3}{2} \\ {} & {} & a_{k +2}=-\frac {k a_{k}+k a_{k +1}-\frac {1}{2} a_{k}+\frac {3}{2} a_{k +1}}{k +\frac {5}{2}} \\ \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 +2}=-\frac {k a_{k}+k a_{k +1}-\frac {1}{2} a_{k}+\frac {3}{2} a_{k +1}}{k +\frac {5}{2}}, a_{1}=-\frac {a_{0}}{3}\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 -2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -\frac {3}{2}}\right ), a_{k +2}=-\frac {k a_{k}+k a_{k +1}-a_{k}+a_{k +1}}{k +2}, a_{1}=0, b_{k +2}=-\frac {k b_{k}+k b_{k +1}-\frac {1}{2} b_{k}+\frac {3}{2} b_{k +1}}{k +\frac {5}{2}}, b_{1}=-\frac {b_{0}}{3}\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
A Liouvillian solution exists
Reducible group (found an exponential solution)
Group is reducible, not completely reducible
Solution has integrals. Trying a special function solution free of integrals...
-> 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
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius
<- Heun successful: received ODE is equivalent to the HeunG ODE, case a <> 0, e <> 0, g <> 0, c = 0
<- Kovacics algorithm successful`
Maple dsolve solution
Solving time : 0.038 (sec)
Leaf size : 45
dsolve(2*x^2*(x^2+x+1)*diff(diff(y(x),x),x)+x*(11*x^2+11*x+9)*diff(y(x),x)+(7*x^2+10*x+6)*y(x) = 0,y(x),
series,x=0)
\[
y = \frac {c_{1} \left (1+\frac {1}{2} x^{2}-\frac {1}{3} x^{3}+\frac {1}{8} x^{4}+\frac {1}{30} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x^{2}}+\frac {c_{2} \left (1-\frac {1}{3} x +\frac {2}{5} x^{2}-\frac {5}{21} x^{3}+\frac {7}{135} x^{4}+\frac {76}{1155} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x^{{3}/{2}}}
\]
Mathematica DSolve solution
Solving time : 0.027 (sec)
Leaf size : 83
AsymptoticDSolveValue[{2*x^2*(1+x+x^2)*D[y[x],{x,2}] + x*(9+11*x+11*x^2)*D[y[x],x] + (6+10*x+7*x^2)*y[x] == 0,{}},
y[x],{x,0,5}]
\[
y(x)\to \frac {c_2 \left (\frac {x^5}{30}+\frac {x^4}{8}-\frac {x^3}{3}+\frac {x^2}{2}+1\right )}{x^2}+\frac {c_1 \left (\frac {76 x^5}{1155}+\frac {7 x^4}{135}-\frac {5 x^3}{21}+\frac {2 x^2}{5}-\frac {x}{3}+1\right )}{x^{3/2}}
\]