2.4.31 problem 27
Internal
problem
ID
[8346]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
27
Date
solved
:
Sunday, November 10, 2024 at 03:38:59 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} x^{2} \left (-x^{2}+2\right ) y^{\prime \prime }-x \left (4 x^{2}+3\right ) y^{\prime }+\left (-2 x^{2}+2\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.
\[ \left (-x^{4}+2 x^{2}\right ) y^{\prime \prime }+\left (-4 x^{3}-3 x \right ) y^{\prime }+\left (-2 x^{2}+2\right ) y = 0 \]
The following is summary of singularities for the above ode. Writing the ode as
\begin{align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end{align*}
Where
\begin{align*} p(x) &= \frac {4 x^{2}+3}{x \left (x^{2}-2\right )}\\ q(x) &= \frac {2 x^{2}-2}{x^{2} \left (x^{2}-2\right )}\\ \end{align*}
Table 2.80: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=\frac {4 x^{2}+3}{x \left (x^{2}-2\right )}\) |
| |
|
singularity |
type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| \(x = \sqrt {2}\) | \(\text {``regular''}\) |
| |
| \(x = -\sqrt {2}\) |
\(\text {``regular''}\) |
| |
| |
| \(q(x)=\frac {2 x^{2}-2}{x^{2} \left (x^{2}-2\right )}\) |
| |
|
singularity |
type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| \(x = \sqrt {2}\) | \(\text {``regular''}\) |
| |
| \(x = -\sqrt {2}\) |
\(\text {``regular''}\) |
| |
Combining everything together gives the following summary of singularities for the ode
as
Regular singular points : \(\left [0, \sqrt {2}, -\sqrt {2}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to
be
\[ -x^{2} \left (x^{2}-2\right ) y^{\prime \prime }+\left (-4 x^{3}-3 x \right ) y^{\prime }+\left (-2 x^{2}+2\right ) y = 0 \]
Let the solution be represented as Frobenius power series of the form
\[
y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}
\]
Then
\begin{align*}
y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\
y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\
\end{align*}
Substituting the above back into the ode gives
\begin{equation}
\tag{1} -x^{2} \left (x^{2}-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 (-4 x^{3}-3 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-2 x^{2}+2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0
\end{equation}
Which simplifies to
\begin{equation}
\tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-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^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n +r +2} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r}\right ) = 0
\end{equation}
The next step is to
make all powers of \(x\) be \(n +r\) 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 }}\left (-4 x^{n +r +2} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-4 a_{n -2} \left (n +r -2\right ) x^{n +r}\right ) \\
\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 a_{n -2} x^{n +r}\right ) \\
\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 }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-4 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r}\right ) = 0
\end{equation}
The indicial
equation is obtained from \(n = 0\). From Eq (2B) this gives
\[ 2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-3 x^{n +r} a_{n} \left (n +r \right )+2 a_{n} x^{n +r} = 0 \]
When \(n = 0\) the above becomes
\[ 2 x^{r} a_{0} r \left (-1+r \right )-3 x^{r} a_{0} r +2 a_{0} x^{r} = 0 \]
Or
\[ \left (2 x^{r} r \left (-1+r \right )-3 x^{r} r +2 x^{r}\right ) a_{0} = 0 \]
Since \(a_{0}\neq 0\) then the above simplifies to
\[ \left (2 r^{2}-5 r +2\right ) x^{r} = 0 \]
Since the above is true for all \(x\) then the
indicial equation becomes
\[ 2 r^{2}-5 r +2 = 0 \]
Solving for \(r\) gives the roots of the indicial equation as
\begin{align*} r_1 &= 2\\ r_2 &= {\frac {1}{2}} \end{align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes
\[ \left (2 r^{2}-5 r +2\right ) x^{r} = 0 \]
Solving for \(r\) gives the roots of the indicial equation
as \(\left [2, {\frac {1}{2}}\right ]\).
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 +2}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{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
\[ a_{1} = 0 \]
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 )+2 a_{n} \left (n +r \right ) \left (n +r -1\right )-4 a_{n -2} \left (n +r -2\right )-3 a_{n} \left (n +r \right )-2 a_{n -2}+2 a_{n} = 0
\end{equation}
Solving for \(a_{n}\) from recursive equation (4) gives
\[ a_{n} = \frac {a_{n -2} \left (n^{2}+2 n r +r^{2}-n -r \right )}{2 n^{2}+4 n r +2 r^{2}-5 n -5 r +2}\tag {4} \]
Which for the root \(r = 2\)
becomes
\[ a_{n} = \frac {a_{n -2} \left (n^{2}+3 n +2\right )}{2 n^{2}+3 n}\tag {5} \]
At this point, it is a good idea to keep track of \(a_{n}\) in a table both before
substituting \(r = 2\) and after as more terms are found using the above recursive equation.
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(0\) | \(0\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ a_{2}=\frac {r^{2}+3 r +2}{2 r^{2}+3 r} \]
Which for the root \(r = 2\) becomes
\[ a_{2}={\frac {6}{7}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(0\) | \(0\) |
| | |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{2 r^{2}+3 r}\) | \(\frac {6}{7}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=0 \]
And the table now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(0\) | \(0\) |
| | |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{2 r^{2}+3 r}\) | \(\frac {6}{7}\) |
| | |
| \(a_{3}\) |
\(0\) |
\(0\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=\frac {r^{3}+8 r^{2}+19 r +12}{4 r^{3}+20 r^{2}+21 r} \]
Which for the root \(r = 2\) becomes
\[ a_{4}={\frac {45}{77}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(0\) | \(0\) |
| | |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{2 r^{2}+3 r}\) | \(\frac {6}{7}\) |
| | |
| \(a_{3}\) | \(0\) | \(0\) |
| | |
| \(a_{4}\) |
\(\frac {r^{3}+8 r^{2}+19 r +12}{4 r^{3}+20 r^{2}+21 r}\) |
\(\frac {45}{77}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=0 \]
And the table now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(0\) |
\(0\) |
| | |
| \(a_{2}\) |
\(\frac {r^{2}+3 r +2}{2 r^{2}+3 r}\) | \(\frac {6}{7}\) |
| | |
| \(a_{3}\) | \(0\) | \(0\) |
| | |
| \(a_{4}\) |
\(\frac {r^{3}+8 r^{2}+19 r +12}{4 r^{3}+20 r^{2}+21 r}\) |
\(\frac {45}{77}\) |
| | |
| \(a_{5}\) |
\(0\) |
\(0\) |
| | |
Using the above table, then the solution \(y_{1}\left (x \right )\) is
\begin{align*} y_{1}\left (x \right )&= x^{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^{2} \left (1+\frac {6 x^{2}}{7}+\frac {45 x^{4}}{77}+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
\[ b_{1} = 0 \]
For \(2\le n\) the recursive
equation is
\begin{equation}
\tag{3} -b_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+2 b_{n} \left (n +r \right ) \left (n +r -1\right )-4 b_{n -2} \left (n +r -2\right )-3 b_{n} \left (n +r \right )-2 b_{n -2}+2 b_{n} = 0
\end{equation}
Solving for \(b_{n}\) from recursive equation (4) gives
\[ b_{n} = \frac {b_{n -2} \left (n^{2}+2 n r +r^{2}-n -r \right )}{2 n^{2}+4 n r +2 r^{2}-5 n -5 r +2}\tag {4} \]
Which for the root \(r = {\frac {1}{2}}\)
becomes
\[ b_{n} = \frac {4 n^{2} b_{n -2}-b_{n -2}}{8 n^{2}-12 n}\tag {5} \]
At this point, it is a good idea to keep track of \(b_{n}\) in a table both before
substituting \(r = {\frac {1}{2}}\) and after as more terms are found using the above recursive equation.
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) | \(1\) |
| | |
| \(b_{1}\) | \(0\) | \(0\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ b_{2}=\frac {r^{2}+3 r +2}{2 r^{2}+3 r} \]
Which for the root \(r = {\frac {1}{2}}\) becomes
\[ b_{2}={\frac {15}{8}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) | \(1\) |
| | |
| \(b_{1}\) | \(0\) | \(0\) |
| | |
| \(b_{2}\) | \(\frac {r^{2}+3 r +2}{2 r^{2}+3 r}\) | \(\frac {15}{8}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ b_{3}=0 \]
And the table now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(0\) | \(0\) |
| | |
| \(b_{2}\) | \(\frac {r^{2}+3 r +2}{2 r^{2}+3 r}\) | \(\frac {15}{8}\) |
| | |
| \(b_{3}\) |
\(0\) |
\(0\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ b_{4}=\frac {r^{3}+8 r^{2}+19 r +12}{4 r^{3}+20 r^{2}+21 r} \]
Which for the root \(r = {\frac {1}{2}}\) becomes
\[ b_{4}={\frac {189}{128}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(0\) | \(0\) |
| | |
| \(b_{2}\) | \(\frac {r^{2}+3 r +2}{2 r^{2}+3 r}\) | \(\frac {15}{8}\) |
| | |
| \(b_{3}\) | \(0\) | \(0\) |
| | |
| \(b_{4}\) |
\(\frac {r^{3}+8 r^{2}+19 r +12}{4 r^{3}+20 r^{2}+21 r}\) |
\(\frac {189}{128}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ b_{5}=0 \]
And the table now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(0\) |
\(0\) |
| | |
| \(b_{2}\) |
\(\frac {r^{2}+3 r +2}{2 r^{2}+3 r}\) | \(\frac {15}{8}\) |
| | |
| \(b_{3}\) | \(0\) | \(0\) |
| | |
| \(b_{4}\) |
\(\frac {r^{3}+8 r^{2}+19 r +12}{4 r^{3}+20 r^{2}+21 r}\) |
\(\frac {189}{128}\) |
| | |
| \(b_{5}\) |
\(0\) |
\(0\) |
| | |
Using the above table, then the solution \(y_{2}\left (x \right )\) is
\begin{align*} y_{2}\left (x \right )&= x^{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 ) \\ &= \sqrt {x}\, \left (1+\frac {15 x^{2}}{8}+\frac {189 x^{4}}{128}+O\left (x^{6}\right )\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^{2} \left (1+\frac {6 x^{2}}{7}+\frac {45 x^{4}}{77}+O\left (x^{6}\right )\right ) + c_2 \sqrt {x}\, \left (1+\frac {15 x^{2}}{8}+\frac {189 x^{4}}{128}+O\left (x^{6}\right )\right ) \\
\end{align*}
Hence the final solution is
\begin{align*}
y &= y_h \\
&= c_1 \,x^{2} \left (1+\frac {6 x^{2}}{7}+\frac {45 x^{4}}{77}+O\left (x^{6}\right )\right )+c_2 \sqrt {x}\, \left (1+\frac {15 x^{2}}{8}+\frac {189 x^{4}}{128}+O\left (x^{6}\right )\right ) \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (-x^{2}+2\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-x \left (4 x^{2}+3\right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (-2 x^{2}+2\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 {2 \left (x^{2}-1\right ) y \left (x \right )}{x^{2} \left (x^{2}-2\right )}-\frac {\left (4 x^{2}+3\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (x^{2}-2\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 (4 x^{2}+3\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (x^{2}-2\right )}+\frac {2 \left (x^{2}-1\right ) y \left (x \right )}{x^{2} \left (x^{2}-2\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 {4 x^{2}+3}{x \left (x^{2}-2\right )}, P_{3}\left (x \right )=\frac {2 \left (x^{2}-1\right )}{x^{2} \left (x^{2}-2\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 {3}{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}}}=1 \\ {} & \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^{2} \left (x^{2}-2\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+x \left (4 x^{2}+3\right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (2 x^{2}-2\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 (-1+2 r \right ) \left (-2+r \right ) x^{r}-a_{1} \left (1+2 r \right ) \left (-1+r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (-a_{k} \left (2 k +2 r -1\right ) \left (k +r -2\right )+a_{k -2} \left (k +r \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 (-1+2 r \right ) \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{2, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & -a_{1} \left (1+2 r \right ) \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 \left (k +r -2\right ) \left (k +r -\frac {1}{2}\right ) a_{k}+a_{k -2} \left (k +r \right ) \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & -2 \left (k +r \right ) \left (k +\frac {3}{2}+r \right ) a_{k +2}+a_{k} \left (k +r +2\right ) \left (k +r +1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {a_{k} \left (k +r +2\right ) \left (k +r +1\right )}{\left (k +r \right ) \left (2 k +3+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +2}=\frac {a_{k} \left (k +4\right ) \left (k +3\right )}{\left (k +2\right ) \left (2 k +7\right )} \\ \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 {a_{k} \left (k +4\right ) \left (k +3\right )}{\left (k +2\right ) \left (2 k +7\right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +2}=\frac {a_{k} \left (k +\frac {5}{2}\right ) \left (k +\frac {3}{2}\right )}{\left (k +\frac {1}{2}\right ) \left (2 k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}, a_{k +2}=\frac {a_{k} \left (k +\frac {5}{2}\right ) \left (k +\frac {3}{2}\right )}{\left (k +\frac {1}{2}\right ) \left (2 k +4\right )}, a_{1}=0\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 {1}{2}}\right ), a_{k +2}=\frac {a_{k} \left (4+k \right ) \left (k +3\right )}{\left (k +2\right ) \left (2 k +7\right )}, a_{1}=0, b_{k +2}=\frac {b_{k} \left (k +\frac {5}{2}\right ) \left (k +\frac {3}{2}\right )}{\left (k +\frac {1}{2}\right ) \left (2 k +4\right )}, b_{1}=0\right ] \end {array} \]
Maple trace
`Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
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
<- heuristic approach successful
<- hypergeometric successful
<- special function solution successful
-> Trying to convert hypergeometric functions to elementary form...
<- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals
<- Kovacics algorithm successful`
Maple dsolve solution
Solving time : 0.018 (sec)
Leaf size : 35
dsolve(x^2*(-x^2+2)*diff(diff(y(x),x),x)-x*(4*x^2+3)*diff(y(x),x)+(-2*x^2+2)*y(x) = 0,y(x),
series,x=0)
\[
y = c_{1} \sqrt {x}\, \left (1+\frac {15}{8} x^{2}+\frac {189}{128} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x^{2} \left (1+\frac {6}{7} x^{2}+\frac {45}{77} x^{4}+\operatorname {O}\left (x^{6}\right )\right )
\]
Mathematica DSolve solution
Solving time : 0.024 (sec)
Leaf size : 50
AsymptoticDSolveValue[{x^2*(2-x^2)*D[y[x],{x,2}] - x*(3+4*x^2)*D[y[x],x] + (2-2*x^2)*y[x] == 0,{}},
y[x],{x,0,5}]
\[
y(x)\to c_1 \left (\frac {45 x^4}{77}+\frac {6 x^2}{7}+1\right ) x^2+c_2 \left (\frac {189 x^4}{128}+\frac {15 x^2}{8}+1\right ) \sqrt {x}
\]