15.60 problem 61
Internal
problem
ID
[2058]
Book
:
Elementary
differential
equations
with
boundary
value
problems.
William
F.
Trench.
Brooks/Cole
2001
Section
:
Chapter
7
Series
Solutions
of
Linear
Second
Equations.
7.6
THE
METHOD
OF
FROBENIUS
II.
Exercises
7.6.
Page
374
Problem
number
:
61
Date
solved
:
Thursday, October 17, 2024 at 02:19:30 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} 16 x^{2} \left (x^{2}+1\right ) y^{\prime \prime }+8 x \left (9 x^{2}+1\right ) y^{\prime }+\left (49 x^{2}+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.
\[ \left (16 x^{4}+16 x^{2}\right ) y^{\prime \prime }+\left (72 x^{3}+8 x \right ) y^{\prime }+\left (49 x^{2}+1\right ) y = 0 \]
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 {9 x^{2}+1}{2 x \left (x^{2}+1\right )}\\ q(x) &= \frac {49 x^{2}+1}{16 x^{2} \left (x^{2}+1\right )}\\ \end{align*}
Table 182: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=\frac {9 x^{2}+1}{2 x \left (x^{2}+1\right )}\) |
| |
|
singularity |
type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| \(x = -i\) | \(\text {``regular''}\) |
| |
| \(x = i\) |
\(\text {``regular''}\) |
| |
| |
| \(q(x)=\frac {49 x^{2}+1}{16 x^{2} \left (x^{2}+1\right )}\) |
| |
|
singularity |
type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| \(x = -i\) | \(\text {``regular''}\) |
| |
| \(x = i\) |
\(\text {``regular''}\) |
| |
Combining everything together gives the following summary of singularities for the ode
as
Regular singular points : \([0, -i, i, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to
be
\[ 16 x^{2} \left (x^{2}+1\right ) y^{\prime \prime }+\left (72 x^{3}+8 x \right ) y^{\prime }+\left (49 x^{2}+1\right ) y = 0 \]
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} 16 x^{2} \left (x^{2}+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 (72 x^{3}+8 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (49 x^{2}+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 }}16 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}16 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}72 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}49 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0
\end{equation}
The next step is to
make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation
term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and
the corresponding index gives
\begin{align*}
\moverset {\infty }{\munderset {n =0}{\sum }}16 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}16 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\
\moverset {\infty }{\munderset {n =0}{\sum }}72 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}72 a_{n -2} \left (n +r -2\right ) x^{n +r} \\
\moverset {\infty }{\munderset {n =0}{\sum }}49 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}49 a_{n -2} 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 }}16 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}16 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}72 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}49 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0
\end{equation}
The indicial
equation is obtained from \(n = 0\). From Eq (2B) this gives
\[ 16 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+8 x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \]
When \(n = 0\) the above becomes
\[ 16 x^{r} a_{0} r \left (-1+r \right )+8 x^{r} a_{0} r +a_{0} x^{r} = 0 \]
Or
\[ \left (16 x^{r} r \left (-1+r \right )+8 x^{r} r +x^{r}\right ) a_{0} = 0 \]
Since \(a_{0}\neq 0\) then the above simplifies to
\[ x^{r} \left (4 r -1\right )^{2} = 0 \]
Since the above is true for all \(x\) then the
indicial equation becomes
\[ \left (4 r -1\right )^{2} = 0 \]
Solving for \(r\) gives the roots of the indicial equation as
\begin{align*} r_1 &= {\frac {1}{4}}\\ r_2 &= {\frac {1}{4}} \end{align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes
\[ x^{r} \left (4 r -1\right )^{2} = 0 \]
Solving for \(r\) gives the roots of the indicial equation
as \(\left [{\frac {1}{4}}, {\frac {1}{4}}\right ]\).
Since the root of the indicial equation is repeated, then we can construct two linearly
independent solutions. The first solution has the form
\begin{align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\tag {1A} \end{align*}
Now the second solution \(y_{2}\) is found using
\begin{align*} y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right )\tag {1B} \end{align*}
Then the general solution will be
\[ y = c_1 y_{1}\left (x \right )+c_2 y_{2}\left (x \right ) \]
In Eq (1B) the sum starts from 1 and not zero. In Eq
(1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_1, c_2\}\) are two arbitray
constants of integration which can be found from initial conditions. Using the value of the
indicial root found earlier, \(r = {\frac {1}{4}}\), Eqs (1A,1B) become
\begin{align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {1}{4}}\\ y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +\frac {1}{4}}\right ) \end{align*}
We start by finding the first solution \(y_{1}\left (x \right )\). Eq (2B) derived above is now used to find all \(a_{n}\)
coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation.
\(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives
\[ a_{1} = 0 \]
For \(2\le n\) the recursive
equation is
\begin{equation}
\tag{3} 16 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+16 a_{n} \left (n +r \right ) \left (n +r -1\right )+72 a_{n -2} \left (n +r -2\right )+8 a_{n} \left (n +r \right )+49 a_{n -2}+a_{n} = 0
\end{equation}
Solving for \(a_{n}\) from recursive equation (4) gives
\[ a_{n} = -a_{n -2}\tag {4} \]
Which for the root \(r = {\frac {1}{4}}\)
becomes
\[ a_{n} = -a_{n -2}\tag {5} \]
At this point, it is a good idea to keep track of \(a_{n}\) in a table both before
substituting \(r = {\frac {1}{4}}\) 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
\[ a_{2}=-1 \]
Which for the root \(r = {\frac {1}{4}}\) becomes
\[ a_{2}=-1 \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(0\) | \(0\) |
| | |
| \(a_{2}\) | \(-1\) | \(-1\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=0 \]
And the table now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(0\) | \(0\) |
| | |
| \(a_{2}\) | \(-1\) | \(-1\) |
| | |
| \(a_{3}\) |
\(0\) |
\(0\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=1 \]
Which for the root \(r = {\frac {1}{4}}\) becomes
\[ a_{4}=1 \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(0\) | \(0\) |
| | |
| \(a_{2}\) | \(-1\) | \(-1\) |
| | |
| \(a_{3}\) | \(0\) | \(0\) |
| | |
| \(a_{4}\) |
\(1\) |
\(1\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=0 \]
And the table now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(0\) |
\(0\) |
| | |
| \(a_{2}\) |
\(-1\) | \(-1\) |
| | |
| \(a_{3}\) | \(0\) | \(0\) |
| | |
| \(a_{4}\) |
\(1\) |
\(1\) |
| | |
| \(a_{5}\) |
\(0\) |
\(0\) |
| | |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is
\begin{align*}
y_{1}\left (x \right )&= x^{{1}/{4}} \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^{{1}/{4}} \left (x^{4}-x^{2}+1+O\left (x^{6}\right )\right ) \\
\end{align*}
Now the second solution is found. The
second solution is given by
\[ y_{2}\left (x \right ) = y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right ) \]
Where \(b_{n}\) is found using
\[ b_{n} = \frac {d}{d r}a_{n ,r} \]
And the above is then evaluated at \(r = {\frac {1}{4}}\). The
above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| | | | |
| \(n\) |
\(b_{n ,r}\) |
\(a_{n}\) |
\(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) |
\(b_{n}\left (r =\frac {1}{4}\right )\) |
| | | | |
| \(b_{0}\) |
\(1\) |
\(1\) |
N/A since \(b_{n}\) starts from 1 |
N/A |
| | | | |
| \(b_{1}\) | \(0\) | \(0\) | \(0\) | \(0\) |
| | | | |
| \(b_{2}\) | \(-1\) | \(-1\) | \(0\) | \(0\) |
| | | | |
| \(b_{3}\) | \(0\) | \(0\) | \(0\) | \(0\) |
| | | | |
| \(b_{4}\) |
\(1\) |
\(1\) |
\(0\) |
\(0\) |
| | | | |
| \(b_{5}\) |
\(0\) |
\(0\) |
\(0\) |
\(0\) |
| | | | |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is
\begin{align*}
y_{2}\left (x \right )&=y_{1}\left (x \right ) \ln \left (x \right )+b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \\
&= x^{{1}/{4}} \left (x^{4}-x^{2}+1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{{1}/{4}} 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^{{1}/{4}} \left (x^{4}-x^{2}+1+O\left (x^{6}\right )\right ) + c_2 \left (x^{{1}/{4}} \left (x^{4}-x^{2}+1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{{1}/{4}} O\left (x^{6}\right )\right ) \\
\end{align*}
Hence the final solution is
\begin{align*}
y &= y_h \\
&= c_1 \,x^{{1}/{4}} \left (x^{4}-x^{2}+1+O\left (x^{6}\right )\right )+c_2 \left (x^{{1}/{4}} \left (x^{4}-x^{2}+1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{{1}/{4}} O\left (x^{6}\right )\right ) \\
\end{align*}
15.60.1 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 16 x^{2} \left (x^{2}+1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+8 x \left (9 x^{2}+1\right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (49 x^{2}+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 (49 x^{2}+1\right ) y \left (x \right )}{16 x^{2} \left (x^{2}+1\right )}-\frac {\left (9 x^{2}+1\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{2 x \left (x^{2}+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 (9 x^{2}+1\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{2 x \left (x^{2}+1\right )}+\frac {\left (49 x^{2}+1\right ) y \left (x \right )}{16 x^{2} \left (x^{2}+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 {9 x^{2}+1}{2 x \left (x^{2}+1\right )}, P_{3}\left (x \right )=\frac {49 x^{2}+1}{16 x^{2} \left (x^{2}+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 {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}}}=\frac {1}{16} \\ {} & \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} \\ {} & {} & 16 x^{2} \left (x^{2}+1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+8 x \left (9 x^{2}+1\right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (49 x^{2}+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 =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+4 r \right )^{2} x^{r}+a_{1} \left (3+4 r \right )^{2} x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (4 k +4 r -1\right )^{2}+a_{k -2} \left (4 k +4 r -1\right )^{2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (-1+4 r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =\frac {1}{4} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (3+4 r \right )^{2}=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} \\ {} & {} & \left (4 k +4 r -1\right )^{2} \left (a_{k}+a_{k -2}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (4 k +4 r +7\right )^{2} \left (a_{k +2}+a_{k}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-a_{k} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{4} \\ {} & {} & a_{k +2}=-a_{k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{4} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{4}}, a_{k +2}=-a_{k}, a_{1}=0\right ] \end {array} \]
15.60.2 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
<- Kovacics algorithm successful`
15.60.3 Maple dsolve solution
Solving time : 0.008 (sec)
Leaf size : 29
dsolve(16*x^2*(x^2+1)*diff(diff(y(x),x),x)+8*x*(9*x^2+1)*diff(y(x),x)+(49*x^2+1)*y(x) = 0,y(x),
series,x=0)
\[
y = x^{{1}/{4}} \left (x^{4}-x^{2}+1\right ) \left (\ln \left (x \right ) c_2 +c_1 \right )+O\left (x^{6}\right )
\]
15.60.4 Mathematica DSolve solution
Solving time : 0.009 (sec)
Leaf size : 42
AsymptoticDSolveValue[{16*x^2*(1+x^2)*D[y[x],{x,2}]+8*x*(1+9*x^2)*D[y[x],x]+(1+49*x^2)*y[x]==0,{}},
y[x],{x,0,5}]
\[
y(x)\to c_1 \sqrt [4]{x} \left (x^4-x^2+1\right )+c_2 \sqrt [4]{x} \left (x^4-x^2+1\right ) \log (x)
\]