2.2.44 problem 44
Internal
problem
ID
[8579]
Book
:
Second
order
enumerated
odes
Section
:
section
2
Problem
number
:
44
Date
solved
:
Sunday, November 10, 2024 at 04:05:29 AM
CAS
classification
:
[_Bessel]
Solve
\begin{align*} x^{2} y^{\prime \prime }+x y^{\prime }+\left (x^{2}-5\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.
\[ x^{2} y^{\prime \prime }+x y^{\prime }+\left (x^{2}-5\right ) y = 0 \]
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}\\ q(x) &= \frac {x^{2}-5}{x^{2}}\\ \end{align*}
Table 2.52: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=\frac {1}{x}\) |
| |
|
singularity | type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| |
| \(q(x)=\frac {x^{2}-5}{x^{2}}\) |
| |
|
singularity | type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
Combining everything together gives the following summary of singularities for the ode
as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to
be
\[ x^{2} y^{\prime \prime }+x y^{\prime }+\left (x^{2}-5\right ) y = 0 \]
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 (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (x^{2}-5\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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 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 }}x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}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 =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 a_{n} x^{n +r}\right ) = 0
\end{equation}
The indicial
equation is obtained from \(n = 0\). From Eq (2B) this gives
\[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )-5 a_{n} x^{n +r} = 0 \]
When \(n = 0\) the above becomes
\[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r -5 a_{0} x^{r} = 0 \]
Or
\[ \left (x^{r} r \left (-1+r \right )+x^{r} r -5 x^{r}\right ) a_{0} = 0 \]
Since \(a_{0}\neq 0\) then the above simplifies to
\[ \left (r^{2}-5\right ) x^{r} = 0 \]
Since the above is true for all \(x\) then the
indicial equation becomes
\[ r^{2}-5 = 0 \]
Solving for \(r\) gives the roots of the indicial equation as
\begin{align*} r_1 &= \sqrt {5}\\ r_2 &= -\sqrt {5} \end{align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes
\[ \left (r^{2}-5\right ) x^{r} = 0 \]
Solving for \(r\) gives the roots of the indicial equation
as \(\left [\sqrt {5}, -\sqrt {5}\right ]\).
Since \(r_1 - r_2 = 2 \sqrt {5}\) 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 +\sqrt {5}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\sqrt {5}} \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} \left (n +r \right ) \left (n +r -1\right )+a_{n} \left (n +r \right )+a_{n -2}-5 a_{n} = 0
\end{equation}
Solving for \(a_{n}\) from recursive equation (4) gives
\[ a_{n} = -\frac {a_{n -2}}{n^{2}+2 n r +r^{2}-5}\tag {4} \]
Which for the root \(r = \sqrt {5}\)
becomes
\[ a_{n} = -\frac {a_{n -2}}{n \left (2 \sqrt {5}+n \right )}\tag {5} \]
At this point, it is a good idea to keep track of \(a_{n}\) in a table both before
substituting \(r = \sqrt {5}\) 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
\[ a_{2}=-\frac {1}{r^{2}+4 r -1} \]
Which for the root \(r = \sqrt {5}\) becomes
\[ a_{2}=-\frac {1}{4+4 \sqrt {5}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(0\) | \(0\) |
| | |
| \(a_{2}\) | \(-\frac {1}{r^{2}+4 r -1}\) | \(-\frac {1}{4+4 \sqrt {5}}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=0 \]
And the table now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(0\) | \(0\) |
| | |
| \(a_{2}\) | \(-\frac {1}{r^{2}+4 r -1}\) | \(-\frac {1}{4+4 \sqrt {5}}\) |
| | |
| \(a_{3}\) |
\(0\) |
\(0\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=\frac {1}{\left (r^{2}+4 r -1\right ) \left (r^{2}+8 r +11\right )} \]
Which for the root \(r = \sqrt {5}\) becomes
\[ a_{4}=\frac {1}{32 \left (\sqrt {5}+1\right ) \left (2+\sqrt {5}\right )} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(0\) | \(0\) |
| | |
| \(a_{2}\) | \(-\frac {1}{r^{2}+4 r -1}\) | \(-\frac {1}{4+4 \sqrt {5}}\) |
| | |
| \(a_{3}\) | \(0\) | \(0\) |
| | |
| \(a_{4}\) |
\(\frac {1}{\left (r^{2}+4 r -1\right ) \left (r^{2}+8 r +11\right )}\) |
\(\frac {1}{32 \left (\sqrt {5}+1\right ) \left (2+\sqrt {5}\right )}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=0 \]
And the table now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(0\) |
\(0\) |
| | |
| \(a_{2}\) |
\(-\frac {1}{r^{2}+4 r -1}\) | \(-\frac {1}{4+4 \sqrt {5}}\) |
| | |
| \(a_{3}\) | \(0\) | \(0\) |
| | |
| \(a_{4}\) |
\(\frac {1}{\left (r^{2}+4 r -1\right ) \left (r^{2}+8 r +11\right )}\) |
\(\frac {1}{32 \left (\sqrt {5}+1\right ) \left (2+\sqrt {5}\right )}\) |
| | |
| \(a_{5}\) |
\(0\) |
\(0\) |
| | |
Using the above table, then the solution \(y_{1}\left (x \right )\) is
\begin{align*} y_{1}\left (x \right )&= x^{\sqrt {5}} \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^{\sqrt {5}} \left (1-\frac {x^{2}}{4+4 \sqrt {5}}+\frac {x^{4}}{32 \left (\sqrt {5}+1\right ) \left (2+\sqrt {5}\right )}+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} \left (n +r \right ) \left (n +r -1\right )+b_{n} \left (n +r \right )+b_{n -2}-5 b_{n} = 0
\end{equation}
Solving for \(b_{n}\) from recursive equation (4) gives
\[ b_{n} = -\frac {b_{n -2}}{n^{2}+2 n r +r^{2}-5}\tag {4} \]
Which for the root \(r = -\sqrt {5}\)
becomes
\[ b_{n} = -\frac {b_{n -2}}{n \left (-2 \sqrt {5}+n \right )}\tag {5} \]
At this point, it is a good idea to keep track of \(b_{n}\) in a table both before
substituting \(r = -\sqrt {5}\) 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
\[ b_{2}=-\frac {1}{r^{2}+4 r -1} \]
Which for the root \(r = -\sqrt {5}\) becomes
\[ b_{2}=\frac {1}{-4+4 \sqrt {5}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) | \(1\) |
| | |
| \(b_{1}\) | \(0\) | \(0\) |
| | |
| \(b_{2}\) | \(-\frac {1}{r^{2}+4 r -1}\) | \(\frac {1}{-4+4 \sqrt {5}}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ b_{3}=0 \]
And the table now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(0\) | \(0\) |
| | |
| \(b_{2}\) | \(-\frac {1}{r^{2}+4 r -1}\) | \(\frac {1}{-4+4 \sqrt {5}}\) |
| | |
| \(b_{3}\) |
\(0\) |
\(0\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ b_{4}=\frac {1}{\left (r^{2}+4 r -1\right ) \left (r^{2}+8 r +11\right )} \]
Which for the root \(r = -\sqrt {5}\) becomes
\[ b_{4}=\frac {1}{32 \left (\sqrt {5}-1\right ) \left (-2+\sqrt {5}\right )} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(0\) | \(0\) |
| | |
| \(b_{2}\) | \(-\frac {1}{r^{2}+4 r -1}\) | \(\frac {1}{-4+4 \sqrt {5}}\) |
| | |
| \(b_{3}\) | \(0\) | \(0\) |
| | |
| \(b_{4}\) |
\(\frac {1}{\left (r^{2}+4 r -1\right ) \left (r^{2}+8 r +11\right )}\) |
\(\frac {1}{32 \left (\sqrt {5}-1\right ) \left (-2+\sqrt {5}\right )}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ b_{5}=0 \]
And the table now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(0\) |
\(0\) |
| | |
| \(b_{2}\) |
\(-\frac {1}{r^{2}+4 r -1}\) | \(\frac {1}{-4+4 \sqrt {5}}\) |
| | |
| \(b_{3}\) | \(0\) | \(0\) |
| | |
| \(b_{4}\) |
\(\frac {1}{\left (r^{2}+4 r -1\right ) \left (r^{2}+8 r +11\right )}\) |
\(\frac {1}{32 \left (\sqrt {5}-1\right ) \left (-2+\sqrt {5}\right )}\) |
| | |
| \(b_{5}\) |
\(0\) |
\(0\) |
| | |
Using the above table, then the solution \(y_{2}\left (x \right )\) is
\begin{align*} y_{2}\left (x \right )&= x^{\sqrt {5}} \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 ) \\ &= x^{-\sqrt {5}} \left (1+\frac {x^{2}}{-4+4 \sqrt {5}}+\frac {x^{4}}{32 \left (\sqrt {5}-1\right ) \left (-2+\sqrt {5}\right )}+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^{\sqrt {5}} \left (1-\frac {x^{2}}{4+4 \sqrt {5}}+\frac {x^{4}}{32 \left (\sqrt {5}+1\right ) \left (2+\sqrt {5}\right )}+O\left (x^{6}\right )\right ) + c_2 \,x^{-\sqrt {5}} \left (1+\frac {x^{2}}{-4+4 \sqrt {5}}+\frac {x^{4}}{32 \left (\sqrt {5}-1\right ) \left (-2+\sqrt {5}\right )}+O\left (x^{6}\right )\right ) \\
\end{align*}
Hence the final solution is
\begin{align*}
y &= y_h \\
&= c_1 \,x^{\sqrt {5}} \left (1-\frac {x^{2}}{4+4 \sqrt {5}}+\frac {x^{4}}{32 \left (\sqrt {5}+1\right ) \left (2+\sqrt {5}\right )}+O\left (x^{6}\right )\right )+c_2 \,x^{-\sqrt {5}} \left (1+\frac {x^{2}}{-4+4 \sqrt {5}}+\frac {x^{4}}{32 \left (\sqrt {5}-1\right ) \left (-2+\sqrt {5}\right )}+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 (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+x \left (\frac {d}{d x}y \left (x \right )\right )+\left (x^{2}-5\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^{2}-5\right ) y \left (x \right )}{x^{2}}-\frac {\frac {d}{d x}y \left (x \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 )}{x}+\frac {\left (x^{2}-5\right ) y \left (x \right )}{x^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=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}, P_{3}\left (x \right )=\frac {x^{2}-5}{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}}}=1 \\ {} & \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}}}=-5 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+x \left (\frac {d}{d x}y \left (x \right )\right )+\left (x^{2}-5\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 \cdot \left (\frac {d}{d x}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \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} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (r^{2}-5\right ) x^{r}+a_{1} \left (r^{2}+2 r -4\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k^{2}+2 k r +r^{2}-5\right )+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^{2}-5=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\sqrt {5}, -\sqrt {5}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (r^{2}+2 r -4\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} \\ {} & {} & a_{k} \left (k^{2}+2 k r +r^{2}-5\right )+a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (\left (k +2\right )^{2}+2 \left (k +2\right ) r +r^{2}-5\right )+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k}}{k^{2}+2 k r +r^{2}+4 k +4 r -1} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\sqrt {5} \\ {} & {} & a_{k +2}=-\frac {a_{k}}{k^{2}+2 k \sqrt {5}+4+4 k +4 \sqrt {5}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\sqrt {5} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\sqrt {5}}, a_{k +2}=-\frac {a_{k}}{k^{2}+2 k \sqrt {5}+4+4 k +4 \sqrt {5}}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\sqrt {5} \\ {} & {} & a_{k +2}=-\frac {a_{k}}{k^{2}-2 k \sqrt {5}+4+4 k -4 \sqrt {5}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\sqrt {5} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\sqrt {5}}, a_{k +2}=-\frac {a_{k}}{k^{2}-2 k \sqrt {5}+4+4 k -4 \sqrt {5}}, 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 +\sqrt {5}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -\sqrt {5}}\right ), a_{k +2}=-\frac {a_{k}}{k^{2}+2 k \sqrt {5}+4+4 k +4 \sqrt {5}}, a_{1}=0, b_{k +2}=-\frac {b_{k}}{k^{2}-2 k \sqrt {5}+4+4 k -4 \sqrt {5}}, 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
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful`
Maple dsolve solution
Solving time : 0.020 (sec)
Leaf size : 89
dsolve(x^2*diff(diff(y(x),x),x)+diff(y(x),x)*x+(x^2-5)*y(x) = 0,y(x),
series,x=0)
\[
y = c_{1} x^{-\sqrt {5}} \left (1+\frac {1}{-4+4 \sqrt {5}} x^{2}+\frac {1}{32} \frac {1}{\left (-2+\sqrt {5}\right ) \left (\sqrt {5}-1\right )} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x^{\sqrt {5}} \left (1-\frac {1}{4+4 \sqrt {5}} x^{2}+\frac {1}{32} \frac {1}{\left (\sqrt {5}+2\right ) \left (\sqrt {5}+1\right )} x^{4}+\operatorname {O}\left (x^{6}\right )\right )
\]
Mathematica DSolve solution
Solving time : 0.007 (sec)
Leaf size : 210
AsymptoticDSolveValue[{x^2*D[y[x],{x,2}]+x*D[y[x],x]+(x^2-5)*y[x]==0,{}},
y[x],{x,0,5}]
\[
y(x)\to c_2 \left (\frac {x^4}{\left (-3-\sqrt {5}+\left (1-\sqrt {5}\right ) \left (2-\sqrt {5}\right )\right ) \left (-1-\sqrt {5}+\left (3-\sqrt {5}\right ) \left (4-\sqrt {5}\right )\right )}-\frac {x^2}{-3-\sqrt {5}+\left (1-\sqrt {5}\right ) \left (2-\sqrt {5}\right )}+1\right ) x^{-\sqrt {5}}+c_1 \left (\frac {x^4}{\left (-3+\sqrt {5}+\left (1+\sqrt {5}\right ) \left (2+\sqrt {5}\right )\right ) \left (-1+\sqrt {5}+\left (3+\sqrt {5}\right ) \left (4+\sqrt {5}\right )\right )}-\frac {x^2}{-3+\sqrt {5}+\left (1+\sqrt {5}\right ) \left (2+\sqrt {5}\right )}+1\right ) x^{\sqrt {5}}
\]