2.4.55 problem 52
Internal
problem
ID
[8370]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
52
Date
solved
:
Sunday, November 10, 2024 at 03:39:35 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} x^{2} y^{\prime \prime }-x y^{\prime }-\left (x^{2}+\frac {5}{4}\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.
\[ x^{2} y^{\prime \prime }-x y^{\prime }+\left (-x^{2}-\frac {5}{4}\right ) y = 0 \]
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 {4 x^{2}+5}{4 x^{2}}\\ \end{align*}
Table 2.101: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=-\frac {1}{x}\) |
| |
|
singularity | type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| |
| \(q(x)=-\frac {4 x^{2}+5}{4 x^{2}}\) |
| |
|
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
\[ x^{2} y^{\prime \prime }-x y^{\prime }+\left (-x^{2}-\frac {5}{4}\right ) y = 0 \]
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}-\frac {5}{4}\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 (-x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {5 a_{n} x^{n +r}}{4}\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}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-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} \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 (-x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {5 a_{n} x^{n +r}}{4}\right ) = 0
\end{equation}
The indicial
equation is obtained from \(n = 0\). From Eq (2B) this gives
\[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-x^{n +r} a_{n} \left (n +r \right )-\frac {5 a_{n} x^{n +r}}{4} = 0 \]
When \(n = 0\) the above becomes
\[ x^{r} a_{0} r \left (-1+r \right )-x^{r} a_{0} r -\frac {5 a_{0} x^{r}}{4} = 0 \]
Or
\[ \left (x^{r} r \left (-1+r \right )-x^{r} r -\frac {5 x^{r}}{4}\right ) a_{0} = 0 \]
Since \(a_{0}\neq 0\) then the above simplifies to
\[ \frac {\left (4 r^{2}-8 r -5\right ) x^{r}}{4} = 0 \]
Since the above is true for all \(x\) then the
indicial equation becomes
\[ r^{2}-2 r -\frac {5}{4} = 0 \]
Solving for \(r\) gives the roots of the indicial equation as
\begin{align*} r_1 &= {\frac {5}{2}}\\ r_2 &= -{\frac {1}{2}} \end{align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes
\[ \frac {\left (4 r^{2}-8 r -5\right ) x^{r}}{4} = 0 \]
Solving for \(r\) gives the roots of the indicial equation
as \(\left [{\frac {5}{2}}, -{\frac {1}{2}}\right ]\).
Since \(r_1 - r_2 = 3\) is 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 ) &= C y_{1}\left (x \right ) \ln \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 ) &= x^{{5}/{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{\sqrt {x}} \end{align*}
Or
\begin{align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {5}{2}}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {1}{2}}\right ) \end{align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). 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}-\frac {5 a_{n}}{4} = 0
\end{equation}
Solving for \(a_{n}\) from recursive equation (4) gives
\[ a_{n} = \frac {4 a_{n -2}}{4 n^{2}+8 n r +4 r^{2}-8 n -8 r -5}\tag {4} \]
Which for the root \(r = {\frac {5}{2}}\)
becomes
\[ a_{n} = \frac {a_{n -2}}{n \left (n +3\right )}\tag {5} \]
At this point, it is a good idea to keep track of \(a_{n}\) in a table both before
substituting \(r = {\frac {5}{2}}\) 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
\[ a_{2}=\frac {4}{4 r^{2}+8 r -5} \]
Which for the root \(r = {\frac {5}{2}}\) becomes
\[ a_{2}={\frac {1}{10}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(0\) | \(0\) |
| | |
| \(a_{2}\) | \(\frac {4}{4 r^{2}+8 r -5}\) | \(\frac {1}{10}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=0 \]
And the table now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(0\) | \(0\) |
| | |
| \(a_{2}\) | \(\frac {4}{4 r^{2}+8 r -5}\) | \(\frac {1}{10}\) |
| | |
| \(a_{3}\) |
\(0\) |
\(0\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=\frac {16}{\left (4 r^{2}+8 r -5\right ) \left (4 r^{2}+24 r +27\right )} \]
Which for the root \(r = {\frac {5}{2}}\) becomes
\[ a_{4}={\frac {1}{280}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(0\) | \(0\) |
| | |
| \(a_{2}\) | \(\frac {4}{4 r^{2}+8 r -5}\) | \(\frac {1}{10}\) |
| | |
| \(a_{3}\) | \(0\) | \(0\) |
| | |
| \(a_{4}\) |
\(\frac {16}{\left (4 r^{2}+8 r -5\right ) \left (4 r^{2}+24 r +27\right )}\) |
\(\frac {1}{280}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=0 \]
And the table now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(0\) |
\(0\) |
| | |
| \(a_{2}\) |
\(\frac {4}{4 r^{2}+8 r -5}\) | \(\frac {1}{10}\) |
| | |
| \(a_{3}\) | \(0\) | \(0\) |
| | |
| \(a_{4}\) |
\(\frac {16}{\left (4 r^{2}+8 r -5\right ) \left (4 r^{2}+24 r +27\right )}\) |
\(\frac {1}{280}\) |
| | |
| \(a_{5}\) |
\(0\) |
\(0\) |
| | |
Using the above table, then the solution \(y_{1}\left (x \right )\) is
\begin{align*} y_{1}\left (x \right )&= x^{{5}/{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^{{5}/{2}} \left (1+\frac {x^{2}}{10}+\frac {x^{4}}{280}+O\left (x^{6}\right )\right ) \end{align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Let
\[ r_{1}-r_{2} = N \]
Where \(N\) is positive integer which is the difference
between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=3\).
Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{3}\left (r \right )\). If this limit
exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that
\begin{align*} a_N &= a_{3} \\ &= 0 \end{align*}
Therefore
\begin{align*} \lim _{r\rightarrow r_{2}}0&= \lim _{r\rightarrow -{\frac {1}{2}}}0\\ &= 0 \end{align*}
The limit is \(0\). Since the limit exists then the log term is not needed and we can set \(C = 0\).
Therefore the second solution has the form
\begin{align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {1}{2}} \end{align*}
Eq (3) derived above is 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(3) gives
\[ b_{1} = 0 \]
For \(2\le n\) the recursive equation is
\begin{equation}
\tag{4} b_{n} \left (n +r \right ) \left (n +r -1\right )-b_{n} \left (n +r \right )-b_{n -2}-\frac {5 b_{n}}{4} = 0
\end{equation}
Which for for the root \(r = -{\frac {1}{2}}\)
becomes
\begin{equation}
\tag{4A} b_{n} \left (n -\frac {1}{2}\right ) \left (n -\frac {3}{2}\right )-b_{n} \left (n -\frac {1}{2}\right )-b_{n -2}-\frac {5 b_{n}}{4} = 0
\end{equation}
Solving for \(b_{n}\) from the recursive equation (4) gives
\[ b_{n} = \frac {4 b_{n -2}}{4 n^{2}+8 n r +4 r^{2}-8 n -8 r -5}\tag {5} \]
Which for the root \(r = -{\frac {1}{2}}\)
becomes
\[ b_{n} = \frac {4 b_{n -2}}{4 n^{2}-12 n}\tag {6} \]
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
\[ b_{2}=\frac {4}{4 r^{2}+8 r -5} \]
Which for the root \(r = -{\frac {1}{2}}\) becomes
\[ b_{2}=-{\frac {1}{2}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) | \(1\) |
| | |
| \(b_{1}\) | \(0\) | \(0\) |
| | |
| \(b_{2}\) | \(\frac {4}{4 r^{2}+8 r -5}\) | \(-{\frac {1}{2}}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ b_{3}=0 \]
And the table now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(0\) | \(0\) |
| | |
| \(b_{2}\) | \(\frac {4}{4 r^{2}+8 r -5}\) | \(-{\frac {1}{2}}\) |
| | |
| \(b_{3}\) |
\(0\) |
\(0\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ b_{4}=\frac {16}{\left (4 r^{2}+8 r -5\right ) \left (4 r^{2}+24 r +27\right )} \]
Which for the root \(r = -{\frac {1}{2}}\) becomes
\[ b_{4}=-{\frac {1}{8}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(0\) | \(0\) |
| | |
| \(b_{2}\) | \(\frac {4}{4 r^{2}+8 r -5}\) | \(-{\frac {1}{2}}\) |
| | |
| \(b_{3}\) | \(0\) | \(0\) |
| | |
| \(b_{4}\) |
\(\frac {16}{\left (4 r^{2}+8 r -5\right ) \left (4 r^{2}+24 r +27\right )}\) |
\(-{\frac {1}{8}}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ b_{5}=0 \]
And the table now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(0\) |
\(0\) |
| | |
| \(b_{2}\) |
\(\frac {4}{4 r^{2}+8 r -5}\) | \(-{\frac {1}{2}}\) |
| | |
| \(b_{3}\) | \(0\) | \(0\) |
| | |
| \(b_{4}\) |
\(\frac {16}{\left (4 r^{2}+8 r -5\right ) \left (4 r^{2}+24 r +27\right )}\) |
\(-{\frac {1}{8}}\) |
| | |
| \(b_{5}\) |
\(0\) |
\(0\) |
| | |
Using the above table, then the solution \(y_{2}\left (x \right )\) is
\begin{align*} y_{2}\left (x \right )&= x^{{5}/{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^{4}}{8}+O\left (x^{6}\right )}{\sqrt {x}} \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^{{5}/{2}} \left (1+\frac {x^{2}}{10}+\frac {x^{4}}{280}+O\left (x^{6}\right )\right ) + \frac {c_2 \left (1-\frac {x^{2}}{2}-\frac {x^{4}}{8}+O\left (x^{6}\right )\right )}{\sqrt {x}} \\
\end{align*}
Hence the final solution is
\begin{align*}
y &= y_h \\
&= c_1 \,x^{{5}/{2}} \left (1+\frac {x^{2}}{10}+\frac {x^{4}}{280}+O\left (x^{6}\right )\right )+\frac {c_2 \left (1-\frac {x^{2}}{2}-\frac {x^{4}}{8}+O\left (x^{6}\right )\right )}{\sqrt {x}} \\
\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}+\frac {5}{4}\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 (4 x^{2}+5\right ) y \left (x \right )}{4 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 (4 x^{2}+5\right ) y \left (x \right )}{4 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 {4 x^{2}+5}{4 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}}}=-\frac {5}{4} \\ {} & \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} \\ {} & {} & 4 x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-4 x \left (\frac {d}{d x}y \left (x \right )\right )+\left (-4 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 (1+2 r \right ) \left (-5+2 r \right ) x^{r}+a_{1} \left (3+2 r \right ) \left (-3+2 r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (2 k +2 r +1\right ) \left (2 k +2 r -5\right )-4 a_{k -2}\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 (-5+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {1}{2}, \frac {5}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (3+2 r \right ) \left (-3+2 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} \\ {} & {} & 4 \left (k +r -\frac {5}{2}\right ) \left (k +\frac {1}{2}+r \right ) a_{k}-4 a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & 4 \left (k -\frac {1}{2}+r \right ) \left (k +\frac {5}{2}+r \right ) a_{k +2}-4 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {4 a_{k}}{\left (2 k -1+2 r \right ) \left (2 k +5+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & a_{k +2}=\frac {4 a_{k}}{\left (2 k -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 {4 a_{k}}{\left (2 k -2\right ) \left (2 k +4\right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {5}{2} \\ {} & {} & a_{k +2}=\frac {4 a_{k}}{\left (2 k +4\right ) \left (2 k +10\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {5}{2} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {5}{2}}, a_{k +2}=\frac {4 a_{k}}{\left (2 k +4\right ) \left (2 k +10\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 -\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {5}{2}}\right ), a_{k +2}=\frac {4 a_{k}}{\left (2 k -2\right ) \left (2 k +4\right )}, a_{1}=0, b_{k +2}=\frac {4 b_{k}}{\left (2 k +4\right ) \left (2 k +10\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)
Reducible group (found another exponential solution)
<- Kovacics algorithm successful`
Maple dsolve solution
Solving time : 0.022 (sec)
Leaf size : 36
dsolve(x^2*diff(diff(y(x),x),x)-diff(y(x),x)*x-(x^2+5/4)*y(x) = 0,y(x),
series,x=0)
\[
y = \frac {c_{1} x^{3} \left (1+\frac {1}{10} x^{2}+\frac {1}{280} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (12-6 x^{2}-\frac {3}{2} x^{4}+\operatorname {O}\left (x^{6}\right )\right )}{\sqrt {x}}
\]
Mathematica DSolve solution
Solving time : 0.014 (sec)
Leaf size : 58
AsymptoticDSolveValue[{x^2*D[y[x],{x,2}]-x*D[y[x],x]-(x^2+5/4)*y[x]==0,{}},
y[x],{x,0,5}]
\[
y(x)\to c_1 \left (-\frac {x^{7/2}}{8}-\frac {x^{3/2}}{2}+\frac {1}{\sqrt {x}}\right )+c_2 \left (\frac {x^{13/2}}{280}+\frac {x^{9/2}}{10}+x^{5/2}\right )
\]