13.10 problem 20.7

13.10.1 Maple step by step solution

Internal problem ID [12079]
Internal file name [OUTPUT/10732_Monday_September_11_2023_12_50_12_AM_11152359/index.tex]

Book: AN INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS by JAMES C. ROBINSON. Cambridge University Press 2004
Section: Chapter 20, Series solutions of second order linear equations. Exercises page 195
Problem number: 20.7.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"

Maple gives the following as the ode type

[_Bessel]

\[ \boxed {x^{2} y^{\prime \prime }+x y^{\prime }+\left (-n^{2}+x^{2}\right ) y=0} \] With the expansion point for the power series method at \(x = 0\).

The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ x^{2} y^{\prime \prime }+x y^{\prime }+\left (-n^{2}+x^{2}\right ) y = 0 \] 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 {1}{x}\\ q(x) &= -\frac {n^{2}-x^{2}}{x^{2}}\\ \end {align*}

Table 89: Table \(p(x),q(x)\) singularites.
\(p(x)=\frac {1}{x}\)
singularity type
\(x = 0\) \(\text {``regular''}\)
\(q(x)=-\frac {n^{2}-x^{2}}{x^{2}}\)
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 \[ x^{2} y^{\prime \prime }+x y^{\prime }+\left (-n^{2}+x^{2}\right ) y = 0 \] 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 (\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 (-n^{2}+x^{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} \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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} n^{2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n}\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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} n^{2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )-x^{n +r} n^{2} a_{n} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r -x^{r} n^{2} a_{0} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r} r -x^{r} n^{2}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (-n^{2}+r^{2}\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ -n^{2}+r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= n\\ r_2 &= -n \end {align*}

Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (-n^{2}+r^{2}\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([n, -n]\).

Assuming the roots differ by non-integer Since \(r_1 - r_2 = 2 n\) 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 +n}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -n} \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} n^{2}+a_{n -2} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -2}}{n^{2}-n^{2}-2 n r -r^{2}}\tag {4} \] Which for the root \(r = n\) becomes \[ a_{n} = -\frac {a_{n -2}}{n \left (2 n +n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = n\) 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 \[ a_{2}=\frac {1}{n^{2}-r^{2}-4 r -4} \] Which for the root \(r = n\) becomes \[ a_{2}=\frac {1}{-4 n -4} \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(0\) \(0\)
\(a_{2}\) \(\frac {1}{n^{2}-r^{2}-4 r -4}\) \(\frac {1}{-4 n -4}\)

For \(n = 3\), using the above recursive equation gives \[ a_{3}=0 \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(0\) \(0\)
\(a_{2}\) \(\frac {1}{n^{2}-r^{2}-4 r -4}\) \(\frac {1}{-4 n -4}\)
\(a_{3}\) \(0\) \(0\)

For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{\left (n^{2}-r^{2}-4 r -4\right ) \left (n^{2}-r^{2}-8 r -16\right )} \] Which for the root \(r = n\) becomes \[ a_{4}=\frac {1}{32 \left (n +1\right ) \left (n +2\right )} \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(0\) \(0\)
\(a_{2}\) \(\frac {1}{n^{2}-r^{2}-4 r -4}\) \(\frac {1}{-4 n -4}\)
\(a_{3}\) \(0\) \(0\)
\(a_{4}\) \(\frac {1}{\left (n^{2}-r^{2}-4 r -4\right ) \left (n^{2}-r^{2}-8 r -16\right )}\) \(\frac {1}{32 \left (n +1\right ) \left (n +2\right )}\)

For \(n = 5\), using the above recursive equation gives \[ a_{5}=0 \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(0\) \(0\)
\(a_{2}\) \(\frac {1}{n^{2}-r^{2}-4 r -4}\) \(\frac {1}{-4 n -4}\)
\(a_{3}\) \(0\) \(0\)
\(a_{4}\) \(\frac {1}{\left (n^{2}-r^{2}-4 r -4\right ) \left (n^{2}-r^{2}-8 r -16\right )}\) \(\frac {1}{32 \left (n +1\right ) \left (n +2\right )}\)
\(a_{5}\) \(0\) \(0\)

Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{n} \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^{n} \left (1+\frac {x^{2}}{-4 n -4}+\frac {x^{4}}{32 \left (n +1\right ) \left (n +2\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 )-n^{2} b_{n}+b_{n -2} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {b_{n -2}}{n^{2}-n^{2}-2 n r -r^{2}}\tag {4} \] Which for the root \(r = -n\) becomes \[ b_{n} = \frac {b_{n -2}}{n \left (2 n -n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -n\) 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 \[ b_{2}=\frac {1}{n^{2}-r^{2}-4 r -4} \] Which for the root \(r = -n\) becomes \[ b_{2}=\frac {1}{4 n -4} \] And the table now becomes

\(n\) \(b_{n ,r}\) \(b_{n}\)
\(b_{0}\) \(1\) \(1\)
\(b_{1}\) \(0\) \(0\)
\(b_{2}\) \(\frac {1}{n^{2}-r^{2}-4 r -4}\) \(\frac {1}{4 n -4}\)

For \(n = 3\), using the above recursive equation gives \[ b_{3}=0 \] And the table now becomes

\(n\) \(b_{n ,r}\) \(b_{n}\)
\(b_{0}\) \(1\) \(1\)
\(b_{1}\) \(0\) \(0\)
\(b_{2}\) \(\frac {1}{n^{2}-r^{2}-4 r -4}\) \(\frac {1}{4 n -4}\)
\(b_{3}\) \(0\) \(0\)

For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {1}{\left (n^{2}-r^{2}-4 r -4\right ) \left (n^{2}-r^{2}-8 r -16\right )} \] Which for the root \(r = -n\) becomes \[ b_{4}=\frac {1}{32 \left (n -1\right ) \left (n -2\right )} \] And the table now becomes

\(n\) \(b_{n ,r}\) \(b_{n}\)
\(b_{0}\) \(1\) \(1\)
\(b_{1}\) \(0\) \(0\)
\(b_{2}\) \(\frac {1}{n^{2}-r^{2}-4 r -4}\) \(\frac {1}{4 n -4}\)
\(b_{3}\) \(0\) \(0\)
\(b_{4}\) \(\frac {1}{\left (n^{2}-r^{2}-4 r -4\right ) \left (n^{2}-r^{2}-8 r -16\right )}\) \(\frac {1}{32 \left (n -1\right ) \left (n -2\right )}\)

For \(n = 5\), using the above recursive equation gives \[ b_{5}=0 \] And the table now becomes

\(n\) \(b_{n ,r}\) \(b_{n}\)
\(b_{0}\) \(1\) \(1\)
\(b_{1}\) \(0\) \(0\)
\(b_{2}\) \(\frac {1}{n^{2}-r^{2}-4 r -4}\) \(\frac {1}{4 n -4}\)
\(b_{3}\) \(0\) \(0\)
\(b_{4}\) \(\frac {1}{\left (n^{2}-r^{2}-4 r -4\right ) \left (n^{2}-r^{2}-8 r -16\right )}\) \(\frac {1}{32 \left (n -1\right ) \left (n -2\right )}\)
\(b_{5}\) \(0\) \(0\)

Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{n} \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^{-n} \left (1+\frac {x^{2}}{4 n -4}+\frac {x^{4}}{32 \left (n -1\right ) \left (n -2\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^{n} \left (1+\frac {x^{2}}{-4 n -4}+\frac {x^{4}}{32 \left (n +1\right ) \left (n +2\right )}+O\left (x^{6}\right )\right ) + c_{2} x^{-n} \left (1+\frac {x^{2}}{4 n -4}+\frac {x^{4}}{32 \left (n -1\right ) \left (n -2\right )}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{n} \left (1+\frac {x^{2}}{-4 n -4}+\frac {x^{4}}{32 \left (n +1\right ) \left (n +2\right )}+O\left (x^{6}\right )\right )+c_{2} x^{-n} \left (1+\frac {x^{2}}{4 n -4}+\frac {x^{4}}{32 \left (n -1\right ) \left (n -2\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{n} \left (1+\frac {x^{2}}{-4 n -4}+\frac {x^{4}}{32 \left (n +1\right ) \left (n +2\right )}+O\left (x^{6}\right )\right )+c_{2} x^{-n} \left (1+\frac {x^{2}}{4 n -4}+\frac {x^{4}}{32 \left (n -1\right ) \left (n -2\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}

Verification of solutions

\[ y = c_{1} x^{n} \left (1+\frac {x^{2}}{-4 n -4}+\frac {x^{4}}{32 \left (n +1\right ) \left (n +2\right )}+O\left (x^{6}\right )\right )+c_{2} x^{-n} \left (1+\frac {x^{2}}{4 n -4}+\frac {x^{4}}{32 \left (n -1\right ) \left (n -2\right )}+O\left (x^{6}\right )\right ) \] Verified OK.

13.10.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+x y^{\prime }+\left (-n^{2}+x^{2}\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {\left (n^{2}-x^{2}\right ) y}{x^{2}}-\frac {y^{\prime }}{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\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}{d x}y^{\prime }+\frac {y^{\prime }}{x}-\frac {\left (n^{2}-x^{2}\right ) y}{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 {n^{2}-x^{2}}{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}}}=-n^{2} \\ {} & \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}{d x}y^{\prime }\right )+x y^{\prime }+\left (-n^{2}+x^{2}\right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\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\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y=\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=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\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}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}y^{\prime }\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 (n +r \right ) \left (-n +r \right ) x^{r}+a_{1} \left (1+n +r \right ) \left (1-n +r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +n +r \right ) \left (k -n +r \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} \\ {} & {} & \left (n +r \right ) \left (-n +r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{n , -n \right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+n +r \right ) \left (1-n +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} \\ {} & {} & a_{k} \left (k +n +r \right ) \left (k -n +r \right )+a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (k +2+n +r \right ) \left (k +2-n +r \right )+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2+n +r \right ) \left (k +2-n +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =n \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2+2 n \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =n \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +n}, a_{k +2}=-\frac {a_{k}}{\left (k +2+2 n \right ) \left (k +2\right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-n \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k +2-2 n \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-n \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -n}, a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k +2-2 n \right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +n}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -n}\right ), a_{k +2}=-\frac {a_{k}}{\left (k +2+2 n \right ) \left (k +2\right )}, a_{1}=0, b_{k +2}=-\frac {b_{k}}{\left (k +2\right ) \left (k +2-2 n \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 
<- No Liouvillian solutions exists 
-> Trying a solution in terms of special functions: 
   -> Bessel 
   <- Bessel successful 
<- special function solution successful`
 

Solution by Maple

Time used: 0.031 (sec). Leaf size: 77

Order:=6; 
dsolve(x^2*diff(y(x),x$2)+x*diff(y(x),x)+(x^2-n^2)*y(x)=0,y(x),type='series',x=0);
 

\[ y \left (x \right ) = x^{-n} \left (1+\frac {1}{4 n -4} x^{2}+\frac {1}{32} \frac {1}{\left (n -2\right ) \left (n -1\right )} x^{4}+\operatorname {O}\left (x^{6}\right )\right ) c_{1} +c_{2} x^{n} \left (1-\frac {1}{4 n +4} x^{2}+\frac {1}{32} \frac {1}{\left (n +2\right ) \left (n +1\right )} x^{4}+\operatorname {O}\left (x^{6}\right )\right ) \]

Solution by Mathematica

Time used: 0.005 (sec). Leaf size: 160

AsymptoticDSolveValue[x^2*y''[x]+x*y'[x]+(x^2-n^2)*y[x]==0,y[x],{x,0,5}]
 

\[ y(x)\to c_2 \left (\frac {x^4}{\left (-n^2-n+(1-n) (2-n)+2\right ) \left (-n^2-n+(3-n) (4-n)+4\right )}-\frac {x^2}{-n^2-n+(1-n) (2-n)+2}+1\right ) x^{-n}+c_1 \left (\frac {x^4}{\left (-n^2+n+(n+1) (n+2)+2\right ) \left (-n^2+n+(n+3) (n+4)+4\right )}-\frac {x^2}{-n^2+n+(n+1) (n+2)+2}+1\right ) x^n \]