5.5 problem 15

5.5.1 Maple step by step solution

Internal problem ID [5673]
Internal file name [OUTPUT/4921_Sunday_June_05_2022_03_10_34_PM_45195600/index.tex]

Book: ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section: Chapter 5. Series Solutions of ODEs. REVIEW QUESTIONS. page 201
Problem number: 15.
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]

\[ \boxed {x^{2} y^{\prime \prime }+y^{\prime } x +\left (x^{2}-5\right ) y=0} \] 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. \[ x^{2} y^{\prime \prime }+y^{\prime } x +\left (x^{2}-5\right ) y = 0 \] 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 26: 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 \[ x^{2} y^{\prime \prime }+y^{\prime } x +\left (x^{2}-5\right ) y = 0 \] 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} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x^{2}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x +\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*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 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*}

Verification of solutions

\[ y = c_{1} x^{\sqrt {5}} \left (1-\frac {x^{2}}{4+4 \sqrt {5}}+\frac {x^{4}}{32 \left (\sqrt {5}+1\right ) \left (2+\sqrt {5}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{-\sqrt {5}} \left (1+\frac {x^{2}}{-4+4 \sqrt {5}}+\frac {x^{4}}{32 \left (\sqrt {5}-1\right ) \left (-2+\sqrt {5}\right )}+O\left (x^{6}\right )\right ) \] Verified OK.

5.5.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x^{2}+y^{\prime } x +\left (x^{2}-5\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 (x^{2}-5\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 (x^{2}-5\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 {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} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x^{2}+y^{\prime } x +\left (x^{2}-5\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 (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=\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=\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 (\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`
 

Solution by Maple

Time used: 0.015 (sec). Leaf size: 97

Order:=6; 
dsolve(x^2*diff(y(x),x$2)+x*diff(y(x),x)+(x^2-5)*y(x)=0,y(x),type='series',x=0);
 

\[ y \left (x \right ) = c_{1} x^{-\sqrt {5}} \left (1+\frac {1}{-4+4 \sqrt {5}} x^{2}+\frac {1}{32} \frac {1}{\left (-2+\sqrt {5}\right ) \left (\sqrt {5}-1\right )} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x^{\sqrt {5}} \left (1-\frac {1}{4+4 \sqrt {5}} x^{2}+\frac {1}{32} \frac {1}{\left (\sqrt {5}+2\right ) \left (\sqrt {5}+1\right )} x^{4}+\operatorname {O}\left (x^{6}\right )\right ) \]

Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 210

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

\[ y(x)\to c_2 \left (\frac {x^4}{\left (-3-\sqrt {5}+\left (1-\sqrt {5}\right ) \left (2-\sqrt {5}\right )\right ) \left (-1-\sqrt {5}+\left (3-\sqrt {5}\right ) \left (4-\sqrt {5}\right )\right )}-\frac {x^2}{-3-\sqrt {5}+\left (1-\sqrt {5}\right ) \left (2-\sqrt {5}\right )}+1\right ) x^{-\sqrt {5}}+c_1 \left (\frac {x^4}{\left (-3+\sqrt {5}+\left (1+\sqrt {5}\right ) \left (2+\sqrt {5}\right )\right ) \left (-1+\sqrt {5}+\left (3+\sqrt {5}\right ) \left (4+\sqrt {5}\right )\right )}-\frac {x^2}{-3+\sqrt {5}+\left (1+\sqrt {5}\right ) \left (2+\sqrt {5}\right )}+1\right ) x^{\sqrt {5}} \]