2.4.53 problem 50

Internal problem ID [8618]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 50
Date solved : Thursday, December 12, 2024 at 09:32:22 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

\begin{align*} x^{2} y^{\prime \prime }-x y^{\prime }+\left (x^{2}-8\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.

\[ x^{2} y^{\prime \prime }-x y^{\prime }+\left (x^{2}-8\right ) y = 0 \]

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}-8}{x^{2}}\\ \end{align*}

Table 2.101: Table \(p(x),q(x)\) singularites.

\(p(x)=-\frac {1}{x}\)
singularity	type
\(x = 0\)	\(\text {``regular''}\)

\(q(x)=\frac {x^{2}-8}{x^{2}}\)
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

\[ x^{2} y^{\prime \prime }-x y^{\prime }+\left (x^{2}-8\right ) y = 0 \]

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}-8\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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-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 (-8 a_{n} x^{n +r}\right ) = 0 \end{equation}

The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives

\[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-x^{n +r} a_{n} \left (n +r \right )-8 a_{n} x^{n +r} = 0 \]

When \(n = 0\) the above becomes

\[ x^{r} a_{0} r \left (-1+r \right )-x^{r} a_{0} r -8 a_{0} x^{r} = 0 \]

Or

\[ \left (x^{r} r \left (-1+r \right )-x^{r} r -8 x^{r}\right ) a_{0} = 0 \]

Since \(a_{0}\neq 0\) then the above simplifies to

\[ \left (r^{2}-2 r -8\right ) x^{r} = 0 \]

Since the above is true for all \(x\) then the indicial equation becomes

\[ r^{2}-2 r -8 = 0 \]

Solving for \(r\) gives the roots of the indicial equation as

\begin{align*} r_1 &= 4\\ r_2 &= -2 \end{align*}

Since \(a_{0}\neq 0\) then the indicial equation becomes

\[ \left (r^{2}-2 r -8\right ) x^{r} = 0 \]

Solving for \(r\) gives the roots of the indicial equation as \([4, -2]\).

Since \(r_1 - r_2 = 6\) 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^{4} \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}}{x^{2}} \end{align*}

Or

\begin{align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +4}\\ 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 -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}-8 a_{n} = 0 \end{equation}

Solving for \(a_{n}\) from recursive equation (4) gives

\[ a_{n} = -\frac {a_{n -2}}{n^{2}+2 n r +r^{2}-2 n -2 r -8}\tag {4} \]

Which for the root \(r = 4\) becomes

\[ a_{n} = -\frac {a_{n -2}}{n \left (n +6\right )}\tag {5} \]

At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 4\) 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

\[ a_{2}=-\frac {1}{r^{2}+2 r -8} \]

Which for the root \(r = 4\) becomes

\[ a_{2}=-{\frac {1}{16}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(-\frac {1}{r^{2}+2 r -8}\)	\(-{\frac {1}{16}}\)

For \(n = 3\), using the above recursive equation gives

\[ a_{3}=0 \]

And the table now becomes

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

For \(n = 4\), using the above recursive equation gives

\[ a_{4}=\frac {1}{\left (r +4\right ) \left (r -2\right ) r \left (r +6\right )} \]

Which for the root \(r = 4\) becomes

\[ a_{4}={\frac {1}{640}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(-\frac {1}{r^{2}+2 r -8}\)	\(-{\frac {1}{16}}\)
\(a_{3}\)	\(0\)	\(0\)
\(a_{4}\)	\(\frac {1}{\left (r +4\right ) \left (r -2\right ) r \left (r +6\right )}\)	\(\frac {1}{640}\)

For \(n = 5\), using the above recursive equation gives

\[ a_{5}=0 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(-\frac {1}{r^{2}+2 r -8}\)	\(-{\frac {1}{16}}\)
\(a_{3}\)	\(0\)	\(0\)
\(a_{4}\)	\(\frac {1}{\left (r +4\right ) \left (r -2\right ) r \left (r +6\right )}\)	\(\frac {1}{640}\)
\(a_{5}\)	\(0\)	\(0\)

For \(n = 6\), using the above recursive equation gives

\[ a_{6}=-\frac {1}{\left (r +4\right ) \left (r -2\right ) r \left (r +6\right ) \left (r +8\right ) \left (r +2\right )} \]

Which for the root \(r = 4\) becomes

\[ a_{6}=-{\frac {1}{46080}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(-\frac {1}{r^{2}+2 r -8}\)	\(-{\frac {1}{16}}\)
\(a_{3}\)	\(0\)	\(0\)
\(a_{4}\)	\(\frac {1}{\left (r +4\right ) \left (r -2\right ) r \left (r +6\right )}\)	\(\frac {1}{640}\)
\(a_{5}\)	\(0\)	\(0\)
\(a_{6}\)	\(-\frac {1}{\left (r +4\right ) \left (r -2\right ) r \left (r +6\right ) \left (r +8\right ) \left (r +2\right )}\)	\(-{\frac {1}{46080}}\)

Using the above table, then the solution \(y_{1}\left (x \right )\) is

\begin{align*} y_{1}\left (x \right )&= x^{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}+a_{7} x^{7}\dots \right ) \\ &= x^{4} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\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 \(N=6\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{6}\left (r \right )\). 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_{6} \\ &= -\frac {1}{\left (r +4\right ) \left (r -2\right ) r \left (r +6\right ) \left (r +8\right ) \left (r +2\right )} \end{align*}

Therefore

\begin{align*} \lim _{r\rightarrow r_{2}}-\frac {1}{\left (r +4\right ) \left (r -2\right ) r \left (r +6\right ) \left (r +8\right ) \left (r +2\right )}&= \lim _{r\rightarrow -2}-\frac {1}{\left (r +4\right ) \left (r -2\right ) r \left (r +6\right ) \left (r +8\right ) \left (r +2\right )}\\ &= \textit {undefined} \end{align*}

Since the limit does not exist then the log term is needed. Therefore the second solution has the form

\[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \]

Therefore

\begin{align*} \frac {d}{d x}y_{2}\left (x \right ) &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) \\ &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) \\ \frac {d^{2}}{d x^{2}}y_{2}\left (x \right ) &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right ) \\ &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) \\ \end{align*}

Substituting these back into the given ode \(x^{2} y^{\prime \prime }-x y^{\prime }+\left (x^{2}-8\right ) y = 0\) gives

\[ x^{2} \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )-x \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+\left (x^{2}-8\right ) \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \]

Which can be written as

\begin{equation} \tag{7} \left (\left (x^{2} y_{1}^{\prime \prime }\left (x \right )-y_{1}^{\prime }\left (x \right ) x +\left (x^{2}-8\right ) y_{1}\left (x \right )\right ) \ln \left (x \right )+x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )-y_{1}\left (x \right )\right ) C +x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )-x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (x^{2}-8\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation}

But since \(y_{1}\left (x \right )\) is a solution to the ode, then

\[ x^{2} y_{1}^{\prime \prime }\left (x \right )-y_{1}^{\prime }\left (x \right ) x +\left (x^{2}-8\right ) y_{1}\left (x \right ) = 0 \]

Eq (7) simplifes to

\begin{equation} \tag{8} \left (x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )-y_{1}\left (x \right )\right ) C +x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )-x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (x^{2}-8\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation}

Substituting \(y_{1} = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\) into the above gives

\begin{equation} \tag{9} \left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right ) x -2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C +\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x^{2}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) x^{2}-\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) x -8 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation}

Since \(r_{1} = 4\) and \(r_{2} = -2\) then the above becomes

\begin{equation} \tag{10} \left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{3+n} a_{n} \left (n +4\right )\right ) x -2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +4}\right )\right ) C +\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -2}\right ) x^{2}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-4+n} b_{n} \left (n -2\right ) \left (-3+n \right )\right ) x^{2}-\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-3+n} b_{n} \left (n -2\right )\right ) x -8 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -2}\right ) = 0 \end{equation}

Which simplifies to

\begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +4} a_{n} \left (n +4\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +4} C \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} \left (n^{2}-5 n +6\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n -2} b_{n} \left (n -2\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 b_{n} x^{n -2}\right ) = 0 \end{equation}

The next step is to make all powers of \(x\) be \(n -2\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -2}\) and adjusting the power and the corresponding index gives

\begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +4} a_{n} \left (n +4\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}2 C a_{n -6} \left (n -2\right ) x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +4} C \right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-2 C a_{n -6} x^{n -2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}b_{n -2} x^{n -2} \\ \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 \(n -2\).

\begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =6}{\sum }}2 C a_{n -6} \left (n -2\right ) x^{n -2}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-2 C a_{n -6} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}b_{n -2} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} \left (n^{2}-5 n +6\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n -2} b_{n} \left (n -2\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 b_{n} x^{n -2}\right ) = 0 \end{equation}

For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=1\), Eq (2B) gives

\[ -5 b_{1} = 0 \]

Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives

\[ -5 b_{1} = 0 \]

Solving the above for \(b_{1}\) gives

\[ b_{1}=0 \]

For \(n=2\), Eq (2B) gives

\[ b_{0}-8 b_{2} = 0 \]

Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives

\[ 1-8 b_{2} = 0 \]

Solving the above for \(b_{2}\) gives

\[ b_{2}={\frac {1}{8}} \]

For \(n=3\), Eq (2B) gives

\[ b_{1}-9 b_{3} = 0 \]

Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives

\[ -9 b_{3} = 0 \]

Solving the above for \(b_{3}\) gives

\[ b_{3}=0 \]

For \(n=4\), Eq (2B) gives

\[ b_{2}-8 b_{4} = 0 \]

Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives

\[ \frac {1}{8}-8 b_{4} = 0 \]

Solving the above for \(b_{4}\) gives

\[ b_{4}={\frac {1}{64}} \]

For \(n=5\), Eq (2B) gives

\[ b_{3}-5 b_{5} = 0 \]

Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives

\[ -5 b_{5} = 0 \]

Solving the above for \(b_{5}\) gives

\[ b_{5}=0 \]

For \(n=N\), where \(N=6\) which is the difference between the two roots, we are free to choose \(b_{6} = 0\). Hence for \(n=6\), Eq (2B) gives

\[ 6 C +\frac {1}{64} = 0 \]

Which is solved for \(C\). Solving for \(C\) gives

\[ C=-{\frac {1}{384}} \]

Now that we found all \(b_{n}\) and \(C\), we can calculate the second solution from

\[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \]

Using the above value found for \(C=-{\frac {1}{384}}\) and all \(b_{n}\), then the second solution becomes

\[ y_{2}\left (x \right )= -\frac {1}{384}\eslowast \left (x^{4} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\right )\right )\right ) \ln \left (x \right )+\frac {1+\frac {x^{2}}{8}+\frac {x^{4}}{64}+O\left (x^{7}\right )}{x^{2}} \]

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^{4} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\right )\right ) + c_2 \left (-\frac {1}{384}\eslowast \left (x^{4} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\right )\right )\right ) \ln \left (x \right )+\frac {1+\frac {x^{2}}{8}+\frac {x^{4}}{64}+O\left (x^{7}\right )}{x^{2}}\right ) \\ \end{align*}

Hence the final solution is

\begin{align*} y &= y_h \\ &= c_1 \,x^{4} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\right )\right )+c_2 \left (-\frac {x^{4} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\right )\right ) \ln \left (x \right )}{384}+\frac {1+\frac {x^{2}}{8}+\frac {x^{4}}{64}+O\left (x^{7}\right )}{x^{2}}\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}-8\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}-8\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}-8\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}-8}{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}}}=-8 \\ {} & \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}-8\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 (2+r \right ) \left (-4+r \right ) x^{r}+a_{1} \left (3+r \right ) \left (-3+r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +r +2\right ) \left (k +r -4\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 (2+r \right ) \left (-4+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-2, 4\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (3+r \right ) \left (-3+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 +r +2\right ) \left (k +r -4\right )+a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (k +4+r \right ) \left (k -2+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 +4+r \right ) \left (k -2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2 \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k -4\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-2\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =4 \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k -4\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =4 \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +8\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =4 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +4}, a_{k +2}=-\frac {a_{k}}{\left (k +8\right ) \left (k +2\right )}, a_{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 : 0.023 (sec)
Leaf size : 35

dsolve(x^2*diff(diff(y(x),x),x)-diff(y(x),x)*x+(x^2-8)*y(x) = 0,y(x), 
       series,x=0)
 

\[ y = c_{1} x^{4} \left (1-\frac {1}{16} x^{2}+\frac {1}{640} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+\frac {c_{2} \left (-86400-10800 x^{2}-1350 x^{4}+\operatorname {O}\left (x^{6}\right )\right )}{x^{2}} \]

Mathematica DSolve solution

Solving time : 0.013 (sec)
Leaf size : 42

AsymptoticDSolveValue[{x^2*D[y[x],{x,2}]-x*D[y[x],x]+(x^2-8)*y[x]==0,{}}, 
       y[x],{x,0,5}]
 

\[ y(x)\to c_1 \left (\frac {x^2}{64}+\frac {1}{x^2}+\frac {1}{8}\right )+c_2 \left (\frac {x^8}{640}-\frac {x^6}{16}+x^4\right ) \]