2.2.43 problem 43
Internal
problem
ID
[8578]
Book
:
Second
order
enumerated
odes
Section
:
section
2
Problem
number
:
43
Date
solved
:
Sunday, November 10, 2024 at 04:05:28 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} x^{2} y^{\prime \prime }-x \left (x +6\right ) y^{\prime }+10 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.
\[ x^{2} y^{\prime \prime }+\left (-x^{2}-6 x \right ) y^{\prime }+10 y = 0 \]
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 {x +6}{x}\\ q(x) &= \frac {10}{x^{2}}\\ \end{align*}
Table 2.51: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=-\frac {x +6}{x}\) |
| |
|
singularity | type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| |
| \(q(x)=\frac {10}{x^{2}}\) |
| |
|
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
\[ x^{2} y^{\prime \prime }+\left (-x^{2}-6 x \right ) y^{\prime }+10 y = 0 \]
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 )+\left (-x^{2}-6 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+10 \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^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 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 }}\left (-x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) 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 =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 a_{n} x^{n +r}\right ) = 0
\end{equation}
The indicial
equation is obtained from \(n = 0\). From Eq (2B) this gives
\[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-6 x^{n +r} a_{n} \left (n +r \right )+10 a_{n} x^{n +r} = 0 \]
When \(n = 0\) the above becomes
\[ x^{r} a_{0} r \left (-1+r \right )-6 x^{r} a_{0} r +10 a_{0} x^{r} = 0 \]
Or
\[ \left (x^{r} r \left (-1+r \right )-6 x^{r} r +10 x^{r}\right ) a_{0} = 0 \]
Since \(a_{0}\neq 0\) then the above simplifies to
\[ \left (r -2\right ) \left (r -5\right ) x^{r} = 0 \]
Since the above is true for all \(x\) then the
indicial equation becomes
\[ \left (r -2\right ) \left (r -5\right ) = 0 \]
Solving for \(r\) gives the roots of the indicial equation as
\begin{align*} r_1 &= 5\\ r_2 &= 2 \end{align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes
\[ \left (r -2\right ) \left (r -5\right ) x^{r} = 0 \]
Solving for \(r\) gives the roots of the indicial equation
as \([5, 2]\).
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} \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^{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 +5}\\ 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 \(a_{0} = 1\). For \(1\le n\) the recursive equation is
\begin{equation}
\tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -1} \left (n +r -1\right )-6 a_{n} \left (n +r \right )+10 a_{n} = 0
\end{equation}
Solving for \(a_{n}\) from
recursive equation (4) gives
\[ a_{n} = \frac {a_{n -1} \left (n +r -1\right )}{n^{2}+2 n r +r^{2}-7 n -7 r +10}\tag {4} \]
Which for the root \(r = 5\) becomes
\[ a_{n} = \frac {a_{n -1} \left (n +4\right )}{n \left (n +3\right )}\tag {5} \]
At this point, it is a good idea to
keep track of \(a_{n}\) in a table both before substituting \(r = 5\) and after as more terms are found using
the above recursive equation.
| | |
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| | |
| \(a_{0}\) | \(1\) | \(1\) |
| | |
For \(n = 1\), using the above recursive equation gives
\[ a_{1}=\frac {r}{r^{2}-5 r +4} \]
Which for the root \(r = 5\) becomes
\[ a_{1}={\frac {5}{4}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {r}{r^{2}-5 r +4}\) | \(\frac {5}{4}\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ a_{2}=\frac {1+r}{r^{3}-8 r^{2}+19 r -12} \]
Which for the root \(r = 5\) becomes
\[ a_{2}={\frac {3}{4}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {r}{r^{2}-5 r +4}\) | \(\frac {5}{4}\) |
| | |
| \(a_{2}\) | \(\frac {1+r}{r^{3}-8 r^{2}+19 r -12}\) | \(\frac {3}{4}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=\frac {2+r}{r^{4}-10 r^{3}+35 r^{2}-50 r +24} \]
Which for the root \(r = 5\) becomes
\[ a_{3}={\frac {7}{24}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {r}{r^{2}-5 r +4}\) | \(\frac {5}{4}\) |
| | |
| \(a_{2}\) | \(\frac {1+r}{r^{3}-8 r^{2}+19 r -12}\) | \(\frac {3}{4}\) |
| | |
| \(a_{3}\) |
\(\frac {2+r}{r^{4}-10 r^{3}+35 r^{2}-50 r +24}\) |
\(\frac {7}{24}\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=\frac {3+r}{\left (-1+r \right )^{2} \left (r -2\right ) \left (r -4\right ) \left (r -3\right )} \]
Which for the root \(r = 5\) becomes
\[ a_{4}={\frac {1}{12}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {r}{r^{2}-5 r +4}\) | \(\frac {5}{4}\) |
| | |
| \(a_{2}\) | \(\frac {1+r}{r^{3}-8 r^{2}+19 r -12}\) | \(\frac {3}{4}\) |
| | |
| \(a_{3}\) | \(\frac {2+r}{r^{4}-10 r^{3}+35 r^{2}-50 r +24}\) | \(\frac {7}{24}\) |
| | |
| \(a_{4}\) |
\(\frac {3+r}{\left (-1+r \right )^{2} \left (r -2\right ) \left (r -4\right ) \left (r -3\right )}\) |
\(\frac {1}{12}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=\frac {4+r}{r \left (-1+r \right )^{2} \left (r -2\right ) \left (r -4\right ) \left (r -3\right )} \]
Which for the root \(r = 5\) becomes
\[ a_{5}={\frac {3}{160}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {r}{r^{2}-5 r +4}\) |
\(\frac {5}{4}\) |
| | |
| \(a_{2}\) |
\(\frac {1+r}{r^{3}-8 r^{2}+19 r -12}\) | \(\frac {3}{4}\) |
| | |
| \(a_{3}\) | \(\frac {2+r}{r^{4}-10 r^{3}+35 r^{2}-50 r +24}\) | \(\frac {7}{24}\) |
| | |
| \(a_{4}\) |
\(\frac {3+r}{\left (-1+r \right )^{2} \left (r -2\right ) \left (r -4\right ) \left (r -3\right )}\) |
\(\frac {1}{12}\) |
| | |
| \(a_{5}\) |
\(\frac {4+r}{r \left (-1+r \right )^{2} \left (r -2\right ) \left (r -4\right ) \left (r -3\right )}\) |
\(\frac {3}{160}\) |
| | |
Using the above table, then the solution \(y_{1}\left (x \right )\) is
\begin{align*} y_{1}\left (x \right )&= x^{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^{5} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+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} \\ &= \frac {2+r}{r^{4}-10 r^{3}+35 r^{2}-50 r +24} \end{align*}
Therefore
\begin{align*} \lim _{r\rightarrow r_{2}}\frac {2+r}{r^{4}-10 r^{3}+35 r^{2}-50 r +24}&= \lim _{r\rightarrow 2}\frac {2+r}{r^{4}-10 r^{3}+35 r^{2}-50 r +24}\\ &= \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 \(x^{2} y^{\prime \prime }+\left (-x^{2}-6 x \right ) y^{\prime }+10 y = 0\) gives
\[
x^{2} \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (-x^{2}-6 x \right ) \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+10 C y_{1}\left (x \right ) \ln \left (x \right )+10 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0
\]
Which
can be written as
\begin{equation}
\tag{7} \left (\left (x^{2} y_{1}^{\prime \prime }\left (x \right )+\left (-x^{2}-6 x \right ) y_{1}^{\prime }\left (x \right )+10 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 )+\frac {\left (-x^{2}-6 x \right ) y_{1}\left (x \right )}{x}\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 )+\left (-x^{2}-6 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+10 \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
\[ x^{2} y_{1}^{\prime \prime }\left (x \right )+\left (-x^{2}-6 x \right ) y_{1}^{\prime }\left (x \right )+10 y_{1}\left (x \right ) = 0 \]
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 )+\frac {\left (-x^{2}-6 x \right ) y_{1}\left (x \right )}{x}\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 )+\left (-x^{2}-6 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+10 \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 -\left (x +7\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C +\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 (-x^{2}-6 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )+10 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0
\end{equation}
Since \(r_{1} = 5\) and \(r_{2} = 2\) then the above becomes
\begin{equation}
\tag{10} \left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +4} a_{n} \left (n +5\right )\right ) x -\left (x +7\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +5}\right )\right ) C +\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} b_{n} \left (n +2\right ) \left (1+n \right )\right ) x^{2}+\left (-x^{2}-6 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} b_{n} \left (n +2\right )\right )+10 \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 +5} a_{n} \left (n +5\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n +6} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-7 C \,x^{n +5} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +2} b_{n} \left (n^{2}+3 n +2\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +3} b_{n} \left (n +2\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{n +2} b_{n} \left (n +2\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 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 +5} a_{n} \left (n +5\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}2 C a_{n -3} \left (n +2\right ) x^{n +2} \\
\moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n +6} a_{n}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-C a_{n -4} x^{n +2}\right ) \\
\moverset {\infty }{\munderset {n =0}{\sum }}\left (-7 C \,x^{n +5} a_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-7 C a_{n -3} x^{n +2}\right ) \\
\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +3} b_{n} \left (n +2\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} \left (1+n \right ) x^{n +2}\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 \(n +2\).
\begin{equation}
\tag{2B} \left (\moverset {\infty }{\munderset {n =3}{\sum }}2 C a_{n -3} \left (n +2\right ) x^{n +2}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-C a_{n -4} x^{n +2}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-7 C a_{n -3} x^{n +2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +2} b_{n} \left (n^{2}+3 n +2\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} \left (1+n \right ) x^{n +2}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{n +2} b_{n} \left (n +2\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 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
\[ -2 b_{1}-2 b_{0} = 0 \]
Which when replacing the above values found already for \(b_{n}\) and the values
found earlier for \(a_{n}\) and for \(C\), gives
\[ -2 b_{1}-2 = 0 \]
Solving the above for \(b_{1}\) gives
\[ b_{1}=-1 \]
For \(n=2\), Eq (2B) gives
\[ -2 b_{2}-3 b_{1} = 0 \]
Which when replacing the above values found already for \(b_{n}\) and the values found
earlier for \(a_{n}\) and for \(C\), gives
\[ -2 b_{2}+3 = 0 \]
Solving the above for \(b_{2}\) gives
\[ b_{2}={\frac {3}{2}} \]
For \(n=N\), where \(N=3\) which is the
difference between the two roots, we are free to choose \(b_{3} = 0\). Hence for \(n=3\), Eq (2B) gives
\[ 3 C -6 = 0 \]
Which is solved for \(C\). Solving for \(C\) gives
\[ C=2 \]
For \(n=4\), Eq (2B) gives
\[ \left (-a_{0}+5 a_{1}\right ) C -5 b_{3}+4 b_{4} = 0 \]
Which when replacing
the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\),
gives
\[ \frac {21}{2}+4 b_{4} = 0 \]
Solving the above for \(b_{4}\) gives
\[ b_{4}=-{\frac {21}{8}} \]
For \(n=5\), Eq (2B) gives
\[ \left (-a_{1}+7 a_{2}\right ) C -6 b_{4}+10 b_{5} = 0 \]
Which when replacing the
above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives
\[ \frac {95}{4}+10 b_{5} = 0 \]
Solving the above for \(b_{5}\) gives
\[ b_{5}=-{\frac {19}{8}} \]
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 \(C=2\) and all \(b_{n}\), then the second
solution becomes
\[
y_{2}\left (x \right )= 2\eslowast \left (x^{5} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+x^{2} \left (1-x +\frac {3 x^{2}}{2}-\frac {21 x^{4}}{8}-\frac {19 x^{5}}{8}+O\left (x^{6}\right )\right )
\]
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} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+O\left (x^{6}\right )\right ) + c_2 \left (2\eslowast \left (x^{5} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+x^{2} \left (1-x +\frac {3 x^{2}}{2}-\frac {21 x^{4}}{8}-\frac {19 x^{5}}{8}+O\left (x^{6}\right )\right )\right ) \\
\end{align*}
Hence the final solution is
\begin{align*}
y &= y_h \\
&= c_1 \,x^{5} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+O\left (x^{6}\right )\right )+c_2 \left (2 x^{5} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{2} \left (1-x +\frac {3 x^{2}}{2}-\frac {21 x^{4}}{8}-\frac {19 x^{5}}{8}+O\left (x^{6}\right )\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 (6+x \right ) \left (\frac {d}{d x}y \left (x \right )\right )+10 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 {10 y \left (x \right )}{x^{2}}+\frac {\left (6+x \right ) \left (\frac {d}{d x}y \left (x \right )\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 {\left (6+x \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x}+\frac {10 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 {6+x}{x}, P_{3}\left (x \right )=\frac {10}{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}}}=-6 \\ {} & \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}}}=10 \\ {} & \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 (6+x \right ) \left (\frac {d}{d x}y \left (x \right )\right )+10 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 \left (\frac {d}{d x}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\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 -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +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 (-5+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k +r -2\right ) \left (k +r -5\right )-a_{k -1} \left (k +r -1\right )\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 (-5+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{2, 5\right \} \\ \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 -5\right )-a_{k -1} \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +1} \left (k +r -1\right ) \left (k -4+r \right )-a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +r \right )}{\left (k +r -1\right ) \left (k -4+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +2\right )}{\left (k +1\right ) \left (k -2\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 =2 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +2\right )}{\left (k +1\right ) \left (k -2\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =5 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +5\right )}{\left (k +4\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =5 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +5}, a_{k +1}=\frac {a_{k} \left (k +5\right )}{\left (k +4\right ) \left (k +1\right )}\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)
Group is reducible, not completely reducible
<- Kovacics algorithm successful`
Maple dsolve solution
Solving time : 0.027 (sec)
Leaf size : 62
dsolve(x^2*diff(diff(y(x),x),x)-x*(6+x)*diff(y(x),x)+10*y(x) = 0,y(x),
series,x=0)
\[
y = \left (c_{1} x^{3} \left (1+\frac {5}{4} x +\frac {3}{4} x^{2}+\frac {7}{24} x^{3}+\frac {1}{12} x^{4}+\frac {3}{160} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (24 x^{3}+30 x^{4}+18 x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (12-12 x +18 x^{2}+26 x^{3}+x^{4}-9 x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right )\right ) x^{2}
\]
Mathematica DSolve solution
Solving time : 0.04 (sec)
Leaf size : 84
AsymptoticDSolveValue[{x^2*D[y[x],{x,2}]-x*(x+6)*D[y[x],x]+10*y[x]==0,{}},
y[x],{x,0,5}]
\[
y(x)\to c_1 \left (\frac {1}{2} x^5 (5 x+4) \log (x)-\frac {1}{4} x^2 \left (3 x^4-6 x^3-6 x^2+4 x-4\right )\right )+c_2 \left (\frac {x^9}{12}+\frac {7 x^8}{24}+\frac {3 x^7}{4}+\frac {5 x^6}{4}+x^5\right )
\]