2.4.51 problem 48
Internal
problem
ID
[8366]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
48
Date
solved
:
Sunday, November 10, 2024 at 03:39:30 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} \left (x^{2}-x \right ) y^{\prime \prime }-x y^{\prime }+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.
\[ \left (x^{2}-x \right ) y^{\prime \prime }-x y^{\prime }+y = 0 \]
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 -1}\\ q(x) &= \frac {1}{x \left (x -1\right )}\\ \end{align*}
Table 2.97: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=-\frac {1}{x -1}\) |
| |
|
singularity | type |
| |
| \(x = 1\) | \(\text {``regular''}\) |
| |
| |
| \(q(x)=\frac {1}{x \left (x -1\right )}\) |
| |
|
singularity | type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| \(x = 1\) |
\(\text {``regular''}\) |
| |
Combining everything together gives the following summary of singularities for the ode
as
Regular singular points : \([1, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to
be
\[ x \left (x -1\right ) y^{\prime \prime }-x y^{\prime }+y = 0 \]
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 \left (x -1\right ) \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 (\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 -1} 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 }}a_{n} x^{n +r}\right ) = 0
\end{equation}
The next step is to
make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation
term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and
the corresponding index gives
\begin{align*}
\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1} \\
\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{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 -1}\right ) \\
\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1} \\
\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 \(n +r -1\).
\begin{equation}
\tag{2B} \left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r -1} 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 -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1}\right ) = 0
\end{equation}
The indicial
equation is obtained from \(n = 0\). From Eq (2B) this gives
\[ -x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right ) = 0 \]
When \(n = 0\) the above becomes
\[ -x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \]
Or
\[ -x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \]
Since \(a_{0}\neq 0\) then the above simplifies to
\[ -x^{-1+r} r \left (-1+r \right ) = 0 \]
Since the above is true for all \(x\) then the
indicial equation becomes
\[ -r \left (-1+r \right ) = 0 \]
Solving for \(r\) gives the roots of the indicial equation as
\begin{align*} r_1 &= 1\\ r_2 &= 0 \end{align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes
\[ -x^{-1+r} r \left (-1+r \right ) = 0 \]
Solving for \(r\) gives the roots of the indicial equation
as \([1, 0]\).
Since \(r_1 - r_2 = 1\) 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 \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 )+\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 +1}\\ 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}\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 -1} \left (n +r -1\right ) \left (n +r -2\right )-a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -1} \left (n +r -1\right )+a_{n -1} = 0
\end{equation}
Solving for \(a_{n}\) from
recursive equation (4) gives
\[ a_{n} = \frac {a_{n -1} \left (n^{2}+2 n r +r^{2}-4 n -4 r +4\right )}{\left (n +r \right ) \left (n +r -1\right )}\tag {4} \]
Which for the root \(r = 1\) becomes
\[ a_{n} = \frac {a_{n -1} \left (n -1\right )^{2}}{\left (n +1\right ) n}\tag {5} \]
At this point, it is a good idea to
keep track of \(a_{n}\) in a table both before substituting \(r = 1\) 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
\[ a_{1}=\frac {\left (-1+r \right )^{2}}{\left (1+r \right ) r} \]
Which for the root \(r = 1\) becomes
\[ a_{1}=0 \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {\left (-1+r \right )^{2}}{\left (1+r \right ) r}\) | \(0\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ a_{2}=\frac {\left (-1+r \right )^{2} r}{\left (1+r \right )^{2} \left (2+r \right )} \]
Which for the root \(r = 1\) becomes
\[ a_{2}=0 \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {\left (-1+r \right )^{2}}{\left (1+r \right ) r}\) | \(0\) |
| | |
| \(a_{2}\) | \(\frac {\left (-1+r \right )^{2} r}{\left (1+r \right )^{2} \left (2+r \right )}\) | \(0\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=\frac {\left (-1+r \right )^{2} r}{\left (3+r \right ) \left (2+r \right )^{2}} \]
Which for the root \(r = 1\) becomes
\[ a_{3}=0 \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {\left (-1+r \right )^{2}}{\left (1+r \right ) r}\) | \(0\) |
| | |
| \(a_{2}\) | \(\frac {\left (-1+r \right )^{2} r}{\left (1+r \right )^{2} \left (2+r \right )}\) | \(0\) |
| | |
| \(a_{3}\) |
\(\frac {\left (-1+r \right )^{2} r}{\left (3+r \right ) \left (2+r \right )^{2}}\) |
\(0\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=\frac {\left (-1+r \right )^{2} r}{\left (4+r \right ) \left (3+r \right )^{2}} \]
Which for the root \(r = 1\) becomes
\[ a_{4}=0 \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {\left (-1+r \right )^{2}}{\left (1+r \right ) r}\) | \(0\) |
| | |
| \(a_{2}\) | \(\frac {\left (-1+r \right )^{2} r}{\left (1+r \right )^{2} \left (2+r \right )}\) | \(0\) |
| | |
| \(a_{3}\) | \(\frac {\left (-1+r \right )^{2} r}{\left (3+r \right ) \left (2+r \right )^{2}}\) | \(0\) |
| | |
| \(a_{4}\) |
\(\frac {\left (-1+r \right )^{2} r}{\left (4+r \right ) \left (3+r \right )^{2}}\) |
\(0\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=\frac {\left (-1+r \right )^{2} r}{\left (5+r \right ) \left (4+r \right )^{2}} \]
Which for the root \(r = 1\) becomes
\[ a_{5}=0 \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {\left (-1+r \right )^{2}}{\left (1+r \right ) r}\) |
\(0\) |
| | |
| \(a_{2}\) |
\(\frac {\left (-1+r \right )^{2} r}{\left (1+r \right )^{2} \left (2+r \right )}\) | \(0\) |
| | |
| \(a_{3}\) | \(\frac {\left (-1+r \right )^{2} r}{\left (3+r \right ) \left (2+r \right )^{2}}\) | \(0\) |
| | |
| \(a_{4}\) |
\(\frac {\left (-1+r \right )^{2} r}{\left (4+r \right ) \left (3+r \right )^{2}}\) |
\(0\) |
| | |
| \(a_{5}\) |
\(\frac {\left (-1+r \right )^{2} r}{\left (5+r \right ) \left (4+r \right )^{2}}\) |
\(0\) |
| | |
Using the above table, then the solution \(y_{1}\left (x \right )\) is
\begin{align*} y_{1}\left (x \right )&= x \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 \left (1+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 \(N=1\).
Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{1}\left (r \right )\). 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_{1} \\ &= \frac {\left (-1+r \right )^{2}}{\left (1+r \right ) r} \end{align*}
Therefore
\begin{align*} \lim _{r\rightarrow r_{2}}\frac {\left (-1+r \right )^{2}}{\left (1+r \right ) r}&= \lim _{r\rightarrow 0}\frac {\left (-1+r \right )^{2}}{\left (1+r \right ) r}\\ &= \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 \(x \left (x -1\right ) y^{\prime \prime }-x y^{\prime }+y = 0\) gives
\[
x \left (x -1\right ) \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )-x \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0
\]
Which can be written
as
\begin{equation}
\tag{7} \left (\left (x \left (x -1\right ) y_{1}^{\prime \prime }\left (x \right )-y_{1}^{\prime }\left (x \right ) x +y_{1}\left (x \right )\right ) \ln \left (x \right )+x \left (x -1\right ) \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 \left (x -1\right ) \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 (\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
\[ x \left (x -1\right ) y_{1}^{\prime \prime }\left (x \right )-y_{1}^{\prime }\left (x \right ) x +y_{1}\left (x \right ) = 0 \]
Eq (7) simplifes to
\begin{equation}
\tag{8} \left (x \left (x -1\right ) \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 \left (x -1\right ) \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 (\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} \frac {\left (2 x \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )+\left (1-2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C}{x}+\frac {x^{2} \left (x -1\right ) \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 )-\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) x^{2}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x}{x} = 0
\end{equation}
Since \(r_{1} = 1\) and \(r_{2} = 0\) then the above becomes
\begin{equation}
\tag{10} \frac {\left (2 x \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} a_{n} \left (n +1\right )\right )+\left (1-2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +1}\right )\right ) C}{x}+\frac {x^{2} \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n} b_{n} n \left (n -1\right )\right )-\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} n \right ) x^{2}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) x}{x} = 0
\end{equation}
Which simplifies to
\begin{equation}
\tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +1} a_{n} \left (n +1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n} a_{n} \left (n +1\right ) C \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n} C \right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{n +1} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} b_{n} n \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-n \,x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n} b_{n} n \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) = 0
\end{equation}
The
next step is to make all powers of \(x\) be \(n -1\) in each summation term. Going over each
summation term above with power of \(x\) in it which is not already \(x^{n -1}\) and adjusting the
power and the corresponding index gives
\begin{align*}
\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +1} a_{n} \left (n +1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{-2+n} \left (n -1\right ) x^{n -1} \\
\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n} a_{n} \left (n +1\right ) C \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 C a_{n -1} n \,x^{n -1}\right ) \\
\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n} C &= \moverset {\infty }{\munderset {n =1}{\sum }}C a_{n -1} x^{n -1} \\
\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{n +1} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 C a_{-2+n} x^{n -1}\right ) \\
\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} b_{n} n \left (n -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} \left (-2+n \right ) x^{n -1} \\
\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n} b_{n} n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-\left (n -1\right ) b_{n -1} x^{n -1}\right ) \\
\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} x^{n -1} \\
\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
\(n -1\).
\begin{equation}
\tag{2B} \left (\moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{-2+n} \left (n -1\right ) x^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 C a_{n -1} n \,x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}C a_{n -1} x^{n -1}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 C a_{-2+n} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} \left (-2+n \right ) x^{n -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-n \,x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-\left (n -1\right ) b_{n -1} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} x^{n -1}\right ) = 0
\end{equation}
For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=N\), where \(N=1\) which is the
difference between the two roots, we are free to choose \(b_{1} = 0\). Hence for \(n=1\), Eq (2B) gives
\[ -C +1 = 0 \]
Which is solved for \(C\). Solving for \(C\) gives
\[ C=1 \]
For \(n=2\), Eq (2B) gives
\[ -3 C a_{1}-2 b_{2} = 0 \]
Which when replacing
the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\),
gives
\[ -2 b_{2} = 0 \]
Solving the above for \(b_{2}\) gives
\[ b_{2}=0 \]
For \(n=3\), Eq (2B) gives
\[ \left (2 a_{1}-5 a_{2}\right ) C +b_{2}-6 b_{3} = 0 \]
Which when replacing
the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\),
gives
\[ -6 b_{3} = 0 \]
Solving the above for \(b_{3}\) gives
\[ b_{3}=0 \]
For \(n=4\), Eq (2B) gives
\[ \left (4 a_{2}-7 a_{3}\right ) C +4 b_{3}-12 b_{4} = 0 \]
Which when replacing
the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\),
gives
\[ -12 b_{4} = 0 \]
Solving the above for \(b_{4}\) gives
\[ b_{4}=0 \]
For \(n=5\), Eq (2B) gives
\[ \left (6 a_{3}-9 a_{4}\right ) C +9 b_{4}-20 b_{5} = 0 \]
Which when replacing the
above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives
\[ -20 b_{5} = 0 \]
Solving the above for \(b_{5}\) gives
\[ b_{5}=0 \]
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 \(C=1\) and all \(b_{n}\), then the second
solution becomes
\[
y_{2}\left (x \right )= 1\eslowast \left (x \left (1+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1+O\left (x^{6}\right )
\]
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 \left (1+O\left (x^{6}\right )\right ) + c_2 \left (1\eslowast \left (x \left (1+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1+O\left (x^{6}\right )\right ) \\
\end{align*}
Hence the final solution is
\begin{align*}
y &= y_h \\
&= c_1 x \left (1+O\left (x^{6}\right )\right )+c_2 \left (x \left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )+1+O\left (x^{6}\right )\right ) \\
\end{align*}
Maple step by step solution
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 : 0.025 (sec)
Leaf size : 34
dsolve((x^2-x)*diff(diff(y(x),x),x)-diff(y(x),x)*x+y(x) = 0,y(x),
series,x=0)
\[
y = c_{1} x \left (1+\operatorname {O}\left (x^{6}\right )\right )+\left (x +\operatorname {O}\left (x^{6}\right )\right ) \ln \left (x \right ) c_{2} +\left (1-x +\operatorname {O}\left (x^{6}\right )\right ) c_{2}
\]
Mathematica DSolve solution
Solving time : 0.049 (sec)
Leaf size : 20
AsymptoticDSolveValue[{(x^2-x)*D[y[x],{x,2}]-x*D[y[x],x]+y[x]==0,{}},
y[x],{x,0,5}]
\[
y(x)\to c_2 x+c_1 (-3 x+x \log (x)+1)
\]