2.4.39 Problem 36
Internal
problem
ID
[10202]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
36
Date
solved
:
Monday, December 29, 2025 at 09:55:10 PM
CAS
classification
:
[[_2nd_order, _exact, _linear, _homogeneous]]
\begin{align*}
x \left (x -1\right ) y^{\prime \prime }+3 y^{\prime } x +y&=0 \\
\end{align*}
Series expansion around \(x=0\).
Entering second order ode series solverThe type of the expansion point is first determined. This is
done on the homogeneous part of the ODE.
\[ \left (x^{2}-x \right ) y^{\prime \prime }+3 y^{\prime } x +y = 0 \]
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 {3}{x -1}\\ q(x) &= \frac {1}{x \left (x -1\right )}\\ \end{align*}
Table 2.93: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=\frac {3}{x -1}\) |
| |
|
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
\[ x \left (x -1\right ) y^{\prime \prime }+3 y^{\prime } x +y = 0 \]
Entering second order ode series frobenius solverLet 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 )+3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x +\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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 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 }}3 x^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\
\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 )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 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 )+3 a_{n -1} \left (n +r -1\right )+a_{n -1} = 0
\end{equation}
Solving for \(a_{n}\) from recursive equation (4)
gives \[ a_{n} = \frac {\left (n +r \right ) a_{n -1}}{n +r -1}\tag {4} \]
Which for the root \(r = 1\) becomes \[ a_{n} = \frac {\left (n +1\right ) a_{n -1}}{n}\tag {5} \]
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
\[ a_{1}=\frac {1+r}{r} \]
Which for the root \(r = 1\) becomes \[ a_{1}=2 \]
And the table now
becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {1+r}{r}\) | \(2\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ a_{2}=\frac {2+r}{r} \]
Which for the root \(r = 1\) becomes \[ a_{2}=3 \]
And the table now
becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {1+r}{r}\) | \(2\) |
| | |
| \(a_{2}\) | \(\frac {2+r}{r}\) | \(3\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=\frac {3+r}{r} \]
Which for the root \(r = 1\) becomes \[ a_{3}=4 \]
And the table now
becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {1+r}{r}\) | \(2\) |
| | |
| \(a_{2}\) | \(\frac {2+r}{r}\) | \(3\) |
| | |
| \(a_{3}\) |
\(\frac {3+r}{r}\) |
\(4\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=\frac {4+r}{r} \]
Which for the root \(r = 1\) becomes \[ a_{4}=5 \]
And the table now
becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {1+r}{r}\) | \(2\) |
| | |
| \(a_{2}\) | \(\frac {2+r}{r}\) | \(3\) |
| | |
| \(a_{3}\) | \(\frac {3+r}{r}\) | \(4\) |
| | |
| \(a_{4}\) |
\(\frac {4+r}{r}\) |
\(5\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=\frac {5+r}{r} \]
Which for the root \(r = 1\) becomes \[ a_{5}=6 \]
And the table now
becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {1+r}{r}\) |
\(2\) |
| | |
| \(a_{2}\) |
\(\frac {2+r}{r}\) | \(3\) |
| | |
| \(a_{3}\) | \(\frac {3+r}{r}\) | \(4\) |
| | |
| \(a_{4}\) |
\(\frac {4+r}{r}\) |
\(5\) |
| | |
| \(a_{5}\) |
\(\frac {5+r}{r}\) |
\(6\) |
| | |
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+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+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 {1+r}{r} \end{align*}
Therefore
\begin{align*} \lim _{r\rightarrow r_{2}}\frac {1+r}{r}&= \lim _{r\rightarrow 0}\frac {1+r}{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 \(\left (x^{2}-x \right ) y^{\prime \prime }+3 y^{\prime } x +y = 0\) gives \[
\left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right ) x^{2}-\left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right ) x +3 \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right ) x +C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0
\]
Which can be written as \begin{equation}
\tag{7} \left (\left (y_{1}^{\prime \prime }\left (x \right ) x^{2}+3 y_{1}^{\prime }\left (x \right ) x -y_{1}^{\prime \prime }\left (x \right ) x +y_{1}\left (x \right )\right ) \ln \left (x \right )+\left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) x^{2}-\left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) x +3 y_{1}\left (x \right )\right ) C +\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^{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 +3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) x +\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 \[ \left (x^{2}-x \right ) y_{1}^{\prime \prime }\left (x \right )+3 y_{1}^{\prime }\left (x \right ) x +y_{1}\left (x \right ) = 0 \]
Eq (7) simplifes to \begin{equation}
\tag{8} \left (\left (y_{1}^{\prime \prime }\left (x \right ) x^{2}+3 y_{1}^{\prime }\left (x \right ) x -y_{1}^{\prime \prime }\left (x \right ) x +y_{1}\left (x \right )\right ) \ln \left (x \right )+\left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) x^{2}-\left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) x +3 y_{1}\left (x \right )\right ) C +\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^{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 +3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) x +\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 (x^{2} \ln \left (x \right ) \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{1}} a_{n} \left (n +r_{1}\right ) \left (-1+n +r_{1}\right )\right )+3 x \left (x \ln \left (x \right )+\frac {2 x}{3}-\frac {2}{3}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )+\left (x \ln \left (x \right )+2 x +1\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 )+3 \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 (x^{2} \ln \left (x \right ) \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} a_{n} \left (n +1\right ) n \right )+3 x \left (x \ln \left (x \right )+\frac {2 x}{3}-\frac {2}{3}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} a_{n} \left (n +1\right )\right )+\left (x \ln \left (x \right )+2 x +1\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 )+3 \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 }}C n \,x^{n +1} a_{n} \ln \left (x \right ) \left (n +1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n} a_{n} n \left (n +1\right ) C \ln \left (x \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 C \,x^{n +1} a_{n} \ln \left (x \right ) \left (n +1\right )\right )+\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 }}C \,x^{n +1} a_{n} \ln \left (x \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +1} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n} C \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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 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 }}C n \,x^{n +1} a_{n} \ln \left (x \right ) \left (n +1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}C \left (-2+n \right ) a_{-2+n} \left (n -1\right ) \ln \left (x \right ) x^{n -1} \\
\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n} a_{n} n \left (n +1\right ) C \ln \left (x \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-C \left (n -1\right ) a_{n -1} n \ln \left (x \right ) x^{n -1}\right ) \\
\moverset {\infty }{\munderset {n =0}{\sum }}3 C \,x^{n +1} a_{n} \ln \left (x \right ) \left (n +1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}3 C a_{-2+n} \left (n -1\right ) \ln \left (x \right ) x^{n -1} \\
\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 }}C \,x^{n +1} a_{n} \ln \left (x \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}C a_{-2+n} \ln \left (x \right ) x^{n -1} \\
\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +1} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{-2+n} x^{n -1} \\
\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 }}x^{n} b_{n} n \left (n -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} \left (n -1\right ) \left (-2+n \right ) x^{n -1} \\
\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n} b_{n} n &= \moverset {\infty }{\munderset {n =1}{\sum }}3 b_{n -1} \left (n -1\right ) x^{n -1} \\
\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 }}C \left (-2+n \right ) a_{-2+n} \left (n -1\right ) \ln \left (x \right ) x^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-C \left (n -1\right ) a_{n -1} n \ln \left (x \right ) x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}3 C a_{-2+n} \left (n -1\right ) \ln \left (x \right ) x^{n -1}\right )+\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 =2}{\sum }}C a_{-2+n} \ln \left (x \right ) x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{-2+n} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}C a_{n -1} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} \left (n -1\right ) \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 )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 b_{n -1} \left (n -1\right ) 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 \[ 4 \left (a_{0}-\frac {a_{1}}{2}\right ) C \ln \left (x \right )+\left (4 a_{0}-3 a_{1}\right ) C +4 b_{1}-2 b_{2} = 0 \]
Which when replacing the above values found already for \(b_{n}\) and
the values found earlier for \(a_{n}\) and for \(C\), gives \[ -2-2 b_{2} = 0 \]
Solving the above for \(b_{2}\) gives \[ b_{2}=-1 \]
For \(n=3\), Eq
(2B) gives \[ 9 \left (a_{1}-\frac {2 a_{2}}{3}\right ) C \ln \left (x \right )+\left (6 a_{1}-5 a_{2}\right ) C +9 b_{2}-6 b_{3} = 0 \]
Which when replacing the above values found already for \(b_{n}\) and the values
found earlier for \(a_{n}\) and for \(C\), gives \[ -12-6 b_{3} = 0 \]
Solving the above for \(b_{3}\) gives \[ b_{3}=-2 \]
For \(n=4\), Eq (2B) gives \[ 16 \left (a_{2}-\frac {3 a_{3}}{4}\right ) C \ln \left (x \right )+\left (8 a_{2}-7 a_{3}\right ) C +16 b_{3}-12 b_{4} = 0 \]
Which when replacing the above values found already for \(b_{n}\) and the values found earlier
for \(a_{n}\) and for \(C\), gives \[ -36-12 b_{4} = 0 \]
Solving the above for \(b_{4}\) gives \[ b_{4}=-3 \]
For \(n=5\), Eq (2B) gives \[ 25 \left (a_{3}-\frac {4 a_{4}}{5}\right ) C \ln \left (x \right )+\left (10 a_{3}-9 a_{4}\right ) C +25 b_{4}-20 b_{5} = 0 \]
Which when
replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for
\(C\), gives \[ -80-20 b_{5} = 0 \]
Solving the above for \(b_{5}\) gives \[ b_{5}=-4 \]
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 \[
y_{2}\left (x \right )= 1 \left (x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-x^{2}-2 x^{3}-3 x^{4}-4 x^{5}+O\left (x^{6}\right )
\]
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+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right ) + c_2 \left (1 \left (x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-x^{2}-2 x^{3}-3 x^{4}-4 x^{5}+O\left (x^{6}\right )\right ) \\
\end{align*}
Hence the final solution is
\begin{align*}
y &= y_h \\
&= c_1 x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right )+c_2 \left (x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-x^{2}-2 x^{3}-3 x^{4}-4 x^{5}+O\left (x^{6}\right )\right ) \\
\end{align*}
2.4.39.1 ✓ Maple. Time used: 0.014 (sec). Leaf size: 60
Order:=6;
ode:=x*(x-1)*diff(diff(y(x),x),x)+3*x*diff(y(x),x)+y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
\[
y = \ln \left (x \right ) \left (x +2 x^{2}+3 x^{3}+4 x^{4}+5 x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_2 +c_1 x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (1+3 x +5 x^{2}+7 x^{3}+9 x^{4}+11 x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_2
\]
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)]
<- linear_1 successful
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (x -1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+3 x \left (\frac {d}{d x}y \left (x \right )\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 {y \left (x \right )}{x \left (x -1\right )}-\frac {3 \left (\frac {d}{d x}y \left (x \right )\right )}{x -1} \\ \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 {3 \left (\frac {d}{d x}y \left (x \right )\right )}{x -1}+\frac {y \left (x \right )}{x \left (x -1\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {3}{x -1}, P_{3}\left (x \right )=\frac {1}{x \left (x -1\right )}\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}}}=0 \\ {} & \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}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (x -1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+3 x \left (\frac {d}{d x}y \left (x \right )\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 \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^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\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 -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (-1+r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +r +1\right ) \left (k +r \right )+a_{k} \left (k +r +1\right )^{2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r +1\right ) \left (-a_{k +1} \left (k +r \right )+a_{k} \left (k +r +1\right )\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +r +1\right )}{k +r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +1\right )}{k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=\frac {a_{k} \left (k +1\right )}{k}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +2\right )}{k +1} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +1}=\frac {a_{k} \left (k +2\right )}{k +1}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +1}\right ), a_{k +1}=\frac {a_{k} \left (k +1\right )}{k}, b_{k +1}=\frac {b_{k} \left (k +2\right )}{k +1}\right ] \end {array} \]
2.4.39.2 ✓ Mathematica. Time used: 0.027 (sec). Leaf size: 63
ode=x*(x-1)*D[y[x],{x,2}] +3*x*D[y[x],x]+y[x] == 0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
\[
y(x)\to c_1 \left (x^4+x^3+x^2+\left (4 x^3+3 x^2+2 x+1\right ) x \log (x)+x+1\right )+c_2 \left (5 x^5+4 x^4+3 x^3+2 x^2+x\right )
\]
2.4.39.3 ✓ Sympy. Time used: 0.297 (sec). Leaf size: 37
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*(x - 1)*Derivative(y(x), (x, 2)) + 3*x*Derivative(y(x), x) + y(x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
\[
y{\left (x \right )} = C_{2} x \left (\frac {27 x^{4}}{40} - \frac {9 x^{3}}{8} + \frac {3 x^{2}}{2} - \frac {3 x}{2} + 1\right ) + C_{1} + O\left (x^{6}\right )
\]