2.4.40 problem 37
Internal
problem
ID
[8355]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
37
Date
solved
:
Sunday, November 10, 2024 at 03:39:15 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} x^{2} \left (x^{2}-2 x +1\right ) y^{\prime \prime }-x \left (3+x \right ) y^{\prime }+\left (4+x \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.
\[ \left (x^{4}-2 x^{3}+x^{2}\right ) y^{\prime \prime }+\left (-x^{2}-3 x \right ) y^{\prime }+\left (4+x \right ) y = 0 \]
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 {3+x}{x \left (x -1\right )^{2}}\\ q(x) &= \frac {4+x}{x^{2} \left (x -1\right )^{2}}\\ \end{align*}
Table 2.87: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=-\frac {3+x}{x \left (x -1\right )^{2}}\) |
| |
|
singularity | type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| \(x = 1\) |
\(\text {``irregular''}\) |
| |
| |
| \(q(x)=\frac {4+x}{x^{2} \left (x -1\right )^{2}}\) |
| |
|
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 : \([0, \infty ]\)
Irregular singular points : \([1]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to
be
\[ x^{2} \left (x^{2}-2 x +1\right ) y^{\prime \prime }+\left (-x^{2}-3 x \right ) y^{\prime }+\left (4+x \right ) y = 0 \]
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 (x^{2}-2 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 )+\left (-x^{2}-3 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (4+x \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 +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \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 (-3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\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} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\
\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right ) \\
\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 ) \\
\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} 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 =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\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 (-3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r}\right ) = 0
\end{equation}
The indicial
equation is obtained from \(n = 0\). From Eq (2B) this gives
\[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-3 x^{n +r} a_{n} \left (n +r \right )+4 a_{n} x^{n +r} = 0 \]
When \(n = 0\) the above becomes
\[ x^{r} a_{0} r \left (-1+r \right )-3 x^{r} a_{0} r +4 a_{0} x^{r} = 0 \]
Or
\[ \left (x^{r} r \left (-1+r \right )-3 x^{r} r +4 x^{r}\right ) a_{0} = 0 \]
Since \(a_{0}\neq 0\) then the above simplifies to
\[ \left (r -2\right )^{2} x^{r} = 0 \]
Since the above is true for all \(x\) then the
indicial equation becomes
\[ \left (r -2\right )^{2} = 0 \]
Solving for \(r\) gives the roots of the indicial equation as
\begin{align*} r_1 &= 2\\ r_2 &= 2 \end{align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes
\[ \left (r -2\right )^{2} x^{r} = 0 \]
Solving for \(r\) gives the roots of the indicial equation
as \([2, 2]\).
Since the root of the indicial equation is repeated, then we can construct two linearly
independent solutions. The first solution has the form
\begin{align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\tag {1A} \end{align*}
Now the second solution \(y_{2}\) is found using
\begin{align*} y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right )\tag {1B} \end{align*}
Then the general solution will be
\[ y = c_1 y_{1}\left (x \right )+c_2 y_{2}\left (x \right ) \]
In Eq (1B) the sum starts from 1 and not zero. In Eq
(1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_1, c_2\}\) are two arbitray
constants of integration which can be found from initial conditions. Using the value of the
indicial root found earlier, \(r = 2\), Eqs (1A,1B) become
\begin{align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +2}\\ y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +2}\right ) \end{align*}
We start by finding the first solution \(y_{1}\left (x \right )\). Eq (2B) derived above is now used to find all \(a_{n}\)
coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation.
\(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives
\[ a_{1} = \frac {2 r +1}{-1+r} \]
For \(2\le n\) the recursive
equation is
\begin{equation}
\tag{3} a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )-2 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 )-3 a_{n} \left (n +r \right )+4 a_{n}+a_{n -1} = 0
\end{equation}
Solving for \(a_{n}\) from recursive equation (4) gives
\[ a_{n} = -\frac {n a_{n -2}-2 n a_{n -1}+r a_{n -2}-2 r a_{n -1}-3 a_{n -2}+a_{n -1}}{n +r -2}\tag {4} \]
Which for the root \(r = 2\)
becomes
\[ a_{n} = \frac {\left (-a_{n -2}+2 a_{n -1}\right ) n +a_{n -2}+3 a_{n -1}}{n}\tag {5} \]
At this point, it is a good idea to keep track of \(a_{n}\) in a table both before
substituting \(r = 2\) and after as more terms are found using the above recursive equation.
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {2 r +1}{-1+r}\) | \(5\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ a_{2}=\frac {3 r^{2}+10 r +2}{r \left (-1+r \right )} \]
Which for the root \(r = 2\) becomes
\[ a_{2}=17 \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {2 r +1}{-1+r}\) | \(5\) |
| | |
| \(a_{2}\) | \(\frac {3 r^{2}+10 r +2}{r \left (-1+r \right )}\) | \(17\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=\frac {4 r^{3}+34 r^{2}+54 r +10}{r^{3}-r} \]
Which for the root \(r = 2\) becomes
\[ a_{3}={\frac {143}{3}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {2 r +1}{-1+r}\) | \(5\) |
| | |
| \(a_{2}\) | \(\frac {3 r^{2}+10 r +2}{r \left (-1+r \right )}\) | \(17\) |
| | |
| \(a_{3}\) |
\(\frac {4 r^{3}+34 r^{2}+54 r +10}{r^{3}-r}\) |
\(\frac {143}{3}\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=\frac {5 r^{4}+80 r^{3}+321 r^{2}+384 r +68}{\left (2+r \right ) r \left (r^{2}-1\right )} \]
Which for the root \(r = 2\) becomes
\[ a_{4}={\frac {355}{3}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {2 r +1}{-1+r}\) | \(5\) |
| | |
| \(a_{2}\) | \(\frac {3 r^{2}+10 r +2}{r \left (-1+r \right )}\) | \(17\) |
| | |
| \(a_{3}\) | \(\frac {4 r^{3}+34 r^{2}+54 r +10}{r^{3}-r}\) | \(\frac {143}{3}\) |
| | |
| \(a_{4}\) |
\(\frac {5 r^{4}+80 r^{3}+321 r^{2}+384 r +68}{\left (2+r \right ) r \left (r^{2}-1\right )}\) |
\(\frac {355}{3}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=\frac {6 r^{5}+155 r^{4}+1156 r^{3}+3295 r^{2}+3336 r +572}{r^{5}+5 r^{4}+5 r^{3}-5 r^{2}-6 r} \]
Which for the root \(r = 2\) becomes
\[ a_{5}={\frac {4043}{15}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {2 r +1}{-1+r}\) |
\(5\) |
| | |
| \(a_{2}\) |
\(\frac {3 r^{2}+10 r +2}{r \left (-1+r \right )}\) | \(17\) |
| | |
| \(a_{3}\) | \(\frac {4 r^{3}+34 r^{2}+54 r +10}{r^{3}-r}\) | \(\frac {143}{3}\) |
| | |
| \(a_{4}\) |
\(\frac {5 r^{4}+80 r^{3}+321 r^{2}+384 r +68}{\left (2+r \right ) r \left (r^{2}-1\right )}\) |
\(\frac {355}{3}\) |
| | |
| \(a_{5}\) |
\(\frac {6 r^{5}+155 r^{4}+1156 r^{3}+3295 r^{2}+3336 r +572}{r^{5}+5 r^{4}+5 r^{3}-5 r^{2}-6 r}\) |
\(\frac {4043}{15}\) |
| | |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is
\begin{align*}
y_{1}\left (x \right )&= x^{2} \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^{2} \left (17 x^{2}+5 x +1+\frac {143 x^{3}}{3}+\frac {355 x^{4}}{3}+\frac {4043 x^{5}}{15}+O\left (x^{6}\right )\right ) \\
\end{align*}
Now the second solution is found. The
second solution is given by
\[ y_{2}\left (x \right ) = y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right ) \]
Where \(b_{n}\) is found using
\[ b_{n} = \frac {d}{d r}a_{n ,r} \]
And the above is then evaluated at \(r = 2\). The
above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| | | | |
| \(n\) |
\(b_{n ,r}\) |
\(a_{n}\) |
\(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) |
\(b_{n}\left (r =2\right )\) |
| | | | |
| \(b_{0}\) |
\(1\) |
\(1\) |
N/A since \(b_{n}\) starts from 1 |
N/A |
| | | | |
| \(b_{1}\) | \(\frac {2 r +1}{-1+r}\) | \(5\) | \(-\frac {3}{\left (-1+r \right )^{2}}\) | \(-3\) |
| | | | |
| \(b_{2}\) | \(\frac {3 r^{2}+10 r +2}{r \left (-1+r \right )}\) | \(17\) | \(\frac {-13 r^{2}-4 r +2}{r^{2} \left (-1+r \right )^{2}}\) | \(-{\frac {29}{2}}\) |
| | | | |
| \(b_{3}\) | \(\frac {4 r^{3}+34 r^{2}+54 r +10}{r^{3}-r}\) | \(\frac {143}{3}\) | \(\frac {-34 r^{4}-116 r^{3}-64 r^{2}+10}{r^{2} \left (r^{2}-1\right )^{2}}\) | \(-{\frac {859}{18}}\) |
| | | | |
| \(b_{4}\) |
\(\frac {5 r^{4}+80 r^{3}+321 r^{2}+384 r +68}{\left (2+r \right ) r \left (r^{2}-1\right )}\) |
\(\frac {355}{3}\) |
\(\frac {-70 r^{6}-652 r^{5}-1904 r^{4}-2128 r^{3}-666 r^{2}+136 r +136}{\left (2+r \right )^{2} r^{2} \left (r^{2}-1\right )^{2}}\) |
\(-{\frac {4693}{36}}\) |
| | | | |
| \(b_{5}\) |
\(\frac {6 r^{5}+155 r^{4}+1156 r^{3}+3295 r^{2}+3336 r +572}{r^{5}+5 r^{4}+5 r^{3}-5 r^{2}-6 r}\) |
\(\frac {4043}{15}\) |
\(\frac {-125 r^{8}-2252 r^{7}-14980 r^{6}-47988 r^{5}-77945 r^{4}-58672 r^{3}-11670 r^{2}+5720 r +3432}{r^{2} \left (r^{4}+5 r^{3}+5 r^{2}-5 r -6\right )^{2}}\) |
\(-{\frac {285181}{900}}\) |
| | | | |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is
\begin{align*}
y_{2}\left (x \right )&=y_{1}\left (x \right ) \ln \left (x \right )+b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \\
&= x^{2} \left (17 x^{2}+5 x +1+\frac {143 x^{3}}{3}+\frac {355 x^{4}}{3}+\frac {4043 x^{5}}{15}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{2} \left (-3 x -\frac {29 x^{2}}{2}-\frac {859 x^{3}}{18}-\frac {4693 x^{4}}{36}-\frac {285181 x^{5}}{900}+O\left (x^{6}\right )\right ) \\
\end{align*}
Therefore the
homogeneous solution is
\begin{align*}
y_h(x) &= c_1 y_{1}\left (x \right )+c_2 y_{2}\left (x \right ) \\
&= c_1 \,x^{2} \left (17 x^{2}+5 x +1+\frac {143 x^{3}}{3}+\frac {355 x^{4}}{3}+\frac {4043 x^{5}}{15}+O\left (x^{6}\right )\right ) + c_2 \left (x^{2} \left (17 x^{2}+5 x +1+\frac {143 x^{3}}{3}+\frac {355 x^{4}}{3}+\frac {4043 x^{5}}{15}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{2} \left (-3 x -\frac {29 x^{2}}{2}-\frac {859 x^{3}}{18}-\frac {4693 x^{4}}{36}-\frac {285181 x^{5}}{900}+O\left (x^{6}\right )\right )\right ) \\
\end{align*}
Hence the final solution is
\begin{align*}
y &= y_h \\
&= c_1 \,x^{2} \left (17 x^{2}+5 x +1+\frac {143 x^{3}}{3}+\frac {355 x^{4}}{3}+\frac {4043 x^{5}}{15}+O\left (x^{6}\right )\right )+c_2 \left (x^{2} \left (17 x^{2}+5 x +1+\frac {143 x^{3}}{3}+\frac {355 x^{4}}{3}+\frac {4043 x^{5}}{15}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{2} \left (-3 x -\frac {29 x^{2}}{2}-\frac {859 x^{3}}{18}-\frac {4693 x^{4}}{36}-\frac {285181 x^{5}}{900}+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 (x^{2}-2 x +1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-x \left (x +3\right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (x +4\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 +4\right ) y \left (x \right )}{x^{2} \left (x^{2}-2 x +1\right )}+\frac {\left (x +3\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (x^{2}-2 x +1\right )} \\ \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 (x +3\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (x^{2}-2 x +1\right )}+\frac {\left (x +4\right ) y \left (x \right )}{x^{2} \left (x^{2}-2 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 {x +3}{x \left (x^{2}-2 x +1\right )}, P_{3}\left (x \right )=\frac {x +4}{x^{2} \left (x^{2}-2 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}}}=-3 \\ {} & \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}}}=4 \\ {} & \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^{2} \left (x^{2}-2 x +1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-x \left (x +3\right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (x +4\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..1 \\ {} & {} & 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^{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^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & 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} \left (-2+r \right )^{2} x^{r}+\left (a_{1} \left (-1+r \right )^{2}-a_{0} \left (1+2 r \right ) \left (-1+r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +r -2\right )^{2}-a_{k -1} \left (2 k -1+2 r \right ) \left (k +r -2\right )+a_{k -2} \left (k +r -2\right ) \left (k -3+r \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 )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =2 \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (-1+r \right )^{2}-a_{0} \left (1+2 r \right ) \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=\frac {a_{0} \left (1+2 r \right )}{-1+r} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r -2\right ) \left (\left (a_{k}+a_{k -2}-2 a_{k -1}\right ) k +\left (a_{k}+a_{k -2}-2 a_{k -1}\right ) r -2 a_{k}-3 a_{k -2}+a_{k -1}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (k +r \right ) \left (\left (a_{k +2}+a_{k}-2 a_{k +1}\right ) \left (k +2\right )+\left (a_{k +2}+a_{k}-2 a_{k +1}\right ) r -2 a_{k +2}-3 a_{k}+a_{k +1}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k a_{k}-2 k a_{k +1}+r a_{k}-2 r a_{k +1}-a_{k}-3 a_{k +1}}{k +r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +2}=-\frac {k a_{k}-2 k a_{k +1}+a_{k}-7 a_{k +1}}{k +2} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +2}=-\frac {k a_{k}-2 k a_{k +1}+a_{k}-7 a_{k +1}}{k +2}, a_{1}=5 a_{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
A Liouvillian solution exists
Reducible group (found an exponential solution)
Group is reducible, not completely reducible
<- Kovacics algorithm successful`
Maple dsolve solution
Solving time : 0.020 (sec)
Leaf size : 48
dsolve(x^2*(x^2-2*x+1)*diff(diff(y(x),x),x)-x*(x+3)*diff(y(x),x)+(x+4)*y(x) = 0,y(x),
series,x=0)
\[
y = \left (\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1+5 x +17 x^{2}+\frac {143}{3} x^{3}+\frac {355}{3} x^{4}+\frac {4043}{15} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (\left (-3\right ) x -\frac {29}{2} x^{2}-\frac {859}{18} x^{3}-\frac {4693}{36} x^{4}-\frac {285181}{900} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} \right ) x^{2}
\]
Mathematica DSolve solution
Solving time : 0.024 (sec)
Leaf size : 118
AsymptoticDSolveValue[{x^2*(1-2*x+x^2)*D[y[x],{x,2}] -x*(3+x)*D[y[x],x]+(4+x)*y[x] == 0,{}},
y[x],{x,0,5}]
\[
y(x)\to c_1 \left (\frac {4043 x^5}{15}+\frac {355 x^4}{3}+\frac {143 x^3}{3}+17 x^2+5 x+1\right ) x^2+c_2 \left (\left (-\frac {285181 x^5}{900}-\frac {4693 x^4}{36}-\frac {859 x^3}{18}-\frac {29 x^2}{2}-3 x\right ) x^2+\left (\frac {4043 x^5}{15}+\frac {355 x^4}{3}+\frac {143 x^3}{3}+17 x^2+5 x+1\right ) x^2 \log (x)\right )
\]