2.4.38 problem 35
Internal
problem
ID
[8353]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
35
Date
solved
:
Sunday, November 10, 2024 at 03:39:12 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} x y^{\prime \prime }+\left (1+x \right ) y^{\prime }+2 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.
\[ x y^{\prime \prime }+\left (1+x \right ) y^{\prime }+2 y = 0 \]
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}{x}\\ q(x) &= \frac {2}{x}\\ \end{align*}
Table 2.85: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=\frac {1+x}{x}\) |
| |
|
singularity | type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| |
| \(q(x)=\frac {2}{x}\) |
| |
|
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
\[ x y^{\prime \prime }+\left (1+x \right ) y^{\prime }+2 y = 0 \]
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 (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (1+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+2 \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 -1} 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 )\right )+\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 }}2 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 ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\
\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}2 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 =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} x^{n +r -1}\right ) = 0
\end{equation}
The indicial
equation is obtained from \(n = 0\). From Eq (2B) this gives
\[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} x^{n +r -1} = 0 \]
When \(n = 0\) the above becomes
\[ x^{-1+r} a_{0} r \left (-1+r \right )+r a_{0} x^{-1+r} = 0 \]
Or
\[ \left (x^{-1+r} r \left (-1+r \right )+r \,x^{-1+r}\right ) a_{0} = 0 \]
Since \(a_{0}\neq 0\) then the above simplifies to
\[ x^{-1+r} r^{2} = 0 \]
Since the above is true for all \(x\) then the
indicial equation becomes
\[ r^{2} = 0 \]
Solving for \(r\) gives the roots of the indicial equation as
\begin{align*} r_1 &= 0\\ r_2 &= 0 \end{align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes
\[ x^{-1+r} r^{2} = 0 \]
Solving for \(r\) gives the roots of the indicial equation
as \([0, 0]\).
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. 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
\(a_{0} = 1\). 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 )+a_{n} \left (n +r \right )+2 a_{n -1} = 0
\end{equation}
Solving for \(a_{n}\) from recursive equation (4) gives
\[ a_{n} = -\frac {a_{n -1} \left (n +r +1\right )}{n^{2}+2 n r +r^{2}}\tag {4} \]
Which for the
root \(r = 0\) becomes
\[ a_{n} = -\frac {a_{n -1} \left (n +1\right )}{n^{2}}\tag {5} \]
At this point, it is a good idea to keep track of \(a_{n}\) in a table both before
substituting \(r = 0\) 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
\[ a_{1}=\frac {-2-r}{\left (r +1\right )^{2}} \]
Which for the root \(r = 0\) becomes
\[ a_{1}=-2 \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {-2-r}{\left (r +1\right )^{2}}\) | \(-2\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ a_{2}=\frac {3+r}{\left (2+r \right ) \left (r +1\right )^{2}} \]
Which for the root \(r = 0\) becomes
\[ a_{2}={\frac {3}{2}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {-2-r}{\left (r +1\right )^{2}}\) | \(-2\) |
| | |
| \(a_{2}\) | \(\frac {3+r}{\left (2+r \right ) \left (r +1\right )^{2}}\) | \(\frac {3}{2}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=\frac {-4-r}{\left (3+r \right ) \left (2+r \right ) \left (r +1\right )^{2}} \]
Which for the root \(r = 0\) becomes
\[ a_{3}=-{\frac {2}{3}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {-2-r}{\left (r +1\right )^{2}}\) | \(-2\) |
| | |
| \(a_{2}\) | \(\frac {3+r}{\left (2+r \right ) \left (r +1\right )^{2}}\) | \(\frac {3}{2}\) |
| | |
| \(a_{3}\) |
\(\frac {-4-r}{\left (3+r \right ) \left (2+r \right ) \left (r +1\right )^{2}}\) |
\(-{\frac {2}{3}}\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=\frac {5+r}{\left (4+r \right ) \left (3+r \right ) \left (2+r \right ) \left (r +1\right )^{2}} \]
Which for the root \(r = 0\) becomes
\[ a_{4}={\frac {5}{24}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {-2-r}{\left (r +1\right )^{2}}\) | \(-2\) |
| | |
| \(a_{2}\) | \(\frac {3+r}{\left (2+r \right ) \left (r +1\right )^{2}}\) | \(\frac {3}{2}\) |
| | |
| \(a_{3}\) | \(\frac {-4-r}{\left (3+r \right ) \left (2+r \right ) \left (r +1\right )^{2}}\) | \(-{\frac {2}{3}}\) |
| | |
| \(a_{4}\) |
\(\frac {5+r}{\left (4+r \right ) \left (3+r \right ) \left (2+r \right ) \left (r +1\right )^{2}}\) |
\(\frac {5}{24}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=\frac {-6-r}{\left (5+r \right ) \left (4+r \right ) \left (3+r \right ) \left (2+r \right ) \left (r +1\right )^{2}} \]
Which for the root \(r = 0\) becomes
\[ a_{5}=-{\frac {1}{20}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {-2-r}{\left (r +1\right )^{2}}\) |
\(-2\) |
| | |
| \(a_{2}\) |
\(\frac {3+r}{\left (2+r \right ) \left (r +1\right )^{2}}\) | \(\frac {3}{2}\) |
| | |
| \(a_{3}\) | \(\frac {-4-r}{\left (3+r \right ) \left (2+r \right ) \left (r +1\right )^{2}}\) | \(-{\frac {2}{3}}\) |
| | |
| \(a_{4}\) |
\(\frac {5+r}{\left (4+r \right ) \left (3+r \right ) \left (2+r \right ) \left (r +1\right )^{2}}\) |
\(\frac {5}{24}\) |
| | |
| \(a_{5}\) |
\(\frac {-6-r}{\left (5+r \right ) \left (4+r \right ) \left (3+r \right ) \left (2+r \right ) \left (r +1\right )^{2}}\) |
\(-{\frac {1}{20}}\) |
| | |
Using the above table, then the first solution \(y_{1}\left (x \right )\) becomes
\begin{align*}
y_{1}\left (x \right )&= 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 \\
&= 1-2 x +\frac {3 x^{2}}{2}-\frac {2 x^{3}}{3}+\frac {5 x^{4}}{24}-\frac {x^{5}}{20}+O\left (x^{6}\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 \(r = 0\).
The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following
table
| | | | |
| \(n\) |
\(b_{n ,r}\) |
\(a_{n}\) |
\(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) |
\(b_{n}\left (r =0\right )\) |
| | | | |
| \(b_{0}\) |
\(1\) |
\(1\) |
N/A since \(b_{n}\) starts from 1 |
N/A |
| | | | |
| \(b_{1}\) | \(\frac {-2-r}{\left (r +1\right )^{2}}\) | \(-2\) | \(\frac {3+r}{\left (r +1\right )^{3}}\) | \(3\) |
| | | | |
| \(b_{2}\) | \(\frac {3+r}{\left (2+r \right ) \left (r +1\right )^{2}}\) | \(\frac {3}{2}\) | \(\frac {-2 r^{2}-11 r -13}{\left (2+r \right )^{2} \left (r +1\right )^{3}}\) | \(-{\frac {13}{4}}\) |
| | | | |
| \(b_{3}\) | \(\frac {-4-r}{\left (3+r \right ) \left (2+r \right ) \left (r +1\right )^{2}}\) | \(-{\frac {2}{3}}\) | \(\frac {3 r^{3}+27 r^{2}+74 r +62}{\left (3+r \right )^{2} \left (2+r \right )^{2} \left (r +1\right )^{3}}\) | \(\frac {31}{18}\) |
| | | | |
| \(b_{4}\) |
\(\frac {5+r}{\left (4+r \right ) \left (3+r \right ) \left (2+r \right ) \left (r +1\right )^{2}}\) |
\(\frac {5}{24}\) |
\(\frac {-4 r^{4}-54 r^{3}-256 r^{2}-504 r -346}{\left (4+r \right )^{2} \left (3+r \right )^{2} \left (2+r \right )^{2} \left (r +1\right )^{3}}\) |
\(-{\frac {173}{288}}\) |
| | | | |
| \(b_{5}\) |
\(\frac {-6-r}{\left (5+r \right ) \left (4+r \right ) \left (3+r \right ) \left (2+r \right ) \left (r +1\right )^{2}}\) |
\(-{\frac {1}{20}}\) |
\(\frac {5 r^{5}+95 r^{4}+685 r^{3}+2335 r^{2}+3744 r +2244}{\left (5+r \right )^{2} \left (4+r \right )^{2} \left (3+r \right )^{2} \left (2+r \right )^{2} \left (r +1\right )^{3}}\) |
\(\frac {187}{1200}\) |
| | | | |
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 \\
&= \left (1-2 x +\frac {3 x^{2}}{2}-\frac {2 x^{3}}{3}+\frac {5 x^{4}}{24}-\frac {x^{5}}{20}+O\left (x^{6}\right )\right ) \ln \left (x \right )+3 x -\frac {13 x^{2}}{4}+\frac {31 x^{3}}{18}-\frac {173 x^{4}}{288}+\frac {187 x^{5}}{1200}+O\left (x^{6}\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 \left (1-2 x +\frac {3 x^{2}}{2}-\frac {2 x^{3}}{3}+\frac {5 x^{4}}{24}-\frac {x^{5}}{20}+O\left (x^{6}\right )\right ) + c_2 \left (\left (1-2 x +\frac {3 x^{2}}{2}-\frac {2 x^{3}}{3}+\frac {5 x^{4}}{24}-\frac {x^{5}}{20}+O\left (x^{6}\right )\right ) \ln \left (x \right )+3 x -\frac {13 x^{2}}{4}+\frac {31 x^{3}}{18}-\frac {173 x^{4}}{288}+\frac {187 x^{5}}{1200}+O\left (x^{6}\right )\right ) \\
\end{align*}
Hence the final solution is
\begin{align*}
y &= y_h \\
&= c_1 \left (1-2 x +\frac {3 x^{2}}{2}-\frac {2 x^{3}}{3}+\frac {5 x^{4}}{24}-\frac {x^{5}}{20}+O\left (x^{6}\right )\right )+c_2 \left (\left (1-2 x +\frac {3 x^{2}}{2}-\frac {2 x^{3}}{3}+\frac {5 x^{4}}{24}-\frac {x^{5}}{20}+O\left (x^{6}\right )\right ) \ln \left (x \right )+3 x -\frac {13 x^{2}}{4}+\frac {31 x^{3}}{18}-\frac {173 x^{4}}{288}+\frac {187 x^{5}}{1200}+O\left (x^{6}\right )\right ) \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right ) x +\left (x +1\right ) \left (\frac {d}{d x}y \left (x \right )\right )+2 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 {2 y \left (x \right )}{x}-\frac {\left (x +1\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 (x +1\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x}+\frac {2 y \left (x \right )}{x}=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 {x +1}{x}, P_{3}\left (x \right )=\frac {2}{x}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=1 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \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} \\ {} & {} & \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right ) x +\left (x +1\right ) \left (\frac {d}{d x}y \left (x \right )\right )+2 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 =0..1 \\ {} & {} & 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 \cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r^{2} x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right )^{2}+a_{k} \left (k +r +2\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1\right )^{2}+a_{k} \left (k +2\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +2\right )}{\left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +2\right )}{\left (k +1\right )^{2}} \\ \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 +2\right )}{\left (k +1\right )^{2}}\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 : 0.017 (sec)
Leaf size : 44
dsolve(x*diff(diff(y(x),x),x)+(x+1)*diff(y(x),x)+2*y(x) = 0,y(x),
series,x=0)
\[
y = \left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-2 x +\frac {3}{2} x^{2}-\frac {2}{3} x^{3}+\frac {5}{24} x^{4}-\frac {1}{20} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (3 x -\frac {13}{4} x^{2}+\frac {31}{18} x^{3}-\frac {173}{288} x^{4}+\frac {187}{1200} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2}
\]
Mathematica DSolve solution
Solving time : 0.007 (sec)
Leaf size : 111
AsymptoticDSolveValue[{x*D[y[x],{x,2}] +(1+x)*D[y[x],x]+2*y[x] == 0,{}},
y[x],{x,0,5}]
\[
y(x)\to c_1 \left (-\frac {x^5}{20}+\frac {5 x^4}{24}-\frac {2 x^3}{3}+\frac {3 x^2}{2}-2 x+1\right )+c_2 \left (\frac {187 x^5}{1200}-\frac {173 x^4}{288}+\frac {31 x^3}{18}-\frac {13 x^2}{4}+\left (-\frac {x^5}{20}+\frac {5 x^4}{24}-\frac {2 x^3}{3}+\frac {3 x^2}{2}-2 x+1\right ) \log (x)+3 x\right )
\]