2.4.17 Problem 17

Internal problem ID [10180]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 17
Date solved : Thursday, November 27, 2025 at 10:24:26 AM
CAS classification : [_Lienard]

\begin{align*} x y^{\prime \prime }+2 y^{\prime }+x y&=0 \\ y \left (0\right ) &= 1 \\ y^{\prime }\left (0\right ) &= 0 \\ \end{align*}

Series expansion around \(x=0\).

The type of the expansion point is first determined. This is done on the homogeneous part of the ODE.

\[ x y^{\prime \prime }+2 y^{\prime }+x y = 0 \]

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 {2}{x}\\ q(x) &= 1\\ \end{align*}

Table 2.74: Table \(p(x),q(x)\) singularites.

\(p(x)=\frac {2}{x}\)
singularity	type
\(x = 0\)	\(\text {``regular''}\)

\(q(x)=1\)
singularity	type

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

\[ x y^{\prime \prime }+2 y^{\prime }+x y = 0 \]

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} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x +2 \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 -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 \left (n +r \right ) a_{n} x^{n +r -1}\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 \(n +r -1\) 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^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} 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 =0}{\sum }}2 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r -1}\right ) = 0 \end{equation}

The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives

\[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \]

When \(n = 0\) the above becomes

\[ x^{-1+r} a_{0} r \left (-1+r \right )+2 r a_{0} x^{-1+r} = 0 \]

Or

\[ \left (x^{-1+r} r \left (-1+r \right )+2 r \,x^{-1+r}\right ) a_{0} = 0 \]

Since \(a_{0}\neq 0\) then the above simplifies to

\[ r \,x^{-1+r} \left (1+r \right ) = 0 \]

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 &= 0\\ r_2 &= -1 \end{align*}

Since \(a_{0}\neq 0\) then the indicial equation becomes

\[ r \,x^{-1+r} \left (1+r \right ) = 0 \]

Solving for \(r\) gives the roots of the indicial equation as \([0, -1]\).

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 ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{x} \end{align*}

Or

\begin{align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ 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 -1}\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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives

\[ a_{1} = 0 \]

For \(2\le n\) the recursive equation is

\begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n} \left (n +r \right )+a_{n -2} = 0 \end{equation}

Solving for \(a_{n}\) from recursive equation (4) gives

\[ a_{n} = -\frac {a_{n -2}}{n^{2}+2 n r +r^{2}+n +r}\tag {4} \]

Which for the root \(r = 0\) becomes

\[ a_{n} = -\frac {a_{n -2}}{n \left (1+n \right )}\tag {5} \]

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.

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)

For \(n = 2\), using the above recursive equation gives

\[ a_{2}=-\frac {1}{r^{2}+5 r +6} \]

Which for the root \(r = 0\) becomes

\[ a_{2}=-{\frac {1}{6}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(-\frac {1}{r^{2}+5 r +6}\)	\(-{\frac {1}{6}}\)

For \(n = 3\), using the above recursive equation gives

\[ a_{3}=0 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(-\frac {1}{r^{2}+5 r +6}\)	\(-{\frac {1}{6}}\)
\(a_{3}\)	\(0\)	\(0\)

For \(n = 4\), using the above recursive equation gives

\[ a_{4}=\frac {1}{r^{4}+14 r^{3}+71 r^{2}+154 r +120} \]

Which for the root \(r = 0\) becomes

\[ a_{4}={\frac {1}{120}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(-\frac {1}{r^{2}+5 r +6}\)	\(-{\frac {1}{6}}\)
\(a_{3}\)	\(0\)	\(0\)
\(a_{4}\)	\(\frac {1}{r^{4}+14 r^{3}+71 r^{2}+154 r +120}\)	\(\frac {1}{120}\)

For \(n = 5\), using the above recursive equation gives

\[ a_{5}=0 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(-\frac {1}{r^{2}+5 r +6}\)	\(-{\frac {1}{6}}\)
\(a_{3}\)	\(0\)	\(0\)
\(a_{4}\)	\(\frac {1}{r^{4}+14 r^{3}+71 r^{2}+154 r +120}\)	\(\frac {1}{120}\)
\(a_{5}\)	\(0\)	\(0\)

Using the above table, then the solution \(y_{1}\left (x \right )\) is

\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-\frac {x^{2}}{6}+\frac {x^{4}}{120}+O\left (x^{6}\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} \\ &= 0 \end{align*}

Therefore

\begin{align*} \lim _{r\rightarrow r_{2}}0&= \lim _{r\rightarrow -1}0\\ &= 0 \end{align*}

The limit is \(0\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form

\begin{align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -1} \end{align*}

Eq (3) derived above is used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq(3) gives

\[ b_{1} = 0 \]

For \(2\le n\) the recursive equation is

\begin{equation} \tag{4} b_{n} \left (n +r \right ) \left (n +r -1\right )+2 \left (n +r \right ) b_{n}+b_{n -2} = 0 \end{equation}

Which for for the root \(r = -1\) becomes

\begin{equation} \tag{4A} b_{n} \left (n -1\right ) \left (n -2\right )+2 \left (n -1\right ) b_{n}+b_{n -2} = 0 \end{equation}

Solving for \(b_{n}\) from the recursive equation (4) gives

\[ b_{n} = -\frac {b_{n -2}}{n^{2}+2 n r +r^{2}+n +r}\tag {5} \]

Which for the root \(r = -1\) becomes

\[ b_{n} = -\frac {b_{n -2}}{n^{2}-n}\tag {6} \]

At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -1\) and after as more terms are found using the above recursive equation.

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)

For \(n = 2\), using the above recursive equation gives

\[ b_{2}=-\frac {1}{r^{2}+5 r +6} \]

Which for the root \(r = -1\) becomes

\[ b_{2}=-{\frac {1}{2}} \]

And the table now becomes

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)
\(b_{2}\)	\(-\frac {1}{r^{2}+5 r +6}\)	\(-{\frac {1}{2}}\)

For \(n = 3\), using the above recursive equation gives

\[ b_{3}=0 \]

And the table now becomes

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)
\(b_{2}\)	\(-\frac {1}{r^{2}+5 r +6}\)	\(-{\frac {1}{2}}\)
\(b_{3}\)	\(0\)	\(0\)

For \(n = 4\), using the above recursive equation gives

\[ b_{4}=\frac {1}{\left (r^{2}+5 r +6\right ) \left (r^{2}+9 r +20\right )} \]

Which for the root \(r = -1\) becomes

\[ b_{4}={\frac {1}{24}} \]

And the table now becomes

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)
\(b_{2}\)	\(-\frac {1}{r^{2}+5 r +6}\)	\(-{\frac {1}{2}}\)
\(b_{3}\)	\(0\)	\(0\)
\(b_{4}\)	\(\frac {1}{r^{4}+14 r^{3}+71 r^{2}+154 r +120}\)	\(\frac {1}{24}\)

For \(n = 5\), using the above recursive equation gives

\[ b_{5}=0 \]

And the table now becomes

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)
\(b_{2}\)	\(-\frac {1}{r^{2}+5 r +6}\)	\(-{\frac {1}{2}}\)
\(b_{3}\)	\(0\)	\(0\)
\(b_{4}\)	\(\frac {1}{r^{4}+14 r^{3}+71 r^{2}+154 r +120}\)	\(\frac {1}{24}\)
\(b_{5}\)	\(0\)	\(0\)

Using the above table, then the solution \(y_{2}\left (x \right )\) is

\begin{align*} y_{2}\left (x \right )&= 1 \left (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 \right ) \\ &= \frac {1-\frac {x^{2}}{2}+\frac {x^{4}}{24}+O\left (x^{6}\right )}{x} \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-\frac {x^{2}}{6}+\frac {x^{4}}{120}+O\left (x^{6}\right )\right ) + \frac {c_2 \left (1-\frac {x^{2}}{2}+\frac {x^{4}}{24}+O\left (x^{6}\right )\right )}{x} \\ \end{align*}

Hence the final solution is

\begin{align*} y &= y_h \\ &= c_1 \left (1-\frac {x^{2}}{6}+\frac {x^{4}}{120}+O\left (x^{6}\right )\right )+\frac {c_2 \left (1-\frac {x^{2}}{2}+\frac {x^{4}}{24}+O\left (x^{6}\right )\right )}{x} \\ \end{align*}

Initial conditions are now used to to solve for the constants of integrations. This results in

\[ y = 1-\frac {x^{2}}{6}+\frac {x^{4}}{120}+O\left (x^{6}\right ) \]

Figure 2.211: Solutions plot

✓ Maple. Time used: 0.003 (sec). Leaf size: 14

Order:=6; 
ode:=x*diff(diff(y(x),x),x)+2*diff(y(x),x)+y(x)*x = 0; 
ic:=[y(0) = 1, D(y)(0) = 0]; 
dsolve([ode,op(ic)],y(x),type='series',x=0);
 

\[ y = 1-\frac {1}{6} x^{2}+\frac {1}{120} x^{4}+\operatorname {O}\left (x^{6}\right ) \]

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 
   Group is reducible or imprimitive 
<- Kovacics algorithm successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+2 \frac {d}{d x}y \left (x \right )+x y \left (x \right )=0, y \left (0\right )=1, \left (\frac {d}{d x}y \left (x \right )\right )\bigg | {\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0\right ] \\ \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 )=-y \left (x \right )-\frac {2 \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 {2 \left (\frac {d}{d x}y \left (x \right )\right )}{x}+y \left (x \right )=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 {2}{x}, P_{3}\left (x \right )=1\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}}}=2 \\ {} & \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} \\ {} & {} & x \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+2 \frac {d}{d x}y \left (x \right )+x 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 y \left (x \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x \cdot y \left (x \right )=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d}{d x}y \left (x \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \frac {d}{d x}y \left (x \right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +r +1\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 +r +1\right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (1+r \right ) x^{-1+r}+a_{1} \left (1+r \right ) \left (2+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +r +1\right ) \left (k +2+r \right )+a_{k -1}\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 \{-1, 0\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (2+r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +r +1\right ) \left (k +2+r \right )+a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (k +3+r \right )+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2+r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +1\right ) \left (k +2\right )} \\ \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 +2}=-\frac {a_{k}}{\left (k +1\right ) \left (k +2\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k +3\right )} \\ \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 +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k +3\right )}, 2 a_{1}=0\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 -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k}\right ), a_{k +2}=-\frac {a_{k}}{\left (k +1\right ) \left (k +2\right )}, 0=0, b_{k +2}=-\frac {b_{k}}{\left (k +2\right ) \left (k +3\right )}, 2 b_{1}=0\right ] \end {array} \]

✓ Mathematica. Time used: 0.127 (sec). Leaf size: 19

ode=x*D[y[x],{x,2}]+2*D[y[x],x]+x*y[x]==0; 
ic={y[0]==1,Derivative[1][y][0] ==0}; 
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
 

\[ y(x)\to \frac {x^4}{120}-\frac {x^2}{6}+1 \]

✓ Sympy. Time used: 0.268 (sec). Leaf size: 39

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x*y(x) + x*Derivative(y(x), (x, 2)) + 2*Derivative(y(x), x),0) 
ics = {y(0): 1, Subs(Derivative(y(x), x), x, 0): 0} 
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
 

\[ y{\left (x \right )} = C_{2} \left (\frac {x^{4}}{120} - \frac {x^{2}}{6} + 1\right ) + \frac {C_{1} \left (- \frac {x^{6}}{720} + \frac {x^{4}}{24} - \frac {x^{2}}{2} + 1\right )}{x} + O\left (x^{6}\right ) \]