2.4.61 Problem 58

Internal problem ID [8949]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 58
Date solved : Friday, April 25, 2025 at 05:25:35 PM
CAS classification : [[_Emden, _Fowler]]

Solve

$$\begin{array}{rl}{x}^2{y}^{\prime \prime }-xy& =0\end{array}$$

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.

$$x^2{y}^{\prime \prime }-xy=0$$

The following is summary of singularities for the above ode. Writing the ode as

$$\begin{array}{rl}{y}^{\prime \prime }+p(x)y^{\prime }+q(x)y& =0\end{array}$$

Where

$$\begin{array}{rl}p(x)& =0\\ q(x)& =-\frac{1}{x}\end{array}$$

Table 2.112: Table $p(x),q(x)$ singularites.

$p(x)=0$
singularity	type

$q(x)=-\frac{1}{x}$
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 : $[\mathrm{\infty }]$

Since $x=0$ is regular singular point, then Frobenius power series is used. The ode is normalized to be

$$x^2{y}^{\prime \prime }-xy=0$$

Let the solution be represented as Frobenius power series of the form

$$y=\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r}$$

Then

$$\begin{array}{rl}{y}^{\prime }& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)a_n{x}^{n+r-1}\\ y^{\prime \prime }& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)(n+r-1)a_n{x}^{n+r-2}\end{array}$$

Substituting the above back into the ode gives

$$\begin{array}{}\text{(1)}& x^2(\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)(n+r-1)a_n{x}^{n+r-2})-x(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r})=0\end{array}$$

Which simplifies to

$$\begin{array}{}\text{(2A)}& (\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{n+r}{a}_n(n+r)(n+r-1))+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-x^{1+n+r}{a}_n)=0\end{array}$$

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

$$\begin{array}{rl}\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-x^{1+n+r}{a}_n)& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}(-a_{n-1}{x}^{n+r})\end{array}$$

Substituting all the above in Eq (2A) gives the following equation where now all powers of $x$ are the same and equal to $n+r$.

$$\begin{array}{}\text{(2B)}& (\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{n+r}{a}_n(n+r)(n+r-1))+\stackrel{\mathrm{\infty }}{\sum _{n=1}}(-a_{n-1}{x}^{n+r})=0\end{array}$$

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

$$x^{n+r}{a}_n(n+r)(n+r-1)=0$$

When $n=0$ the above becomes

$$x^r{a}_{0}r(-1+r)=0$$

Or

$$x^r{a}_{0}r(-1+r)=0$$

Since $a_0\ne 0$ then the above simplifies to

$$x^{r}r(-1+r)=0$$

Since the above is true for all $x$ then the indicial equation becomes

$$r(-1+r)=0$$

Solving for $r$ gives the roots of the indicial equation as

$$\begin{array}{rl}{r}_1& =1\\ r_2& =0\end{array}$$

Since $a_0\ne 0$ then the indicial equation becomes

$$x^{r}r(-1+r)=0$$

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

$$\begin{array}{rl}{y}_1\left(x\right)& =x^{r_1}(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^n)\\ y_2\left(x\right)& =C{y}_1\left(x\right)\ln \left(x\right)+x^{r_2}(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^n)\end{array}$$

Or

$$\begin{array}{rl}{y}_1\left(x\right)& =x(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^n)\\ y_2\left(x\right)& =C{y}_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^n)\end{array}$$

Or

$$\begin{array}{rl}{y}_1\left(x\right)& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{1+n}\\ y_2\left(x\right)& =C{y}_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^n)\end{array}$$

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

$$\begin{array}{}\text{(3)}& a_n(n+r)(n+r-1)-a_{n-1}=0\end{array}$$

Solving for $a_n$ from recursive equation (4) gives

$$\begin{array}{}\text{(4)}& a_n=\frac{a_{n-1}}{(n+r)(n+r-1)}\end{array}$$

Which for the root $r=1$ becomes

$$\begin{array}{}\text{(5)}& a_n=\frac{a_{n-1}}{(1+n)n}\end{array}$$

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

$$a_1=\frac{1}{(1+r)r}$$

Which for the root $r=1$ becomes

$$a_1=\frac{1}{2}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{1}{(1+r)r}$	$\frac{1}{2}$

For $n=2$, using the above recursive equation gives

$$a_2=\frac{1}{{(1+r)}^{2}r(2+r)}$$

Which for the root $r=1$ becomes

$$a_2=\frac{1}{12}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{1}{(1+r)r}$	$\frac{1}{2}$
$a_2$	$\frac{1}{{(1+r)}^{2}r(2+r)}$	$\frac{1}{12}$

For $n=3$, using the above recursive equation gives

$$a_3=\frac{1}{{(1+r)}^{2}r{(2+r)}^2(3+r)}$$

Which for the root $r=1$ becomes

$$a_3=\frac{1}{144}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{1}{(1+r)r}$	$\frac{1}{2}$
$a_2$	$\frac{1}{{(1+r)}^{2}r(2+r)}$	$\frac{1}{12}$
$a_3$	$\frac{1}{{(1+r)}^{2}r{(2+r)}^2(3+r)}$	$\frac{1}{144}$

For $n=4$, using the above recursive equation gives

$$a_4=\frac{1}{{(1+r)}^{2}r{(2+r)}^2{(3+r)}^2(4+r)}$$

Which for the root $r=1$ becomes

$$a_4=\frac{1}{2880}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{1}{(1+r)r}$	$\frac{1}{2}$
$a_2$	$\frac{1}{{(1+r)}^{2}r(2+r)}$	$\frac{1}{12}$
$a_3$	$\frac{1}{{(1+r)}^{2}r{(2+r)}^2(3+r)}$	$\frac{1}{144}$
$a_4$	$\frac{1}{{(1+r)}^{2}r{(2+r)}^2{(3+r)}^2(4+r)}$	$\frac{1}{2880}$

For $n=5$, using the above recursive equation gives

$$a_5=\frac{1}{{(1+r)}^{2}r{(2+r)}^2{(3+r)}^2{(4+r)}^2(5+r)}$$

Which for the root $r=1$ becomes

$$a_5=\frac{1}{86400}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{1}{(1+r)r}$	$\frac{1}{2}$
$a_2$	$\frac{1}{{(1+r)}^{2}r(2+r)}$	$\frac{1}{12}$
$a_3$	$\frac{1}{{(1+r)}^{2}r{(2+r)}^2(3+r)}$	$\frac{1}{144}$
$a_4$	$\frac{1}{{(1+r)}^{2}r{(2+r)}^2{(3+r)}^2(4+r)}$	$\frac{1}{2880}$
$a_5$	$\frac{1}{{(1+r)}^{2}r{(2+r)}^2{(3+r)}^2{(4+r)}^2(5+r)}$	$\frac{1}{86400}$

Using the above table, then the solution $y_1\left(x\right)$ is

$$\begin{array}{rl}{y}_1\left(x\right)& =x(a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+a_5{x}^5+a_6{x}^6\dots )\\ & =x(1+\frac{x}{2}+\frac{x^2}{12}+\frac{x^3}{144}+\frac{x^4}{2880}+\frac{x^5}{86400}+O\left(x^6\right))\end{array}$$

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 $\underset{r\to r_2}{lim}{a}_1\left(r\right)$. If this limit exists, then $C=0$, else we need to keep the log term and $C\ne 0$. The above table shows that

$$\begin{array}{rl}{a}_N& =a_1\\ & =\frac{1}{(1+r)r}\end{array}$$

Therefore

$$\begin{array}{rl}\underset{r\to r_2}{lim}\frac{1}{(1+r)r}& =\underset{r\to 0}{lim}\frac{1}{(1+r)r}\\ & =\mathit{\text{undefined}}\end{array}$$

Since the limit does not exist then the log term is needed. Therefore the second solution has the form

$$y_2\left(x\right)=C{y}_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n+r_2})$$

Therefore

$$\begin{array}{rl}\frac{d}{dx}{y}_2\left(x\right)& =C{y}_1^{\prime }\left(x\right)\ln \left(x\right)+\frac{C{y}_1\left(x\right)}{x}+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}\frac{b_n{x}^{n+r_2}(n+r_2)}{x})\\ & =C{y}_1^{\prime }\left(x\right)\ln \left(x\right)+\frac{C{y}_1\left(x\right)}{x}+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{-1+n+r_2}{b}_n(n+r_2))\\ \frac{d^2}{d{x}^2}{y}_2\left(x\right)& =C{y}_1^{\prime \prime }\left(x\right)\ln \left(x\right)+\frac{2C{y}_1^{\prime }\left(x\right)}{x}-\frac{C{y}_1\left(x\right)}{x^2}+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(\frac{b_n{x}^{n+r_2}{(n+r_2)}^2}{x^2}-\frac{b_n{x}^{n+r_2}(n+r_2)}{x^2})\\ & =C{y}_1^{\prime \prime }\left(x\right)\ln \left(x\right)+\frac{2C{y}_1^{\prime }\left(x\right)}{x}-\frac{C{y}_1\left(x\right)}{x^2}+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{-2+n+r_2}{b}_n(n+r_2)(-1+n+r_2))\end{array}$$

Substituting these back into the given ode $x^2{y}^{\prime \prime }-xy=0$ gives

$$x^2(C{y}_1^{\prime \prime }\left(x\right)\ln \left(x\right)+\frac{2C{y}_1^{\prime }\left(x\right)}{x}-\frac{C{y}_1\left(x\right)}{x^2}+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(\frac{b_n{x}^{n+r_2}{(n+r_2)}^2}{x^2}-\frac{b_n{x}^{n+r_2}(n+r_2)}{x^2}))-x(C{y}_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n+r_2}))=0$$

Which can be written as

$$\begin{array}{}\text{(7)}& ((y_1^{\prime \prime }\left(x\right)x^2-y_1\left(x\right)x)\ln \left(x\right)+x^2(\frac{2y_1^{\prime }\left(x\right)}{x}-\frac{y_1\left(x\right)}{x^2}))C+x^2(\stackrel{\mathrm{\infty }}{\sum _{n=0}}(\frac{b_n{x}^{n+r_2}{(n+r_2)}^2}{x^2}-\frac{b_n{x}^{n+r_2}(n+r_2)}{x^2}))-x(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n+r_2})=0\end{array}$$

But since $y_1\left(x\right)$ is a solution to the ode, then

$$y_1^{\prime \prime }\left(x\right)x^2-y_1\left(x\right)x=0$$

Eq (7) simplifes to

$$\begin{array}{}\text{(8)}& x^2(\frac{2y_1^{\prime }\left(x\right)}{x}-\frac{y_1\left(x\right)}{x^2})C+x^2(\stackrel{\mathrm{\infty }}{\sum _{n=0}}(\frac{b_n{x}^{n+r_2}{(n+r_2)}^2}{x^2}-\frac{b_n{x}^{n+r_2}(n+r_2)}{x^2}))-x(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n+r_2})=0\end{array}$$

Substituting $y_1=\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r_1}$ into the above gives

$$\begin{array}{}\text{(9)}& (2(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{-1+n+r_1}{a}_n(n+r_1))x-(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r_1}))C+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{-2+n+r_2}{b}_n(n+r_2)(-1+n+r_2))x^2-x(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n+r_2})=0\end{array}$$

Since $r_1=1$ and $r_2=0$ then the above becomes

$$\begin{array}{}\text{(10)}& (2(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^n{a}_n(1+n))x-(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{1+n}))C+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{n-2}{b}_{n}n(n-1))x^2-x(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^n)=0\end{array}$$

Which simplifies to

$$\begin{array}{}\text{(2A)}& (\stackrel{\mathrm{\infty }}{\sum _{n=0}}2C{x}^{1+n}{a}_n(1+n))+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-a_n{x}^{1+n}C)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}n{x}^n{b}_n(n-1))+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-x^{1+n}{b}_n)=0\end{array}$$

The next step is to make all powers of $x$ be $n$ in each summation term. Going over each summation term above with power of $x$ in it which is not already $x^n$ and adjusting the power and the corresponding index gives

$$\begin{array}{rl}\stackrel{\mathrm{\infty }}{\sum _{n=0}}2C{x}^{1+n}{a}_n(1+n)& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}2C{a}_{n-1}n{x}^n\\ \stackrel{\mathrm{\infty }}{\sum _{n=0}}(-a_n{x}^{1+n}C)& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}(-C{a}_{n-1}{x}^n)\\ \stackrel{\mathrm{\infty }}{\sum _{n=0}}(-x^{1+n}{b}_n)& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}(-b_{n-1}{x}^n)\end{array}$$

Substituting all the above in Eq (2A) gives the following equation where now all powers of $x$ are the same and equal to $n$.

$$\begin{array}{}\text{(2B)}& (\stackrel{\mathrm{\infty }}{\sum _{n=1}}2C{a}_{n-1}n{x}^n)+\stackrel{\mathrm{\infty }}{\sum _{n=1}}(-C{a}_{n-1}{x}^n)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}n{x}^n{b}_n(n-1))+\stackrel{\mathrm{\infty }}{\sum _{n=1}}(-b_{n-1}{x}^n)=0\end{array}$$

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

$$3C{a}_1-b_1+2b_2=0$$

Which when replacing the above values found already for $b_n$ and the values found earlier for $a_n$ and for $C$, gives

$$2b_2+\frac{3}{2}=0$$

Solving the above for $b_2$ gives

$$b_2=-\frac{3}{4}$$

For $n=3$, Eq (2B) gives

$$5C{a}_2-b_2+6b_3=0$$

Which when replacing the above values found already for $b_n$ and the values found earlier for $a_n$ and for $C$, gives

$$6b_3+\frac{7}{6}=0$$

Solving the above for $b_3$ gives

$$b_3=-\frac{7}{36}$$

For $n=4$, Eq (2B) gives

$$7C{a}_3-b_3+12b_4=0$$

Which when replacing the above values found already for $b_n$ and the values found earlier for $a_n$ and for $C$, gives

$$12b_4+\frac{35}{144}=0$$

Solving the above for $b_4$ gives

$$b_4=-\frac{35}{1728}$$

For $n=5$, Eq (2B) gives

$$9C{a}_4-b_4+20b_5=0$$

Which when replacing the above values found already for $b_n$ and the values found earlier for $a_n$ and for $C$, gives

$$20b_5+\frac{101}{4320}=0$$

Solving the above for $b_5$ gives

$$b_5=-\frac{101}{86400}$$

Now that we found all $b_n$ and $C$, we can calculate the second solution from

$$y_2\left(x\right)=C{y}_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n+r_2})$$

Using the above value found for $C=1$ and all $b_n$, then the second solution becomes

$$y_2\left(x\right)=1\left(x(1+\frac{x}{2}+\frac{x^2}{12}+\frac{x^3}{144}+\frac{x^4}{2880}+\frac{x^5}{86400}+O\left(x^6\right))\right)\ln \left(x\right)+1-\frac{3x^2}{4}-\frac{7x^3}{36}-\frac{35x^4}{1728}-\frac{101x^5}{86400}+O\left(x^6\right)$$

Therefore the homogeneous solution is

$$\begin{array}{rl}{y}_h(x)& =c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)\\ & =c_{1}x(1+\frac{x}{2}+\frac{x^2}{12}+\frac{x^3}{144}+\frac{x^4}{2880}+\frac{x^5}{86400}+O\left(x^6\right))+c_2(1\left(x(1+\frac{x}{2}+\frac{x^2}{12}+\frac{x^3}{144}+\frac{x^4}{2880}+\frac{x^5}{86400}+O\left(x^6\right))\right)\ln \left(x\right)+1-\frac{3x^2}{4}-\frac{7x^3}{36}-\frac{35x^4}{1728}-\frac{101x^5}{86400}+O\left(x^6\right))\end{array}$$

Hence the final solution is

$$\begin{array}{rl}y& =y_h\\ & =c_{1}x(1+\frac{x}{2}+\frac{x^2}{12}+\frac{x^3}{144}+\frac{x^4}{2880}+\frac{x^5}{86400}+O\left(x^6\right))+c_2(x(1+\frac{x}{2}+\frac{x^2}{12}+\frac{x^3}{144}+\frac{x^4}{2880}+\frac{x^5}{86400}+O\left(x^6\right))\ln \left(x\right)+1-\frac{3x^2}{4}-\frac{7x^3}{36}-\frac{35x^4}{1728}-\frac{101x^5}{86400}+O\left(x^6\right))\end{array}$$

✓ Maple. Time used: 0.024 (sec). Leaf size: 58

Order:=6; 
ode:=x^2*diff(diff(y(x),x),x)-x*y(x) = 0; 
dsolve(ode,y(x),type='series',x=0);
 

$$y=c_{1}x(1+\frac{1}{2}x+\frac{1}{12}{x}^2+\frac{1}{144}{x}^3+\frac{1}{2880}{x}^4+\frac{1}{86400}{x}^5+\mathrm{O}\left(x^6\right))+c_2(\ln \left(x\right)(x+\frac{1}{2}{x}^2+\frac{1}{12}{x}^3+\frac{1}{144}{x}^4+\frac{1}{2880}{x}^5+\mathrm{O}\left(x^6\right))+(1-\frac{3}{4}{x}^2-\frac{7}{36}{x}^3-\frac{35}{1728}{x}^4-\frac{101}{86400}{x}^5+\mathrm{O}\left(x^6\right)))$$

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 
<- No Liouvillian solutions exists 
-> Trying a solution in terms of special functions: 
   -> Bessel 
   <- Bessel successful 
<- special function solution successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)-xy\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=\frac{y\left(x\right)}{x}\\ \bullet & & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)-\frac{y\left(x\right)}{x}=0\\ ◻& & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=0,P_3\left(x\right)=-\frac{1}{x}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=0\\ & \circ & x^2\cdot P_3\left(x\right)\text{is analytic at}x=0\\ & & (x^2\cdot P_3\left(x\right))|{}_{x=0}=0\\ & \circ & x=0\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & & x_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)x-y\left(x\right)=0\\ \bullet & & \text{Assume series solution for}y\left(x\right)\\ & & y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}x\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)\text{to series expansion}\\ & & x\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)x^{k+r-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & x\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+1+r)(k+r)x^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_{0}r(-1+r)x^{-1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}(a_{k+1}(k+1+r)(k+r)-a_k)x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & r(-1+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,1\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_{k+1}(k+1+r)(k+r)-a_k=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=\frac{a_k}{(k+1+r)(k+r)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+1}=\frac{a_k}{(k+1)k}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+1}=\frac{a_k}{(k+1)k}]\\ \bullet & & \text{Recursion relation for}r=1\\ & & a_{k+1}=\frac{a_k}{(k+2)(k+1)}\\ \bullet & & \text{Solution for}r=1\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+1},a_{k+1}=\frac{a_k}{(k+2)(k+1)}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k)+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{x}^{k+1}),a_{k+1}=\frac{a_k}{(k+1)k},b_{k+1}=\frac{b_k}{(k+2)(k+1)}]\end{array}$$

✓ Mathematica. Time used: 0.019 (sec). Leaf size: 85

ode=x^2*D[y[x],{x,2}]-x*y[x]==0; 
ic={}; 
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
 

$$y(x)\to c_1(\frac{1}{144}x(x^3+12x^2+72x+144)\log (x)+\frac{-47x^4-480x^3-2160x^2-1728x+1728}{1728})+c_2(\frac{x^5}{2880}+\frac{x^4}{144}+\frac{x^3}{12}+\frac{x^2}{2}+x)$$

✓ Sympy. Time used: 0.823 (sec). Leaf size: 29

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

$$y\left(x\right)=C_{1}x(\frac{x^4}{2880}+\frac{x^3}{144}+\frac{x^2}{12}+\frac{x}{2}+1)+O\left(x^6\right)$$