2.4.40 Problem 37

Internal problem ID [8929]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 37
Date solved : Sunday, March 30, 2025 at 01:55:35 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}{x}^2(x^2-2x+1)y^{\prime \prime }-x(3+x)y^{\prime }+(4+x)y& =0\end{array}$$

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.

$$(x^4-2x^3+x^2)y^{\prime \prime }+(-x^2-3x)y^{\prime }+(4+x)y=0$$

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

$$\begin{array}{rl}p(x)& =-\frac{3+x}{x{(x-1)}^2}\\ q(x)& =\frac{4+x}{x^2{(x-1)}^2}\end{array}$$

Table 2.93: Table $p(x),q(x)$ singularites.

$p(x)=-\frac{3+x}{x{(x-1)}^2}$
singularity	type
$x=0$	$\text{“regular”}$
$x=1$	$\text{“irregular”}$

$q(x)=\frac{4+x}{x^2{(x-1)}^2}$
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 : $[0,\mathrm{\infty }]$

Irregular singular points : $[1]$

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

$$x^2(x^2-2x+1)y^{\prime \prime }+(-x^2-3x)y^{\prime }+(4+x)y=0$$

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

$$\begin{array}{}\text{(1)}& x^2(x^2-2x+1)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)(n+r-1)a_n{x}^{n+r-2})+(-x^2-3x)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)a_n{x}^{n+r-1})+(4+x)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r})=0\end{array}$$

Which simplifies to

$$\begin{array}{}\text{(2A)}& (\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{n+r+2}{a}_n(n+r)(n+r-1))+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-2x^{1+n+r}{a}_n(n+r)(n+r-1))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{n+r}{a}_n(n+r)(n+r-1))+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-x^{1+n+r}{a}_n(n+r))+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-3x^{n+r}{a}_n(n+r))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}4a_n{x}^{n+r})+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{1+n+r}{a}_n)=0\end{array}$$

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

$$\begin{array}{rl}\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{n+r+2}{a}_n(n+r)(n+r-1)& =\stackrel{\mathrm{\infty }}{\sum _{n=2}}{a}_{n-2}(n+r-2)(n-3+r)x^{n+r}\\ \stackrel{\mathrm{\infty }}{\sum _{n=0}}(-2x^{1+n+r}{a}_n(n+r)(n+r-1))& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}(-2a_{n-1}(n+r-1)(n+r-2)x^{n+r})\\ \stackrel{\mathrm{\infty }}{\sum _{n=0}}(-x^{1+n+r}{a}_n(n+r))& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}(-a_{n-1}(n+r-1)x^{n+r})\\ \stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{1+n+r}{a}_n& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}{a}_{n-1}{x}^{n+r}\end{array}$$

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

$$\begin{array}{}\text{(2B)}& (\stackrel{\mathrm{\infty }}{\sum _{n=2}}{a}_{n-2}(n+r-2)(n-3+r)x^{n+r})+\stackrel{\mathrm{\infty }}{\sum _{n=1}}(-2a_{n-1}(n+r-1)(n+r-2)x^{n+r})+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{n+r}{a}_n(n+r)(n+r-1))+\stackrel{\mathrm{\infty }}{\sum _{n=1}}(-a_{n-1}(n+r-1)x^{n+r})+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-3x^{n+r}{a}_n(n+r))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}4a_n{x}^{n+r})+(\stackrel{\mathrm{\infty }}{\sum _{n=1}}{a}_{n-1}{x}^{n+r})=0\end{array}$$

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

$$x^{n+r}{a}_n(n+r)(n+r-1)-3x^{n+r}{a}_n(n+r)+4a_n{x}^{n+r}=0$$

When $n=0$ the above becomes

$$x^r{a}_{0}r(-1+r)-3x^r{a}_{0}r+4a_0{x}^r=0$$

Or

$$(x^{r}r(-1+r)-3x^{r}r+4x^r)a_0=0$$

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

$${(r-2)}^2{x}^r=0$$

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

$${(r-2)}^2=0$$

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

$$\begin{array}{rl}{r}_1& =2\\ r_2& =2\end{array}$$

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

$${(r-2)}^2{x}^r=0$$

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

$$\begin{array}{}\text{(1A)}& y_1\left(x\right)& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r}\end{array}$$

Now the second solution $y_2$ is found using

$$\begin{array}{}\text{(1B)}& y_2\left(x\right)& =y_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=1}}{b}_n{x}^{n+r})\end{array}$$

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

$$\begin{array}{rl}{y}_1\left(x\right)& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+2}\\ y_2\left(x\right)& =y_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=1}}{b}_n{x}^{n+2})\end{array}$$

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

$$a_1=\frac{2r+1}{-1+r}$$

For $2\le n$ the recursive equation is

$$\begin{array}{}\text{(3)}& a_{n-2}(n+r-2)(n-3+r)-2a_{n-1}(n+r-1)(n+r-2)+a_n(n+r)(n+r-1)-a_{n-1}(n+r-1)-3a_n(n+r)+4a_n+a_{n-1}=0\end{array}$$

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

$$\begin{array}{}\text{(4)}& a_n=-\frac{n{a}_{n-2}-2n{a}_{n-1}+r{a}_{n-2}-2r{a}_{n-1}-3a_{n-2}+a_{n-1}}{n+r-2}\end{array}$$

Which for the root $r=2$ becomes

$$\begin{array}{}\text{(5)}& a_n=\frac{(-a_{n-2}+2a_{n-1})n+a_{n-2}+3a_{n-1}}{n}\end{array}$$

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.

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{2r+1}{-1+r}$	$5$

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

$$a_2=\frac{3r^2+10r+2}{r(-1+r)}$$

Which for the root $r=2$ becomes

$$a_2=17$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{2r+1}{-1+r}$	$5$
$a_2$	$\frac{3r^2+10r+2}{r(-1+r)}$	$17$

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

$$a_3=\frac{4r^3+34r^2+54r+10}{r^3-r}$$

Which for the root $r=2$ becomes

$$a_3=\frac{143}{3}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{2r+1}{-1+r}$	$5$
$a_2$	$\frac{3r^2+10r+2}{r(-1+r)}$	$17$
$a_3$	$\frac{4r^3+34r^2+54r+10}{r^3-r}$	$\frac{143}{3}$

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

$$a_4=\frac{5r^4+80r^3+321r^2+384r+68}{(2+r)r(r^2-1)}$$

Which for the root $r=2$ becomes

$$a_4=\frac{355}{3}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{2r+1}{-1+r}$	$5$
$a_2$	$\frac{3r^2+10r+2}{r(-1+r)}$	$17$
$a_3$	$\frac{4r^3+34r^2+54r+10}{r^3-r}$	$\frac{143}{3}$
$a_4$	$\frac{5r^4+80r^3+321r^2+384r+68}{(2+r)r(r^2-1)}$	$\frac{355}{3}$

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

$$a_5=\frac{6r^5+155r^4+1156r^3+3295r^2+3336r+572}{r^5+5r^4+5r^3-5r^2-6r}$$

Which for the root $r=2$ becomes

$$a_5=\frac{4043}{15}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{2r+1}{-1+r}$	$5$
$a_2$	$\frac{3r^2+10r+2}{r(-1+r)}$	$17$
$a_3$	$\frac{4r^3+34r^2+54r+10}{r^3-r}$	$\frac{143}{3}$
$a_4$	$\frac{5r^4+80r^3+321r^2+384r+68}{(2+r)r(r^2-1)}$	$\frac{355}{3}$
$a_5$	$\frac{6r^5+155r^4+1156r^3+3295r^2+3336r+572}{r^5+5r^4+5r^3-5r^2-6r}$	$\frac{4043}{15}$

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

$$\begin{array}{rl}{y}_1\left(x\right)& =x^2(a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+a_5{x}^5+a_6{x}^6\dots )\\ & =x^2(17x^2+5x+1+\frac{143x^3}{3}+\frac{355x^4}{3}+\frac{4043x^5}{15}+O\left(x^6\right))\end{array}$$

Now the second solution is found. The second solution is given by

$$y_2\left(x\right)=y_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=1}}{b}_n{x}^{n+r})$$

Where $b_n$ is found using

$$b_n=\frac{d}{dr}{a}_{n,r}$$

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

$n$	$b_{n,r}$	$a_n$	$b_{n,r}=\frac{d}{dr}{a}_{n,r}$	$b_n(r=2)$
$b_0$	$1$	$1$	N/A since $b_n$ starts from 1	N/A
$b_1$	$\frac{2r+1}{-1+r}$	$5$	$-\frac{3}{{(-1+r)}^2}$	$-3$
$b_2$	$\frac{3r^2+10r+2}{r(-1+r)}$	$17$	$\frac{-13r^2-4r+2}{r^2{(-1+r)}^2}$	$-\frac{29}{2}$
$b_3$	$\frac{4r^3+34r^2+54r+10}{r^3-r}$	$\frac{143}{3}$	$\frac{-34r^4-116r^3-64r^2+10}{r^2{(r^2-1)}^2}$	$-\frac{859}{18}$
$b_4$	$\frac{5r^4+80r^3+321r^2+384r+68}{(2+r)r(r^2-1)}$	$\frac{355}{3}$	$\frac{-70r^6-652r^5-1904r^4-2128r^3-666r^2+136r+136}{{(2+r)}^2{r}^2{(r^2-1)}^2}$	$-\frac{4693}{36}$
$b_5$	$\frac{6r^5+155r^4+1156r^3+3295r^2+3336r+572}{r^5+5r^4+5r^3-5r^2-6r}$	$\frac{4043}{15}$	$\frac{-125r^8-2252r^7-14980r^6-47988r^5-77945r^4-58672r^3-11670r^2+5720r+3432}{r^2{(r^4+5r^3+5r^2-5r-6)}^2}$	$-\frac{285181}{900}$

The above table gives all values of $b_n$ needed. Hence the second solution is

$$\begin{array}{rl}{y}_2\left(x\right)& =y_1\left(x\right)\ln \left(x\right)+b_0+b_{1}x+b_2{x}^2+b_3{x}^3+b_4{x}^4+b_5{x}^5+b_6{x}^6\dots \\ & =x^2(17x^2+5x+1+\frac{143x^3}{3}+\frac{355x^4}{3}+\frac{4043x^5}{15}+O\left(x^6\right))\ln \left(x\right)+x^2(-3x-\frac{29x^2}{2}-\frac{859x^3}{18}-\frac{4693x^4}{36}-\frac{285181x^5}{900}+O\left(x^6\right))\end{array}$$

Therefore the homogeneous solution is

$$\begin{array}{rl}{y}_h(x)& =c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)\\ & =c_1{x}^2(17x^2+5x+1+\frac{143x^3}{3}+\frac{355x^4}{3}+\frac{4043x^5}{15}+O\left(x^6\right))+c_2(x^2(17x^2+5x+1+\frac{143x^3}{3}+\frac{355x^4}{3}+\frac{4043x^5}{15}+O\left(x^6\right))\ln \left(x\right)+x^2(-3x-\frac{29x^2}{2}-\frac{859x^3}{18}-\frac{4693x^4}{36}-\frac{285181x^5}{900}+O\left(x^6\right)))\end{array}$$

Hence the final solution is

$$\begin{array}{rl}y& =y_h\\ & =c_1{x}^2(17x^2+5x+1+\frac{143x^3}{3}+\frac{355x^4}{3}+\frac{4043x^5}{15}+O\left(x^6\right))+c_2(x^2(17x^2+5x+1+\frac{143x^3}{3}+\frac{355x^4}{3}+\frac{4043x^5}{15}+O\left(x^6\right))\ln \left(x\right)+x^2(-3x-\frac{29x^2}{2}-\frac{859x^3}{18}-\frac{4693x^4}{36}-\frac{285181x^5}{900}+O\left(x^6\right)))\end{array}$$

✓ Maple. Time used: 0.026 (sec). Leaf size: 48

Order:=6; 
ode:=x^2*(x^2-2*x+1)*diff(diff(y(x),x),x)-(3+x)*x*diff(y(x),x)+(x+4)*y(x) = 0; 
dsolve(ode,y(x),type='series',x=0);
 

$$y=x^2((c_2\ln \left(x\right)+c_1)(1+5x+17x^2+\frac{143}{3}{x}^3+\frac{355}{3}{x}^4+\frac{4043}{15}{x}^5+\mathrm{O}\left(x^6\right))+((-3)x-\frac{29}{2}{x}^2-\frac{859}{18}{x}^3-\frac{4693}{36}{x}^4-\frac{285181}{900}{x}^5+\mathrm{O}\left(x^6\right))c_2)$$

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)] 
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 step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x^2(x^2-2x+1)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)-x(3+x)\left(\frac{d}{dx}y\left(x\right)\right)+(4+x)y\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{(4+x)y\left(x\right)}{x^2(x^2-2x+1)}+\frac{(3+x)\left(\frac{d}{dx}y\left(x\right)\right)}{x(x^2-2x+1)}\\ \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{(3+x)\left(\frac{d}{dx}y\left(x\right)\right)}{x(x^2-2x+1)}+\frac{(4+x)y\left(x\right)}{x^2(x^2-2x+1)}=0\\ ◻& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=-\frac{3+x}{x(x^2-2x+1)},P_3\left(x\right)=\frac{4+x}{x^2(x^2-2x+1)}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=-3\\ & \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}=4\\ & \circ & x=0\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & & x_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & x^2(x^2-2x+1)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)-x(3+x)\left(\frac{d}{dx}y\left(x\right)\right)+(4+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}^m\cdot y\left(x\right)\text{to series expansion for}m=0..1\\ & & x^m\cdot y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+r+m}\\ & \circ & \text{Shift index using}k->k-m\\ & & x^m\cdot y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=m}}{a}_{k-m}{x}^{k+r}\\ & \circ & \text{Convert}{x}^m\cdot \left(\frac{d}{dx}y\left(x\right)\right)\text{to series expansion for}m=1..2\\ & & x^m\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)x^{k+r-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & x^m\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1+m}}{a}_{k+1-m}(k+1-m+r)x^{k+r}\\ & \circ & \text{Convert}{x}^m\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)\text{to series expansion for}m=2..4\\ & & x^m\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-2+m}\\ & \circ & \text{Shift index using}k->k+2-m\\ & & x^m\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-2+m}}{a}_{k+2-m}(k+2-m+r)(k+1-m+r)x^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_0{(-2+r)}^2{x}^r+(a_1{(-1+r)}^2-a_0(1+2r)(-1+r))x^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(a_k{(k+r-2)}^2-a_{k-1}(2k-1+2r)(k+r-2)+a_{k-2}(k+r-2)(k-3+r))x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & {(-2+r)}^2=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r=2\\ \bullet & & \text{Each term must be 0}\\ & & a_1{(-1+r)}^2-a_0(1+2r)(-1+r)=0\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & a_1=\frac{a_0(1+2r)}{-1+r}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & (k+r-2)((a_k+a_{k-2}-2a_{k-1})k+(a_k+a_{k-2}-2a_{k-1})r-2a_k-3a_{k-2}+a_{k-1})=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & (k+r)((a_{k+2}+a_k-2a_{k+1})(k+2)+(a_{k+2}+a_k-2a_{k+1})r-2a_{k+2}-3a_k+a_{k+1})=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=-\frac{k{a}_k-2k{a}_{k+1}+r{a}_k-2r{a}_{k+1}-a_k-3a_{k+1}}{k+r}\\ \bullet & & \text{Recursion relation for}r=2\\ & & a_{k+2}=-\frac{k{a}_k-2k{a}_{k+1}+a_k-7a_{k+1}}{k+2}\\ \bullet & & \text{Solution for}r=2\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+2},a_{k+2}=-\frac{k{a}_k-2k{a}_{k+1}+a_k-7a_{k+1}}{k+2},a_1=5a_0]\end{array}$$

✓ Mathematica. Time used: 0.023 (sec). Leaf size: 118

ode=x^2*(1-2*x+x^2)*D[y[x],{x,2}] -x*(3+x)*D[y[x],x]+(4+x)*y[x] == 0; 
ic={}; 
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
 

$$y(x)\to c_1(\frac{4043x^5}{15}+\frac{355x^4}{3}+\frac{143x^3}{3}+17x^2+5x+1)x^2+c_2((-\frac{285181x^5}{900}-\frac{4693x^4}{36}-\frac{859x^3}{18}-\frac{29x^2}{2}-3x)x^2+(\frac{4043x^5}{15}+\frac{355x^4}{3}+\frac{143x^3}{3}+17x^2+5x+1)x^2\log (x))$$

✓ Sympy. Time used: 1.133 (sec). Leaf size: 10

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x**2*(x**2 - 2*x + 1)*Derivative(y(x), (x, 2)) - x*(x + 3)*Derivative(y(x), x) + (x + 4)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
 

$$y\left(x\right)=C_1{x}^2+O\left(x^6\right)$$