2.1.546 Problem 562

Internal problem ID [9718]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 562
Date solved : Sunday, March 30, 2025 at 02:45:04 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order ode using Kovacic algorithm

Time used: 1.045 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =x^2(x^2+x+1)\\ \text{(3)}& B& =2x^3+4x^2-xC& =1\end{array}$$

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(x)& =y{e}^{\int \frac{B}{2A}dx}\end{array}$$

Then (2) becomes

$$\begin{array}{}\text{(4)}& z^{″}(x)=rz(x)\end{array}$$

Where $r$ is given by

$$\begin{array}{rl}\text{(5)}& r& =\frac{s}{t}& =\frac{2A{B}^{\prime }-2B{A}^{\prime }+B^2-4AC}{4A^2}\end{array}$$

Substituting the values of $A,B,C$ from (3) in the above and simplifying gives

$$\begin{array}{}\text{(6)}& r& =\frac{10x^2-8x-1}{4{(x^3+x^2+x)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =10x^2-8x-1\\ t& =4{(x^3+x^2+x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{10x^2-8x-1}{4{(x^3+x^2+x)}^2}\right)z(x)\end{array}$$

Equation (7) is now solved. After finding $z(x)$ then $y$ is found using the inverse transformation

$$\begin{array}{rl}y& =z\left(x\right)e^{-\int \frac{B}{2A}dx}\end{array}$$

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of $r$ and the order of $r$ at $\mathrm{\infty }$. The following table summarizes these cases.

Case	Allowed pole order for $r$	Allowed value for $\mathcal{O}(\mathrm{\infty })$
1	$\{0,1,2,4,6,8,\cdots \}$	$\{\cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \}$
2	Need to have at least one pole that is either order $2$ or odd order greater than $2$. Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. $\{1,2\}$,$\{1,3\}$,$\{2\}$,$\{3\}$,$\{3,4\}$,$\{1,2,5\}$.	no condition
3	$\{1,2\}$	$\{2,3,4,5,6,7,\cdots \}$

Table 2.546: Necessary conditions for each Kovacic case

The order of $r$ at $\mathrm{\infty }$ is the degree of $t$ minus the degree of $s$. Therefore

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =6-2\\ & =4\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=4{(x^3+x^2+x)}^2$. There is a pole at $x=0$ of order $2$. There is a pole at $x=-\frac{1}{2}+\frac{i\sqrt{3}}{2}$ of order $2$. There is a pole at $x=-\frac{1}{2}-\frac{i\sqrt{3}}{2}$ of order $2$. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $4$ then the necessary conditions for case one are met. Since there is a pole of order $2$ then necessary conditions for case two are met. Since pole order is not larger than $2$ and the order at $\mathrm{\infty }$ is $4$ then the necessary conditions for case three are met. Therefore

$$\begin{array}{rl}L& =[1,2,4,6,12]\end{array}$$

Attempting to find a solution using case $n=1$.

Looking at poles of order 2. The partial fractions decomposition of $r$ is

$$r=\frac{-\frac{29}{24}-\frac{7i\sqrt{3}}{24}}{{(x+\frac{1}{2}-\frac{i\sqrt{3}}{2})}^2}+\frac{-\frac{29}{24}+\frac{7i\sqrt{3}}{24}}{{(x+\frac{1}{2}+\frac{i\sqrt{3}}{2})}^2}+\frac{\frac{3}{4}-\frac{41i\sqrt{3}}{36}}{x+\frac{1}{2}-\frac{i\sqrt{3}}{2}}+\frac{\frac{3}{4}+\frac{41i\sqrt{3}}{36}}{x+\frac{1}{2}+\frac{i\sqrt{3}}{2}}-\frac{3}{2x}-\frac{1}{4x^2}$$

For the pole at $x=0$ let $b$ be the coefficient of $\frac{1}{x^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=-\frac{1}{4}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{1}{2}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{1}{2}\end{array}$$

For the pole at $x=-\frac{1}{2}+\frac{i\sqrt{3}}{2}$ let $b$ be the coefficient of $\frac{1}{{(x+\frac{1}{2}-\frac{i\sqrt{3}}{2})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=-\frac{29}{24}-\frac{7i\sqrt{3}}{24}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{1}{2}+\frac{\sqrt{-138-42i\sqrt{3}}}{12}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{1}{2}-\frac{\sqrt{-138-42i\sqrt{3}}}{12}\end{array}$$

For the pole at $x=-\frac{1}{2}-\frac{i\sqrt{3}}{2}$ let $b$ be the coefficient of $\frac{1}{{(x+\frac{1}{2}+\frac{i\sqrt{3}}{2})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=-\frac{29}{24}+\frac{7i\sqrt{3}}{24}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{1}{2}+\frac{\sqrt{-138+42i\sqrt{3}}}{12}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{1}{2}-\frac{\sqrt{-138+42i\sqrt{3}}}{12}\end{array}$$

Since the order of $r$ at $\mathrm{\infty }$ is $4>2$ then

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =0\\ {\alpha }_{\mathrm{\infty }}^{+}& =0\\ {\alpha }_{\mathrm{\infty }}^{-}& =1\end{array}$$

The following table summarizes the findings so far for poles and for the order of $r$ at $\mathrm{\infty }$ where $r$ is

$$r=\frac{10x^2-8x-1}{4{(x^3+x^2+x)}^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$2$	$0$	$\frac{1}{2}$	$\frac{1}{2}$
$-\frac{1}{2}+\frac{i\sqrt{3}}{2}$	$2$	$0$	$\frac{1}{2}+\frac{\sqrt{-138-42i\sqrt{3}}}{12}$	$\frac{1}{2}-\frac{\sqrt{-138-42i\sqrt{3}}}{12}$
$-\frac{1}{2}-\frac{i\sqrt{3}}{2}$	$2$	$0$	$\frac{1}{2}+\frac{\sqrt{-138+42i\sqrt{3}}}{12}$	$\frac{1}{2}-\frac{\sqrt{-138+42i\sqrt{3}}}{12}$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$4$	$0$	$0$	$1$

Now that the all $[\sqrt{r}{]}_c$ and its associated ${\alpha }_c^{\pm }$ have been determined for all the poles in the set $\mathrm{\Gamma }$ and $[\sqrt{r}{]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$ have also been found, the next step is to determine possible non negative integer $d$ from these using

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{s(\mathrm{\infty })}-\sum _{c\in \mathrm{\Gamma }}{\alpha }_c^{s(c)}\end{array}$$

Where $s(c)$ is either $+$ or $-$ and $s(\mathrm{\infty })$ is the sign of ${\alpha }_{\mathrm{\infty }}^{\pm }$. This is done by trial over all set of families $s=(s(c){)}_{c\in \mathrm{\Gamma }\cup \mathrm{\infty }}$ until such $d$ is found to work in finding candidate $\omega$. Trying ${\alpha }_{\mathrm{\infty }}^{-}=1$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{-}+{\alpha }_{c_3}^{-})\\ & =1-\left(1\right)\\ & =0\end{array}$$

Since $d$ an integer and $d\ge 0$ then it can be used to find $\omega$ using

$$\begin{array}{rl}\omega & =\sum _{c\in \mathrm{\Gamma }}(s(c)[\sqrt{r}{]}_c+\frac{{\alpha }_c^{s(c)}}{x-c})+s(\mathrm{\infty })[\sqrt{r}{]}_{\mathrm{\infty }}\end{array}$$

The above gives

$$\begin{array}{rl}\omega & =((+)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{+}}{x-c_1})+((-)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{-}}{x-c_2})+((-)[\sqrt{r}{]}_{c_3}+\frac{{\alpha }_{c_3}^{-}}{x-c_3})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =\frac{1}{2x}+\frac{\frac{1}{2}-\frac{\sqrt{-138-42i\sqrt{3}}}{12}}{x+\frac{1}{2}-\frac{i\sqrt{3}}{2}}+\frac{\frac{1}{2}-\frac{\sqrt{-138+42i\sqrt{3}}}{12}}{x+\frac{1}{2}+\frac{i\sqrt{3}}{2}}+(-)\left(0\right)\\ & =\frac{1}{2x}+\frac{\frac{1}{2}-\frac{\sqrt{-138-42i\sqrt{3}}}{12}}{x+\frac{1}{2}-\frac{i\sqrt{3}}{2}}+\frac{\frac{1}{2}-\frac{\sqrt{-138+42i\sqrt{3}}}{12}}{x+\frac{1}{2}+\frac{i\sqrt{3}}{2}}\\ & =\frac{2x^2-2x+1}{2x(x^2+x+1)}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=0$ to solve the ode. The polynomial $p(x)$ needs to satisfy the equation

$$\begin{array}{}\text{(1A)}& p^{″}+2\omega p^{\prime }+({\omega }^{\prime }+{\omega }^2-r)p=0\end{array}$$

Let

$$\begin{array}{}\text{(2A)}& p(x)& =1\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(\frac{1}{2x}+\frac{\frac{1}{2}-\frac{\sqrt{-138-42i\sqrt{3}}}{12}}{x+\frac{1}{2}-\frac{i\sqrt{3}}{2}}+\frac{\frac{1}{2}-\frac{\sqrt{-138+42i\sqrt{3}}}{12}}{x+\frac{1}{2}+\frac{i\sqrt{3}}{2}})\left(0\right)+((-\frac{1}{2x^2}-\frac{\frac{1}{2}-\frac{\sqrt{-138-42i\sqrt{3}}}{12}}{{(x+\frac{1}{2}-\frac{i\sqrt{3}}{2})}^2}-\frac{\frac{1}{2}-\frac{\sqrt{-138+42i\sqrt{3}}}{12}}{{(x+\frac{1}{2}+\frac{i\sqrt{3}}{2})}^2})+{(\frac{1}{2x}+\frac{\frac{1}{2}-\frac{\sqrt{-138-42i\sqrt{3}}}{12}}{x+\frac{1}{2}-\frac{i\sqrt{3}}{2}}+\frac{\frac{1}{2}-\frac{\sqrt{-138+42i\sqrt{3}}}{12}}{x+\frac{1}{2}+\frac{i\sqrt{3}}{2}})}^2-\left(\frac{10x^2-8x-1}{4{(x^3+x^2+x)}^2}\right))& =0\\ 0=0\end{array}$$

The equation is satisfied since both sides are zero. Therefore the first solution to the ode $z^{″}=rz$ is

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & ={\mathrm{e}}^{\int (\frac{1}{2x}+\frac{\frac{1}{2}-\frac{\sqrt{-138-42i\sqrt{3}}}{12}}{x+\frac{1}{2}-\frac{i\sqrt{3}}{2}}+\frac{\frac{1}{2}-\frac{\sqrt{-138+42i\sqrt{3}}}{12}}{x+\frac{1}{2}+\frac{i\sqrt{3}}{2}})dx}\\ & ={(x^2+x+1)}^{1/4}\sqrt{2}\sqrt{x}{\mathrm{e}}^{-\frac{7\sqrt{3}\arctan \left(\frac{(2x+1)\sqrt{3}}{3}\right)}{6}}\end{array}$$

The first solution to the original ode in $y$ is found from

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{2x^3+4x^2-x}{x^2(x^2+x+1)}dx}\\ & =z_1{e}^{-\frac{3\ln (x^2+x+1)}{4}-\frac{7\sqrt{3}\arctan \left(\frac{(2x+1)\sqrt{3}}{3}\right)}{6}+\frac{\ln \left(x\right)}{2}}\\ & =z_1\left(\frac{\sqrt{x}{\mathrm{e}}^{-\frac{7\sqrt{3}\arctan \left(\frac{(2x+1)\sqrt{3}}{3}\right)}{6}}}{{(x^2+x+1)}^{3/4}}\right)\end{array}$$

Which simplifies to

$$y_1=\frac{x{\mathrm{e}}^{-\frac{7\sqrt{3}\arctan \left(\frac{(2x+1)\sqrt{3}}{3}\right)}{3}}\sqrt{2}}{\sqrt{x^2+x+1}}$$

The second solution $y_2$ to the original ode is found using reduction of order

$$y_2=y_1\int \frac{e^{\int -\frac{B}{A}dx}}{y_1^2}dx$$

Substituting gives

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{2x^3+4x^2-x}{x^2(x^2+x+1)}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-\frac{3\ln (x^2+x+1)}{2}-\frac{7\sqrt{3}\arctan \left(\frac{(2x+1)\sqrt{3}}{3}\right)}{3}+\ln \left(x\right)}}{{\left(y_1\right)}^2}dx\\ & =y_1(\int \frac{{\mathrm{e}}^{-\frac{3\ln (x^2+x+1)}{2}-\frac{7\sqrt{3}\arctan \left(\frac{(2x+1)\sqrt{3}}{3}\right)}{3}+\ln \left(x\right)}(x^2+x+1){\mathrm{e}}^{\frac{14\sqrt{3}\arctan \left(\frac{(2x+1)\sqrt{3}}{3}\right)}{3}}}{2x^2}dx)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{x{\mathrm{e}}^{-\frac{7\sqrt{3}\arctan \left(\frac{(2x+1)\sqrt{3}}{3}\right)}{3}}\sqrt{2}}{\sqrt{x^2+x+1}}\right)+c_2\left(\frac{x{\mathrm{e}}^{-\frac{7\sqrt{3}\arctan \left(\frac{(2x+1)\sqrt{3}}{3}\right)}{3}}\sqrt{2}}{\sqrt{x^2+x+1}}(\int \frac{{\mathrm{e}}^{-\frac{3\ln (x^2+x+1)}{2}-\frac{7\sqrt{3}\arctan \left(\frac{(2x+1)\sqrt{3}}{3}\right)}{3}+\ln \left(x\right)}(x^2+x+1){\mathrm{e}}^{\frac{14\sqrt{3}\arctan \left(\frac{(2x+1)\sqrt{3}}{3}\right)}{3}}}{2x^2}dx)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.441 (sec). Leaf size: 147

ode:=x^2*(x^2+x+1)*diff(diff(y(x),x),x)-x*(-2*x^2-4*x+1)*diff(y(x),x)+y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{{\mathrm{e}}^{-\frac{7\sqrt{3}\arctan \left(\frac{(2x+1)\sqrt{3}}{3}\right)}{6}}x(c_1{(i\sqrt{3}-2x-1)}^{\frac{1}{4}+\frac{7i\sqrt{3}}{12}}{(i\sqrt{3}+2x+1)}^{\frac{1}{4}-\frac{7i\sqrt{3}}{12}}+c_2{(i\sqrt{3}-2x-1)}^{-\frac{1}{4}-\frac{7i\sqrt{3}}{12}}{(i\sqrt{3}+2x+1)}^{\frac{3}{4}+\frac{7i\sqrt{3}}{12}}\mathrm{hypergeom}([1,\frac{1}{2}+\frac{7i\sqrt{3}}{6}],[\frac{3}{2}+\frac{7i\sqrt{3}}{6}],\frac{-i\sqrt{3}x+x+2}{i\sqrt{3}x+x+2}))}{{(x^2+x+1)}^{3/4}}$$

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 
   Solution has integrals. Trying a special function solution free of integrals\ 
... 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      -> elliptic 
      -> Legendre 
      -> Kummer 
         -> hyper3: Equivalence to 1F1 under a power @ Moebius 
      -> hypergeometric 
         -> heuristic approach 
         -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
         <- hyper3 successful: received ODE is equivalent to the 2F1 ODE 
      <- hypergeometric successful 
   <- special function solution successful 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form is not straightforward to achieve - returning special \ 
function solution free of uncomputed integrals 
   <- Kovacics algorithm successful
 

Maple step by step

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

✓ Mathematica. Time used: 0.552 (sec). Leaf size: 130

ode=x^2*(1+x+x^2)*D[y[x],{x,2}]-x*(1-4*x-2*x^2)*D[y[x],x]+y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \exp ({\int }_1^x\frac{1}{2}(\frac{K[1]-3}{K[1{]}^2+K[1]+1}+\frac{1}{K[1]})dK[1]-\frac{1}{2}{\int }_1^x(\frac{3K[2]+5}{K[2{]}^2+K[2]+1}-\frac{1}{K[2]})dK[2])(c_2{\int }_1^x\exp (-2{\int }_1^{K[3]}\frac{2K[1{]}^2-2K[1]+1}{2K[1](K[1{]}^2+K[1]+1)}dK[1])dK[3]+c_1)$$

✗ Sympy

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

False