2.1.83 Problem 85

Internal problem ID [9253]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 85
Date solved : Friday, April 25, 2025 at 06:03:36 PM
CAS classification : [[_2nd_order, _with_linear_symmetries], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]

Solve

$$\begin{array}{rl}3x^2{y}^{\prime \prime }+2x(-2x^2+x+1)y^{\prime }+(-8x^2+2x)y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.570 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& 3x^2{y}^{\prime \prime }+(-4x^3+2x^2+2x)y^{\prime }+(-8x^2+2x)y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =3x^2\\ \text{(3)}& B& =-4x^3+2x^2+2xC& =-8x^2+2x\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{4x^4-4x^3+15x^2-4x-2}{9x^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =4x^4-4x^3+15x^2-4x-2\\ t& =9x^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{4x^4-4x^3+15x^2-4x-2}{9x^2}\right)z(x)\end{array}$$

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 2.83: Necessary conditions for each Kovacic case

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

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

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=9x^2$. There is a pole at $x=0$ of order $2$. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $-2$ 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. Therefore

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

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

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

$$r=\frac{4x^2}{9}-\frac{4x}{9}+\frac{5}{3}-\frac{2}{9x^2}-\frac{4}{9x}$$

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 $b=-\frac{2}{9}$. Hence

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

Since the order of $r$ at $\mathrm{\infty }$ is $O_r(\mathrm{\infty })=-2$ then

$$\begin{array}{rlrlrl}v& =\frac{-O_r(\mathrm{\infty })}{2}& & =\frac{2}{2}& & =1\end{array}$$

$[\sqrt{r}{]}_{\mathrm{\infty }}$ is the sum of terms involving $x^i$ for $0\le i\le v$ in the Laurent series for $\sqrt{r}$ at $\mathrm{\infty }$. Therefore

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^v{a}_i{x}^i\\ \text{(8)}& & =\sum _{i=0}^1{a}_i{x}^i\end{array}$$

Let $a$ be the coefficient of $x^v=x^1$ in the above sum. The Laurent series of $\sqrt{r}$ at $\mathrm{\infty }$ is

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx \frac{2x}{3}-\frac{1}{3}+\frac{7}{6x}+\frac{1}{4x^2}-\frac{17}{16x^3}-\frac{31}{32x^4}+\frac{85}{64x^5}+\frac{353}{128x^6}+\dots \end{array}$$

Comparing Eq. (9) with Eq. (8) shows that

$$a=\frac{2}{3}$$

From Eq. (9) the sum up to $v=1$ gives

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^1{a}_i{x}^i\\ \text{(10)}& & =\frac{2x}{3}-\frac{1}{3}\end{array}$$

Now we need to find $b$, where $b$ be the coefficient of $x^{v-1}=x^0=1$ in $r$ minus the coefficient of same term but in ${([\sqrt{r}{]}_{\mathrm{\infty }})}^2$ where $[\sqrt{r}{]}_{\mathrm{\infty }}$ was found above in Eq (10). Hence

$${([\sqrt{r}{]}_{\mathrm{\infty }})}^2=\frac{4}{9}{x}^2-\frac{4}{9}x+\frac{1}{9}$$

This shows that the coefficient of $1$ in the above is $\frac{1}{9}$. Now we need to find the coefficient of $1$ in $r$. How this is done depends on if $v=0$ or not. Since $v=1$ which is not zero, then starting $r=\frac{s}{t}$, we do long division and write this in the form

$$r=Q+\frac{R}{t}$$

Where $Q$ is the quotient and $R$ is the remainder. Then the coefficient of $1$ in $r$ will be the coefficient this term in the quotient. Doing long division gives

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{4x^4-4x^3+15x^2-4x-2}{9x^2}\\ & =Q+\frac{R}{9x^2}\\ & =(\frac{4}{9}{x}^2-\frac{4}{9}x+\frac{5}{3})+\left(\frac{-4x-2}{9x^2}\right)\\ & =\frac{4x^2}{9}-\frac{4x}{9}+\frac{5}{3}+\frac{-4x-2}{9x^2}\end{array}$$

We see that the coefficient of the term $x$ in the quotient is $\frac{5}{3}$. Now $b$ can be found.

$$\begin{array}{rl}b& =\left(\frac{5}{3}\right)-\left(\frac{1}{9}\right)\\ & =\frac{14}{9}\end{array}$$

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\frac{2x}{3}-\frac{1}{3}\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)& & =\frac{1}{2}(\frac{\frac{14}{9}}{\frac{2}{3}}-1)& & =\frac{2}{3}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)& & =\frac{1}{2}(-\frac{\frac{14}{9}}{\frac{2}{3}}-1)& & =-\frac{5}{3}\end{array}$$

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

$$r=\frac{4x^4-4x^3+15x^2-4x-2}{9x^2}$$

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

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$-2$	$\frac{2x}{3}-\frac{1}{3}$	$\frac{2}{3}$	$-\frac{5}{3}$

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 ${\alpha }_{\mathrm{\infty }}^{+}=\frac{2}{3}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_{c_1}^{+}\right)\\ & =\frac{2}{3}-\left(\frac{2}{3}\right)\\ & =0\end{array}$$

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

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

Substituting the above values in the above results in

$$\begin{array}{rl}\omega & =((+)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{+}}{x-c_1})+(+)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =\frac{2}{3x}+(\frac{2x}{3}-\frac{1}{3})\\ & =\frac{2}{3x}+\frac{2x}{3}-\frac{1}{3}\\ & =\frac{2}{3x}+\frac{2x}{3}-\frac{1}{3}\end{array}$$

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

$$\begin{array}{rl}\left(0\right)+2(\frac{2}{3x}+\frac{2x}{3}-\frac{1}{3})\left(0\right)+((-\frac{2}{3x^2}+\frac{2}{3})+{(\frac{2}{3x}+\frac{2x}{3}-\frac{1}{3})}^2-\left(\frac{4x^4-4x^3+15x^2-4x-2}{9x^2}\right))& =0\\ 0=0\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & ={\mathrm{e}}^{\int (\frac{2}{3x}+\frac{2x}{3}-\frac{1}{3})dx}\\ & =x^{2/3}{\mathrm{e}}^{\frac{x(x-1)}{3}}\end{array}$$

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

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

Which simplifies to

$$y_1=x^{1/3}{\mathrm{e}}^{\frac{2x(x-1)}{3}}$$

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

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

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(x^{1/3}{\mathrm{e}}^{\frac{2x(x-1)}{3}}\right)+c_2\left(x^{1/3}{\mathrm{e}}^{\frac{2x(x-1)}{3}}(\int \frac{{\mathrm{e}}^{\frac{2x^2}{3}-\frac{2x}{3}-\frac{2\ln \left(x\right)}{3}}{\mathrm{e}}^{-\frac{4x(x-1)}{3}}}{x^{2/3}}dx)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.055 (sec). Leaf size: 38

ode:=3*x^2*diff(diff(y(x),x),x)+2*x*(-2*x^2+x+1)*diff(y(x),x)+(-8*x^2+2*x)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=c_1{x}^{1/3}{\mathrm{e}}^{\frac{2x(x-1)}{3}}+c_2\mathrm{HeunB}(-\frac{1}{3},\frac{\sqrt{6}}{3},-\frac{7}{3},\frac{4\sqrt{6}}{9},-\frac{\sqrt{6}x}{3})$$

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 
      -> Mathieu 
         -> Equivalence to the rational form of Mathieu ODE under a power @ Mo\ 
ebius 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a \ 
power @ Moebius 
      <- Heun successful: received ODE is equivalent to the  HeunB  ODE, case  \ 
c = 0 
   <- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & 3x^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+2x(-2x^2+x+1)\left(\frac{d}{dx}y\left(x\right)\right)+(-8x^2+2x)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{2(4x-1)y\left(x\right)}{3x}+\frac{2(2x^2-x-1)\left(\frac{d}{dx}y\left(x\right)\right)}{3x}\\ \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{2(2x^2-x-1)\left(\frac{d}{dx}y\left(x\right)\right)}{3x}-\frac{2(4x-1)y\left(x\right)}{3x}=0\\ ◻& & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=-\frac{2(2x^2-x-1)}{3x},P_3\left(x\right)=-\frac{2(4x-1)}{3x}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=\frac{2}{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}=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}\\ & & 3\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)x+(-4x^2+2x+2)\left(\frac{d}{dx}y\left(x\right)\right)+(-8x+2)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=0..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\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+3r)x^{-1+r}+(a_1(1+r)(2+3r)+2a_0(1+r))x^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(a_{k+1}(k+1+r)(3k+2+3r)+2a_k(k+1+r)-4a_{k-1}(k+1+r))x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & r(-1+3r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,\frac{1}{3}\}\\ \bullet & & \text{Each term must be 0}\\ & & a_1(1+r)(2+3r)+2a_0(1+r)=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & (k+1+r)(3k{a}_{k+1}+3r{a}_{k+1}+2a_k-4a_{k-1}+2a_{k+1})=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & (k+r+2)(3(k+1)a_{k+2}+3r{a}_{k+2}+2a_{k+1}-4a_k+2a_{k+2})=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=\frac{2(-a_{k+1}+2a_k)}{3k+5+3r}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+2}=\frac{2(-a_{k+1}+2a_k)}{3k+5}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+2}=\frac{2(-a_{k+1}+2a_k)}{3k+5},2a_1+2a_0=0]\\ \bullet & & \text{Recursion relation for}r=\frac{1}{3}\\ & & a_{k+2}=\frac{2(-a_{k+1}+2a_k)}{3k+6}\\ \bullet & & \text{Solution for}r=\frac{1}{3}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+\frac{1}{3}},a_{k+2}=\frac{2(-a_{k+1}+2a_k)}{3k+6},4a_1+\frac{8a_0}{3}=0]\\ \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+\frac{1}{3}}),a_{k+2}=\frac{2(-a_{k+1}+2a_k)}{3k+5},2a_1+2a_0=0,b_{k+2}=\frac{2(-b_{k+1}+2b_k)}{3k+6},4b_1+\frac{8b_0}{3}=0]\end{array}$$

✓ Mathematica. Time used: 0.943 (sec). Leaf size: 53

ode=3*x^2*D[y[x],{x,2}]+2*x*(1+x-2*x^2)*D[y[x],x]+(2*x-8*x^2)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to e^{\frac{2}{3}(x-1)x}\sqrt[3]{x}(c_2{\int }_1^x\frac{e^{-\frac{2}{3}(K[1]-1)K[1]}}{K[1{]}^{4/3}}dK[1]+c_1)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(3*x**2*Derivative(y(x), (x, 2)) + 2*x*(-2*x**2 + x + 1)*Derivative(y(x), x) + (-8*x**2 + 2*x)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False