2.1.259 Problem 262

Internal problem ID [9431]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 262
Date solved : Wednesday, March 05, 2025 at 07:49:09 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}(4x^3-14x^2-2x)y^{\prime \prime }-(6x^2-7x+1)y^{\prime }+(6x-1)y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.416 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& (4x^3-14x^2-2x)y^{\prime \prime }+(-6x^2+7x-1)y^{\prime }+(6x-1)y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =4x^3-14x^2-2x\\ \text{(3)}& B& =-6x^2+7x-1C& =6x-1\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{-12x^4+156x^3+297x^2-78x-3}{16{(2x^3-7x^2-x)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-12x^4+156x^3+297x^2-78x-3\\ t& =16{(2x^3-7x^2-x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{-12x^4+156x^3+297x^2-78x-3}{16{(2x^3-7x^2-x)}^2}\right)z(x)\end{array}$$

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 2.259: Necessary conditions for each Kovacic case

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

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

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=16{(2x^3-7x^2-x)}^2$. There is a pole at $x=0$ of order $2$. There is a pole at $x=\frac{7}{4}+\frac{\sqrt{57}}{4}$ of order $2$. There is a pole at $x=\frac{7}{4}-\frac{\sqrt{57}}{4}$ 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. Since pole order is not larger than $2$ and the order at $\mathrm{\infty }$ is $2$ 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

$$r=-\frac{9}{4x}-\frac{3}{16x^2}+\frac{3}{4{(x-\frac{7}{4}-\frac{\sqrt{57}}{4})}^2}+\frac{3}{4{(x-\frac{7}{4}+\frac{\sqrt{57}}{4})}^2}+\frac{\frac{9}{8}-\frac{29\sqrt{57}}{152}}{x-\frac{7}{4}-\frac{\sqrt{57}}{4}}+\frac{\frac{9}{8}+\frac{29\sqrt{57}}{152}}{x-\frac{7}{4}+\frac{\sqrt{57}}{4}}$$

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 $b=-\frac{3}{16}$. Hence

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

For the pole at $x=\frac{7}{4}+\frac{\sqrt{57}}{4}$ let $b$ be the coefficient of $\frac{1}{{(x-\frac{7}{4}-\frac{\sqrt{57}}{4})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{3}{4}$. Hence

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

For the pole at $x=\frac{7}{4}-\frac{\sqrt{57}}{4}$ let $b$ be the coefficient of $\frac{1}{{(x-\frac{7}{4}+\frac{\sqrt{57}}{4})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{3}{4}$. Hence

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

Since the order of $r$ at $\mathrm{\infty }$ is 2 then $[\sqrt{r}{]}_{\mathrm{\infty }}=0$. Let $b$ be the coefficient of $\frac{1}{x^2}$ in the Laurent series expansion of $r$ at $\mathrm{\infty }$. which can be found by dividing the leading coefficient of $s$ by the leading coefficient of $t$ from

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{-12x^4+156x^3+297x^2-78x-3}{16{(2x^3-7x^2-x)}^2}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=-\frac{3}{16}$. Hence

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

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

$$r=\frac{-12x^4+156x^3+297x^2-78x-3}{16{(2x^3-7x^2-x)}^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$2$	$0$	$\frac{3}{4}$	$\frac{1}{4}$
$\frac{7}{4}+\frac{\sqrt{57}}{4}$	$2$	$0$	$\frac{3}{2}$	$-\frac{1}{2}$
$\frac{7}{4}-\frac{\sqrt{57}}{4}$	$2$	$0$	$\frac{3}{2}$	$-\frac{1}{2}$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$2$	$0$	$\frac{3}{4}$	$\frac{1}{4}$

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 ${\alpha }_{\mathrm{\infty }}^{-}=\frac{1}{4}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{-}+{\alpha }_{c_3}^{-})\\ & =\frac{1}{4}-(-\frac{3}{4})\\ & =1\end{array}$$

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

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{x-c_1})+((-)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{-}}{x-c_2})+((-)[\sqrt{r}{]}_{c_3}+\frac{{\alpha }_{c_3}^{-}}{x-c_3})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =\frac{1}{4x}-\frac{1}{2(x-\frac{7}{4}-\frac{\sqrt{57}}{4})}-\frac{1}{2(x-\frac{7}{4}+\frac{\sqrt{57}}{4})}+(-)\left(0\right)\\ & =\frac{1}{4x}-\frac{1}{2(x-\frac{7}{4}-\frac{\sqrt{57}}{4})}-\frac{1}{2(x-\frac{7}{4}+\frac{\sqrt{57}}{4})}\\ & =\frac{-6x^2+7x-1}{8x^3-28x^2-4x}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=1$ 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

$$\begin{array}{}\text{(2A)}& p(x)& =x+a_0\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(\frac{1}{4x}-\frac{1}{2(x-\frac{7}{4}-\frac{\sqrt{57}}{4})}-\frac{1}{2(x-\frac{7}{4}+\frac{\sqrt{57}}{4})})\left(1\right)+((-\frac{1}{4x^2}+\frac{1}{2{(x-\frac{7}{4}-\frac{\sqrt{57}}{4})}^2}+\frac{1}{2{(x-\frac{7}{4}+\frac{\sqrt{57}}{4})}^2})+{(\frac{1}{4x}-\frac{1}{2(x-\frac{7}{4}-\frac{\sqrt{57}}{4})}-\frac{1}{2(x-\frac{7}{4}+\frac{\sqrt{57}}{4})})}^2-\left(\frac{-12x^4+156x^3+297x^2-78x-3}{16{(2x^3-7x^2-x)}^2}\right))& =0\\ \frac{32(a_0+1)(6x-1)(2x^2-7x-1)}{{(4x-7+\sqrt{57})}^2{(-4x+7+\sqrt{57})}^{2}x}=0\end{array}$$

Solving for the coefficients $a_i$ in the above using method of undetermined coefficients gives

$$\{a_0=-1\}$$

Substituting these coefficients in $p(x)$ in eq. (2A) results in

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

Therefore the first solution to the ode $z^{″}=rz$ is

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & =(x-1){\mathrm{e}}^{\int (\frac{1}{4x}-\frac{1}{2(x-\frac{7}{4}-\frac{\sqrt{57}}{4})}-\frac{1}{2(x-\frac{7}{4}+\frac{\sqrt{57}}{4})})dx}\\ & =(x-1){\mathrm{e}}^{\frac{2\ln \left(x\right)}{(7+\sqrt{57})(-7+\sqrt{57})}-\frac{(57+7\sqrt{57})\sqrt{57}\ln (4x-7-\sqrt{57})}{2(399+57\sqrt{57})}+\frac{(-57+7\sqrt{57})\sqrt{57}\ln (4x-7+\sqrt{57})}{-798+114\sqrt{57}}}\\ & =\frac{(x-1)x^{1/4}}{\sqrt{4x-7-\sqrt{57}}\sqrt{4x-7+\sqrt{57}}}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-6x^2+7x-1}{4x^3-14x^2-2x}dx}\\ & =z_1{e}^{-\frac{\ln \left(x\right)}{4}+\frac{\ln (2x^2-7x-1)}{2}}\\ & =z_1\left(\frac{\sqrt{2x^2-7x-1}}{x^{1/4}}\right)\end{array}$$

Which simplifies to

$$y_1=\frac{(x-1)\sqrt{2}}{4}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-6x^2+7x-1}{4x^3-14x^2-2x}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-\frac{\ln \left(x\right)}{2}+\ln (2x^2-7x-1)}}{{\left(y_1\right)}^2}dx\\ & =y_1\left(\frac{16x(2x+1){\mathrm{e}}^{-\frac{\ln \left(x\right)}{2}+\ln (2x^2-7x-1)}}{(x-1)(2x^2-7x-1)}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{(x-1)\sqrt{2}}{4}\right)+c_2\left(\frac{(x-1)\sqrt{2}}{4}\left(\frac{16x(2x+1){\mathrm{e}}^{-\frac{\ln \left(x\right)}{2}+\ln (2x^2-7x-1)}}{(x-1)(2x^2-7x-1)}\right)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.007 (sec). Leaf size: 21

ode:=(4*x^3-14*x^2-2*x)*diff(diff(y(x),x),x)-(6*x^2-7*x+1)*diff(y(x),x)+(6*x-1)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=c_2\sqrt{x}+c_1(x-1)+2c_2{x}^{3/2}$$

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) 
   Reducible group (found another exponential solution) 
<- Kovacics algorithm successful`
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & (4x^3-14x^2-2x)\left(\frac{d^2}{d{x}^2}y\left(x\right)\right)-(6x^2-7x+1)\left(\frac{d}{dx}y\left(x\right)\right)+(6x-1)y\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d^2}{d{x}^2}y\left(x\right)=-\frac{(6x-1)y\left(x\right)}{2x(2x^2-7x-1)}+\frac{(6x^2-7x+1)\left(\frac{d}{dx}y\left(x\right)\right)}{2x(2x^2-7x-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^2}{d{x}^2}y\left(x\right)-\frac{(6x^2-7x+1)\left(\frac{d}{dx}y\left(x\right)\right)}{2x(2x^2-7x-1)}+\frac{(6x-1)y\left(x\right)}{2x(2x^2-7x-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{6x^2-7x+1}{2x(2x^2-7x-1)},P_3\left(x\right)=\frac{6x-1}{2x(2x^2-7x-1)}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=\frac{1}{2}\\ & \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\text{is a regular singular point}\\ & & x_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & 2x(2x^2-7x-1)\left(\frac{d^2}{d{x}^2}y\left(x\right)\right)+(-6x^2+7x-1)\left(\frac{d}{dx}y\left(x\right)\right)+(6x-1)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}^m\cdot \left(\frac{d^2}{d{x}^2}y\left(x\right)\right)\text{to series expansion for}m=1..3\\ & & x^m\cdot \left(\frac{d^2}{d{x}^2}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^2}{d{x}^2}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}r(-1+2r)x^{-1+r}+(-a_1(1+r)(1+2r)-a_0(14r^2-21r+1))x^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(-a_{k+1}(k+1+r)(2k+1+2r)-a_k(14k^2+28kr+14r^2-21k-21r+1)+2a_{k-1}(k-2+r)(2k-5+2r))x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & -r(-1+2r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,\frac{1}{2}\}\\ \bullet & & \text{Each term must be 0}\\ & & -a_1(1+r)(1+2r)-a_0(14r^2-21r+1)=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & (-14a_k+4a_{k-1}-2a_{k+1})k^2+((-28a_k+8a_{k-1}-4a_{k+1})r+21a_k-18a_{k-1}-3a_{k+1})k+(-14a_k+4a_{k-1}-2a_{k+1})r^2+(21a_k-18a_{k-1}-3a_{k+1})r-a_k+20a_{k-1}-a_{k+1}=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & (-14a_{k+1}+4a_k-2a_{k+2}){(k+1)}^2+((-28a_{k+1}+8a_k-4a_{k+2})r+21a_{k+1}-18a_k-3a_{k+2})(k+1)+(-14a_{k+1}+4a_k-2a_{k+2})r^2+(21a_{k+1}-18a_k-3a_{k+2})r-a_{k+1}+20a_k-a_{k+2}=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=\frac{4k^2{a}_k-14k^2{a}_{k+1}+8kr{a}_k-28kr{a}_{k+1}+4r^2{a}_k-14r^2{a}_{k+1}-10k{a}_k-7k{a}_{k+1}-10r{a}_k-7r{a}_{k+1}+6a_k+6a_{k+1}}{2k^2+4kr+2r^2+7k+7r+6}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+2}=\frac{4k^2{a}_k-14k^2{a}_{k+1}-10k{a}_k-7k{a}_{k+1}+6a_k+6a_{k+1}}{2k^2+7k+6}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+2}=\frac{4k^2{a}_k-14k^2{a}_{k+1}-10k{a}_k-7k{a}_{k+1}+6a_k+6a_{k+1}}{2k^2+7k+6},-a_1-a_0=0]\\ \bullet & & \text{Recursion relation for}r=\frac{1}{2}\\ & & a_{k+2}=\frac{4k^2{a}_k-14k^2{a}_{k+1}-6k{a}_k-21k{a}_{k+1}+2a_k-a_{k+1}}{2k^2+9k+10}\\ \bullet & & \text{Solution for}r=\frac{1}{2}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+\frac{1}{2}},a_{k+2}=\frac{4k^2{a}_k-14k^2{a}_{k+1}-6k{a}_k-21k{a}_{k+1}+2a_k-a_{k+1}}{2k^2+9k+10},-3a_1+6a_0=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}{2}}),a_{k+2}=\frac{4k^2{a}_k-14k^2{a}_{k+1}-10k{a}_k-7k{a}_{k+1}+6a_k+6a_{k+1}}{2k^2+7k+6},-a_1-a_0=0,b_{k+2}=\frac{4k^2{b}_k-14k^2{b}_{k+1}-6k{b}_k-21k{b}_{k+1}+2b_k-b_{k+1}}{2k^2+9k+10},-3b_1+6b_0=0]\end{array}$$

✓ Mathematica. Time used: 1.932 (sec). Leaf size: 155

ode=(4*x^3-14*x^2-2*x)*D[y[x],{x,2}]-(6*x^2-7*x+1)*D[y[x],x]+(6*x-1)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to (x-1)\exp ({\int }_1^x\frac{6K[1{]}^2-7K[1]+1}{-8K[1{]}^3+28K[1{]}^2+4K[1]}dK[1]-\frac{1}{2}{\int }_1^x(\frac{7-4K[2]}{K[2](2K[2]-7)-1}+\frac{1}{2K[2]})dK[2])(c_2{\int }_1^x\frac{\exp (-2{\int }_1^{K[3]}\frac{6K[1{]}^2-7K[1]+1}{-8K[1{]}^3+28K[1{]}^2+4K[1]}dK[1])}{(K[3]-1{)}^2}dK[3]+c_1)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq((6*x - 1)*y(x) - (6*x**2 - 7*x + 1)*Derivative(y(x), x) + (4*x**3 - 14*x**2 - 2*x)*Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False