2.2.1 Problem 1

Internal problem ID [9994]
Book : Collection of Kovacic problems
Section : section 2. Solution found using all possible Kovacic cases
Problem number : 1
Date solved : Sunday, March 30, 2025 at 02:51:10 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order ode using Kovacic algorithm

Time used: 1.217 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& y^{\prime \prime }+\frac{(32x^2-27x+27)y}{144x^2{(x-1)}^2}& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =0C& =\frac{32x^2-27x+27}{144x^2{(x-1)}^2}\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{-32x^2+27x-27}{144{(x^2-x)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-32x^2+27x-27\\ t& =144{(x^2-x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{-32x^2+27x-27}{144{(x^2-x)}^2}\right)z(x)\end{array}$$

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 2.822: Necessary conditions for each Kovacic case

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

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

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=144{(x^2-x)}^2$. There is a pole at $x=0$ of order $2$. There is a pole at $x=1$ 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 $n=1$.

Unable to find solution using case one

Attempting to find a solution using case $n=2$.

Unable to find solution using case two.

Attempting to find a solution using $n=4$.

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

$$r=-\frac{3}{16x^2}-\frac{3}{16x}-\frac{2}{9{(x-1)}^2}+\frac{3}{16(x-1)}$$

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

$$\begin{array}{rl}{E}_c& =\{6+\frac{12k}{n}\sqrt{1+4b}|k=0,\pm 1,\pm 2,\dots ,\pm \frac{n}{2}\}\cap \mathbb{Z}\end{array}$$

Where $n$ for case $3$ is $4,6$ or $12$. For the current case $n=4$. Hence the above becomes

$$\begin{array}{rl}{E}_c& =\{3,6,9\}\end{array}$$

For the pole at $x=1$ let $b$ be the coefficient of $\frac{1}{{(x-1)}^2}$ in the partial fractions decomposition of $r$ given above. This shows that $b=-\frac{2}{9}$. Hence

$$\begin{array}{rl}{E}_c& =\{6+\frac{12k}{n}\sqrt{1+4b}|k=0,\pm 1,\pm 2,\dots ,\pm \frac{n}{2}\}\cap \mathbb{Z}\end{array}$$

Where $n$ for case $3$ is $4,6$ or $12$. For the current case $n=4$. Hence the above becomes

$$\begin{array}{rl}{E}_c& =\{4,5,6,7,8\}\end{array}$$

Let

$$\begin{array}{}\text{(B1)}& E_{\mathrm{\infty }}& =\{6+\frac{12k}{n}\sqrt{1+4b}|k=0,\pm 1,\pm 2,\dots ,\pm \frac{n}{2}\}\cap \mathbb{Z}\end{array}$$

Where $b$ is the coefficient of $\frac{1}{x^2}$ in the Laurent series for $r$ at $\mathrm{\infty }$ given by

$$r\approx -\frac{2}{9x^2}-\frac{37}{144x^3}-\frac{23}{48x^4}-\frac{101}{144x^5}-\frac{133}{144x^6}-\frac{55}{48x^7}+\cdots$$

The above shows that

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

The value of $n$ in eq. (B1) for case $3$ is $4,6$ or $2$.For the current case $n=4$. eq. (B1) simplifies to the following, after removing any duplicate and non integer entries in the set.

$$\begin{array}{rl}{E}_{\mathrm{\infty }}& =\{4,5,6,7,8\}\end{array}$$

The following table summarizes the results found so far for poles and for the order of $r$ at $\mathrm{\infty }$ for case 3 of Kovacic algorithm using $n=4$.

pole $c$ location	pole order	set $\{E_c\}$
$0$	$2$	$\{3,6,9\}$
$1$	$2$	$\{4,5,6,7,8\}$

Order of $r$ at $\mathrm{\infty }$	set $\{E_{\mathrm{\infty }}\}$
$2$	$\{4,5,6,7,8\}$

Now that $E_c$ sets for all poles are found and $E_{\mathrm{\infty }}$ set is found, the next step is to determine a non negative integer $d$ using the following

$$\begin{array}{rl}d& =\frac{n}{12}(e_{\mathrm{\infty }}-\sum _{c\in \mathrm{\Gamma }}{e}_c)\end{array}$$

Where in the above $e_c$ is a distinct element from each corresponding $E_c$. This means all possible tuples $\{e_{c_1},e_{c_2},\dots ,e_{c_n}\}$ are tried in the sum above, where $e_{c_i}$ is one element of each $E_c$ found earlier. Using the following family $\{e_1,e_2,\dots ,e_{\mathrm{\infty }}\}$ given by

$$e_1=3,e_2=4,e_{\mathrm{\infty }}=7$$

Gives a non negative integer $d$ (the degree of the polynomial $p(x)$), which is generated using

$$\begin{array}{rl}d& =\frac{n}{12}(e_{\mathrm{\infty }}-\sum _{c\in \mathrm{\Gamma }}{e}_c)\\ & =\frac{4}{12}(7-(3+\left(4\right)))\\ & =0\end{array}$$

The following rational function is

$$\begin{array}{rl}\theta & =\frac{n}{12}\sum _{c\in \mathrm{\Gamma }}\frac{e_c}{x-c}\\ & =\frac{4}{12}(\frac{3}{(x-\left(0\right))}+\frac{4}{(x-\left(1\right))})\\ & =\frac{1}{x}+\frac{4}{3x-3}\end{array}$$

And

$$\begin{array}{rl}S& =\prod _{c\in \mathrm{\Gamma }}(x-c)\\ & =x(x-1)\end{array}$$

The polynomial $p(x)$ is now determined. Since the degree of the polynomial is $d=0$, then let

$$p(x)=1$$

The following set of equations are set up in order to determine the coefficients $a_i$ (if any) of the above polynomial

$$\begin{array}{rl}{P}_n& =-p(x)\\ & =-1\\ \text{(1A)}& P_{i-1}& =-S{p}_i^{\prime }+((n-i)S^{\prime }-S\theta )P_i-(n-1)(i+1)S^{2}r{P}_{i+1}i=n,n-1,\dots ,0\end{array}$$

The coefficients $a_i$ are solved for from

$$\begin{array}{}\text{(2A)}& P_{-1}=0\end{array}$$

By using method of undetermined coefficients. Carrying the above computation in eq. (1A) gives the following sequence of polynomials $P_i$ (noting that $n=4$ and $r=\frac{-32x^2+27x-27}{144{(x^2-x)}^2}$).

$$\begin{array}{rl}{P}_4& =-p=-1\\ P_3& =\frac{7x}{3}-1\\ P_2& =-4x^2+\frac{41}{12}x-\frac{3}{4}\\ P_1& =\frac{40}{9}{x}^3-\frac{409}{72}{x}^2+\frac{5}{2}x-\frac{3}{8}\\ P_0& =-\frac{64}{27}{x}^4+\frac{871}{216}{x}^3-\frac{257}{96}{x}^2+\frac{13}{16}x-\frac{3}{32}\\ P_{-1}& =0\end{array}$$

Because $P_{-1}=0$ then $z=e^{\int \omega }$ is a solution. $\omega$ is found by finding a solution to the equation generated by the following sum

$$\begin{array}{rl}\sum _{i=0}^n{S}^i\frac{P_i}{(n-i)!}{\omega }^i& =0\\ \sum _{i=0}^4{S}^i\frac{P_i}{(4-i)!}{\omega }^i& =0\end{array}$$

Where the $P_i$ are the polynomials found earlier. Computing the above sum gives

$$\begin{array}{}\text{(3A)}& -\frac{8x^4}{81}+\frac{871x^3}{5184}-\frac{257x^2}{2304}+\frac{13x}{384}-\frac{1}{256}+\frac{x(x-1)(320x^3-409x^2+180x-27)\omega }{432}-\frac{x^2{(x-1)}^2(48x^2-41x+9){\omega }^2}{24}+x^3{(x-1)}^3(\frac{7x}{3}-1){\omega }^3-x^4{(x-1)}^4{\omega }^4=0\end{array}$$

The solution $\omega$ of eq. 3A is found as

$$\begin{array}{}\text{(4A)}& \omega =\frac{1}{12x(x-1)}(7x-3+\sqrt{x^2+{\left({(x-1)}^2{x}^3\right)}^{1/3}-x}+\sqrt{-\frac{2((-x^2+x+\frac{{\left({(x-1)}^2{x}^3\right)}^{1/3}}{2})\sqrt{x^2+{\left({(x-1)}^2{x}^3\right)}^{1/3}-x}+x^2(x-1))}{\sqrt{x^2+{\left({(x-1)}^2{x}^3\right)}^{1/3}-x}}})\end{array}$$

This $\omega$ is used to find a solution to $z^{″}=rz$.

$$\begin{array}{}\text{(5A)}& z_1(x)=e^{\int \omega dx}\end{array}$$

Unable to integrate $\int \omega dx$. Leaving the integral unevaluated. The first solution to the original ode in $y$ is found from

$$y_1=z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}$$

Since $B=0$ then the above reduces to

$$\begin{array}{rl}{y}_1& =z_1\\ & =e^{\int \omega dx}\end{array}$$

Where $\omega$ given above. The second solution $y_2$ to the original ode is found using reduction of order

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{B}{A}dx}}{y_1^2}dx\\ & =e^{\int \omega dx}\int \frac{e^{\int -\frac{B}{A}dx}}{{\left(e^{\int \omega dx}\right)}^2}dx\end{array}$$

Since $B=0$ then the above reduces to

$$y_2=e^{\int \omega dx}\int {\left(e^{\int \omega dx}\right)}^{-2}dx$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(e^{\int \omega dx}\right)+c_2(e^{\int \omega dx}\int {\left(e^{\int \omega dx}\right)}^{-2}dx)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.007 (sec). Leaf size: 30

ode:=diff(diff(y(x),x),x) = (-3/16/x^2-2/9/(x-1)^2+3/16/x/(x-1))*y(x); 
dsolve(ode,y(x), singsol=all);
 

$$y=x^{1/4}\sqrt{x-1}(c_1\mathrm{LegendreP}(-\frac{1}{6},\frac{1}{3},\sqrt{x})+c_2\mathrm{LegendreQ}(-\frac{1}{6},\frac{1}{3},\sqrt{x}))$$

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 
   Tetrahedral Galois group A4_SL2. 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=(-\frac{3}{16x^2}-\frac{2}{9{(x-1)}^2}+\frac{3}{16x(x-1)})y\left(x\right)\\ \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{(32x^2-27x+27)y\left(x\right)}{144x^2{(x-1)}^2}\\ \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{(32x^2-27x+27)y\left(x\right)}{144x^2{(x-1)}^2}=0\\ ◻& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=0,P_3\left(x\right)=\frac{32x^2-27x+27}{144x^2{(x-1)}^2}]\\ & \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}=\frac{3}{16}\\ & \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}\\ & & 144x^2{(x-1)}^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+(32x^2-27x+27)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..2\\ & & 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}\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}\\ & & 9a_0(-1+4r)(-3+4r)x^r+(9a_1(3+4r)(1+4r)-9a_0(32r^2-32r+3))x^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(9a_k(4k+4r-1)(4k+4r-3)-9a_{k-1}(32{(k-1)}^2+64(k-1)r+32r^2-32k+35-32r)+16a_{k-2}(3k-7+3r)(3k-8+3r))x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & 9(-1+4r)(-3+4r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{\frac{1}{4},\frac{3}{4}\}\\ \bullet & & \text{Each term must be 0}\\ & & 9a_1(3+4r)(1+4r)-9a_0(32r^2-32r+3)=0\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & a_1=\frac{a_0(32r^2-32r+3)}{16r^2+16r+3}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & 144(a_k+a_{k-2}-2a_{k-1})k^2+144(2(a_k+a_{k-2}-2a_{k-1})r-a_k-5a_{k-2}+6a_{k-1})k+144(a_k+a_{k-2}-2a_{k-1})r^2+144(-a_k-5a_{k-2}+6a_{k-1})r+27a_k+896a_{k-2}-603a_{k-1}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & 144(a_{k+2}+a_k-2a_{k+1}){(k+2)}^2+144(2(a_{k+2}+a_k-2a_{k+1})r-a_{k+2}-5a_k+6a_{k+1})(k+2)+144(a_{k+2}+a_k-2a_{k+1})r^2+144(-a_{k+2}-5a_k+6a_{k+1})r+27a_{k+2}+896a_k-603a_{k+1}=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=-\frac{144k^2{a}_k-288k^2{a}_{k+1}+288kr{a}_k-576kr{a}_{k+1}+144r^2{a}_k-288r^2{a}_{k+1}-144k{a}_k-288k{a}_{k+1}-144r{a}_k-288r{a}_{k+1}+32a_k-27a_{k+1}}{9(16k^2+32kr+16r^2+48k+48r+35)}\\ \bullet & & \text{Recursion relation for}r=\frac{1}{4}\\ & & a_{k+2}=-\frac{144k^2{a}_k-288k^2{a}_{k+1}-72k{a}_k-432k{a}_{k+1}+5a_k-117a_{k+1}}{9(16k^2+56k+48)}\\ \bullet & & \text{Solution for}r=\frac{1}{4}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+\frac{1}{4}},a_{k+2}=-\frac{144k^2{a}_k-288k^2{a}_{k+1}-72k{a}_k-432k{a}_{k+1}+5a_k-117a_{k+1}}{9(16k^2+56k+48)},a_1=-\frac{3a_0}{8}]\\ \bullet & & \text{Recursion relation for}r=\frac{3}{4}\\ & & a_{k+2}=-\frac{144k^2{a}_k-288k^2{a}_{k+1}+72k{a}_k-720k{a}_{k+1}+5a_k-405a_{k+1}}{9(16k^2+72k+80)}\\ \bullet & & \text{Solution for}r=\frac{3}{4}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+\frac{3}{4}},a_{k+2}=-\frac{144k^2{a}_k-288k^2{a}_{k+1}+72k{a}_k-720k{a}_{k+1}+5a_k-405a_{k+1}}{9(16k^2+72k+80)},a_1=-\frac{a_0}{8}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+\frac{1}{4}})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{x}^{k+\frac{3}{4}}),a_{k+2}=-\frac{144k^2{a}_k-288k^2{a}_{k+1}-72k{a}_k-432k{a}_{k+1}+5a_k-117a_{k+1}}{9(16k^2+56k+48)},a_1=-\frac{3a_0}{8},b_{k+2}=-\frac{144k^2{b}_k-288k^2{b}_{k+1}+72k{b}_k-720k{b}_{k+1}+5b_k-405b_{k+1}}{9(16k^2+72k+80)},b_1=-\frac{b_0}{8}]\end{array}$$

✓ Mathematica. Time used: 0.237 (sec). Leaf size: 550

ode=D[y[x],{x,2}]== ( -3/(16*x^2)- 2/(9*(x-1)^2) + 3/(16*x*(x-1))) *y[x]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to c_1\exp ({\int }_1^x\text{Root}[2048K[1{]}^4-3484K[1{]}^3+2313K[1{]}^2-702K[1]+(20736K[1{]}^8-82944K[1{]}^7+124416K[1{]}^6-82944K[1{]}^5+20736K[1{]}^4){\mathrm{\#}\text{1}}^4+(-48384K[1{]}^7+165888K[1{]}^6-207360K[1{]}^5+110592K[1{]}^4-20736K[1{]}^3){\mathrm{\#}\text{1}}^3+(41472K[1{]}^6-118368K[1{]}^5+120096K[1{]}^4-50976K[1{]}^3+7776K[1{]}^2){\mathrm{\#}\text{1}}^2+(-15360K[1{]}^5+34992K[1{]}^4-28272K[1{]}^3+9936K[1{]}^2-1296K[1])\mathrm{\#}\text{1}+81\mathrm{\&},1]dK[1])+c_2\exp ({\int }_1^x\text{Root}[2048K[1{]}^4-3484K[1{]}^3+2313K[1{]}^2-702K[1]+(20736K[1{]}^8-82944K[1{]}^7+124416K[1{]}^6-82944K[1{]}^5+20736K[1{]}^4){\mathrm{\#}\text{1}}^4+(-48384K[1{]}^7+165888K[1{]}^6-207360K[1{]}^5+110592K[1{]}^4-20736K[1{]}^3){\mathrm{\#}\text{1}}^3+(41472K[1{]}^6-118368K[1{]}^5+120096K[1{]}^4-50976K[1{]}^3+7776K[1{]}^2){\mathrm{\#}\text{1}}^2+(-15360K[1{]}^5+34992K[1{]}^4-28272K[1{]}^3+9936K[1{]}^2-1296K[1])\mathrm{\#}\text{1}+81\mathrm{\&},1]dK[1]){\int }_1^x\exp (-2{\int }_1^{K[2]}\text{Root}[2048K[1{]}^4-3484K[1{]}^3+2313K[1{]}^2-702K[1]+(20736K[1{]}^8-82944K[1{]}^7+124416K[1{]}^6-82944K[1{]}^5+20736K[1{]}^4){\mathrm{\#}\text{1}}^4+(-48384K[1{]}^7+165888K[1{]}^6-207360K[1{]}^5+110592K[1{]}^4-20736K[1{]}^3){\mathrm{\#}\text{1}}^3+(41472K[1{]}^6-118368K[1{]}^5+120096K[1{]}^4-50976K[1{]}^3+7776K[1{]}^2){\mathrm{\#}\text{1}}^2+(-15360K[1{]}^5+34992K[1{]}^4-28272K[1{]}^3+9936K[1{]}^2-1296K[1])\mathrm{\#}\text{1}+81\mathrm{\&},1]dK[1])dK[2]$$

✓ Sympy. Time used: 0.355 (sec). Leaf size: 41

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq((2/(9*(x - 1)**2) - 3/(16*x*(x - 1)) + 3/(16*x**2))*y(x) + Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

$$y\left(x\right)=\sqrt[4]{x}\sqrt[3]{x-1}(C_1\sqrt{x}{}_2{F}_1\left(\begin{array}{c}\frac{5}{12},\frac{3}{4}\\ \frac{3}{2}\end{array}\right|x)+C_2{}_2{F}_1\left(\begin{array}{c}-\frac{1}{12},\frac{1}{4}\\ \frac{1}{2}\end{array}\right|x))$$