2.1.102 Problem 104

Internal problem ID [9272]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 104
Date solved : Friday, April 25, 2025 at 06:04:07 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}4x^2(-x^2+1)y^{\prime \prime }+x(-19x^2+7)y^{\prime }-(14x^2+1)y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.515 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& (-4x^4+4x^2)y^{\prime \prime }+(-19x^3+7x)y^{\prime }+(-14x^2-1)y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =-4x^4+4x^2\\ \text{(3)}& B& =-19x^3+7xC& =-14x^2-1\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{-15x^4-42x^2+9}{64{(x^3-x)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-15x^4-42x^2+9\\ t& =64{(x^3-x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{-15x^4-42x^2+9}{64{(x^3-x)}^2}\right)z(x)\end{array}$$

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 2.102: 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 $t=64{(x^3-x)}^2$. There is a pole at $x=0$ of order $2$. There is a pole at $x=1$ 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 $n=1$.

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

$$r=-\frac{3}{16{(x-1)}^2}-\frac{3}{16{(x+1)}^2}+\frac{9}{64x^2}$$

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 $b=\frac{9}{64}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{9}{8}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{1}{8}\end{array}$$

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. 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 $x=-1$ let $b$ be the coefficient of $\frac{1}{{(x+1)}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=-\frac{3}{16}$. Hence

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

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

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{-15x^4-42x^2+9}{64{(x^3-x)}^2}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=-\frac{15}{64}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =0\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{5}{8}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{3}{8}\end{array}$$

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

$$r=\frac{-15x^4-42x^2+9}{64{(x^3-x)}^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$2$	$0$	$\frac{9}{8}$	$-\frac{1}{8}$
$1$	$2$	$0$	$\frac{3}{4}$	$\frac{1}{4}$
$-1$	$2$	$0$	$\frac{3}{4}$	$\frac{1}{4}$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$2$	$0$	$\frac{5}{8}$	$\frac{3}{8}$

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 ${\alpha }_{\mathrm{\infty }}^{-}=\frac{3}{8}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{-}+{\alpha }_{c_3}^{-})\\ & =\frac{3}{8}-\left(\frac{3}{8}\right)\\ & =0\end{array}$$

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

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

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

$$\begin{array}{rl}\left(0\right)+2(-\frac{1}{8x}+\frac{1}{4x-4}+\frac{1}{4x+4})\left(0\right)+((\frac{1}{8x^2}-\frac{1}{4{(x-1)}^2}-\frac{1}{4{(x+1)}^2})+{(-\frac{1}{8x}+\frac{1}{4x-4}+\frac{1}{4x+4})}^2-\left(\frac{-15x^4-42x^2+9}{64{(x^3-x)}^2}\right))& =0\\ 0=0\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & ={\mathrm{e}}^{\int (-\frac{1}{8x}+\frac{1}{4x-4}+\frac{1}{4x+4})dx}\\ & =\frac{{(x-1)}^{1/4}{(x+1)}^{1/4}}{x^{1/8}}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-19x^3+7x}{-4x^4+4x^2}dx}\\ & =z_1{e}^{-\frac{3\ln (x-1)}{4}-\frac{3\ln (x+1)}{4}-\frac{7\ln \left(x\right)}{8}}\\ & =z_1\left(\frac{1}{{(x-1)}^{3/4}{(x+1)}^{3/4}{x}^{7/8}}\right)\end{array}$$

Which simplifies to

$$y_1=\frac{{(x^2-1)}^{1/4}}{{(x-1)}^{3/4}{(x+1)}^{3/4}x}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-19x^3+7x}{-4x^4+4x^2}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-\frac{3\ln (x-1)}{2}-\frac{3\ln (x+1)}{2}-\frac{7\ln \left(x\right)}{4}}}{{\left(y_1\right)}^2}dx\\ & =y_1(\int \frac{{\mathrm{e}}^{-\frac{3\ln (x-1)}{2}-\frac{3\ln (x+1)}{2}-\frac{7\ln \left(x\right)}{4}}{(x-1)}^{3/2}{(x+1)}^{3/2}{x}^2}{\sqrt{x^2-1}}dx)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{{(x^2-1)}^{1/4}}{{(x-1)}^{3/4}{(x+1)}^{3/4}x}\right)+c_2\left(\frac{{(x^2-1)}^{1/4}}{{(x-1)}^{3/4}{(x+1)}^{3/4}x}(\int \frac{{\mathrm{e}}^{-\frac{3\ln (x-1)}{2}-\frac{3\ln (x+1)}{2}-\frac{7\ln \left(x\right)}{4}}{(x-1)}^{3/2}{(x+1)}^{3/2}{x}^2}{\sqrt{x^2-1}}dx)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.177 (sec). Leaf size: 44

ode:=4*x^2*(-x^2+1)*diff(diff(y(x),x),x)+x*(-19*x^2+7)*diff(y(x),x)-(14*x^2+1)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{c_1\mathrm{LegendreP}(-\frac{3}{8},\frac{5}{8},\sqrt{-x^2+1})+c_2\mathrm{LegendreQ}(-\frac{3}{8},\frac{5}{8},\sqrt{-x^2+1})}{x^{3/8}\sqrt{x^2-1}}$$

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 
         <- heuristic approach successful 
      <- 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}\\ & & 4x^2(-x^2+1)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+x(-19x^2+7)\left(\frac{d}{dx}y\left(x\right)\right)-(14x^2+1)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{(14x^2+1)y\left(x\right)}{4x^2(x^2-1)}-\frac{(19x^2-7)\left(\frac{d}{dx}y\left(x\right)\right)}{4x(x^2-1)}\\ \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{(19x^2-7)\left(\frac{d}{dx}y\left(x\right)\right)}{4x(x^2-1)}+\frac{(14x^2+1)y\left(x\right)}{4x^2(x^2-1)}=0\\ ◻& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{19x^2-7}{4x(x^2-1)},P_3\left(x\right)=\frac{14x^2+1}{4x^2(x^2-1)}]\\ & \circ & (x+1)\cdot P_2\left(x\right)\text{is analytic at}x=-1\\ & & ((x+1)\cdot P_2\left(x\right))|{}_{x=-1}=\frac{3}{2}\\ & \circ & {(x+1)}^2\cdot P_3\left(x\right)\text{is analytic at}x=-1\\ & & ({(x+1)}^2\cdot P_3\left(x\right))|{}_{x=-1}=0\\ & \circ & x=-1\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & & x_0=-1\\ \bullet & & \text{Multiply by denominators}\\ & & 4x^2(x^2-1)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+x(19x^2-7)\left(\frac{d}{dx}y\left(x\right)\right)+(14x^2+1)y\left(x\right)=0\\ \bullet & & \text{Change variables using}x=u-1\text{so that the regular singular point is at}u=0\\ & & (4u^4-16u^3+20u^2-8u)\left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)+(19u^3-57u^2+50u-12)\left(\frac{d}{du}y\left(u\right)\right)+(14u^2-28u+15)y\left(u\right)=0\\ \bullet & & \text{Assume series solution for}y\left(u\right)\\ & & y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{u}^m\cdot y\left(u\right)\text{to series expansion for}m=0..2\\ & & u^m\cdot y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+r+m}\\ & \circ & \text{Shift index using}k->k-m\\ & & u^m\cdot y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=m}}{a}_{k-m}{u}^{k+r}\\ & \circ & \text{Convert}{u}^m\cdot \left(\frac{d}{du}y\left(u\right)\right)\text{to series expansion for}m=0..3\\ & & u^m\cdot \left(\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)u^{k+r-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & u^m\cdot \left(\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1+m}}{a}_{k+1-m}(k+1-m+r)u^{k+r}\\ & \circ & \text{Convert}{u}^m\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)\text{to series expansion for}m=1..4\\ & & u^m\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)u^{k+r-2+m}\\ & \circ & \text{Shift index using}k->k+2-m\\ & & u^m\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-2+m}}{a}_{k+2-m}(k+2-m+r)(k+1-m+r)u^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & -4a_{0}r(1+2r)u^{-1+r}+(-4a_1(1+r)(3+2r)+5a_0(4r^2+6r+3))u^r+(-4a_2(2+r)(5+2r)+5a_1(4r^2+14r+13)-a_0(16r^2+41r+28))u^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(-4a_{k+1}(k+1+r)(2k+3+2r)+5a_k(4k^2+8kr+4r^2+6k+6r+3)-a_{k-1}(16{(k-1)}^2+32(k-1)r+16r^2+41k-13+41r)+a_{k-2}(k+r)(4k-1+4r))u^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & -4r(1+2r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,-\frac{1}{2}\}\\ \bullet & & \text{The coefficients of each power of}u\text{must be 0}\\ & & [-4a_1(1+r)(3+2r)+5a_0(4r^2+6r+3)=0,-4a_2(2+r)(5+2r)+5a_1(4r^2+14r+13)-a_0(16r^2+41r+28)=0]\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & \{a_1=\frac{5a_0(4r^2+6r+3)}{4(2r^2+5r+3)},a_2=\frac{a_0(272r^4+1352r^3+2464r^2+1948r+639)}{16(4r^4+28r^3+71r^2+77r+30)}\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & 4(5a_k+a_{k-2}-4a_{k-1}-2a_{k+1})k^2+(8(5a_k+a_{k-2}-4a_{k-1}-2a_{k+1})r+30a_k-a_{k-2}-9a_{k-1}-20a_{k+1})k+4(5a_k+a_{k-2}-4a_{k-1}-2a_{k+1})r^2+(30a_k-a_{k-2}-9a_{k-1}-20a_{k+1})r+15a_k-3a_{k-1}-12a_{k+1}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & 4(5a_{k+2}+a_k-4a_{k+1}-2a_{k+3}){(k+2)}^2+(8(5a_{k+2}+a_k-4a_{k+1}-2a_{k+3})r+30a_{k+2}-a_k-9a_{k+1}-20a_{k+3})(k+2)+4(5a_{k+2}+a_k-4a_{k+1}-2a_{k+3})r^2+(30a_{k+2}-a_k-9a_{k+1}-20a_{k+3})r+15a_{k+2}-3a_{k+1}-12a_{k+3}=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+3}=\frac{4k^2{a}_k-16k^2{a}_{k+1}+20k^2{a}_{k+2}+8kr{a}_k-32kr{a}_{k+1}+40kr{a}_{k+2}+4r^2{a}_k-16r^2{a}_{k+1}+20r^2{a}_{k+2}+15k{a}_k-73k{a}_{k+1}+110k{a}_{k+2}+15r{a}_k-73r{a}_{k+1}+110r{a}_{k+2}+14a_k-85a_{k+1}+155a_{k+2}}{4(2k^2+4kr+2r^2+13k+13r+21)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+3}=\frac{4k^2{a}_k-16k^2{a}_{k+1}+20k^2{a}_{k+2}+15k{a}_k-73k{a}_{k+1}+110k{a}_{k+2}+14a_k-85a_{k+1}+155a_{k+2}}{4(2k^2+13k+21)}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^k,a_{k+3}=\frac{4k^2{a}_k-16k^2{a}_{k+1}+20k^2{a}_{k+2}+15k{a}_k-73k{a}_{k+1}+110k{a}_{k+2}+14a_k-85a_{k+1}+155a_{k+2}}{4(2k^2+13k+21)},a_1=\frac{5a_0}{4},a_2=\frac{213a_0}{160}]\\ \bullet & & \text{Revert the change of variables}u=x+1\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(x+1)}^k,a_{k+3}=\frac{4k^2{a}_k-16k^2{a}_{k+1}+20k^2{a}_{k+2}+15k{a}_k-73k{a}_{k+1}+110k{a}_{k+2}+14a_k-85a_{k+1}+155a_{k+2}}{4(2k^2+13k+21)},a_1=\frac{5a_0}{4},a_2=\frac{213a_0}{160}]\\ \bullet & & \text{Recursion relation for}r=-\frac{1}{2}\\ & & a_{k+3}=\frac{4k^2{a}_k-16k^2{a}_{k+1}+20k^2{a}_{k+2}+11k{a}_k-57k{a}_{k+1}+90k{a}_{k+2}+\frac{15}{2}{a}_k-\frac{105}{2}{a}_{k+1}+105a_{k+2}}{4(2k^2+11k+15)}\\ \bullet & & \text{Solution for}r=-\frac{1}{2}\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k-\frac{1}{2}},a_{k+3}=\frac{4k^2{a}_k-16k^2{a}_{k+1}+20k^2{a}_{k+2}+11k{a}_k-57k{a}_{k+1}+90k{a}_{k+2}+\frac{15}{2}{a}_k-\frac{105}{2}{a}_{k+1}+105a_{k+2}}{4(2k^2+11k+15)},a_1=\frac{5a_0}{4},a_2=\frac{43a_0}{32}]\\ \bullet & & \text{Revert the change of variables}u=x+1\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(x+1)}^{k-\frac{1}{2}},a_{k+3}=\frac{4k^2{a}_k-16k^2{a}_{k+1}+20k^2{a}_{k+2}+11k{a}_k-57k{a}_{k+1}+90k{a}_{k+2}+\frac{15}{2}{a}_k-\frac{105}{2}{a}_{k+1}+105a_{k+2}}{4(2k^2+11k+15)},a_1=\frac{5a_0}{4},a_2=\frac{43a_0}{32}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(x+1)}^k)+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{(x+1)}^{k-\frac{1}{2}}),a_{k+3}=\frac{4k^2{a}_k-16k^2{a}_{k+1}+20k^2{a}_{k+2}+15k{a}_k-73k{a}_{k+1}+110k{a}_{k+2}+14a_k-85a_{k+1}+155a_{k+2}}{4(2k^2+13k+21)},a_1=\frac{5a_0}{4},a_2=\frac{213a_0}{160},b_{k+3}=\frac{4k^2{b}_k-16k^2{b}_{k+1}+20k^2{b}_{k+2}+11k{b}_k-57k{b}_{k+1}+90k{b}_{k+2}+\frac{15}{2}{b}_k-\frac{105}{2}{b}_{k+1}+105b_{k+2}}{4(2k^2+11k+15)},b_1=\frac{5b_0}{4},b_2=\frac{43b_0}{32}]\end{array}$$

✓ Mathematica. Time used: 0.308 (sec). Leaf size: 120

ode=4*x^2*(1-x^2)*D[y[x],{x,2}]+x*(7-19*x^2)*D[y[x],x]-(1+14*x^2)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \exp ({\int }_1^x-\frac{3K[1{]}^2+1}{8K[1]-8K[1{]}^3}dK[1]-\frac{1}{2}{\int }_1^x\frac{7-19K[2{]}^2}{4K[2]-4K[2{]}^3}dK[2])(c_2{\int }_1^x\exp (-2{\int }_1^{K[3]}-\frac{3K[1{]}^2+1}{8K[1]-8K[1{]}^3}dK[1])dK[3]+c_1)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(4*x**2*(1 - x**2)*Derivative(y(x), (x, 2)) + x*(7 - 19*x**2)*Derivative(y(x), x) - (14*x**2 + 1)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False