2.1.92 Problem 94

Internal problem ID [9264]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 94
Date solved : Sunday, March 30, 2025 at 02:31:00 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order ode using Kovacic algorithm

Time used: 0.379 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& (2x^3+6x^2)y^{\prime \prime }+(5x^2+x)y^{\prime }+(1+x)y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =2x^3+6x^2\\ \text{(3)}& B& =5x^2+xC& =1+x\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{-3x^2-30x-35}{16{(x^2+3x)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-3x^2-30x-35\\ t& =16{(x^2+3x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{-3x^2-30x-35}{16{(x^2+3x)}^2}\right)z(x)\end{array}$$

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 2.92: 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 $t=16{(x^2+3x)}^2$. There is a pole at $x=0$ of order $2$. There is a pole at $x=-3$ 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

$$r=-\frac{5}{108x}+\frac{5}{108(3+x)}+\frac{7}{36{(3+x)}^2}-\frac{35}{144x^2}$$

For the pole at $x=-3$ let $b$ be the coefficient of $\frac{1}{{(3+x)}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{7}{36}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{7}{6}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{1}{6}\end{array}$$

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 $b=-\frac{35}{144}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{7}{12}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{5}{12}\end{array}$$

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

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{-3x^2-30x-35}{16{(x^2+3x)}^2}\end{array}$$

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

$$r=\frac{-3x^2-30x-35}{16{(x^2+3x)}^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$-3$	$2$	$0$	$\frac{7}{6}$	$-\frac{1}{6}$
$0$	$2$	$0$	$\frac{7}{12}$	$\frac{5}{12}$

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

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{-})\\ & =\frac{1}{4}-\left(\frac{1}{4}\right)\\ & =0\end{array}$$

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

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{x-c_1})+((-)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{-}}{x-c_2})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =-\frac{1}{6(3+x)}+\frac{5}{12x}+(-)\left(0\right)\\ & =-\frac{1}{6(3+x)}+\frac{5}{12x}\\ & =\frac{x+5}{4x(3+x)}\end{array}$$

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

$$\begin{array}{rl}\left(0\right)+2(-\frac{1}{6(3+x)}+\frac{5}{12x})\left(0\right)+((\frac{1}{6{(3+x)}^2}-\frac{5}{12x^2})+{(-\frac{1}{6(3+x)}+\frac{5}{12x})}^2-\left(\frac{-3x^2-30x-35}{16{(x^2+3x)}^2}\right))& =0\\ 0=0\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & ={\mathrm{e}}^{\int (-\frac{1}{6(3+x)}+\frac{5}{12x})dx}\\ & =\frac{x^{5/12}}{{(3+x)}^{1/6}}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{5x^2+x}{2x^3+6x^2}dx}\\ & =z_1{e}^{-\frac{\ln \left(x\right)}{12}-\frac{7\ln (3+x)}{6}}\\ & =z_1\left(\frac{1}{x^{1/12}{(3+x)}^{7/6}}\right)\end{array}$$

Which simplifies to

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

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{5x^2+x}{2x^3+6x^2}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-\frac{\ln \left(x\right)}{6}-\frac{7\ln (3+x)}{3}}}{{\left(y_1\right)}^2}dx\\ & =y_1(\int \frac{{\mathrm{e}}^{-\frac{\ln \left(x\right)}{6}-\frac{7\ln (3+x)}{3}}{(3+x)}^{8/3}}{x^{2/3}}dx)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{x^{1/3}}{{(3+x)}^{4/3}}\right)+c_2\left(\frac{x^{1/3}}{{(3+x)}^{4/3}}(\int \frac{{\mathrm{e}}^{-\frac{\ln \left(x\right)}{6}-\frac{7\ln (3+x)}{3}}{(3+x)}^{8/3}}{x^{2/3}}dx)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.047 (sec). Leaf size: 36

ode:=2*(3+x)*x^2*diff(diff(y(x),x),x)+x*(1+5*x)*diff(y(x),x)+(x+1)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

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

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 
         <- heuristic approach successful 
      <- hypergeometric successful 
   <- special function solution successful 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form for at least one hypergeometric solution is achieved -\ 
 returning with no uncomputed integrals 
   <- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & 2x^2(3+x)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+x(1+5x)\left(\frac{d}{dx}y\left(x\right)\right)+(1+x)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{(1+x)y\left(x\right)}{2x^2(3+x)}-\frac{(1+5x)\left(\frac{d}{dx}y\left(x\right)\right)}{2x(3+x)}\\ \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{(1+5x)\left(\frac{d}{dx}y\left(x\right)\right)}{2x(3+x)}+\frac{(1+x)y\left(x\right)}{2x^2(3+x)}=0\\ ◻& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{1+5x}{2x(3+x)},P_3\left(x\right)=\frac{1+x}{2x^2(3+x)}]\\ & \circ & (3+x)\cdot P_2\left(x\right)\text{is analytic at}x=-3\\ & & ((3+x)\cdot P_2\left(x\right))|{}_{x=-3}=\frac{7}{3}\\ & \circ & {(3+x)}^2\cdot P_3\left(x\right)\text{is analytic at}x=-3\\ & & ({(3+x)}^2\cdot P_3\left(x\right))|{}_{x=-3}=0\\ & \circ & x=-3\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & & x_0=-3\\ \bullet & & \text{Multiply by denominators}\\ & & 2x^2(3+x)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+x(1+5x)\left(\frac{d}{dx}y\left(x\right)\right)+(1+x)y\left(x\right)=0\\ \bullet & & \text{Change variables using}x=u-3\text{so that the regular singular point is at}u=0\\ & & (2u^3-12u^2+18u)\left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)+(5u^2-29u+42)\left(\frac{d}{du}y\left(u\right)\right)+(-2+u)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..1\\ & & 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..2\\ & & 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..3\\ & & 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}\\ & & 6a_{0}r(4+3r)u^{-1+r}+(6a_1(1+r)(7+3r)-a_0(12r^2+17r+2))u^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(6a_{k+1}(k+r+1)(3k+7+3r)-a_k(12k^2+24kr+12r^2+17k+17r+2)+a_{k-1}(k+r)(2k-1+2r))u^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & 6r(4+3r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,-\frac{4}{3}\}\\ \bullet & & \text{Each term must be 0}\\ & & 6a_1(1+r)(7+3r)-a_0(12r^2+17r+2)=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & 2(-6a_k+a_{k-1}+9a_{k+1})k^2+(4(-6a_k+a_{k-1}+9a_{k+1})r-17a_k-a_{k-1}+60a_{k+1})k+2(-6a_k+a_{k-1}+9a_{k+1})r^2+(-17a_k-a_{k-1}+60a_{k+1})r-2a_k+42a_{k+1}=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & 2(-6a_{k+1}+a_k+9a_{k+2}){(k+1)}^2+(4(-6a_{k+1}+a_k+9a_{k+2})r-17a_{k+1}-a_k+60a_{k+2})(k+1)+2(-6a_{k+1}+a_k+9a_{k+2})r^2+(-17a_{k+1}-a_k+60a_{k+2})r-2a_{k+1}+42a_{k+2}=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=-\frac{2k^2{a}_k-12k^2{a}_{k+1}+4kr{a}_k-24kr{a}_{k+1}+2r^2{a}_k-12r^2{a}_{k+1}+3k{a}_k-41k{a}_{k+1}+3r{a}_k-41r{a}_{k+1}+a_k-31a_{k+1}}{6(3k^2+6kr+3r^2+16k+16r+20)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+2}=-\frac{2k^2{a}_k-12k^2{a}_{k+1}+3k{a}_k-41k{a}_{k+1}+a_k-31a_{k+1}}{6(3k^2+16k+20)}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^k,a_{k+2}=-\frac{2k^2{a}_k-12k^2{a}_{k+1}+3k{a}_k-41k{a}_{k+1}+a_k-31a_{k+1}}{6(3k^2+16k+20)},42a_1-2a_0=0]\\ \bullet & & \text{Revert the change of variables}u=3+x\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(3+x)}^k,a_{k+2}=-\frac{2k^2{a}_k-12k^2{a}_{k+1}+3k{a}_k-41k{a}_{k+1}+a_k-31a_{k+1}}{6(3k^2+16k+20)},42a_1-2a_0=0]\\ \bullet & & \text{Recursion relation for}r=-\frac{4}{3}\\ & & a_{k+2}=-\frac{2k^2{a}_k-12k^2{a}_{k+1}-\frac{7}{3}k{a}_k-9k{a}_{k+1}+\frac{5}{9}{a}_k+\frac{7}{3}{a}_{k+1}}{6(3k^2+8k+4)}\\ \bullet & & \text{Solution for}r=-\frac{4}{3}\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k-\frac{4}{3}},a_{k+2}=-\frac{2k^2{a}_k-12k^2{a}_{k+1}-\frac{7}{3}k{a}_k-9k{a}_{k+1}+\frac{5}{9}{a}_k+\frac{7}{3}{a}_{k+1}}{6(3k^2+8k+4)},-6a_1-\frac{2a_0}{3}=0]\\ \bullet & & \text{Revert the change of variables}u=3+x\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(3+x)}^{k-\frac{4}{3}},a_{k+2}=-\frac{2k^2{a}_k-12k^2{a}_{k+1}-\frac{7}{3}k{a}_k-9k{a}_{k+1}+\frac{5}{9}{a}_k+\frac{7}{3}{a}_{k+1}}{6(3k^2+8k+4)},-6a_1-\frac{2a_0}{3}=0]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(3+x)}^k)+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{(3+x)}^{k-\frac{4}{3}}),a_{k+2}=-\frac{2k^2{a}_k-12k^2{a}_{k+1}+3k{a}_k-41k{a}_{k+1}+a_k-31a_{k+1}}{6(3k^2+16k+20)},42a_1-2a_0=0,b_{k+2}=-\frac{2k^2{b}_k-12k^2{b}_{k+1}-\frac{7}{3}k{b}_k-9k{b}_{k+1}+\frac{5}{9}{b}_k+\frac{7}{3}{b}_{k+1}}{6(3k^2+8k+4)},-6b_1-\frac{2b_0}{3}=0]\end{array}$$

✓ Mathematica. Time used: 0.256 (sec). Leaf size: 108

ode=2*x^2*(3+x)*D[y[x],{x,2}]+x*(1+5*x)*D[y[x],x]+(1+x)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \exp ({\int }_1^x\frac{K[1]+5}{4K[1{]}^2+12K[1]}dK[1]-\frac{1}{2}{\int }_1^x\frac{5K[2]+1}{2K[2{]}^2+6K[2]}dK[2])(c_2{\int }_1^x\exp (-2{\int }_1^{K[3]}\frac{K[1]+5}{4K[1{]}^2+12K[1]}dK[1])dK[3]+c_1)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(2*x**2*(x + 3)*Derivative(y(x), (x, 2)) + x*(5*x + 1)*Derivative(y(x), x) + (x + 1)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False