2.1.130 Problem 132

Internal problem ID [9302]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 132
Date solved : Wednesday, March 05, 2025 at 07:47:11 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}9x^2(1+x)y^{\prime \prime }+3x(-x^2+11x+5)y^{\prime }+(-7x^2+16x+1)y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.399 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& (9x^3+9x^2)y^{\prime \prime }+(-3x^3+33x^2+15x)y^{\prime }+(-7x^2+16x+1)y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =9x^3+9x^2\\ \text{(3)}& B& =-3x^3+33x^2+15xC& =-7x^2+16x+1\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{x^4+6x^3+3x^2-18x-9}{36{(x^2+x)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =x^4+6x^3+3x^2-18x-9\\ t& =36{(x^2+x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{x^4+6x^3+3x^2-18x-9}{36{(x^2+x)}^2}\right)z(x)\end{array}$$

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 2.130: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =4-4\\ & =0\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=36{(x^2+x)}^2$. 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 $0$ 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

$$r=\frac{1}{36}+\frac{7}{36{(1+x)}^2}+\frac{1}{9+9x}-\frac{1}{4x^2}$$

For the pole at $x=-1$ let $b$ be the coefficient of $\frac{1}{{(1+x)}^2}$ 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 $b=-\frac{1}{4}$. Hence

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

Since the order of $r$ at $\mathrm{\infty }$ is $O_r(\mathrm{\infty })=0$ then

$$\begin{array}{rlrlrl}v& =\frac{-O_r(\mathrm{\infty })}{2}& & =\frac{0}{2}& & =0\end{array}$$

$[\sqrt{r}{]}_{\mathrm{\infty }}$ 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

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

Let $a$ be the coefficient of $x^v=x^0$ in the above sum. The Laurent series of $\sqrt{r}$ at $\mathrm{\infty }$ is

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx \frac{1}{6}+\frac{1}{3x}-\frac{5}{6x^2}+\frac{5}{6x^3}-\frac{7}{3x^4}+\frac{41}{6x^5}-\frac{149}{6x^6}+\frac{277}{3x^7}+\dots \end{array}$$

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

$$a=\frac{1}{6}$$

From Eq. (9) the sum up to $v=0$ gives

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^0{a}_i{x}^i\\ \text{(10)}& & =\frac{1}{6}\end{array}$$

Now we need to find $b$, where $b$ be the coefficient of $x^{v-1}=x^{-1}=\frac{1}{x}$ 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

$${([\sqrt{r}{]}_{\mathrm{\infty }})}^2=\frac{1}{36}$$

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

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

Where $Q$ is the quotient and $R$ is the remainder. Then the coefficient of $\frac{1}{x}$ in $r$ will be the coefficient in $R$ of the term in $x$ of degree of $t$ minus one, divided by the leading coefficient in $t$. Doing long division gives

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{x^4+6x^3+3x^2-18x-9}{36x^4+72x^3+36x^2}\\ & =Q+\frac{R}{36x^4+72x^3+36x^2}\\ & =\left(\frac{1}{36}\right)+\left(\frac{4x^3+2x^2-18x-9}{36x^4+72x^3+36x^2}\right)\\ & =\frac{1}{36}+\frac{4x^3+2x^2-18x-9}{36x^4+72x^3+36x^2}\end{array}$$

Since the degree of $t$ is $4$, then we see that the coefficient of the term $x^3$ in the remainder $R$ is $4$. Dividing this by leading coefficient in $t$ which is $36$ gives $\frac{1}{9}$. Now $b$ can be found.

$$\begin{array}{rl}b& =\left(\frac{1}{9}\right)-\left(0\right)\\ & =\frac{1}{9}\end{array}$$

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\frac{1}{6}\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)& & =\frac{1}{2}(\frac{\frac{1}{9}}{\frac{1}{6}}-0)& & =\frac{1}{3}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)& & =\frac{1}{2}(-\frac{\frac{1}{9}}{\frac{1}{6}}-0)& & =-\frac{1}{3}\end{array}$$

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

$$r=\frac{x^4+6x^3+3x^2-18x-9}{36{(x^2+x)}^2}$$

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

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$0$	$\frac{1}{6}$	$\frac{1}{3}$	$-\frac{1}{3}$

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 ${\alpha }_{\mathrm{\infty }}^{+}=\frac{1}{3}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{+})\\ & =\frac{1}{3}-\left(\frac{1}{3}\right)\\ & =0\end{array}$$

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

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

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

$$\begin{array}{rl}\left(0\right)+2(-\frac{1}{6(1+x)}+\frac{1}{2x}+\frac{1}{6})\left(0\right)+((\frac{1}{6{(1+x)}^2}-\frac{1}{2x^2})+{(-\frac{1}{6(1+x)}+\frac{1}{2x}+\frac{1}{6})}^2-\left(\frac{x^4+6x^3+3x^2-18x-9}{36{(x^2+x)}^2}\right))& =0\\ 0=0\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & ={\mathrm{e}}^{\int (-\frac{1}{6(1+x)}+\frac{1}{2x}+\frac{1}{6})dx}\\ & =\frac{\sqrt{x}{\mathrm{e}}^{\frac{x}{6}}}{{(1+x)}^{1/6}}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-3x^3+33x^2+15x}{9x^3+9x^2}dx}\\ & =z_1{e}^{\frac{x}{6}-\frac{7\ln (1+x)}{6}-\frac{5\ln \left(x\right)}{6}}\\ & =z_1\left(\frac{{\mathrm{e}}^{\frac{x}{6}}}{{(1+x)}^{7/6}{x}^{5/6}}\right)\end{array}$$

Which simplifies to

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

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-3x^3+33x^2+15x}{9x^3+9x^2}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{\frac{x}{3}-\frac{7\ln (1+x)}{3}-\frac{5\ln \left(x\right)}{3}}}{{\left(y_1\right)}^2}dx\\ & =y_1(\int {\mathrm{e}}^{\frac{x}{3}-\frac{7\ln (1+x)}{3}-\frac{5\ln \left(x\right)}{3}}{(1+x)}^{8/3}{x}^{2/3}{\mathrm{e}}^{-\frac{2x}{3}}dx)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{{\mathrm{e}}^{\frac{x}{3}}}{{(1+x)}^{4/3}{x}^{1/3}}\right)+c_2\left(\frac{{\mathrm{e}}^{\frac{x}{3}}}{{(1+x)}^{4/3}{x}^{1/3}}(\int {\mathrm{e}}^{\frac{x}{3}-\frac{7\ln (1+x)}{3}-\frac{5\ln \left(x\right)}{3}}{(1+x)}^{8/3}{x}^{2/3}{\mathrm{e}}^{-\frac{2x}{3}}dx)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.388 (sec). Leaf size: 36

ode:=9*x^2*(x+1)*diff(diff(y(x),x),x)+3*x*(-x^2+11*x+5)*diff(y(x),x)+(-7*x^2+16*x+1)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{\frac{c_1\mathrm{HeunC}(-\frac{1}{3},-\frac{4}{3},0,-\frac{1}{9},\frac{11}{18},x+1)}{{(x+1)}^{4/3}}+c_2\mathrm{HeunC}(-\frac{1}{3},\frac{4}{3},0,-\frac{1}{9},\frac{11}{18},x+1)}{x^{1/3}}$$

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 
         -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
      -> Mathieu 
         -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
      <- Heun successful: received ODE is equivalent to the  HeunC  ODE, case  a <> 0, e <> 0, c = 0 
   <- Kovacics algorithm successful`
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & 9x^2(x+1)\left(\frac{d^2}{d{x}^2}y\left(x\right)\right)+3x(-x^2+11x+5)\left(\frac{d}{dx}y\left(x\right)\right)+(-7x^2+16x+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{(7x^2-16x-1)y\left(x\right)}{9x^2(x+1)}+\frac{(x^2-11x-5)\left(\frac{d}{dx}y\left(x\right)\right)}{3x(x+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{(x^2-11x-5)\left(\frac{d}{dx}y\left(x\right)\right)}{3x(x+1)}-\frac{(7x^2-16x-1)y\left(x\right)}{9x^2(x+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{x^2-11x-5}{3x(x+1)},P_3\left(x\right)=-\frac{7x^2-16x-1}{9x^2(x+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{7}{3}\\ & \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}\\ & & 9x^2(x+1)\left(\frac{d^2}{d{x}^2}y\left(x\right)\right)-3(x^2-11x-5)x\left(\frac{d}{dx}y\left(x\right)\right)+(-7x^2+16x+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\\ & & (9u^3-18u^2+9u)\left(\frac{d^2}{d{u}^2}y\left(u\right)\right)+(-3u^3+42u^2-60u+21)\left(\frac{d}{du}y\left(u\right)\right)+(-7u^2+30u-22)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^2}{d{u}^2}y\left(u\right)\right)\text{to series expansion for}m=1..3\\ & & u^m\cdot \left(\frac{d^2}{d{u}^2}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^2}{d{u}^2}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}\\ & & 3a_{0}r(4+3r)u^{-1+r}+(3a_1(1+r)(7+3r)-2a_0(9r^2+21r+11))u^r+(3a_2(2+r)(10+3r)-2a_1(9r^2+39r+41)+3a_0(2+r)(5+3r))u^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(3a_{k+1}(k+1+r)(3k+3r+7)-2a_k(9k^2+18kr+9r^2+21k+21r+11)+3a_{k-1}(k+1+r)(3k+2+3r)-a_{k-2}(3k+1+3r))u^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & 3r(4+3r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,-\frac{4}{3}\}\\ \bullet & & \text{The coefficients of each power of}u\text{must be 0}\\ & & [3a_1(1+r)(7+3r)-2a_0(9r^2+21r+11)=0,3a_2(2+r)(10+3r)-2a_1(9r^2+39r+41)+3a_0(2+r)(5+3r)=0]\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & \{a_1=\frac{2a_0(9r^2+21r+11)}{3(3r^2+10r+7)},a_2=\frac{a_0(243r^4+1593r^3+3699r^2+3567r+1174)}{9(9r^4+78r^3+241r^2+312r+140)}\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & 9(-2a_k+a_{k-1}+a_{k+1})k^2+3(6(-2a_k+a_{k-1}+a_{k+1})r-14a_k-a_{k-2}+5a_{k-1}+10a_{k+1})k+9(-2a_k+a_{k-1}+a_{k+1})r^2+3(-14a_k-a_{k-2}+5a_{k-1}+10a_{k+1})r-22a_k-a_{k-2}+6a_{k-1}+21a_{k+1}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & 9(-2a_{k+2}+a_{k+1}+a_{k+3}){(k+2)}^2+3(6(-2a_{k+2}+a_{k+1}+a_{k+3})r-14a_{k+2}-a_k+5a_{k+1}+10a_{k+3})(k+2)+9(-2a_{k+2}+a_{k+1}+a_{k+3})r^2+3(-14a_{k+2}-a_k+5a_{k+1}+10a_{k+3})r-22a_{k+2}-a_k+6a_{k+1}+21a_{k+3}=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+3}=-\frac{9k^2{a}_{k+1}-18k^2{a}_{k+2}+18kr{a}_{k+1}-36kr{a}_{k+2}+9r^2{a}_{k+1}-18r^2{a}_{k+2}-3k{a}_k+51k{a}_{k+1}-114k{a}_{k+2}-3r{a}_k+51r{a}_{k+1}-114r{a}_{k+2}-7a_k+72a_{k+1}-178a_{k+2}}{3(3k^2+6kr+3r^2+22k+22r+39)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+3}=-\frac{9k^2{a}_{k+1}-18k^2{a}_{k+2}-3k{a}_k+51k{a}_{k+1}-114k{a}_{k+2}-7a_k+72a_{k+1}-178a_{k+2}}{3(3k^2+22k+39)}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^k,a_{k+3}=-\frac{9k^2{a}_{k+1}-18k^2{a}_{k+2}-3k{a}_k+51k{a}_{k+1}-114k{a}_{k+2}-7a_k+72a_{k+1}-178a_{k+2}}{3(3k^2+22k+39)},a_1=\frac{22a_0}{21},a_2=\frac{587a_0}{630}]\\ \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{9k^2{a}_{k+1}-18k^2{a}_{k+2}-3k{a}_k+51k{a}_{k+1}-114k{a}_{k+2}-7a_k+72a_{k+1}-178a_{k+2}}{3(3k^2+22k+39)},a_1=\frac{22a_0}{21},a_2=\frac{587a_0}{630}]\\ \bullet & & \text{Recursion relation for}r=-\frac{4}{3}\\ & & a_{k+3}=-\frac{9k^2{a}_{k+1}-18k^2{a}_{k+2}-3k{a}_k+27k{a}_{k+1}-66k{a}_{k+2}-3a_k+20a_{k+1}-58a_{k+2}}{3(3k^2+14k+15)}\\ \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+3}=-\frac{9k^2{a}_{k+1}-18k^2{a}_{k+2}-3k{a}_k+27k{a}_{k+1}-66k{a}_{k+2}-3a_k+20a_{k+1}-58a_{k+2}}{3(3k^2+14k+15)},a_1=\frac{2a_0}{3},a_2=\frac{7a_0}{18}]\\ \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{4}{3}},a_{k+3}=-\frac{9k^2{a}_{k+1}-18k^2{a}_{k+2}-3k{a}_k+27k{a}_{k+1}-66k{a}_{k+2}-3a_k+20a_{k+1}-58a_{k+2}}{3(3k^2+14k+15)},a_1=\frac{2a_0}{3},a_2=\frac{7a_0}{18}]\\ \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{4}{3}}),a_{k+3}=-\frac{9k^2{a}_{k+1}-18k^2{a}_{k+2}-3k{a}_k+51k{a}_{k+1}-114k{a}_{k+2}-7a_k+72a_{k+1}-178a_{k+2}}{3(3k^2+22k+39)},a_1=\frac{22a_0}{21},a_2=\frac{587a_0}{630},b_{k+3}=-\frac{9k^2{b}_{k+1}-18k^2{b}_{k+2}-3k{b}_k+27k{b}_{k+1}-66k{b}_{k+2}-3b_k+20b_{k+1}-58b_{k+2}}{3(3k^2+14k+15)},b_1=\frac{2b_0}{3},b_2=\frac{7b_0}{18}]\end{array}$$

✓ Mathematica. Time used: 0.346 (sec). Leaf size: 120

ode=9*x^2*(1+x)*D[y[x],{x,2}]+3*x*(5+11*x-x^2)*D[y[x],x]+(1+16*x-7*x^2)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \exp ({\int }_1^x\frac{K[1{]}^2+3K[1]+3}{6K[1{]}^2+6K[1]}dK[1]-\frac{1}{2}{\int }_1^x\frac{1}{3}(\frac{7}{K[2]+1}-1+\frac{5}{K[2]})dK[2])(c_2{\int }_1^x\exp (-2{\int }_1^{K[3]}\frac{K[1{]}^2+3K[1]+3}{6K[1{]}^2+6K[1]}dK[1])dK[3]+c_1)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(9*x**2*(x + 1)*Derivative(y(x), (x, 2)) + 3*x*(-x**2 + 11*x + 5)*Derivative(y(x), x) + (-7*x**2 + 16*x + 1)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False