2.1.85 Problem 87

Internal problem ID [9255]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 87
Date solved : Friday, April 25, 2025 at 06:03:39 PM
CAS classification : [[_2nd_order, _with_linear_symmetries], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]

Solve

$$\begin{array}{rl}{x}^2(10x^2+x+5)y^{\prime \prime }+x(48x^2+3x+4)y^{\prime }+(36x^2+x)y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 1.496 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& (10x^4+x^3+5x^2)y^{\prime \prime }+(48x^3+3x^2+4x)y^{\prime }+(36x^2+x)y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =10x^4+x^3+5x^2\\ \text{(3)}& B& =48x^3+3x^2+4xC& =36x^2+x\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{-96x^4-16x^3-97x^2-12x-24}{4{(10x^3+x^2+5x)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-96x^4-16x^3-97x^2-12x-24\\ t& =4{(10x^3+x^2+5x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{-96x^4-16x^3-97x^2-12x-24}{4{(10x^3+x^2+5x)}^2}\right)z(x)\end{array}$$

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 2.85: 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 $t=4{(10x^3+x^2+5x)}^2$. There is a pole at $x=0$ of order $2$. There is a pole at $x=-\frac{1}{20}+\frac{i\sqrt{199}}{20}$ of order $2$. There is a pole at $x=-\frac{1}{20}-\frac{i\sqrt{199}}{20}$ 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

$$r=\frac{-\frac{1}{19900}-\frac{i\sqrt{199}}{1990}}{{(x+\frac{1}{20}-\frac{i\sqrt{199}}{20})}^2}+\frac{-\frac{1}{19900}+\frac{i\sqrt{199}}{1990}}{{(x+\frac{1}{20}+\frac{i\sqrt{199}}{20})}^2}+\frac{\frac{3}{250}-\frac{647i\sqrt{199}}{9900250}}{x+\frac{1}{20}-\frac{i\sqrt{199}}{20}}+\frac{\frac{3}{250}+\frac{647i\sqrt{199}}{9900250}}{x+\frac{1}{20}+\frac{i\sqrt{199}}{20}}-\frac{6}{25x^2}-\frac{3}{125x}$$

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 $b=-\frac{6}{25}$. Hence

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

For the pole at $x=-\frac{1}{20}+\frac{i\sqrt{199}}{20}$ let $b$ be the coefficient of $\frac{1}{{(x+\frac{1}{20}-\frac{i\sqrt{199}}{20})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=-\frac{1}{19900}-\frac{i\sqrt{199}}{1990}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{1}{2}+\frac{\sqrt{989826-1990i\sqrt{199}}}{1990}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{1}{2}-\frac{\sqrt{989826-1990i\sqrt{199}}}{1990}\end{array}$$

For the pole at $x=-\frac{1}{20}-\frac{i\sqrt{199}}{20}$ let $b$ be the coefficient of $\frac{1}{{(x+\frac{1}{20}+\frac{i\sqrt{199}}{20})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=-\frac{1}{19900}+\frac{i\sqrt{199}}{1990}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{1}{2}+\frac{\sqrt{989826+1990i\sqrt{199}}}{1990}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{1}{2}-\frac{\sqrt{989826+1990i\sqrt{199}}}{1990}\end{array}$$

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

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{-96x^4-16x^3-97x^2-12x-24}{4{(10x^3+x^2+5x)}^2}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=-\frac{6}{25}$. Hence

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

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

$$r=\frac{-96x^4-16x^3-97x^2-12x-24}{4{(10x^3+x^2+5x)}^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$2$	$0$	$\frac{3}{5}$	$\frac{2}{5}$
$-\frac{1}{20}+\frac{i\sqrt{199}}{20}$	$2$	$0$	$\frac{1}{2}+\frac{\sqrt{989826-1990i\sqrt{199}}}{1990}$	$\frac{1}{2}-\frac{\sqrt{989826-1990i\sqrt{199}}}{1990}$
$-\frac{1}{20}-\frac{i\sqrt{199}}{20}$	$2$	$0$	$\frac{1}{2}+\frac{\sqrt{989826+1990i\sqrt{199}}}{1990}$	$\frac{1}{2}-\frac{\sqrt{989826+1990i\sqrt{199}}}{1990}$

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

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 ${\alpha }_{\mathrm{\infty }}^{+}=\frac{3}{5}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{-}+{\alpha }_{c_3}^{-})\\ & =\frac{3}{5}-\left(\frac{3}{5}\right)\\ & =0\end{array}$$

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

$$\begin{array}{rl}\omega & =((+)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{+}}{x-c_1})+((-)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{-}}{x-c_2})+((-)[\sqrt{r}{]}_{c_3}+\frac{{\alpha }_{c_3}^{-}}{x-c_3})+(+)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =\frac{3}{5x}+\frac{\frac{1}{2}-\frac{\sqrt{989826-1990i\sqrt{199}}}{1990}}{x+\frac{1}{20}-\frac{i\sqrt{199}}{20}}+\frac{\frac{1}{2}-\frac{\sqrt{989826+1990i\sqrt{199}}}{1990}}{x+\frac{1}{20}+\frac{i\sqrt{199}}{20}}+\left(0\right)\\ & =\frac{3}{5x}+\frac{\frac{1}{2}-\frac{\sqrt{989826-1990i\sqrt{199}}}{1990}}{x+\frac{1}{20}-\frac{i\sqrt{199}}{20}}+\frac{\frac{1}{2}-\frac{\sqrt{989826+1990i\sqrt{199}}}{1990}}{x+\frac{1}{20}+\frac{i\sqrt{199}}{20}}\\ & =\frac{12x^2+x+6}{20x^3+2x^2+10x}\end{array}$$

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

$$\begin{array}{rl}\left(0\right)+2(\frac{3}{5x}+\frac{\frac{1}{2}-\frac{\sqrt{989826-1990i\sqrt{199}}}{1990}}{x+\frac{1}{20}-\frac{i\sqrt{199}}{20}}+\frac{\frac{1}{2}-\frac{\sqrt{989826+1990i\sqrt{199}}}{1990}}{x+\frac{1}{20}+\frac{i\sqrt{199}}{20}})\left(0\right)+((-\frac{3}{5x^2}-\frac{\frac{1}{2}-\frac{\sqrt{989826-1990i\sqrt{199}}}{1990}}{{(x+\frac{1}{20}-\frac{i\sqrt{199}}{20})}^2}-\frac{\frac{1}{2}-\frac{\sqrt{989826+1990i\sqrt{199}}}{1990}}{{(x+\frac{1}{20}+\frac{i\sqrt{199}}{20})}^2})+{(\frac{3}{5x}+\frac{\frac{1}{2}-\frac{\sqrt{989826-1990i\sqrt{199}}}{1990}}{x+\frac{1}{20}-\frac{i\sqrt{199}}{20}}+\frac{\frac{1}{2}-\frac{\sqrt{989826+1990i\sqrt{199}}}{1990}}{x+\frac{1}{20}+\frac{i\sqrt{199}}{20}})}^2-\left(\frac{-96x^4-16x^3-97x^2-12x-24}{4{(10x^3+x^2+5x)}^2}\right))& =0\\ 0=0\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & ={\mathrm{e}}^{\int (\frac{3}{5x}+\frac{\frac{1}{2}-\frac{\sqrt{989826-1990i\sqrt{199}}}{1990}}{x+\frac{1}{20}-\frac{i\sqrt{199}}{20}}+\frac{\frac{1}{2}-\frac{\sqrt{989826+1990i\sqrt{199}}}{1990}}{x+\frac{1}{20}+\frac{i\sqrt{199}}{20}})dx}\\ & =x^{3/5}{\mathrm{e}}^{-\frac{\sqrt{199}\arctan \left(\frac{(20x+1)\sqrt{199}}{199}\right)}{995}}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{48x^3+3x^2+4x}{10x^4+x^3+5x^2}dx}\\ & =z_1{e}^{-\ln (10x^2+x+5)-\frac{\sqrt{199}\arctan \left(\frac{(20x+1)\sqrt{199}}{199}\right)}{995}-\frac{2\ln \left(x\right)}{5}}\\ & =z_1\left(\frac{{\mathrm{e}}^{-\frac{\sqrt{199}\arctan \left(\frac{(20x+1)\sqrt{199}}{199}\right)}{995}}}{(10x^2+x+5)x^{2/5}}\right)\end{array}$$

Which simplifies to

$$y_1=\frac{x^{1/5}{\mathrm{e}}^{-\frac{2\sqrt{199}\arctan \left(\frac{(20x+1)\sqrt{199}}{199}\right)}{995}}}{10x^2+x+5}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{48x^3+3x^2+4x}{10x^4+x^3+5x^2}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-2\ln (10x^2+x+5)-\frac{2\sqrt{199}\arctan \left(\frac{(20x+1)\sqrt{199}}{199}\right)}{995}-\frac{4\ln \left(x\right)}{5}}}{{\left(y_1\right)}^2}dx\\ & =y_1(\int \frac{{\mathrm{e}}^{-2\ln (10x^2+x+5)-\frac{2\sqrt{199}\arctan \left(\frac{(20x+1)\sqrt{199}}{199}\right)}{995}-\frac{4\ln \left(x\right)}{5}}{(10x^2+x+5)}^2{\mathrm{e}}^{\frac{4\sqrt{199}\arctan \left(\frac{(20x+1)\sqrt{199}}{199}\right)}{995}}}{x^{2/5}}dx)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{x^{1/5}{\mathrm{e}}^{-\frac{2\sqrt{199}\arctan \left(\frac{(20x+1)\sqrt{199}}{199}\right)}{995}}}{10x^2+x+5}\right)+c_2\left(\frac{x^{1/5}{\mathrm{e}}^{-\frac{2\sqrt{199}\arctan \left(\frac{(20x+1)\sqrt{199}}{199}\right)}{995}}}{10x^2+x+5}(\int \frac{{\mathrm{e}}^{-2\ln (10x^2+x+5)-\frac{2\sqrt{199}\arctan \left(\frac{(20x+1)\sqrt{199}}{199}\right)}{995}-\frac{4\ln \left(x\right)}{5}}{(10x^2+x+5)}^2{\mathrm{e}}^{\frac{4\sqrt{199}\arctan \left(\frac{(20x+1)\sqrt{199}}{199}\right)}{995}}}{x^{2/5}}dx)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.284 (sec). Leaf size: 162

ode:=x^2*(10*x^2+x+5)*diff(diff(y(x),x),x)+x*(48*x^2+3*x+4)*diff(y(x),x)+(36*x^2+x)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{{(i\sqrt{199}+20x+1)}^{-\frac{i\sqrt{199}}{1990}}{(i\sqrt{199}-20x-1)}^{\frac{i\sqrt{199}}{1990}}{\mathrm{e}}^{-\frac{\sqrt{199}\arctan \left(\frac{(20x+1)\sqrt{199}}{199}\right)}{995}}(x^{1/5}\mathrm{HeunG}(\frac{\sqrt{199}+i}{i-\sqrt{199}},0,0,\frac{1}{5},\frac{6}{5},-\frac{i\sqrt{199}}{995},-\frac{20x}{1+i\sqrt{199}})c_2+\mathrm{HeunG}(\frac{\sqrt{199}+i}{i-\sqrt{199}},\frac{15721-179i\sqrt{199}}{194275i\sqrt{199}+641775},-\frac{1}{5},0,\frac{4}{5},-\frac{i\sqrt{199}}{995},-\frac{20x}{1+i\sqrt{199}})c_1)}{10x^2+x+5}$$

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 @ Mo\ 
ebius 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a \ 
power @ Moebius 
      <- Heun successful: received ODE is equivalent to the  HeunG  ODE, case  \ 
a <> 0, e <> 0, g <> 0, c = 0 
   <- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x^2(10x^2+x+5)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+x(48x^2+3x+4)\left(\frac{d}{dx}y\left(x\right)\right)+(36x^2+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{(36x+1)y\left(x\right)}{x(10x^2+x+5)}-\frac{(48x^2+3x+4)\left(\frac{d}{dx}y\left(x\right)\right)}{x(10x^2+x+5)}\\ \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{(48x^2+3x+4)\left(\frac{d}{dx}y\left(x\right)\right)}{x(10x^2+x+5)}+\frac{(36x+1)y\left(x\right)}{x(10x^2+x+5)}=0\\ ◻& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{48x^2+3x+4}{x(10x^2+x+5)},P_3\left(x\right)=\frac{36x+1}{x(10x^2+x+5)}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=\frac{4}{5}\\ & \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}\\ & & x(10x^2+x+5)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+(48x^2+3x+4)\left(\frac{d}{dx}y\left(x\right)\right)+(36x+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}{dx}\frac{d}{dx}y\left(x\right)\right)\text{to series expansion for}m=1..3\\ & & 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}\\ & & a_{0}r(-1+5r)x^{-1+r}+(a_1(1+r)(4+5r)+a_0{(1+r)}^2)x^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(a_{k+1}(k+r+1)(5k+4+5r)+a_k{(k+r+1)}^2+2a_{k-1}(k+r+1)(5k+4+5r))x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & r(-1+5r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,\frac{1}{5}\}\\ \bullet & & \text{Each term must be 0}\\ & & a_1(1+r)(4+5r)+a_0{(1+r)}^2=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & (k+r+1)((a_k+10a_{k-1}+5a_{k+1})k+(a_k+10a_{k-1}+5a_{k+1})r+a_k+8a_{k-1}+4a_{k+1})=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & (k+r+2)((a_{k+1}+10a_k+5a_{k+2})(k+1)+(a_{k+1}+10a_k+5a_{k+2})r+a_{k+1}+8a_k+4a_{k+2})=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=-\frac{10k{a}_k+k{a}_{k+1}+10r{a}_k+r{a}_{k+1}+18a_k+2a_{k+1}}{5k+5r+9}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+2}=-\frac{10k{a}_k+k{a}_{k+1}+18a_k+2a_{k+1}}{5k+9}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+2}=-\frac{10k{a}_k+k{a}_{k+1}+18a_k+2a_{k+1}}{5k+9},4a_1+a_0=0]\\ \bullet & & \text{Recursion relation for}r=\frac{1}{5}\\ & & a_{k+2}=-\frac{10k{a}_k+k{a}_{k+1}+20a_k+\frac{11}{5}{a}_{k+1}}{5k+10}\\ \bullet & & \text{Solution for}r=\frac{1}{5}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+\frac{1}{5}},a_{k+2}=-\frac{10k{a}_k+k{a}_{k+1}+20a_k+\frac{11}{5}{a}_{k+1}}{5k+10},6a_1+\frac{36a_0}{25}=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}{5}}),a_{k+2}=-\frac{10k{a}_k+k{a}_{k+1}+18a_k+2a_{k+1}}{5k+9},4a_1+a_0=0,b_{k+2}=-\frac{10k{b}_k+k{b}_{k+1}+20b_k+\frac{11}{5}{b}_{k+1}}{5k+10},6b_1+\frac{36b_0}{25}=0]\end{array}$$

✓ Mathematica. Time used: 1.139 (sec). Leaf size: 132

ode=x^2*(5+x+10*x^2)*D[y[x],{x,2}]+x*(4+3*x+48*x^2)*D[y[x],x]+(x+36*x^2)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \exp ({\int }_1^x(\frac{3}{5K[1]}-\frac{1}{10(10K[1{]}^2+K[1]+5)})dK[1]-\frac{1}{2}{\int }_1^x\frac{48K[2{]}^2+3K[2]+4}{10K[2{]}^3+K[2{]}^2+5K[2]}dK[2])(c_2{\int }_1^x\exp (-2{\int }_1^{K[3]}(\frac{3}{5K[1]}-\frac{1}{10(10K[1{]}^2+K[1]+5)})dK[1])dK[3]+c_1)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x**2*(10*x**2 + x + 5)*Derivative(y(x), (x, 2)) + x*(48*x**2 + 3*x + 4)*Derivative(y(x), x) + (36*x**2 + x)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False