2.1.98 Problem 100

Internal problem ID [9268]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 100
Date solved : Friday, April 25, 2025 at 06:04:00 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}{x}^2(2+x)y^{\prime \prime }+5x(1-x)y^{\prime }-(2-8x)y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 1.069 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& x^2(2+x)y^{\prime \prime }+(-5x^2+5x)y^{\prime }+(8x-2)y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =x^2(2+x)\\ \text{(3)}& B& =-5x^2+5xC& =8x-2\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{3x^2-126x+21}{4{(x^2+2x)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =3x^2-126x+21\\ t& =4{(x^2+2x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{3x^2-126x+21}{4{(x^2+2x)}^2}\right)z(x)\end{array}$$

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 2.98: 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 $t=4{(x^2+2x)}^2$. There is a pole at $x=0$ of order $2$. There is a pole at $x=-2$ 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

$$r=\frac{21}{16x^2}+\frac{147}{16(2+x)}-\frac{147}{16x}+\frac{285}{16{(2+x)}^2}$$

For the pole at $x=-2$ let $b$ be the coefficient of $\frac{1}{{(2+x)}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{285}{16}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{19}{4}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{15}{4}\end{array}$$

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 $b=\frac{21}{16}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{7}{4}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{3}{4}\end{array}$$

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

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{3x^2-126x+21}{4{(x^2+2x)}^2}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=\frac{3}{4}$. Hence

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

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

$$r=\frac{3x^2-126x+21}{4{(x^2+2x)}^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$-2$	$2$	$0$	$\frac{19}{4}$	$-\frac{15}{4}$
$0$	$2$	$0$	$\frac{7}{4}$	$-\frac{3}{4}$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$2$	$0$	$\frac{3}{2}$	$-\frac{1}{2}$

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 ${\alpha }_{\mathrm{\infty }}^{-}=-\frac{1}{2}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{-})\\ & =-\frac{1}{2}-(-\frac{9}{2})\\ & =4\end{array}$$

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

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{x-c_1})+((-)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{-}}{x-c_2})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =-\frac{15}{4(2+x)}-\frac{3}{4x}+(-)\left(0\right)\\ & =-\frac{15}{4(2+x)}-\frac{3}{4x}\\ & =-\frac{3(3x+1)}{2x(2+x)}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=4$ 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

$$\begin{array}{}\text{(2A)}& p(x)& =x^4+a_3{x}^3+a_2{x}^2+a_{1}x+a_0\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}(12x^2+6x{a}_3+2a_2)+2(-\frac{15}{4(2+x)}-\frac{3}{4x})(4x^3+3x^2{a}_3+2a_{2}x+a_1)+((\frac{15}{4{(2+x)}^2}+\frac{3}{4x^2})+{(-\frac{15}{4(2+x)}-\frac{3}{4x})}^2-\left(\frac{3x^2-126x+21}{4{(x^2+2x)}^2}\right))& =0\\ \frac{3(4+a_3)x^3+(8a_2+3a_3)x^2+(15a_1-2a_2)x+24a_0-3a_1}{x(2+x)}=0\end{array}$$

Solving for the coefficients $a_i$ in the above using method of undetermined coefficients gives

$$\{a_0=\frac{1}{40},a_1=\frac{1}{5},a_2=\frac{3}{2},a_3=-4\}$$

Substituting these coefficients in $p(x)$ in eq. (2A) results in

$$\begin{array}{rl}p(x)& =x^4-4x^3+\frac{3}{2}{x}^2+\frac{1}{5}x+\frac{1}{40}\end{array}$$

Therefore the first solution to the ode $z^{″}=rz$ is

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & =(x^4-4x^3+\frac{3}{2}{x}^2+\frac{1}{5}x+\frac{1}{40}){\mathrm{e}}^{\int (-\frac{15}{4(2+x)}-\frac{3}{4x})dx}\\ & =(x^4-4x^3+\frac{3}{2}{x}^2+\frac{1}{5}x+\frac{1}{40}){\mathrm{e}}^{-\frac{3\ln \left(x\right)}{4}-\frac{15\ln (2+x)}{4}}\\ & =\frac{40x^4-160x^3+60x^2+8x+1}{40x^{3/4}{(2+x)}^{15/4}}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-5x^2+5x}{x^2(2+x)}dx}\\ & =z_1{e}^{-\frac{5\ln \left(x\right)}{4}+\frac{15\ln (2+x)}{4}}\\ & =z_1\left(\frac{{(2+x)}^{15/4}}{x^{5/4}}\right)\end{array}$$

Which simplifies to

$$y_1=\frac{40x^4-160x^3+60x^2+8x+1}{40x^2}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-5x^2+5x}{x^2(2+x)}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-\frac{5\ln \left(x\right)}{2}+\frac{15\ln (2+x)}{2}}}{{\left(y_1\right)}^2}dx\\ & =y_1(-\frac{10x^{5/2}\sqrt{2+x}(-8x^5\sqrt{x(2+x)}+4200\ln \left(\frac{\sqrt{x(2+x)}-x}{x}\right)x^4-4200\ln \left(\frac{x+\sqrt{x(2+x)}}{x}\right)x^4-328x^4\sqrt{x(2+x)}-16800\ln \left(\frac{\sqrt{x(2+x)}-x}{x}\right)x^3+16800\ln \left(\frac{x+\sqrt{x(2+x)}}{x}\right)x^3+13974\sqrt{x(2+x)}{x}^3+6300\ln \left(\frac{\sqrt{x(2+x)}-x}{x}\right)x^2-6300\ln \left(\frac{x+\sqrt{x(2+x)}}{x}\right)x^2-26734x^2\sqrt{x(2+x)}+840\ln \left(\frac{\sqrt{x(2+x)}-x}{x}\right)x-840\ln \left(\frac{x+\sqrt{x(2+x)}}{x}\right)x+805x\sqrt{x(2+x)}+105\ln \left(\frac{\sqrt{x(2+x)}-x}{x}\right)-105\ln \left(\frac{x+\sqrt{x(2+x)}}{x}\right)+105\sqrt{x(2+x)})}{\sqrt{x(2+x)}{(-\sqrt{x(2+x)}+x)}^2{(x+\sqrt{x(2+x)})}^2(40x^4-160x^3+60x^2+8x+1)})\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{40x^4-160x^3+60x^2+8x+1}{40x^2}\right)+c_2\left(\frac{40x^4-160x^3+60x^2+8x+1}{40x^2}(-\frac{10x^{5/2}\sqrt{2+x}(-8x^5\sqrt{x(2+x)}+4200\ln \left(\frac{\sqrt{x(2+x)}-x}{x}\right)x^4-4200\ln \left(\frac{x+\sqrt{x(2+x)}}{x}\right)x^4-328x^4\sqrt{x(2+x)}-16800\ln \left(\frac{\sqrt{x(2+x)}-x}{x}\right)x^3+16800\ln \left(\frac{x+\sqrt{x(2+x)}}{x}\right)x^3+13974\sqrt{x(2+x)}{x}^3+6300\ln \left(\frac{\sqrt{x(2+x)}-x}{x}\right)x^2-6300\ln \left(\frac{x+\sqrt{x(2+x)}}{x}\right)x^2-26734x^2\sqrt{x(2+x)}+840\ln \left(\frac{\sqrt{x(2+x)}-x}{x}\right)x-840\ln \left(\frac{x+\sqrt{x(2+x)}}{x}\right)x+805x\sqrt{x(2+x)}+105\ln \left(\frac{\sqrt{x(2+x)}-x}{x}\right)-105\ln \left(\frac{x+\sqrt{x(2+x)}}{x}\right)+105\sqrt{x(2+x)})}{\sqrt{x(2+x)}{(-\sqrt{x(2+x)}+x)}^2{(x+\sqrt{x(2+x)})}^2(40x^4-160x^3+60x^2+8x+1)})\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.185 (sec). Leaf size: 113

ode:=x^2*(x+2)*diff(diff(y(x),x),x)+5*x*(1-x)*diff(y(x),x)-(2-8*x)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{c_1(40x^4-160x^3+60x^2+8x+1)}{x^2}+\frac{4(1050(x^4-4x^3+\frac{3}{2}{x}^2+\frac{1}{5}x+\frac{1}{40})x^{3/2}\mathrm{arcsinh}\left(\frac{\sqrt{2}\sqrt{x}}{2}\right)+\sqrt{2+x}{x}^2(x^5+41x^4-\frac{6987}{4}{x}^3+\frac{13367}{4}{x}^2-\frac{805}{8}x-\frac{105}{8})){(-2-x)}^{3/4}{c}_2}{{(2+x)}^{3/4}{x}^{7/2}}$$

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 
         <- hyper3 successful: received ODE is equivalent to the 2F1 ODE 
      <- 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}\\ & & x^2(2+x)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+5x(1-x)\left(\frac{d}{dx}y\left(x\right)\right)-(2-8x)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{2(-1+4x)y\left(x\right)}{x^2(2+x)}+\frac{5(-1+x)\left(\frac{d}{dx}y\left(x\right)\right)}{x(2+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{5(-1+x)\left(\frac{d}{dx}y\left(x\right)\right)}{x(2+x)}+\frac{2(-1+4x)y\left(x\right)}{x^2(2+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{5(-1+x)}{x(2+x)},P_3\left(x\right)=\frac{2(-1+4x)}{x^2(2+x)}]\\ & \circ & (2+x)\cdot P_2\left(x\right)\text{is analytic at}x=-2\\ & & ((2+x)\cdot P_2\left(x\right))|{}_{x=-2}=-\frac{15}{2}\\ & \circ & {(2+x)}^2\cdot P_3\left(x\right)\text{is analytic at}x=-2\\ & & ({(2+x)}^2\cdot P_3\left(x\right))|{}_{x=-2}=0\\ & \circ & x=-2\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & & x_0=-2\\ \bullet & & \text{Multiply by denominators}\\ & & x^2(2+x)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)-5x(-1+x)\left(\frac{d}{dx}y\left(x\right)\right)+(8x-2)y\left(x\right)=0\\ \bullet & & \text{Change variables using}x=u-2\text{so that the regular singular point is at}u=0\\ & & (u^3-4u^2+4u)\left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)+(-5u^2+25u-30)\left(\frac{d}{du}y\left(u\right)\right)+(8u-18)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}\\ & & 2a_{0}r(-17+2r)u^{-1+r}+(2a_1(1+r)(-15+2r)-a_0(4r^2-29r+18))u^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(2a_{k+1}(k+1+r)(2k-15+2r)-a_k(4k^2+8kr+4r^2-29k-29r+18)+a_{k-1}(k-3+r)(k-5+r))u^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & 2r(-17+2r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,\frac{17}{2}\}\\ \bullet & & \text{Each term must be 0}\\ & & 2a_1(1+r)(-15+2r)-a_0(4r^2-29r+18)=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & (-4a_k+a_{k-1}+4a_{k+1})k^2+((-8a_k+2a_{k-1}+8a_{k+1})r+29a_k-8a_{k-1}-26a_{k+1})k+(-4a_k+a_{k-1}+4a_{k+1})r^2+(29a_k-8a_{k-1}-26a_{k+1})r-18a_k+15a_{k-1}-30a_{k+1}=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & (-4a_{k+1}+a_k+4a_{k+2}){(k+1)}^2+((-8a_{k+1}+2a_k+8a_{k+2})r+29a_{k+1}-8a_k-26a_{k+2})(k+1)+(-4a_{k+1}+a_k+4a_{k+2})r^2+(29a_{k+1}-8a_k-26a_{k+2})r-18a_{k+1}+15a_k-30a_{k+2}=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=-\frac{k^2{a}_k-4k^2{a}_{k+1}+2kr{a}_k-8kr{a}_{k+1}+r^2{a}_k-4r^2{a}_{k+1}-6k{a}_k+21k{a}_{k+1}-6r{a}_k+21r{a}_{k+1}+8a_k+7a_{k+1}}{2(2k^2+4kr+2r^2-9k-9r-26)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+2}=-\frac{k^2{a}_k-4k^2{a}_{k+1}-6k{a}_k+21k{a}_{k+1}+8a_k+7a_{k+1}}{2(2k^2-9k-26)}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^k,a_{k+2}=-\frac{k^2{a}_k-4k^2{a}_{k+1}-6k{a}_k+21k{a}_{k+1}+8a_k+7a_{k+1}}{2(2k^2-9k-26)},-30a_1-18a_0=0]\\ \bullet & & \text{Revert the change of variables}u=2+x\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(2+x)}^k,a_{k+2}=-\frac{k^2{a}_k-4k^2{a}_{k+1}-6k{a}_k+21k{a}_{k+1}+8a_k+7a_{k+1}}{2(2k^2-9k-26)},-30a_1-18a_0=0]\\ \bullet & & \text{Recursion relation for}r=\frac{17}{2}\\ & & a_{k+2}=-\frac{k^2{a}_k-4k^2{a}_{k+1}+11k{a}_k-47k{a}_{k+1}+\frac{117}{4}{a}_k-\frac{207}{2}{a}_{k+1}}{2(2k^2+25k+42)}\\ \bullet & & \text{Solution for}r=\frac{17}{2}\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+\frac{17}{2}},a_{k+2}=-\frac{k^2{a}_k-4k^2{a}_{k+1}+11k{a}_k-47k{a}_{k+1}+\frac{117}{4}{a}_k-\frac{207}{2}{a}_{k+1}}{2(2k^2+25k+42)},38a_1-\frac{121a_0}{2}=0]\\ \bullet & & \text{Revert the change of variables}u=2+x\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(2+x)}^{k+\frac{17}{2}},a_{k+2}=-\frac{k^2{a}_k-4k^2{a}_{k+1}+11k{a}_k-47k{a}_{k+1}+\frac{117}{4}{a}_k-\frac{207}{2}{a}_{k+1}}{2(2k^2+25k+42)},38a_1-\frac{121a_0}{2}=0]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(2+x)}^k)+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{(2+x)}^{k+\frac{17}{2}}),a_{k+2}=-\frac{k^2{a}_k-4k^2{a}_{k+1}-6k{a}_k+21k{a}_{k+1}+8a_k+7a_{k+1}}{2(2k^2-9k-26)},-30a_1-18a_0=0,b_{k+2}=-\frac{k^2{b}_k-4k^2{b}_{k+1}+11k{b}_k-47k{b}_{k+1}+\frac{117}{4}{b}_k-\frac{207}{2}{b}_{k+1}}{2(2k^2+25k+42)},38b_1-\frac{121b_0}{2}=0]\end{array}$$

✓ Mathematica. Time used: 0.773 (sec). Leaf size: 163

ode=x^2*(2+x)*D[y[x],{x,2}]+5*x*(1-x)*D[y[x],x]-(2-8*x)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \frac{1}{40}(40x^4-160x^3+60x^2+8x+1)\exp ({\int }_1^x-\frac{9K[1]+3}{2K[1{]}^2+4K[1]}dK[1]-\frac{1}{2}{\int }_1^x\frac{5-5K[2]}{K[2{]}^2+2K[2]}dK[2])(c_2{\int }_1^x\frac{1600\exp (-2{\int }_1^{K[3]}-\frac{9K[1]+3}{2K[1{]}^2+4K[1]}dK[1])}{{(40K[3{]}^4-160K[3{]}^3+60K[3{]}^2+8K[3]+1)}^2}dK[3]+c_1)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x**2*(x + 2)*Derivative(y(x), (x, 2)) + 5*x*(1 - x)*Derivative(y(x), x) - (2 - 8*x)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False