2.1.300 Problem 303

Internal problem ID [9470]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 303
Date solved : Friday, April 25, 2025 at 06:09:08 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

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

Solved as second order ode using Kovacic algorithm

Time used: 0.708 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =-x^3+2x\\ \text{(3)}& B& =x^3+3x^2-2x-2C& =-x^2-4x-2\end{array}$$

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

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

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =x^6+2x^5-5x^4-16x^3+24x^2+24x+12\\ t& =4{(x^3-2x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{x^6+2x^5-5x^4-16x^3+24x^2+24x+12}{4{(x^3-2x)}^2}\right)z(x)\end{array}$$

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 2.300: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =6-6\\ & =0\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=4{(x^3-2x)}^2$. There is a pole at $x=0$ of order $2$. There is a pole at $x=\sqrt{2}$ of order $2$. There is a pole at $x=-\sqrt{2}$ 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

$$r=\frac{1}{4}+\frac{3}{4{(x-\sqrt{2})}^2}+\frac{3}{4{(x+\sqrt{2})}^2}+\frac{-\frac{5\sqrt{2}}{8}-\frac{1}{2}}{x-\sqrt{2}}+\frac{\frac{5\sqrt{2}}{8}-\frac{1}{2}}{x+\sqrt{2}}+\frac{3}{2x}+\frac{3}{4x^2}$$

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 $b=\frac{3}{4}$. Hence

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

For the pole at $x=\sqrt{2}$ let $b$ be the coefficient of $\frac{1}{{(x-\sqrt{2})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{3}{4}$. Hence

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

For the pole at $x=-\sqrt{2}$ let $b$ be the coefficient of $\frac{1}{{(x+\sqrt{2})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{3}{4}$. Hence

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

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

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx \frac{1}{2}+\frac{1}{2x}-\frac{1}{2x^2}-\frac{3}{2x^3}+\frac{21}{4x^4}-\frac{43}{4x^5}+\frac{135}{4x^6}-\frac{147}{4x^7}+\dots \end{array}$$

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

$$a=\frac{1}{2}$$

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

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^0{a}_i{x}^i\\ \text{(10)}& & =\frac{1}{2}\end{array}$$

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

$${([\sqrt{r}{]}_{\mathrm{\infty }})}^2=\frac{1}{4}$$

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

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{x^6+2x^5-5x^4-16x^3+24x^2+24x+12}{4x^6-16x^4+16x^2}\\ & =Q+\frac{R}{4x^6-16x^4+16x^2}\\ & =\left(\frac{1}{4}\right)+\left(\frac{2x^5-x^4-16x^3+20x^2+24x+12}{4x^6-16x^4+16x^2}\right)\\ & =\frac{1}{4}+\frac{2x^5-x^4-16x^3+20x^2+24x+12}{4x^6-16x^4+16x^2}\end{array}$$

Since the degree of $t$ is $6$, then we see that the coefficient of the term $x^5$ in the remainder $R$ is $2$. Dividing this by leading coefficient in $t$ which is $4$ gives $\frac{1}{2}$. Now $b$ can be found.

$$\begin{array}{rl}b& =\left(\frac{1}{2}\right)-\left(0\right)\\ & =\frac{1}{2}\end{array}$$

Hence

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

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

$$r=\frac{x^6+2x^5-5x^4-16x^3+24x^2+24x+12}{4{(x^3-2x)}^2}$$

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

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

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

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{-}+{\alpha }_{c_3}^{-})\\ & =\frac{1}{2}-\left(\frac{1}{2}\right)\\ & =0\end{array}$$

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

$$\begin{array}{rl}\omega & =((+)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{+}}{x-c_1})+((-)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{-}}{x-c_2})+((-)[\sqrt{r}{]}_{c_3}+\frac{{\alpha }_{c_3}^{-}}{x-c_3})+(+)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =\frac{3}{2x}-\frac{1}{2(x-\sqrt{2})}-\frac{1}{2(x+\sqrt{2})}+\left(\frac{1}{2}\right)\\ & =\frac{3}{2x}-\frac{1}{2(x-\sqrt{2})}-\frac{1}{2(x+\sqrt{2})}+\frac{1}{2}\\ & =\frac{x^3+x^2-2x-6}{2x^3-4x}\end{array}$$

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

$$\begin{array}{rl}\left(0\right)+2(\frac{3}{2x}-\frac{1}{2(x-\sqrt{2})}-\frac{1}{2(x+\sqrt{2})}+\frac{1}{2})\left(0\right)+((-\frac{3}{2x^2}+\frac{1}{2{(x-\sqrt{2})}^2}+\frac{1}{2{(x+\sqrt{2})}^2})+{(\frac{3}{2x}-\frac{1}{2(x-\sqrt{2})}-\frac{1}{2(x+\sqrt{2})}+\frac{1}{2})}^2-\left(\frac{x^6+2x^5-5x^4-16x^3+24x^2+24x+12}{4{(x^3-2x)}^2}\right))& =0\\ 0=0\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & ={\mathrm{e}}^{\int (\frac{3}{2x}-\frac{1}{2(x-\sqrt{2})}-\frac{1}{2(x+\sqrt{2})}+\frac{1}{2})dx}\\ & =\frac{x^{3/2}{\mathrm{e}}^{\frac{x}{2}}}{\sqrt{x+\sqrt{2}}\sqrt{x-\sqrt{2}}}\end{array}$$

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

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

Which simplifies to

$$y_1=x^2{\mathrm{e}}^x$$

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

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

Therefore the solution is

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

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.006 (sec). Leaf size: 17

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

$$y=c_1(-1+x)+c_2{\mathrm{e}}^x{x}^2$$

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) 
   Reducible group (found another exponential solution) 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x(-x^2+2)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)-(x^2+4x+2)((1-x)\left(\frac{d}{dx}y\left(x\right)\right)+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{(x^2+4x+2)y\left(x\right)}{x(x^2-2)}+\frac{(x^2+4x+2)(-1+x)\left(\frac{d}{dx}y\left(x\right)\right)}{x(x^2-2)}\\ \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{(x^2+4x+2)(-1+x)\left(\frac{d}{dx}y\left(x\right)\right)}{x(x^2-2)}+\frac{(x^2+4x+2)y\left(x\right)}{x(x^2-2)}=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+x)(x^2+4x+2)}{x(x^2-2)},P_3\left(x\right)=\frac{x^2+4x+2}{x(x^2-2)}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=-1\\ & \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(x^2-2)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)-(-1+x)(x^2+4x+2)\left(\frac{d}{dx}y\left(x\right)\right)+(x^2+4x+2)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..2\\ & & 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..3\\ & & 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}\\ & & -2a_{0}r(-2+r)x^{-1+r}+(-2a_1(1+r)(-1+r)+2a_0(1+r))x^r+(-2a_2(2+r)r+2a_1(2+r)+a_0{(-2+r)}^2)x^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(-2a_{k+1}(k+r+1)(k+r-1)+2a_k(k+r+1)+a_{k-1}{(k-3+r)}^2-a_{k-2}(k-3+r))x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & -2r(-2+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,2\}\\ \bullet & & \text{The coefficients of each power of}x\text{must be 0}\\ & & [-2a_1(1+r)(-1+r)+2a_0(1+r)=0,-2a_2(2+r)r+2a_1(2+r)+a_0{(-2+r)}^2=0]\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & \{a_1=\frac{a_0}{-1+r},a_2=\frac{a_0(r^2-5r+10)}{2(r^2+r-2)}\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_{k-1}{(k-3+r)}^2-2a_{k+1}(k+r+1)(k+r-1)+(2a_k-a_{k-2})k+(2a_k-a_{k-2})r+2a_k+3a_{k-2}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & a_{k+1}{(k+r-1)}^2-2a_{k+3}(k+3+r)(k+r+1)+(2a_{k+2}-a_k)(k+2)+(2a_{k+2}-a_k)r+2a_{k+2}+3a_k=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+3}=\frac{k^2{a}_{k+1}+2kr{a}_{k+1}+r^2{a}_{k+1}-k{a}_k-2k{a}_{k+1}+2k{a}_{k+2}-r{a}_k-2r{a}_{k+1}+2r{a}_{k+2}+a_k+a_{k+1}+6a_{k+2}}{2(k+3+r)(k+r+1)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+3}=\frac{k^2{a}_{k+1}-k{a}_k-2k{a}_{k+1}+2k{a}_{k+2}+a_k+a_{k+1}+6a_{k+2}}{2(k+3)(k+1)}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+3}=\frac{k^2{a}_{k+1}-k{a}_k-2k{a}_{k+1}+2k{a}_{k+2}+a_k+a_{k+1}+6a_{k+2}}{2(k+3)(k+1)},a_1=-a_0,a_2=-\frac{5a_0}{2}]\\ \bullet & & \text{Recursion relation for}r=2\\ & & a_{k+3}=\frac{k^2{a}_{k+1}-k{a}_k+2k{a}_{k+1}+2k{a}_{k+2}-a_k+a_{k+1}+10a_{k+2}}{2(k+5)(k+3)}\\ \bullet & & \text{Solution for}r=2\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+2},a_{k+3}=\frac{k^2{a}_{k+1}-k{a}_k+2k{a}_{k+1}+2k{a}_{k+2}-a_k+a_{k+1}+10a_{k+2}}{2(k+5)(k+3)},a_1=a_0,a_2=\frac{a_0}{2}]\\ \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+2}),a_{k+3}=\frac{k^2{a}_{k+1}-k{a}_k-2k{a}_{k+1}+2k{a}_{k+2}+a_k+a_{k+1}+6a_{k+2}}{2(k+3)(k+1)},a_1=-a_0,a_2=-\frac{5a_0}{2},b_{k+3}=\frac{k^2{b}_{k+1}-k{b}_k+2k{b}_{k+1}+2k{b}_{k+2}-b_k+b_{k+1}+10b_{k+2}}{2(k+5)(k+3)},b_1=b_0,b_2=\frac{b_0}{2}]\end{array}$$

✓ Mathematica. Time used: 0.338 (sec). Leaf size: 126

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

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

✗ Sympy

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

False