2.32.11 Problem 192

Internal problem ID [13853]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.2-6
Problem number : 192
Date solved : Thursday, January 01, 2026 at 03:01:36 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

2.32.11.1 second order change of variable on y method 1

0.938 (sec)

\begin{align*} x^{2} \left (a x +b \right ) y^{\prime \prime }-2 x \left (a x +2 b \right ) y^{\prime }+2 \left (a x +3 b \right ) y&=0 \\ \end{align*}

Entering second order change of variable on \(y\) method 1 solverIn normal form the given ode is written as

\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=\frac {-2 a \,x^{2}-4 b x}{x^{2} \left (a x +b \right )}\\ q \left (x \right )&=\frac {2 a x +6 b}{\left (a x +b \right ) x^{2}} \end{align*}

Calculating the Liouville ode invariant \(Q\) given by

\begin{align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= \frac {2 a x +6 b}{\left (a x +b \right ) x^{2}} - \frac {\left (\frac {-2 a \,x^{2}-4 b x}{x^{2} \left (a x +b \right )}\right )'}{2}- \frac {\left (\frac {-2 a \,x^{2}-4 b x}{x^{2} \left (a x +b \right )}\right )^2}{4} \\ &= \frac {2 a x +6 b}{\left (a x +b \right ) x^{2}} - \frac {\left (\frac {-4 a x -4 b}{x^{2} \left (a x +b \right )}-\frac {2 \left (-2 a \,x^{2}-4 b x \right )}{x^{3} \left (a x +b \right )}-\frac {\left (-2 a \,x^{2}-4 b x \right ) a}{x^{2} \left (a x +b \right )^{2}}\right )}{2}- \frac {\left (\frac {\left (-2 a \,x^{2}-4 b x \right )^{2}}{x^{4} \left (a x +b \right )^{2}}\right )}{4} \\ &= \frac {2 a x +6 b}{\left (a x +b \right ) x^{2}} - \left (\frac {-4 a x -4 b}{2 x^{2} \left (a x +b \right )}-\frac {-2 a \,x^{2}-4 b x}{x^{3} \left (a x +b \right )}-\frac {\left (-2 a \,x^{2}-4 b x \right ) a}{2 x^{2} \left (a x +b \right )^{2}}\right )-\frac {\left (-2 a \,x^{2}-4 b x \right )^{2}}{4 x^{4} \left (a x +b \right )^{2}}\\ &= 0 \end{align*}

Since the Liouville ode invariant does not depend on the independent variable \(x\) then the transformation

\begin{align*} y = v \left (x \right ) z \left (x \right )\tag {3} \end{align*}

is used to change the original ode to a constant coefficients ode in \(v\). In (3) the term \(z \left (x \right )\) is given by

\begin{align*} z \left (x \right )&={\mathrm e}^{-\int \frac {p \left (x \right )}{2}d x}\\ &= e^{-\int \frac {\frac {-2 a \,x^{2}-4 b x}{x^{2} \left (a x +b \right )}}{2} }\\ &= \frac {x^{2}}{a x +b}\tag {5} \end{align*}

Hence (3) becomes

\begin{align*} y = \frac {v \left (x \right ) x^{2}}{a x +b}\tag {4} \end{align*}

Applying this change of variable to the original ode results in

\begin{align*} x^{4} v^{\prime \prime }\left (x \right ) = 0 \end{align*}

Which is now solved for \(v \left (x \right )\).

The above ode simplifies to

\begin{align*} v^{\prime \prime }\left (x \right ) = 0 \end{align*}

Entering second order ode quadrature solverIntegrating twice gives the solution

\[ v \left (x \right )= c_1 x + c_2 \]

Now that \(v \left (x \right )\) is known, then

\begin{align*} y&= v \left (x \right ) z \left (x \right )\\ &= \left (c_1 x +c_2\right ) \left (z \left (x \right )\right )\tag {7} \end{align*}

But from (5)

\begin{align*} z \left (x \right )&= \frac {x^{2}}{a x +b} \end{align*}

Hence (7) becomes

\begin{align*} y = \frac {\left (c_1 x +c_2 \right ) x^{2}}{a x +b} \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {\left (c_1 x +c_2 \right ) x^{2}}{a x +b} \\ \end{align*}

2.32.11.2 second order change of variable on y method 2

1.119 (sec)

\begin{align*} x^{2} \left (a x +b \right ) y^{\prime \prime }-2 x \left (a x +2 b \right ) y^{\prime }+2 \left (a x +3 b \right ) y&=0 \\ \end{align*}

Entering second order change of variable on \(y\) method 2 solverIn normal form the ode

\begin{align*} x^{2} \left (a x +b \right ) y^{\prime \prime }-2 x \left (a x +2 b \right ) y^{\prime }+2 \left (a x +3 b \right ) y = 0\tag {1} \end{align*}

Becomes

\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=\frac {-2 a x -4 b}{x \left (a x +b \right )}\\ q \left (x \right )&=\frac {2 a x +6 b}{\left (a x +b \right ) x^{2}} \end{align*}

Applying change of variables on the depndent variable \(y = v \left (x \right ) x^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (x \right )\) and not \(y\).

\begin{align*} v^{\prime \prime }\left (x \right )+\left (\frac {2 n}{x}+p \right ) v^{\prime }\left (x \right )+\left (\frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q \right ) v \left (x \right )&=0 \tag {3} \end{align*}

Let the coefficient of \(v \left (x \right )\) above be zero. Hence

\begin{align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q&=0 \tag {4} \end{align*}

Substituting the earlier values found for \(p \left (x \right )\) and \(q \left (x \right )\) into (4) gives

\begin{align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n \left (-2 a x -4 b \right )}{x^{2} \left (a x +b \right )}+\frac {2 a x +6 b}{\left (a x +b \right ) x^{2}}&=0 \tag {5} \end{align*}

Solving (5) for \(n\) gives

\begin{align*} n&=2 \tag {6} \end{align*}

Substituting this value in (3) gives

\begin{align*} v^{\prime \prime }\left (x \right )+\left (\frac {4}{x}+\frac {-2 a x -4 b}{x \left (a x +b \right )}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\frac {2 a v^{\prime }\left (x \right )}{a x +b}&=0 \tag {7} \\ \end{align*}

Using the substitution

\begin{align*} u \left (x \right ) = v^{\prime }\left (x \right ) \end{align*}

Then (7) becomes

\begin{align*} u^{\prime }\left (x \right )+\frac {2 a u \left (x \right )}{a x +b} = 0 \tag {8} \\ \end{align*}

The above is now solved for \(u \left (x \right )\). Entering first order ode linear solverIn canonical form a linear first order is

\begin{align*} u^{\prime }\left (x \right ) + q(x)u \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=\frac {2 a}{a x +b}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {2 a}{a x +b}d x}\\ &= \left (a x +b \right )^{2} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu u &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (u \left (a x +b \right )^{2}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \left (a x +b \right )^{2}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \(\left (a x +b \right )^{2}\) gives the final solution

\[ u \left (x \right ) = \frac {c_1}{\left (a x +b \right )^{2}} \]

Now that \(u \left (x \right )\) is known, then

\begin{align*} v^{\prime }\left (x \right )&= u \left (x \right )\\ v \left (x \right )&= \int u \left (x \right )d x +c_2\\ &= -\frac {c_1}{\left (a x +b \right ) a}+c_2 \end{align*}

Hence

\begin{align*} y&= v \left (x \right ) x^{n}\\ &= \left (-\frac {c_1}{\left (a x +b \right ) a}+c_2 \right ) x^{2}\\ &= \left (-\frac {c_1}{\left (a x +b \right ) a}+c_2 \right ) x^{2}\\ \end{align*}

Summary of solutions found

\begin{align*} y &= \left (-\frac {c_1}{\left (a x +b \right ) a}+c_2 \right ) x^{2} \\ \end{align*}

2.32.11.3 second order kovacic

0.421 (sec)

\begin{align*} x^{2} \left (a x +b \right ) y^{\prime \prime }-2 x \left (a x +2 b \right ) y^{\prime }+2 \left (a x +3 b \right ) y&=0 \\ \end{align*}

Entering kovacic solverWriting the ode as

\begin{align*} x^{2} \left (a x +b \right ) y^{\prime \prime }+\left (-2 a \,x^{2}-4 b x \right ) y^{\prime }+\left (2 a x +6 b \right ) y &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end{align*}

Comparing (1) and (2) shows that

\begin{align*} A &= x^{2} \left (a x +b \right ) \\ B &= -2 a \,x^{2}-4 b x\tag {3} \\ C &= 2 a x +6 b \end{align*}

Applying the Liouville transformation on the dependent variable gives

\begin{align*} z(x) &= y e^{\int \frac {B}{2 A} \,dx} \end{align*}

Then (2) becomes

\begin{align*} z''(x) = r z(x)\tag {4} \end{align*}

Where \(r\) is given by

\begin{align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end{align*}

Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives

\begin{align*} r &= \frac {0}{1}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= 0\\ t &= 1 \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(x) &= 0 \tag {7} \end{align*}

Equation (7) is now solved. After finding \(z(x)\) then \(y\) is found using the inverse transformation

\begin{align*} y &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \end{align*}

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 \(\infty \). The following table summarizes these cases.

Case	Allowed pole order for \(r\)	Allowed value for \(\mathcal {O}(\infty )\)
1	\(\left \{ 0,1,2,4,6,8,\cdots \right \} \)	\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \)
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	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)

Table 2.72: Necessary conditions for each Kovacic case

The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore

\begin{align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 0 - -\infty \\ &= \infty \end{align*}

There are no poles in \(r\). Therefore the set of poles \(\Gamma \) is empty. Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(infinity\) then the necessary conditions for case one are met. Therefore

\begin{align*} L &= [1] \end{align*}

Since \(r = 0\) is not a function of \(x\), then there is no need run Kovacic algorithm to obtain a solution for transformed ode \(z''=r z\) as one solution is

\[ z_1(x) = 1 \]

Using the above, the solution for the original ode can now be found. The first solution to the original ode in \(y\) is found from

\begin{align*} y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {-2 a \,x^{2}-4 b x}{x^{2} \left (a x +b \right )} \,dx} \\ &= z_1 e^{2 \ln \left (x \right )-\ln \left (a x +b \right )} \\ &= z_1 \left (\frac {x^{2}}{a x +b}\right ) \\ \end{align*}

Which simplifies to

\[ y_1 = \frac {x^{2}}{a x +b} \]

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{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {-2 a \,x^{2}-4 b x}{x^{2} \left (a x +b \right )} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{4 \ln \left (x \right )-2 \ln \left (a x +b \right )}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (x\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left (\frac {x^{2}}{a x +b}\right ) + c_2 \left (\frac {x^{2}}{a x +b}\left (x\right )\right ) \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {c_1 \,x^{2}}{a x +b}+\frac {c_2 \,x^{3}}{a x +b} \\ \end{align*}

2.32.11.4 ✓ Maple. Time used: 0.004 (sec). Leaf size: 20

ode:=x^2*(a*x+b)*diff(diff(y(x),x),x)-2*x*(a*x+2*b)*diff(y(x),x)+2*(a*x+3*b)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = \frac {x^{2} \left (c_2 x +c_1 \right )}{a x +b} \]

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) 
<- Kovacics algorithm successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (a x +b \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-2 x \left (a x +2 b \right ) \left (\frac {d}{d x}y \left (x \right )\right )+2 \left (a x +3 b \right ) y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=-\frac {2 \left (a x +3 b \right ) y \left (x \right )}{x^{2} \left (a x +b \right )}+\frac {2 \left (a x +2 b \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (a x +b \right )} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )-\frac {2 \left (a x +2 b \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (a x +b \right )}+\frac {2 \left (a x +3 b \right ) y \left (x \right )}{x^{2} \left (a x +b \right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {2 \left (a x +2 b \right )}{x \left (a x +b \right )}, P_{3}\left (x \right )=\frac {2 \left (a x +3 b \right )}{x^{2} \left (a x +b \right )}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-4 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=6 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (a x +b \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-2 x \left (a x +2 b \right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (2 a x +6 b \right ) y \left (x \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y \left (x \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..3 \\ {} & {} & x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & b a_{0} \left (-2+r \right ) \left (-3+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (b a_{k} \left (k +r -2\right ) \left (k +r -3\right )+a a_{k -1} \left (k +r -2\right ) \left (k +r -3\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & b \left (-2+r \right ) \left (-3+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{2, 3\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r -2\right ) \left (k +r -3\right ) \left (a a_{k -1}+b a_{k}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (k +r -1\right ) \left (k +r -2\right ) \left (a a_{k}+b a_{k +1}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a a_{k}}{b} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +1}=-\frac {a a_{k}}{b} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +1}=-\frac {a a_{k}}{b}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =3 \\ {} & {} & a_{k +1}=-\frac {a a_{k}}{b} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =3 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +3}, a_{k +1}=-\frac {a a_{k}}{b}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +3}\right ), c_{k +1}=-\frac {a c_{k}}{b}, d_{k +1}=-\frac {a d_{k}}{b}\right ] \end {array} \]

2.32.11.5 ✓ Mathematica. Time used: 0.032 (sec). Leaf size: 23

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

\begin{align*} y(x)&\to \frac {x^2 (c_2 x+c_1)}{a x+b} \end{align*}

2.32.11.6 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
y = Function("y") 
ode = Eq(x**2*(a*x + b)*Derivative(y(x), (x, 2)) - 2*x*(a*x + 2*b)*Derivative(y(x), x) + (2*a*x + 6*b)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False