Problem 52

2.2.49 Problem 52

2.2.49.1 Solved using ﬁrst_order_ode_riccati
2.2.49.2 ✓Maple
2.2.49.3 ✓Mathematica
2.2.49.4 ✗Sympy

Internal problem ID [13255]
Book : Handbook of exact solutions for ordinary diﬀerential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. 1.2.2. Equations Containing Power Functions
Problem number : 52
Date solved : Sunday, January 18, 2026 at 06:52:41 PM
CAS classiﬁcation : [_rational, _Riccati]

2.2.49.1 Solved using ﬁrst_order_ode_riccati

1.096 (sec)

Entering ﬁrst order ode riccati solver

\begin{align*} x^{2} y^{\prime }&=a \,x^{2} y^{2}+b x y+c \,x^{2 n}+s \,x^{n} \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {a \,x^{2} y^{2}+b x y+c \,x^{2 n}+s \,x^{n}}{x^{2}} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = a y^{2}+\frac {b y}{x}+\frac {c \,x^{2 n}}{x^{2}}+\frac {s \,x^{n}}{x^{2}} \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=\frac {c \,x^{2 n}}{x^{2}}+\frac {s \,x^{n}}{x^{2}}\), \(f_1(x)=\frac {b}{x}\) and \(f_2(x)=a\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u a} \tag {1} \end{align*}

Using the above substitution in the given ODE results (after some simpliﬁcation) in a second order ODE to solve for \(u(x)\) which is

\begin{align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end{align*}

But

\begin{align*} f_2' &=0\\ f_1 f_2 &=\frac {b a}{x}\\ f_2^2 f_0 &=a^{2} \left (\frac {c \,x^{2 n}}{x^{2}}+\frac {s \,x^{n}}{x^{2}}\right ) \end{align*}

Substituting the above terms back in equation (2) gives

\[ a u^{\prime \prime }\left (x \right )-\frac {b a u^{\prime }\left (x \right )}{x}+a^{2} \left (\frac {c \,x^{2 n}}{x^{2}}+\frac {s \,x^{n}}{x^{2}}\right ) u \left (x \right ) = 0 \]

Unable to solve. Will ask Maple to solve this ode now.

The solution for \(u \left (x \right )\) is

\begin{equation} \tag{3} u \left (x \right ) = c_1 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )+c_2 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = \frac {c_1 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \left (\frac {b}{2}+\frac {1}{2}-\frac {n}{2}\right ) \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )}{x}+\frac {2 i c_1 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \left (\left (\frac {1}{2}+\frac {s \,x^{-n}}{4 c}\right ) \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )-\frac {i \left (\frac {1}{2}+\frac {b +1}{2 n}-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}\right ) n \,x^{-n} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}+1, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )}{2 \sqrt {a}\, \sqrt {c}}\right ) \sqrt {a}\, \sqrt {c}\, x^{n}}{x}+\frac {c_2 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \left (\frac {b}{2}+\frac {1}{2}-\frac {n}{2}\right ) \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )}{x}+\frac {2 i c_2 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \left (\left (\frac {1}{2}+\frac {s \,x^{-n}}{4 c}\right ) \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )+\frac {i n \,x^{-n} \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}+1, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )}{2 \sqrt {a}\, \sqrt {c}}\right ) \sqrt {a}\, \sqrt {c}\, x^{n}}{x} \end{equation}

Substituting equations (3,4) into (1) results in

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{u a} \\ y &= -\frac {\frac {c_1 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \left (\frac {b}{2}+\frac {1}{2}-\frac {n}{2}\right ) \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )}{x}+\frac {2 i c_1 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \left (\left (\frac {1}{2}+\frac {s \,x^{-n}}{4 c}\right ) \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )-\frac {i \left (\frac {1}{2}+\frac {b +1}{2 n}-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}\right ) n \,x^{-n} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}+1, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )}{2 \sqrt {a}\, \sqrt {c}}\right ) \sqrt {a}\, \sqrt {c}\, x^{n}}{x}+\frac {c_2 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \left (\frac {b}{2}+\frac {1}{2}-\frac {n}{2}\right ) \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )}{x}+\frac {2 i c_2 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \left (\left (\frac {1}{2}+\frac {s \,x^{-n}}{4 c}\right ) \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )+\frac {i n \,x^{-n} \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}+1, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )}{2 \sqrt {a}\, \sqrt {c}}\right ) \sqrt {a}\, \sqrt {c}\, x^{n}}{x}}{a \left (c_1 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )+c_2 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )\right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\frac {x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \left (\frac {b}{2}+\frac {1}{2}-\frac {n}{2}\right ) \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )}{x}+\frac {2 i x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \left (\left (\frac {1}{2}+\frac {s \,x^{-n}}{4 c}\right ) \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )-\frac {i \left (\frac {1}{2}+\frac {b +1}{2 n}-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}\right ) n \,x^{-n} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}+1, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )}{2 \sqrt {a}\, \sqrt {c}}\right ) \sqrt {a}\, \sqrt {c}\, x^{n}}{x}+\frac {c_3 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \left (\frac {b}{2}+\frac {1}{2}-\frac {n}{2}\right ) \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )}{x}+\frac {2 i c_3 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \left (\left (\frac {1}{2}+\frac {s \,x^{-n}}{4 c}\right ) \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )+\frac {i n \,x^{-n} \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}+1, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )}{2 \sqrt {a}\, \sqrt {c}}\right ) \sqrt {a}\, \sqrt {c}\, x^{n}}{x}}{a \left (x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )+c_3 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )\right )} \]

Simplifying the above gives

\begin{align*} y &= \frac {\frac {\left (-\left (b +1+n \right ) \sqrt {c}+i \sqrt {a}\, s \right ) \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}+1, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )}{2}+\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}+1, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right ) c_3 n \sqrt {c}-\left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right ) c_3 +\operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )\right ) \left (\frac {\left (b -n +1\right ) \sqrt {c}}{2}+i \sqrt {a}\, \left (c \,x^{n}+\frac {s}{2}\right )\right )}{\sqrt {c}\, x a \left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right ) c_3 +\operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {b +1}{2 n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )\right )} \\ \end{align*}

Summary of solutions found

2.2.49.2 ✓ Maple. Time used: 0.003 (sec). Leaf size: 369

ode:=x^2*diff(y(x),x) = a*x^2*y(x)^2+b*x*y(x)+c*x^(2*n)+s*x^n; 
dsolve(ode,y(x), singsol=all);

\[ y = \frac {\frac {\left (-\left (b +n +1\right ) \sqrt {c}+i \sqrt {a}\, s \right ) \operatorname {KummerM}\left (\frac {\left (b -n +1\right ) \sqrt {c}+i \sqrt {a}\, s}{2 \sqrt {c}\, n}, \frac {b +n +1}{n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )}{2}+\operatorname {KummerU}\left (\frac {\left (b -n +1\right ) \sqrt {c}+i \sqrt {a}\, s}{2 \sqrt {c}\, n}, \frac {b +n +1}{n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right ) \sqrt {c}\, c_1 n -\left (\operatorname {KummerU}\left (\frac {\left (b +n +1\right ) \sqrt {c}+i \sqrt {a}\, s}{2 \sqrt {c}\, n}, \frac {b +n +1}{n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right ) c_1 +\operatorname {KummerM}\left (\frac {\left (b +n +1\right ) \sqrt {c}+i \sqrt {a}\, s}{2 \sqrt {c}\, n}, \frac {b +n +1}{n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )\right ) \left (\frac {\left (b -n +1\right ) \sqrt {c}}{2}+i \sqrt {a}\, \left (x^{n} c +\frac {s}{2}\right )\right )}{\sqrt {c}\, x a \left (\operatorname {KummerU}\left (\frac {\left (b +n +1\right ) \sqrt {c}+i \sqrt {a}\, s}{2 \sqrt {c}\, n}, \frac {b +n +1}{n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right ) c_1 +\operatorname {KummerM}\left (\frac {\left (b +n +1\right ) \sqrt {c}+i \sqrt {a}\, s}{2 \sqrt {c}\, n}, \frac {b +n +1}{n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )\right )} \]

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
trying inverse linear 
trying homogeneous types: 
trying Chini 
differential order: 1; looking for linear symmetries 
trying exact 
Looking for potential symmetries 
trying Riccati 
trying Riccati sub-methods: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = 1/x*b*diff(y(x),x)-a 
*(c*x^(-2+2*n)+x^(-2+n)*s)*y(x), y(x) 
      *** Sublevel 2 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying a symmetry of the form [xi=0, eta=F(x)] 
      checking if the LODE is missing y 
      -> Trying an equivalence, under non-integer power transformations, 
         to LODEs admitting Liouvillian solutions. 
         -> Trying a Liouvillian solution using Kovacics algorithm 
         <- No Liouvillian solutions exists 
      -> Trying a solution in terms of special functions: 
         -> Bessel 
         -> elliptic 
         -> Legendre 
         -> Kummer 
            -> hyper3: Equivalence to 1F1 under a power @ Moebius 
            <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
         <- Kummer successful 
      <- special function solution successful 
   <- Riccati to 2nd Order successful

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y \left (x \right )\right )=a \,x^{2} y \left (x \right )^{2}+b x y \left (x \right )+c \,x^{26510}+s \,x^{13255} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {a \,x^{2} y \left (x \right )^{2}+b x y \left (x \right )+c \,x^{26510}+s \,x^{13255}}{x^{2}} \end {array} \]

2.2.49.3 ✓ Mathematica. Time used: 0.634 (sec). Leaf size: 819

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

\begin{align*} y(x)&\to -\frac {i \sqrt {a} c_1 x^n \left (\sqrt {c} (b+n+1)-i \sqrt {a} s\right ) \operatorname {HypergeometricU}\left (\frac {b+3 n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n},\frac {b+2 n+1}{n},-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )+c_1 n \left (i \sqrt {a} \sqrt {c} x^n+b+1\right ) \operatorname {HypergeometricU}\left (\frac {b+n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n},\frac {b+n+1}{n},-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )+n \left (2 i \sqrt {a} \sqrt {c} x^n L_{-\frac {b+3 n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n}}^{\frac {b+n+1}{n}}\left (-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )+\left (i \sqrt {a} \sqrt {c} x^n+b+1\right ) L_{-\frac {b+n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n}}^{\frac {b+1}{n}}\left (-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )\right )}{a n x \left (c_1 \operatorname {HypergeometricU}\left (\frac {b+n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n},\frac {b+n+1}{n},-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )+L_{-\frac {b+n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n}}^{\frac {b+1}{n}}\left (-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )\right )}\\ y(x)&\to -\frac {\frac {\sqrt {a} x^n \left (\sqrt {a} s+i \sqrt {c} (b+n+1)\right ) \operatorname {HypergeometricU}\left (\frac {b+3 n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n},\frac {b+2 n+1}{n},-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )}{n \operatorname {HypergeometricU}\left (\frac {b+n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n},\frac {b+n+1}{n},-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )}+i \sqrt {a} \sqrt {c} x^n+b+1}{a x}\\ y(x)&\to -\frac {\frac {\sqrt {a} x^n \left (\sqrt {a} s+i \sqrt {c} (b+n+1)\right ) \operatorname {HypergeometricU}\left (\frac {b+3 n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n},\frac {b+2 n+1}{n},-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )}{n \operatorname {HypergeometricU}\left (\frac {b+n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n},\frac {b+n+1}{n},-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )}+i \sqrt {a} \sqrt {c} x^n+b+1}{a x} \end{align*}

2.2.49.4 ✗ Sympy

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

NotImplementedError : The given ODE Derivative(y(x), x) - (a*x**2*y(x)**2 + b*x*y(x) + x**n*(c*x**n

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0

classify_ode(ode,func=y(x)) 
 
('factorable', 'lie_group')

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