2.2.50 Problem 53

Internal problem ID [13256]
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 : 53
Date solved : Friday, December 19, 2025 at 02:08:50 AM
CAS classification : [_rational, _Riccati]

\begin{align*} y^{\prime } x^{2}&=c \,x^{2} y^{2}+\left (x^{n} a +b \right ) x y+\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \\ \end{align*}

Entering first order ode riccati solver

\begin{align*} y^{\prime } x^{2}&=c \,x^{2} y^{2}+\left (x^{n} a +b \right ) x y+\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \\ \end{align*}

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

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

With Riccati ODE standard form

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

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{c u} \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 {\left (x^{n} a x +b x \right ) c}{x^{2}}\\ f_2^2 f_0 &=\frac {c^{2} \left (\beta \,x^{n}+\alpha \,x^{2 n}+\gamma \right )}{x^{2}} \end{align*}

Substituting the above terms back in equation (2) gives

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

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}} {\mathrm e}^{\frac {x^{n} a}{2 n}} \operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, x^{n}}{n}\right )+c_2 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} {\mathrm e}^{\frac {x^{n} a}{2 n}} \operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, 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 ) {\mathrm e}^{\frac {x^{n} a}{2 n}} \operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, x^{n}}{n}\right )}{x}+\frac {c_1 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} a \,x^{n} {\mathrm e}^{\frac {x^{n} a}{2 n}} \operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, x^{n}}{n}\right )}{2 x}+\frac {c_1 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} {\mathrm e}^{\frac {x^{n} a}{2 n}} \left (\left (\frac {1}{2}+\frac {\left (\left (b -n +1\right ) a -2 c \beta \right ) x^{-n}}{2 a^{2}-8 c \alpha }\right ) \operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, x^{n}}{n}\right )+\frac {\left (\frac {1}{2}+\frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}-\frac {\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}\right ) n \,x^{-n} \operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}+1, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, x^{n}}{n}\right )}{\sqrt {a^{2}-4 c \alpha }}\right ) \sqrt {a^{2}-4 c \alpha }\, 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 ) {\mathrm e}^{\frac {x^{n} a}{2 n}} \operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, x^{n}}{n}\right )}{x}+\frac {c_2 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} a \,x^{n} {\mathrm e}^{\frac {x^{n} a}{2 n}} \operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, x^{n}}{n}\right )}{2 x}+\frac {c_2 \,x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} {\mathrm e}^{\frac {x^{n} a}{2 n}} \left (\left (\frac {1}{2}+\frac {\left (\left (b -n +1\right ) a -2 c \beta \right ) x^{-n}}{2 a^{2}-8 c \alpha }\right ) \operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, x^{n}}{n}\right )-\frac {n \,x^{-n} \operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}+1, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, x^{n}}{n}\right )}{\sqrt {a^{2}-4 c \alpha }}\right ) \sqrt {a^{2}-4 c \alpha }\, x^{n}}{x} \end{equation}

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

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

Doing change of constants, the above solution becomes

\[ \text {Expression too large to display} \]

Simplifying the above gives

\begin{align*} y &= -\frac {\left (-x^{\frac {n}{2}} x^{-n} \left (-\sqrt {b^{2}-4 c \gamma +2 b +1}\, \sqrt {a^{2}-4 c \alpha }-\sqrt {a^{2}-4 c \alpha }\, n +\left (b -n +1\right ) a -2 c \beta \right ) \operatorname {WhittakerM}\left (-\frac {-2 \sqrt {a^{2}-4 c \alpha }\, n +\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, x^{n}}{n}\right )-2 x^{-n} x^{\frac {n}{2}} \sqrt {a^{2}-4 c \alpha }\, \operatorname {WhittakerW}\left (-\frac {-2 \sqrt {a^{2}-4 c \alpha }\, n +\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, x^{n}}{n}\right ) n c_3 +\left (\operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, x^{n}}{n}\right ) c_3 +\operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, x^{n}}{n}\right )\right ) \left (\left (a \,x^{\frac {n}{2}}+x^{-\frac {n}{2}} \left (b -n +1\right )\right ) \sqrt {a^{2}-4 c \alpha }+x^{\frac {n}{2}} \left (\left (\left (b -n +1\right ) a -2 c \beta \right ) x^{-n}+a^{2}-4 c \alpha \right )\right )\right ) x^{-1+\frac {n}{2}}}{2 \sqrt {a^{2}-4 c \alpha }\, c \left (\operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, x^{n}}{n}\right ) c_3 +\operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 c \beta }{2 \sqrt {a^{2}-4 c \alpha }\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 c \alpha }\, x^{n}}{n}\right )\right )} \\ \end{align*}

The solution

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

was found not to satisfy the ode or the IC. Hence it is removed.

2.2.50.1 ✓ Maple. Time used: 0.004 (sec). Leaf size: 560

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

\[ y = -\frac {\left (\sqrt {b^{2}-4 c \gamma +2 b +1}\, \sqrt {a^{2}-4 \alpha c}+\sqrt {a^{2}-4 \alpha c}\, n +\left (-b +n -1\right ) a +2 c \beta \right ) \operatorname {WhittakerM}\left (-\frac {-2 \sqrt {a^{2}-4 \alpha c}\, n +a \left (b -n +1\right )-2 c \beta }{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )-2 \operatorname {WhittakerW}\left (-\frac {-2 \sqrt {a^{2}-4 \alpha c}\, n +a \left (b -n +1\right )-2 c \beta }{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right ) c_1 n \sqrt {a^{2}-4 \alpha c}+\left (\operatorname {WhittakerW}\left (-\frac {a \left (b -n +1\right )-2 c \beta }{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right ) c_1 +\operatorname {WhittakerM}\left (-\frac {a \left (b -n +1\right )-2 c \beta }{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )\right ) \left (\left (a \,x^{n}+b -n +1\right ) \sqrt {a^{2}-4 \alpha c}+\left (a^{2}-4 \alpha c \right ) x^{n}+a \left (b -n +1\right )-2 c \beta \right )}{2 \sqrt {a^{2}-4 \alpha c}\, \left (\operatorname {WhittakerW}\left (-\frac {a \left (b -n +1\right )-2 c \beta }{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right ) c_1 +\operatorname {WhittakerM}\left (-\frac {a \left (b -n +1\right )-2 c \beta }{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )\right ) c x} \]

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*(x^(-1+n)*a*x+b) 
*diff(y(x),x)-c*(x^(-2+2*n)*alpha*x^2+x^(-2+n)*beta*x^2+gamma)/x^2*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 
         -> Whittaker 
            -> hyper3: Equivalence to 1F1 under a power @ Moebius 
            <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
         <- Whittaker successful 
      <- special function solution successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y \left (x \right )\right )=c \,x^{2} y \left (x \right )^{2}+\left (a \,x^{13256}+b \right ) x y \left (x \right )+\alpha \,x^{26512}+\beta \,x^{13256}+\gamma \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {c \,x^{2} y \left (x \right )^{2}+\left (a \,x^{13256}+b \right ) x y \left (x \right )+\alpha \,x^{26512}+\beta \,x^{13256}+\gamma }{x^{2}} \end {array} \]

2.2.50.2 ✓ Mathematica. Time used: 1.348 (sec). Leaf size: 1837

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

\begin{align*} \text {Solution too large to show}\end{align*}

2.2.50.3 ✗ Sympy

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

Timed Out
 

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