Internal problem ID [10382]
Internal file name [OUTPUT/9330_Monday_June_06_2022_01_51_58_PM_89123123/index.tex]
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.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_rational, _Riccati]
\[ \boxed {x^{2} y^{\prime }-y^{2} c \,x^{2}-\left (x^{n} a +b \right ) x y=\alpha \,x^{2 n}+\beta \,x^{n}+\gamma } \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y^{2} c \,x^{2}+x^{n} a x y +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 \[ y' = c \,y^{2}+\frac {x^{n} a y}{x}+\frac {y b}{x}+\frac {\beta \,x^{n}}{x^{2}}+\frac {\alpha \,x^{2 n}}{x^{2}}+\frac {\gamma }{x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {\alpha \,x^{2 n}+\beta \,x^{n}+\gamma }{x^{2}}\), \(f_1(x)=\frac {x^{n} a x +b x}{x^{2}}\) 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 (\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \right )}{x^{2}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} 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 (\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \right ) u \left (x \right )}{x^{2}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} {\mathrm e}^{\frac {x^{n} a}{2 n}} \left (c_{1} \operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )+c_{2} \operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {{\mathrm e}^{\frac {x^{n} a}{2 n}} x^{-n -\frac {1}{2}+\frac {b}{2}} \left (-x^{\frac {n}{2}} c_{1} \left (-\sqrt {b^{2}-4 c \gamma +2 b +1}\, \sqrt {a^{2}-4 \alpha c}-\sqrt {a^{2}-4 \alpha c}\, n +\left (b -n +1\right ) a -2 \beta c \right ) \operatorname {WhittakerM}\left (-\frac {-2 \sqrt {a^{2}-4 \alpha c}\, n +\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )-2 n c_{2} x^{\frac {n}{2}} \sqrt {a^{2}-4 \alpha c}\, \operatorname {WhittakerW}\left (-\frac {-2 \sqrt {a^{2}-4 \alpha c}\, n +\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )+\left (c_{1} \operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )+c_{2} \operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )\right ) \left (\left (\left (b -n +1\right ) x^{\frac {n}{2}}+a \,x^{\frac {3 n}{2}}\right ) \sqrt {a^{2}-4 \alpha c}+\left (\left (b -n +1\right ) a -2 \beta c \right ) x^{\frac {n}{2}}+x^{\frac {3 n}{2}} \left (a^{2}-4 \alpha c \right )\right )\right )}{2 \sqrt {a^{2}-4 \alpha c}} \] Using the above in (1) gives the solution \[ y = -\frac {x^{-n -\frac {1}{2}+\frac {b}{2}} \left (-x^{\frac {n}{2}} c_{1} \left (-\sqrt {b^{2}-4 c \gamma +2 b +1}\, \sqrt {a^{2}-4 \alpha c}-\sqrt {a^{2}-4 \alpha c}\, n +\left (b -n +1\right ) a -2 \beta c \right ) \operatorname {WhittakerM}\left (-\frac {-2 \sqrt {a^{2}-4 \alpha c}\, n +\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )-2 n c_{2} x^{\frac {n}{2}} \sqrt {a^{2}-4 \alpha c}\, \operatorname {WhittakerW}\left (-\frac {-2 \sqrt {a^{2}-4 \alpha c}\, n +\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )+\left (c_{1} \operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )+c_{2} \operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )\right ) \left (\left (\left (b -n +1\right ) x^{\frac {n}{2}}+a \,x^{\frac {3 n}{2}}\right ) \sqrt {a^{2}-4 \alpha c}+\left (\left (b -n +1\right ) a -2 \beta c \right ) x^{\frac {n}{2}}+x^{\frac {3 n}{2}} \left (a^{2}-4 \alpha c \right )\right )\right ) x^{\frac {n}{2}-\frac {1}{2}-\frac {b}{2}}}{2 \sqrt {a^{2}-4 \alpha c}\, c \left (c_{1} \operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )+c_{2} \operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\left (-x^{\frac {n}{2}} c_{3} \left (-\sqrt {b^{2}-4 c \gamma +2 b +1}\, \sqrt {a^{2}-4 \alpha c}-\sqrt {a^{2}-4 \alpha c}\, n +\left (b -n +1\right ) a -2 \beta c \right ) \operatorname {WhittakerM}\left (-\frac {-2 \sqrt {a^{2}-4 \alpha c}\, n +\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )-2 n \,x^{\frac {n}{2}} \sqrt {a^{2}-4 \alpha c}\, \operatorname {WhittakerW}\left (-\frac {-2 \sqrt {a^{2}-4 \alpha c}\, n +\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )+\left (c_{3} \operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )+\operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )\right ) \left (\left (\left (b -n +1\right ) x^{\frac {n}{2}}+a \,x^{\frac {3 n}{2}}\right ) \sqrt {a^{2}-4 \alpha c}+\left (\left (b -n +1\right ) a -2 \beta c \right ) x^{\frac {n}{2}}+x^{\frac {3 n}{2}} \left (a^{2}-4 \alpha c \right )\right )\right ) x^{-\frac {n}{2}-1}}{2 \sqrt {a^{2}-4 \alpha c}\, c \left (c_{3} \operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )+\operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (-x^{\frac {n}{2}} c_{3} \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 \beta c \right ) \operatorname {WhittakerM}\left (-\frac {-2 \sqrt {a^{2}-4 \alpha c}\, n +\left (b -n +1\right ) a -2 \beta c}{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 n \,x^{\frac {n}{2}} \sqrt {a^{2}-4 \alpha c}\, \operatorname {WhittakerW}\left (-\frac {-2 \sqrt {a^{2}-4 \alpha c}\, n +\left (b -n +1\right ) a -2 \beta c}{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 )+\left (c_{3} \operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{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 )+\operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{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 (\left (b -n +1\right ) x^{\frac {n}{2}}+a \,x^{\frac {3 n}{2}}\right ) \sqrt {a^{2}-4 \alpha c}+\left (\left (b -n +1\right ) a -2 \beta c \right ) x^{\frac {n}{2}}+x^{\frac {3 n}{2}} \left (a^{2}-4 \alpha c \right )\right )\right ) x^{-\frac {n}{2}-1}}{2 \sqrt {a^{2}-4 \alpha c}\, c \left (c_{3} \operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{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 )+\operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{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 )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left (-x^{\frac {n}{2}} c_{3} \left (-\sqrt {b^{2}-4 c \gamma +2 b +1}\, \sqrt {a^{2}-4 \alpha c}-\sqrt {a^{2}-4 \alpha c}\, n +\left (b -n +1\right ) a -2 \beta c \right ) \operatorname {WhittakerM}\left (-\frac {-2 \sqrt {a^{2}-4 \alpha c}\, n +\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )-2 n \,x^{\frac {n}{2}} \sqrt {a^{2}-4 \alpha c}\, \operatorname {WhittakerW}\left (-\frac {-2 \sqrt {a^{2}-4 \alpha c}\, n +\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )+\left (c_{3} \operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )+\operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )\right ) \left (\left (\left (b -n +1\right ) x^{\frac {n}{2}}+a \,x^{\frac {3 n}{2}}\right ) \sqrt {a^{2}-4 \alpha c}+\left (\left (b -n +1\right ) a -2 \beta c \right ) x^{\frac {n}{2}}+x^{\frac {3 n}{2}} \left (a^{2}-4 \alpha c \right )\right )\right ) x^{-\frac {n}{2}-1}}{2 \sqrt {a^{2}-4 \alpha c}\, c \left (c_{3} \operatorname {WhittakerM}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )+\operatorname {WhittakerW}\left (-\frac {\left (b -n +1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime }-y^{2} c \,x^{2}-\left (x^{n} a +b \right ) x y=\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{2} c \,x^{2}+\left (x^{n} a +b \right ) x y+\alpha \,x^{2 n}+\beta \,x^{n}+\gamma }{x^{2}} \end {array} \]
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) = (x^(n-1)*a*x+b)*(diff(y(x), x))/x-c*(x^(2*n-2)*alpha*x^2+x^(n-2)*beta* 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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 560
dsolve(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,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\left (\sqrt {b^{2}-4 c \gamma +2 b +1}\, \sqrt {a^{2}-4 \alpha c}+\sqrt {a^{2}-4 \alpha c}\, n +\left (n -b -1\right ) a +2 \beta c \right ) \operatorname {WhittakerM}\left (-\frac {-2 \sqrt {a^{2}-4 \alpha c}\, n +a \left (b -n +1\right )-2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )-2 \operatorname {WhittakerW}\left (-\frac {-2 \sqrt {a^{2}-4 \alpha c}\, n +a \left (b -n +1\right )-2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right ) c_{1} n \sqrt {a^{2}-4 \alpha c}+\left (\operatorname {WhittakerW}\left (-\frac {a \left (b -n +1\right )-2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right ) c_{1} +\operatorname {WhittakerM}\left (-\frac {a \left (b -n +1\right )-2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )\right ) \left (\left (a \,x^{n}+b -n +1\right ) \sqrt {a^{2}-4 \alpha c}+\left (a^{2}-4 \alpha c \right ) x^{n}+a \left (b -n +1\right )-2 \beta c \right )}{2 \sqrt {a^{2}-4 \alpha c}\, \left (\operatorname {WhittakerW}\left (-\frac {a \left (b -n +1\right )-2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right ) c_{1} +\operatorname {WhittakerM}\left (-\frac {a \left (b -n +1\right )-2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}, \frac {\sqrt {b^{2}-4 c \gamma +2 b +1}}{2 n}, \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )\right ) c x} \]
✓ Solution by Mathematica
Time used: 3.672 (sec). Leaf size: 1837
DSolve[x^2*y'[x]==c*x^2*y[x]^2+(a*x^n+b)*x*y[x]+\[Alpha]*x^(2*n)+\[Beta]*x^n+\[Gamma],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {-\left (\left (-\left (\left (n^2+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right ) a^2\right )+n (-b+n-1) \sqrt {a^2-4 c \alpha } a+2 c \left (2 \alpha n^2+\sqrt {a^2-4 c \alpha } \beta n+2 \alpha \sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right )\right ) c_1 \operatorname {HypergeometricU}\left (\frac {\left (3 n^2+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right ) a^2+(b-n+1) n \sqrt {a^2-4 c \alpha } a-2 c \left (6 \alpha n^2+\sqrt {a^2-4 c \alpha } \beta n+2 \alpha \sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right )}{2 n^2 \left (a^2-4 c \alpha \right )},\frac {2 n^2+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}}{n^2},\frac {x^n \sqrt {a^2-4 c \alpha }}{n}\right ) x^n\right )-n \left (-a^2 n x^n+4 c n \alpha x^n+a n \sqrt {a^2-4 c \alpha } x^n+\sqrt {a^2-4 c \alpha } \left (b n+n+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right )\right ) c_1 \operatorname {HypergeometricU}\left (\frac {\left (n^2+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right ) a^2+(b-n+1) n \sqrt {a^2-4 c \alpha } a-2 c \left (2 \alpha n^2+\sqrt {a^2-4 c \alpha } \beta n+2 \alpha \sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right )}{2 n^2 \left (a^2-4 c \alpha \right )},\frac {n^2+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}}{n^2},\frac {x^n \sqrt {a^2-4 c \alpha }}{n}\right )-n \left (\left (-a^2 n x^n+4 c n \alpha x^n+a n \sqrt {a^2-4 c \alpha } x^n+\sqrt {a^2-4 c \alpha } \left (b n+n+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right )\right ) L_{\frac {-\left (\left (n^2+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right ) a^2\right )+n (-b+n-1) \sqrt {a^2-4 c \alpha } a+2 c \left (2 \alpha n^2+\sqrt {a^2-4 c \alpha } \beta n+2 \alpha \sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right )}{2 n^2 \left (a^2-4 c \alpha \right )}}^{\frac {\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}}{n^2}}\left (\frac {x^n \sqrt {a^2-4 c \alpha }}{n}\right )-2 n x^n \left (a^2-4 c \alpha \right ) L_{\frac {-\left (\left (3 n^2+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right ) a^2\right )+n (-b+n-1) \sqrt {a^2-4 c \alpha } a+2 c \left (6 \alpha n^2+\sqrt {a^2-4 c \alpha } \beta n+2 \alpha \sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right )}{2 n^2 \left (a^2-4 c \alpha \right )}}^{\frac {n^2+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}}{n^2}}\left (\frac {x^n \sqrt {a^2-4 c \alpha }}{n}\right )\right )}{2 c n^2 x \sqrt {a^2-4 c \alpha } \left (c_1 \operatorname {HypergeometricU}\left (\frac {\left (n^2+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right ) a^2+(b-n+1) n \sqrt {a^2-4 c \alpha } a-2 c \left (2 \alpha n^2+\sqrt {a^2-4 c \alpha } \beta n+2 \alpha \sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right )}{2 n^2 \left (a^2-4 c \alpha \right )},\frac {n^2+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}}{n^2},\frac {x^n \sqrt {a^2-4 c \alpha }}{n}\right )+L_{\frac {-\left (\left (n^2+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right ) a^2\right )+n (-b+n-1) \sqrt {a^2-4 c \alpha } a+2 c \left (2 \alpha n^2+\sqrt {a^2-4 c \alpha } \beta n+2 \alpha \sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right )}{2 n^2 \left (a^2-4 c \alpha \right )}}^{\frac {\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}}{n^2}}\left (\frac {x^n \sqrt {a^2-4 c \alpha }}{n}\right )\right )} \\ y(x)\to \frac {-\frac {x^n \left (2 c \left (\beta n \sqrt {a^2-4 \alpha c}+2 \alpha \sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}+2 \alpha n^2\right )-\left (a^2 \left (\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}+n^2\right )\right )+a n (-b+n-1) \sqrt {a^2-4 \alpha c}\right ) \operatorname {HypergeometricU}\left (\frac {\left (3 n^2+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right ) a^2+(b-n+1) n \sqrt {a^2-4 c \alpha } a-2 c \left (6 \alpha n^2+\sqrt {a^2-4 c \alpha } \beta n+2 \alpha \sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right )}{2 n^2 \left (a^2-4 c \alpha \right )},\frac {2 n^2+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}}{n^2},\frac {x^n \sqrt {a^2-4 c \alpha }}{n}\right )}{\operatorname {HypergeometricU}\left (\frac {\left (n^2+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right ) a^2+(b-n+1) n \sqrt {a^2-4 c \alpha } a-2 c \left (2 \alpha n^2+\sqrt {a^2-4 c \alpha } \beta n+2 \alpha \sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}\right )}{2 n^2 \left (a^2-4 c \alpha \right )},\frac {n^2+\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}}{n^2},\frac {x^n \sqrt {a^2-4 c \alpha }}{n}\right )}-n \left (\sqrt {a^2-4 \alpha c} \left (\sqrt {n^2 \left (b^2+2 b-4 c \gamma +1\right )}+b n+n\right )+a n x^n \sqrt {a^2-4 \alpha c}-a^2 n x^n+4 \alpha c n x^n\right )}{2 c n^2 x \sqrt {a^2-4 \alpha c}} \\ \end{align*}