Internal problem ID [10340]
Internal file name [OUTPUT/9288_Monday_June_06_2022_01_48_19_PM_41829524/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: 11.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_Riccati]
\[ \boxed {y^{\prime }-\left (a \,x^{2 n}+b \,x^{-1+n}\right ) y^{2}=c} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= x^{2 n} a \,y^{2}+x^{-1+n} b \,y^{2}+c \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = x^{2 n} a \,y^{2}+\frac {x^{n} b \,y^{2}}{x}+c \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=c\), \(f_1(x)=0\) and \(f_2(x)=a \,x^{2 n}+b \,x^{-1+n}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\left (a \,x^{2 n}+b \,x^{-1+n}\right ) 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' &=\frac {2 a \,x^{2 n} n}{x}+\frac {b \,x^{-1+n} \left (-1+n \right )}{x}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\left (a \,x^{2 n}+b \,x^{-1+n}\right )^{2} c \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \left (a \,x^{2 n}+b \,x^{-1+n}\right ) u^{\prime \prime }\left (x \right )-\left (\frac {2 a \,x^{2 n} n}{x}+\frac {b \,x^{-1+n} \left (-1+n \right )}{x}\right ) u^{\prime }\left (x \right )+\left (a \,x^{2 n}+b \,x^{-1+n}\right )^{2} c u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \frac {-\left (\left (2 i a^{\frac {3}{2}} b n -2 a \sqrt {c}\, b^{2}\right ) x^{1+2 n}+\left (-a^{2} \sqrt {c}\, b +i a^{\frac {5}{2}} n \right ) x^{2+3 n}+b^{2} \left (-\sqrt {c}\, b +i \sqrt {a}\, n \right ) x^{n}\right ) {\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}} c_{1} \left (2+n \right ) \operatorname {hypergeom}\left (\left [\frac {\left (2+3 n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {1+2 n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+\left (-\left (2 b \left (i c \left (2+n \right ) a^{\frac {3}{2}}-a b \,c^{\frac {3}{2}}\right ) x^{2+2 n}+\left (i c \left (2+n \right ) a^{\frac {5}{2}}-c^{\frac {3}{2}} a^{2} b \right ) x^{3+3 n}+x^{1+n} \left (-c^{\frac {3}{2}} b +i \sqrt {a}\, c \left (2+n \right )\right ) b^{2}\right ) c_{2} {\mathrm e}^{\frac {i \left (-4 \sqrt {c}\, \sqrt {a}\, x^{1+n}+\pi \left (2+n \right )\right )}{4+4 n}} \operatorname {hypergeom}\left (\left [\frac {\left (4+3 n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {3+2 n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+\left (2+n \right ) \left (c_{2} {\mathrm e}^{\frac {i \left (-4 \sqrt {c}\, \sqrt {a}\, x^{1+n}+\pi \left (2+n \right )\right )}{4+4 n}} \left (\left (2 i a^{\frac {3}{2}} b c -\sqrt {c}\, a^{2}\right ) x^{2+2 n}+i a^{\frac {5}{2}} x^{3+3 n} c +b \left (\left (i \sqrt {a}\, b c -2 a \sqrt {c}\right ) x^{1+n}-\sqrt {c}\, b \right )\right ) \operatorname {hypergeom}\left (\left [\frac {\left (2+n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {2+n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+i \left (\sqrt {a}\, b^{2} x^{n}+2 a^{\frac {3}{2}} x^{1+2 n} b +a^{\frac {5}{2}} x^{2+3 n}\right ) \operatorname {hypergeom}\left (\left [\frac {i \sqrt {c}\, b +\sqrt {a}\, n}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right ) {\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}} c_{1} \right )\right ) n}{\sqrt {c}\, \left (2+n \right ) n \left (a \,x^{1+n}+b \right )^{2}} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {{\mathrm e}^{\frac {i \left (-4 \sqrt {c}\, \sqrt {a}\, x^{1+n}+\pi \left (2+n \right )\right )}{4+4 n}} c c_{2} \left (x^{1+n} b^{4}+4 x^{2+2 n} a \,b^{3}+6 x^{3+3 n} a^{2} b^{2}+4 x^{4+4 n} a^{3} b +x^{5+5 n} a^{4}\right ) \operatorname {hypergeom}\left (\left [\frac {\left (2+n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {2+n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+{\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}} \operatorname {hypergeom}\left (\left [\frac {i \sqrt {c}\, b +\sqrt {a}\, n}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right ) c_{1} \left (x^{n} b^{4}+4 x^{1+2 n} a \,b^{3}+6 x^{2+3 n} a^{2} b^{2}+x^{4+5 n} a^{4}+4 x^{4 n +3} a^{3} b \right )}{x \left (a \,x^{1+n}+b \right )^{3}} \] Using the above in (1) gives the solution \[ y = -\frac {\left ({\mathrm e}^{\frac {i \left (-4 \sqrt {c}\, \sqrt {a}\, x^{1+n}+\pi \left (2+n \right )\right )}{4+4 n}} c c_{2} \left (x^{1+n} b^{4}+4 x^{2+2 n} a \,b^{3}+6 x^{3+3 n} a^{2} b^{2}+4 x^{4+4 n} a^{3} b +x^{5+5 n} a^{4}\right ) \operatorname {hypergeom}\left (\left [\frac {\left (2+n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {2+n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+{\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}} \operatorname {hypergeom}\left (\left [\frac {i \sqrt {c}\, b +\sqrt {a}\, n}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right ) c_{1} \left (x^{n} b^{4}+4 x^{1+2 n} a \,b^{3}+6 x^{2+3 n} a^{2} b^{2}+x^{4+5 n} a^{4}+4 x^{4 n +3} a^{3} b \right )\right ) \sqrt {c}\, \left (2+n \right ) n}{x \left (a \,x^{1+n}+b \right ) \left (a \,x^{2 n}+b \,x^{-1+n}\right ) \left (-\left (\left (2 i a^{\frac {3}{2}} b n -2 a \sqrt {c}\, b^{2}\right ) x^{1+2 n}+\left (-a^{2} \sqrt {c}\, b +i a^{\frac {5}{2}} n \right ) x^{2+3 n}+b^{2} \left (-\sqrt {c}\, b +i \sqrt {a}\, n \right ) x^{n}\right ) {\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}} c_{1} \left (2+n \right ) \operatorname {hypergeom}\left (\left [\frac {\left (2+3 n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {1+2 n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+\left (-\left (2 b \left (i c \left (2+n \right ) a^{\frac {3}{2}}-a b \,c^{\frac {3}{2}}\right ) x^{2+2 n}+\left (i c \left (2+n \right ) a^{\frac {5}{2}}-c^{\frac {3}{2}} a^{2} b \right ) x^{3+3 n}+x^{1+n} \left (-c^{\frac {3}{2}} b +i \sqrt {a}\, c \left (2+n \right )\right ) b^{2}\right ) c_{2} {\mathrm e}^{\frac {i \left (-4 \sqrt {c}\, \sqrt {a}\, x^{1+n}+\pi \left (2+n \right )\right )}{4+4 n}} \operatorname {hypergeom}\left (\left [\frac {\left (4+3 n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {3+2 n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+\left (2+n \right ) \left (c_{2} {\mathrm e}^{\frac {i \left (-4 \sqrt {c}\, \sqrt {a}\, x^{1+n}+\pi \left (2+n \right )\right )}{4+4 n}} \left (\left (2 i a^{\frac {3}{2}} b c -\sqrt {c}\, a^{2}\right ) x^{2+2 n}+i a^{\frac {5}{2}} x^{3+3 n} c +b \left (\left (i \sqrt {a}\, b c -2 a \sqrt {c}\right ) x^{1+n}-\sqrt {c}\, b \right )\right ) \operatorname {hypergeom}\left (\left [\frac {\left (2+n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {2+n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+i \left (\sqrt {a}\, b^{2} x^{n}+2 a^{\frac {3}{2}} x^{1+2 n} b +a^{\frac {5}{2}} x^{2+3 n}\right ) \operatorname {hypergeom}\left (\left [\frac {i \sqrt {c}\, b +\sqrt {a}\, n}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right ) {\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}} c_{1} \right )\right ) n \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 ({\mathrm e}^{\frac {i \left (-4 \sqrt {c}\, \sqrt {a}\, x^{1+n}+\pi \left (2+n \right )\right )}{4+4 n}} c \left (a^{4} x^{5+4 n}+4 a^{3} b \,x^{4+3 n}+6 a^{2} b^{2} x^{3+2 n}+4 a \,x^{2+n} b^{3}+x \,b^{4}\right ) \operatorname {hypergeom}\left (\left [\frac {\left (2+n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {2+n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+{\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}} \operatorname {hypergeom}\left (\left [\frac {i \sqrt {c}\, b +\sqrt {a}\, n}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right ) c_{3} \left (a^{4} x^{4+4 n}+4 a^{3} b \,x^{3+3 n}+6 a^{2} x^{2+2 n} b^{2}+4 a \,x^{1+n} b^{3}+b^{4}\right )\right ) \left (2+n \right ) n \sqrt {c}}{\left (\left (2 \left (i a^{\frac {3}{2}} b n -a \sqrt {c}\, b^{2}\right ) x^{1+2 n}+\left (-a^{2} \sqrt {c}\, b +i a^{\frac {5}{2}} n \right ) x^{2+3 n}+b^{2} \left (-\sqrt {c}\, b +i \sqrt {a}\, n \right ) x^{n}\right ) {\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}} c_{3} \left (2+n \right ) \operatorname {hypergeom}\left (\left [\frac {\left (2+3 n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {1+2 n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+\left (\left (2 b \left (i c \left (2+n \right ) a^{\frac {3}{2}}-a b \,c^{\frac {3}{2}}\right ) x^{2+2 n}+\left (i c \left (2+n \right ) a^{\frac {5}{2}}-c^{\frac {3}{2}} a^{2} b \right ) x^{3+3 n}+x^{1+n} \left (-c^{\frac {3}{2}} b +i \sqrt {a}\, c \left (2+n \right )\right ) b^{2}\right ) {\mathrm e}^{\frac {i \left (-4 \sqrt {c}\, \sqrt {a}\, x^{1+n}+\pi \left (2+n \right )\right )}{4+4 n}} \operatorname {hypergeom}\left (\left [\frac {\left (4+3 n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {3+2 n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+\left (\left (\left (-2 i a^{\frac {3}{2}} b c +\sqrt {c}\, a^{2}\right ) x^{2+2 n}-i a^{\frac {5}{2}} x^{3+3 n} c +\left (\left (-i \sqrt {a}\, b c +2 a \sqrt {c}\right ) x^{1+n}+\sqrt {c}\, b \right ) b \right ) {\mathrm e}^{\frac {i \left (-4 \sqrt {c}\, \sqrt {a}\, x^{1+n}+\pi \left (2+n \right )\right )}{4+4 n}} \operatorname {hypergeom}\left (\left [\frac {\left (2+n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {2+n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )-i {\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}} \left (\sqrt {a}\, b^{2} x^{n}+2 a^{\frac {3}{2}} x^{1+2 n} b +a^{\frac {5}{2}} x^{2+3 n}\right ) c_{3} \operatorname {hypergeom}\left (\left [\frac {i \sqrt {c}\, b +\sqrt {a}\, n}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )\right ) \left (2+n \right )\right ) n \right ) \left (a \,x^{1+n}+b \right )^{2}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left ({\mathrm e}^{\frac {i \left (-4 \sqrt {c}\, \sqrt {a}\, x^{1+n}+\pi \left (2+n \right )\right )}{4+4 n}} c \left (a^{4} x^{5+4 n}+4 a^{3} b \,x^{4+3 n}+6 a^{2} b^{2} x^{3+2 n}+4 a \,x^{2+n} b^{3}+x \,b^{4}\right ) \operatorname {hypergeom}\left (\left [\frac {\left (2+n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {2+n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+{\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}} \operatorname {hypergeom}\left (\left [\frac {i \sqrt {c}\, b +\sqrt {a}\, n}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right ) c_{3} \left (a^{4} x^{4+4 n}+4 a^{3} b \,x^{3+3 n}+6 a^{2} x^{2+2 n} b^{2}+4 a \,x^{1+n} b^{3}+b^{4}\right )\right ) \left (2+n \right ) n \sqrt {c}}{\left (\left (2 \left (i a^{\frac {3}{2}} b n -a \sqrt {c}\, b^{2}\right ) x^{1+2 n}+\left (-a^{2} \sqrt {c}\, b +i a^{\frac {5}{2}} n \right ) x^{2+3 n}+b^{2} \left (-\sqrt {c}\, b +i \sqrt {a}\, n \right ) x^{n}\right ) {\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}} c_{3} \left (2+n \right ) \operatorname {hypergeom}\left (\left [\frac {\left (2+3 n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {1+2 n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+\left (\left (2 b \left (i c \left (2+n \right ) a^{\frac {3}{2}}-a b \,c^{\frac {3}{2}}\right ) x^{2+2 n}+\left (i c \left (2+n \right ) a^{\frac {5}{2}}-c^{\frac {3}{2}} a^{2} b \right ) x^{3+3 n}+x^{1+n} \left (-c^{\frac {3}{2}} b +i \sqrt {a}\, c \left (2+n \right )\right ) b^{2}\right ) {\mathrm e}^{\frac {i \left (-4 \sqrt {c}\, \sqrt {a}\, x^{1+n}+\pi \left (2+n \right )\right )}{4+4 n}} \operatorname {hypergeom}\left (\left [\frac {\left (4+3 n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {3+2 n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+\left (\left (\left (-2 i a^{\frac {3}{2}} b c +\sqrt {c}\, a^{2}\right ) x^{2+2 n}-i a^{\frac {5}{2}} x^{3+3 n} c +\left (\left (-i \sqrt {a}\, b c +2 a \sqrt {c}\right ) x^{1+n}+\sqrt {c}\, b \right ) b \right ) {\mathrm e}^{\frac {i \left (-4 \sqrt {c}\, \sqrt {a}\, x^{1+n}+\pi \left (2+n \right )\right )}{4+4 n}} \operatorname {hypergeom}\left (\left [\frac {\left (2+n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {2+n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )-i {\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}} \left (\sqrt {a}\, b^{2} x^{n}+2 a^{\frac {3}{2}} x^{1+2 n} b +a^{\frac {5}{2}} x^{2+3 n}\right ) c_{3} \operatorname {hypergeom}\left (\left [\frac {i \sqrt {c}\, b +\sqrt {a}\, n}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )\right ) \left (2+n \right )\right ) n \right ) \left (a \,x^{1+n}+b \right )^{2}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left ({\mathrm e}^{\frac {i \left (-4 \sqrt {c}\, \sqrt {a}\, x^{1+n}+\pi \left (2+n \right )\right )}{4+4 n}} c \left (a^{4} x^{5+4 n}+4 a^{3} b \,x^{4+3 n}+6 a^{2} b^{2} x^{3+2 n}+4 a \,x^{2+n} b^{3}+x \,b^{4}\right ) \operatorname {hypergeom}\left (\left [\frac {\left (2+n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {2+n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+{\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}} \operatorname {hypergeom}\left (\left [\frac {i \sqrt {c}\, b +\sqrt {a}\, n}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right ) c_{3} \left (a^{4} x^{4+4 n}+4 a^{3} b \,x^{3+3 n}+6 a^{2} x^{2+2 n} b^{2}+4 a \,x^{1+n} b^{3}+b^{4}\right )\right ) \left (2+n \right ) n \sqrt {c}}{\left (\left (2 \left (i a^{\frac {3}{2}} b n -a \sqrt {c}\, b^{2}\right ) x^{1+2 n}+\left (-a^{2} \sqrt {c}\, b +i a^{\frac {5}{2}} n \right ) x^{2+3 n}+b^{2} \left (-\sqrt {c}\, b +i \sqrt {a}\, n \right ) x^{n}\right ) {\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}} c_{3} \left (2+n \right ) \operatorname {hypergeom}\left (\left [\frac {\left (2+3 n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {1+2 n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+\left (\left (2 b \left (i c \left (2+n \right ) a^{\frac {3}{2}}-a b \,c^{\frac {3}{2}}\right ) x^{2+2 n}+\left (i c \left (2+n \right ) a^{\frac {5}{2}}-c^{\frac {3}{2}} a^{2} b \right ) x^{3+3 n}+x^{1+n} \left (-c^{\frac {3}{2}} b +i \sqrt {a}\, c \left (2+n \right )\right ) b^{2}\right ) {\mathrm e}^{\frac {i \left (-4 \sqrt {c}\, \sqrt {a}\, x^{1+n}+\pi \left (2+n \right )\right )}{4+4 n}} \operatorname {hypergeom}\left (\left [\frac {\left (4+3 n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {3+2 n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )+\left (\left (\left (-2 i a^{\frac {3}{2}} b c +\sqrt {c}\, a^{2}\right ) x^{2+2 n}-i a^{\frac {5}{2}} x^{3+3 n} c +\left (\left (-i \sqrt {a}\, b c +2 a \sqrt {c}\right ) x^{1+n}+\sqrt {c}\, b \right ) b \right ) {\mathrm e}^{\frac {i \left (-4 \sqrt {c}\, \sqrt {a}\, x^{1+n}+\pi \left (2+n \right )\right )}{4+4 n}} \operatorname {hypergeom}\left (\left [\frac {\left (2+n \right ) \sqrt {a}+i \sqrt {c}\, b}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {2+n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )-i {\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}} \left (\sqrt {a}\, b^{2} x^{n}+2 a^{\frac {3}{2}} x^{1+2 n} b +a^{\frac {5}{2}} x^{2+3 n}\right ) c_{3} \operatorname {hypergeom}\left (\left [\frac {i \sqrt {c}\, b +\sqrt {a}\, n}{\sqrt {a}\, \left (2+2 n \right )}\right ], \left [\frac {n}{1+n}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{1+n}}{1+n}\right )\right ) \left (2+n \right )\right ) n \right ) \left (a \,x^{1+n}+b \right )^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\left (a \,x^{2 n}+b \,x^{-1+n}\right ) y^{2}=c \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (a \,x^{2 n}+b \,x^{-1+n}\right ) y^{2}+c \end {array} \]
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) = (b*x^(n-1)*n+2*x^(2*n)*n*a-b*x^(n-1))*(diff(y(x), x))/(x*(a*x^(2*n)+b* 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 -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- hypergeometric successful <- special function solution successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 1237
dsolve(diff(y(x),x)=(a*x^(2*n)+b*x^(n-1))*y(x)^2+c,y(x), singsol=all)
\[ \text {Expression too large to display} \]
✓ Solution by Mathematica
Time used: 2.128 (sec). Leaf size: 1384
DSolve[y'[x]==(a*x^(2*n)+b*x^(n-1))*y[x]^2+c,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\sqrt {c} (n+1)^2 x^{-n} \left (L_{-\frac {\sqrt {c} b}{2 \sqrt {a} \sqrt {-(n+1)^2}}-\frac {n}{2 (n+1)}}^{-\frac {1}{n+1}}\left (\frac {2 \sqrt {a} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )+c_1 \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} b}{\sqrt {a} \sqrt {-(n+1)^2}}+\frac {n}{n+1}\right ),\frac {n}{n+1},\frac {2 \sqrt {a} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )\right )}{\sqrt {a} c_1 (n+1) \sqrt {-(n+1)^2} \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} b}{\sqrt {a} \sqrt {-(n+1)^2}}+\frac {n}{n+1}\right ),\frac {n}{n+1},\frac {2 \sqrt {a} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )+c_1 \left (\sqrt {a} \sqrt {-(n+1)^2} n+b \sqrt {c} (n+1)\right ) \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} b}{\sqrt {a} \sqrt {-(n+1)^2}}+\frac {3 n+2}{n+1}\right ),\frac {n}{n+1}+1,\frac {2 \sqrt {a} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )+\sqrt {a} (n+1) \sqrt {-(n+1)^2} \left (L_{-\frac {\sqrt {c} b}{2 \sqrt {a} \sqrt {-(n+1)^2}}-\frac {n}{2 (n+1)}}^{-\frac {1}{n+1}}\left (\frac {2 \sqrt {a} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )+2 L_{-\frac {\sqrt {c} b}{2 \sqrt {a} \sqrt {-(n+1)^2}}-\frac {3 n+2}{2 n+2}}^{\frac {n}{n+1}}\left (\frac {2 \sqrt {a} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )\right )} \\ y(x)\to \frac {\sqrt {c} (n+1)^2 x^{-n} \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} b}{\sqrt {a} \sqrt {-(n+1)^2}}+\frac {n}{n+1}\right ),\frac {n}{n+1},\frac {2 \sqrt {a} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )}{\sqrt {a} (n+1) \sqrt {-(n+1)^2} \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} b}{\sqrt {a} \sqrt {-(n+1)^2}}+\frac {n}{n+1}\right ),\frac {n}{n+1},\frac {2 \sqrt {a} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )+\left (\sqrt {a} \sqrt {-(n+1)^2} n+b \sqrt {c} (n+1)\right ) \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} b}{\sqrt {a} \sqrt {-(n+1)^2}}+\frac {n}{n+1}+2\right ),\frac {n}{n+1}+1,\frac {2 \sqrt {a} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )} \\ y(x)\to \frac {\sqrt {c} (n+1)^2 x^{-n} \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} b}{\sqrt {a} \sqrt {-(n+1)^2}}+\frac {n}{n+1}\right ),\frac {n}{n+1},\frac {2 \sqrt {a} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )}{\sqrt {a} (n+1) \sqrt {-(n+1)^2} \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} b}{\sqrt {a} \sqrt {-(n+1)^2}}+\frac {n}{n+1}\right ),\frac {n}{n+1},\frac {2 \sqrt {a} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )+\left (\sqrt {a} \sqrt {-(n+1)^2} n+b \sqrt {c} (n+1)\right ) \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} b}{\sqrt {a} \sqrt {-(n+1)^2}}+\frac {n}{n+1}+2\right ),\frac {n}{n+1}+1,\frac {2 \sqrt {a} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )} \\ y(x)\to \frac {\sqrt {c} (n+1) x^{-n} L_{-\frac {\sqrt {c} b}{2 \sqrt {a} \sqrt {-(n+1)^2}}-\frac {n}{2 (n+1)}}^{-\frac {1}{n+1}}\left (\frac {2 \sqrt {a} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )}{\sqrt {a} \sqrt {-(n+1)^2} \left (2 L_{-\frac {\sqrt {c} b}{2 \sqrt {a} \sqrt {-(n+1)^2}}-\frac {n}{2 (n+1)}-1}^{\frac {n}{n+1}}\left (\frac {2 \sqrt {a} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )+L_{-\frac {\sqrt {c} b}{2 \sqrt {a} \sqrt {-(n+1)^2}}-\frac {n}{2 (n+1)}}^{-\frac {1}{n+1}}\left (\frac {2 \sqrt {a} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )\right )} \\ \end{align*}