Internal problem ID [9590]
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: 3.
ODE order: 1.
ODE degree: 1.
CAS Maple gives this as type [_Riccati]
Solve \begin {gather*} \boxed {y^{\prime }-y^{2}-a^{2} x^{2}-b x -c=0} \end {gather*}
✓ Solution by Maple
Time used: 0.003 (sec). Leaf size: 688
dsolve(diff(y(x),x)=y(x)^2+a^2*x^2+b*x+c,y(x), singsol=all)
\[ y \relax (x ) = -\frac {\left (48 i c_{1} a^{7} x^{2}-16 c_{1} a^{6} c \,x^{2}+48 i c_{1} a^{5} b x +4 c_{1} a^{4} b^{2} x^{2}-16 c_{1} a^{4} b c x +12 i c_{1} a^{3} b^{2}+4 c_{1} a^{2} b^{3} x -4 c_{1} a^{2} b^{2} c +c_{1} b^{4}\right ) \hypergeom \left (\left [\frac {4 i a^{2} c +28 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )}{24 a^{4} \left (\left (2 a^{2} c_{1} x +b c_{1}\right ) \hypergeom \left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\hypergeom \left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )\right )}-\frac {\left (24 i a^{5} x -24 a^{4} c x +12 i a^{3} b +6 a^{2} b^{2} x -12 a^{2} b c +3 b^{3}\right ) \hypergeom \left (\left [\frac {4 i a^{2} c +20 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )}{24 a^{4} \left (\left (2 a^{2} c_{1} x +b c_{1}\right ) \hypergeom \left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\hypergeom \left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )\right )}-\frac {\left (-48 i c_{1} a^{7} x^{2}-48 i c_{1} a^{5} b x +48 c_{1} a^{6}-12 i c_{1} a^{3} b^{2}\right ) \hypergeom \left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\left (-24 i a^{5} x -12 i a^{3} b \right ) \hypergeom \left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )}{24 a^{4} \left (\left (2 a^{2} c_{1} x +b c_{1}\right ) \hypergeom \left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\hypergeom \left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.646 (sec). Leaf size: 602
DSolve[y'[x]==y[x]^2+a^2*x^2+b*x+c,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {4 (-1)^{3/4} a^{3/2} D_{\frac {1}{8} \left (4-\frac {i \left (b^2-4 a^2 c\right )}{a^3}\right )}\left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )+i \sqrt {2} \left (2 a^2 x+b\right ) D_{\frac {1}{8} \left (-\frac {i \left (b^2-4 a^2 c\right )}{a^3}-4\right )}\left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )+c_1 \left (4 \sqrt [4]{-1} a^{3/2} D_{\frac {1}{8} \left (\frac {i \left (b^2-4 a^2 c\right )}{a^3}+4\right )}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )-i \sqrt {2} \left (2 a^2 x+b\right ) D_{\frac {1}{8} \left (\frac {i \left (b^2-4 a^2 c\right )}{a^3}-4\right )}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )\right )}{2 \sqrt {2} a \left (D_{\frac {1}{8} \left (-\frac {i \left (b^2-4 a^2 c\right )}{a^3}-4\right )}\left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )+c_1 D_{\frac {1}{8} \left (\frac {i \left (b^2-4 a^2 c\right )}{a^3}-4\right )}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )\right )} \\ y(x)\to \frac {(1+i) \sqrt {a} D_{\frac {1}{8} \left (\frac {i \left (b^2-4 a^2 c\right )}{a^3}+4\right )}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )}{D_{\frac {1}{8} \left (\frac {i \left (b^2-4 a^2 c\right )}{a^3}-4\right )}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )}-\frac {i \left (2 a^2 x+b\right )}{2 a} \\ y(x)\to \frac {(1+i) \sqrt {a} D_{\frac {1}{8} \left (\frac {i \left (b^2-4 a^2 c\right )}{a^3}+4\right )}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )}{D_{\frac {1}{8} \left (\frac {i \left (b^2-4 a^2 c\right )}{a^3}-4\right )}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )}-\frac {i \left (2 a^2 x+b\right )}{2 a} \\ \end{align*}