61.2.3 problem 3
Internal
problem
ID
[12009]
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
Date
solved
:
Monday, January 27, 2025 at 11:48:14 PM
CAS
classification
:
[_Riccati]
\begin{align*} y^{\prime }&=y^{2}+a^{2} x^{2}+b x +c \end{align*}
✓ Solution by Maple
Time used: 0.003 (sec). Leaf size: 369
dsolve(diff(y(x),x)=y(x)^2+a^2*x^2+b*x+c,y(x), singsol=all)
\[
y = \frac {-48 \left (a^{2} x +\frac {b}{2}\right )^{2} c_{1} \left (i a^{3}-\frac {1}{3} a^{2} c +\frac {1}{12} b^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +28 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+48 a^{3} \left (i x^{2} a^{4}+i a^{2} b x +\frac {1}{4} i b^{2}-a^{3}\right ) c_{1} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+24 \left (a^{2} x +\frac {b}{2}\right ) \left (\left (-i a^{3}+a^{2} c -\frac {1}{4} b^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +20 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+i a^{3} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )\right )}{48 a^{4} \left (\left (a^{2} x +\frac {b}{2}\right ) c_{1} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\frac {\operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )}{2}\right )}
\]
✓ Solution by Mathematica
Time used: 0.729 (sec). Leaf size: 664
DSolve[D[y[x],x]==y[x]^2+a^2*x^2+b*x+c,y[x],x,IncludeSingularSolutions -> True]
\begin{align*}
y(x)\to \frac {2 i \sqrt {2} a^2 x \operatorname {ParabolicCylinderD}\left (\frac {1}{8} \left (-\frac {i b^2}{a^3}+\frac {4 i c}{a}-4\right ),-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )+4 (-1)^{3/4} a^{3/2} \operatorname {ParabolicCylinderD}\left (\frac {1}{8} \left (-\frac {i b^2}{a^3}+\frac {4 i c}{a}+4\right ),-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )+i \sqrt {2} b \operatorname {ParabolicCylinderD}\left (\frac {1}{8} \left (-\frac {i b^2}{a^3}+\frac {4 i c}{a}-4\right ),-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )+4 \sqrt [4]{-1} a^{3/2} c_1 \operatorname {ParabolicCylinderD}\left (\frac {1}{8} \left (\frac {i b^2}{a^3}-\frac {4 i c}{a}+4\right ),\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )-i \sqrt {2} c_1 \left (2 a^2 x+b\right ) \operatorname {ParabolicCylinderD}\left (\frac {1}{8} \left (\frac {i b^2}{a^3}-\frac {4 i c}{a}-4\right ),\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )}{2 \sqrt {2} a \left (\operatorname {ParabolicCylinderD}\left (\frac {1}{8} \left (-\frac {i b^2}{a^3}+\frac {4 i c}{a}-4\right ),-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )+c_1 \operatorname {ParabolicCylinderD}\left (\frac {1}{8} \left (\frac {i b^2}{a^3}-\frac {4 i c}{a}-4\right ),\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} \operatorname {ParabolicCylinderD}\left (\frac {1}{8} \left (\frac {i b^2}{a^3}-\frac {4 i c}{a}+4\right ),\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )}{\operatorname {ParabolicCylinderD}\left (\frac {1}{8} \left (\frac {i b^2}{a^3}-\frac {4 i c}{a}-4\right ),\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} \operatorname {ParabolicCylinderD}\left (\frac {1}{8} \left (\frac {i b^2}{a^3}-\frac {4 i c}{a}+4\right ),\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (2 x a^2+b\right )}{a^{3/2}}\right )}{\operatorname {ParabolicCylinderD}\left (\frac {1}{8} \left (\frac {i b^2}{a^3}-\frac {4 i c}{a}-4\right ),\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*}