Internal problem ID [10332]
Internal file name [OUTPUT/9280_Monday_June_06_2022_01_45_25_PM_48071204/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: 3.
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]
\[ \boxed {y^{\prime }-y^{2}=x^{2} a^{2}+b x +c} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= x^{2} a^{2}+b x +y^{2}+c \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = x^{2} a^{2}+b x +y^{2}+c \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=x^{2} a^{2}+b x +c\), \(f_1(x)=0\) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{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 &=0\\ f_2^2 f_0 &=x^{2} a^{2}+b x +c \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\left (x^{2} a^{2}+b x +c \right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{-\frac {i x \left (a^{2} x +b \right )}{2 a}} \left (2 x \,a^{2} c_{2} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+b c_{2} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right ) c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {2 \left (\left (i a^{3}-\frac {1}{3} a^{2} c +\frac {1}{12} b^{2}\right ) \left (a^{2} x +\frac {b}{2}\right )^{2} c_{2} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +28 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\frac {\left (a^{2} x +\frac {b}{2}\right ) c_{1} \left (-a^{2} c +\frac {1}{4} b^{2}+i a^{3}\right ) \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +20 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )}{2}-\left (\left (i a^{4} x^{2}+i a^{2} b x +\frac {1}{4} i b^{2}-a^{3}\right ) c_{2} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\frac {i \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right ) \left (a^{2} x +\frac {b}{2}\right ) c_{1}}{2}\right ) a^{3}\right ) {\mathrm e}^{-\frac {i x \left (a^{2} x +b \right )}{2 a}}}{a^{4}} \] Using the above in (1) gives the solution \[ y = -\frac {2 \left (\left (i a^{3}-\frac {1}{3} a^{2} c +\frac {1}{12} b^{2}\right ) \left (a^{2} x +\frac {b}{2}\right )^{2} c_{2} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +28 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\frac {\left (a^{2} x +\frac {b}{2}\right ) c_{1} \left (-a^{2} c +\frac {1}{4} b^{2}+i a^{3}\right ) \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +20 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )}{2}-\left (\left (i a^{4} x^{2}+i a^{2} b x +\frac {1}{4} i b^{2}-a^{3}\right ) c_{2} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\frac {i \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right ) \left (a^{2} x +\frac {b}{2}\right ) c_{1}}{2}\right ) a^{3}\right )}{a^{4} \left (2 x \,a^{2} c_{2} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+b c_{2} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right ) c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {4 \left (-i a^{3}+\frac {1}{3} a^{2} c -\frac {1}{12} b^{2}\right ) \left (a^{2} x +\frac {b}{2}\right )^{2} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +28 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\left (4 i a^{7} x^{2}+4 i a^{5} b x -4 a^{6}+i a^{3} b^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+2 \left (a^{2} x +\frac {b}{2}\right ) \left (\left (a^{2} c -\frac {1}{4} b^{2}-i a^{3}\right ) \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +20 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+i a^{3} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )\right ) c_{3}}{4 \left (\left (a^{2} x +\frac {b}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\frac {\operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right ) c_{3}}{2}\right ) a^{4}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {4 \left (-i a^{3}+\frac {1}{3} a^{2} c -\frac {1}{12} b^{2}\right ) \left (a^{2} x +\frac {b}{2}\right )^{2} \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 )+\left (4 i a^{7} x^{2}+4 i a^{5} b x -4 a^{6}+i a^{3} b^{2}\right ) \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 )+2 \left (a^{2} x +\frac {b}{2}\right ) \left (\left (a^{2} c -\frac {1}{4} b^{2}-i a^{3}\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 ) c_{3}}{4 \left (\left (a^{2} x +\frac {b}{2}\right ) \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 ) c_{3}}{2}\right ) a^{4}} \\ \end{align*}
Verification of solutions
\[ y = \frac {4 \left (-i a^{3}+\frac {1}{3} a^{2} c -\frac {1}{12} b^{2}\right ) \left (a^{2} x +\frac {b}{2}\right )^{2} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +28 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\left (4 i a^{7} x^{2}+4 i a^{5} b x -4 a^{6}+i a^{3} b^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+2 \left (a^{2} x +\frac {b}{2}\right ) \left (\left (a^{2} c -\frac {1}{4} b^{2}-i a^{3}\right ) \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +20 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+i a^{3} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )\right ) c_{3}}{4 \left (\left (a^{2} x +\frac {b}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\frac {\operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right ) c_{3}}{2}\right ) a^{4}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}=x^{2} a^{2}+b x +c \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+x^{2} a^{2}+b x +c \end {array} \]
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 Special trying Riccati sub-methods: trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (-a^2*x^2-b*x-c)*y(x), y(x)` *** Sublevel 2 *** Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> 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: indirect Equivalence to 0F1 under \`\`^ @ Moebius\`\` is resolved <- hypergeometric successful <- special function solution successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 393
dsolve(diff(y(x),x)=y(x)^2+a^2*x^2+b*x+c,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-48 \left (i a^{3}-\frac {1}{3} a^{2} c +\frac {1}{12} b^{2}\right ) \left (a^{2} x +\frac {b}{2}\right )^{2} c_{1} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +28 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+48 c_{1} a^{3} \left (i a^{4} x^{2}+i a^{2} b x +\frac {1}{4} i b^{2}-a^{3}\right ) \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+24 \left (a^{2} x +\frac {b}{2}\right ) \left (\left (-i a^{3}+a^{2} c -\frac {1}{4} b^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +20 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+i a^{3} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )\right )}{48 \left (\left (a^{2} x +\frac {b}{2}\right ) c_{1} \operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +12 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )+\frac {\operatorname {hypergeom}\left (\left [\frac {4 i a^{2} c +4 a^{3}-i b^{2}}{16 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a^{2} x +b \right )^{2}}{4 a^{3}}\right )}{2}\right ) a^{4}} \]
✓ Solution by Mathematica
Time used: 1.582 (sec). Leaf size: 664
DSolve[y'[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*}