Internal problem ID [10343]
Internal file name [OUTPUT/9291_Monday_June_06_2022_01_49_10_PM_88481665/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: 14.
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]
\[ \boxed {x^{2} y^{\prime }-x^{2} y^{2}=-a^{2} x^{4}+a \left (1-2 b \right ) x^{2}-b \left (1+b \right )} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {-a^{2} x^{4}-2 a b \,x^{2}+x^{2} y^{2}+x^{2} a -b^{2}-b}{x^{2}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -x^{2} a^{2}-2 a b +y^{2}+a -\frac {b^{2}}{x^{2}}-\frac {b}{x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {-a^{2} x^{4}-2 a b \,x^{2}+x^{2} a -b^{2}-b}{x^{2}}\), \(f_1(x)=0\) 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 &=\frac {-a^{2} x^{4}-2 a b \,x^{2}+x^{2} a -b^{2}-b}{x^{2}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\frac {\left (-a^{2} x^{4}-2 a b \,x^{2}+x^{2} a -b^{2}-b \right ) u \left (x \right )}{x^{2}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{-b} {\mathrm e}^{-\frac {x^{2} a}{2}} \left (c_{2} \Gamma \left (b +\frac {1}{2}\right )-c_{2} \Gamma \left (b +\frac {1}{2}, -x^{2} a \right )+c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \left (\left (x^{2} a +b \right ) \left (-c_{2} \Gamma \left (b +\frac {1}{2}\right )+c_{2} \Gamma \left (b +\frac {1}{2}, -x^{2} a \right )-c_{1} \right ) {\mathrm e}^{-\frac {x^{2} a}{2}}-2 \,{\mathrm e}^{\frac {x^{2} a}{2}} \left (-x^{2} a \right )^{b -\frac {1}{2}} c_{2} a \,x^{2}\right ) x^{-1-b} \] Using the above in (1) gives the solution \[ y = -\frac {\left (\left (x^{2} a +b \right ) \left (-c_{2} \Gamma \left (b +\frac {1}{2}\right )+c_{2} \Gamma \left (b +\frac {1}{2}, -x^{2} a \right )-c_{1} \right ) {\mathrm e}^{-\frac {x^{2} a}{2}}-2 \,{\mathrm e}^{\frac {x^{2} a}{2}} \left (-x^{2} a \right )^{b -\frac {1}{2}} c_{2} a \,x^{2}\right ) x^{-1-b} x^{b} {\mathrm e}^{\frac {x^{2} a}{2}}}{c_{2} \Gamma \left (b +\frac {1}{2}\right )-c_{2} \Gamma \left (b +\frac {1}{2}, -x^{2} a \right )+c_{1}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-2 x^{2} a \left (-x^{2} a \right )^{b -\frac {1}{2}} {\mathrm e}^{x^{2} a}+\left (x^{2} a +b \right ) \left (-\Gamma \left (b +\frac {1}{2}\right )+\Gamma \left (b +\frac {1}{2}, -x^{2} a \right )-c_{3} \right )}{x \left (-\Gamma \left (b +\frac {1}{2}\right )+\Gamma \left (b +\frac {1}{2}, -x^{2} a \right )-c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-2 x^{2} a \left (-x^{2} a \right )^{b -\frac {1}{2}} {\mathrm e}^{x^{2} a}+\left (x^{2} a +b \right ) \left (-\Gamma \left (b +\frac {1}{2}\right )+\Gamma \left (b +\frac {1}{2}, -x^{2} a \right )-c_{3} \right )}{x \left (-\Gamma \left (b +\frac {1}{2}\right )+\Gamma \left (b +\frac {1}{2}, -x^{2} a \right )-c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {-2 x^{2} a \left (-x^{2} a \right )^{b -\frac {1}{2}} {\mathrm e}^{x^{2} a}+\left (x^{2} a +b \right ) \left (-\Gamma \left (b +\frac {1}{2}\right )+\Gamma \left (b +\frac {1}{2}, -x^{2} a \right )-c_{3} \right )}{x \left (-\Gamma \left (b +\frac {1}{2}\right )+\Gamma \left (b +\frac {1}{2}, -x^{2} a \right )-c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime }-x^{2} y^{2}=-a^{2} x^{4}+a \left (1-2 b \right ) x^{2}-b \left (1+b \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x^{2} y^{2}-a^{2} x^{4}+a \left (1-2 b \right ) x^{2}-b \left (1+b \right )}{x^{2}} \end {array} \]
Maple trace Kovacic algorithm successful
`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^4+2*a*b*x^2-a*x^2+b^2+b)*y(x)/x^2, 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 A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible <- Kovacics algorithm successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 84
dsolve(x^2*diff(y(x),x)=x^2*y(x)^2-a^2*x^4+a*(1-2*b)*x^2-b*(b+1),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-2 \left (-a \,x^{2}\right )^{b -\frac {1}{2}} {\mathrm e}^{a \,x^{2}} c_{1} a \,x^{2}+\left (-c_{1} \Gamma \left (b +\frac {1}{2}\right )+c_{1} \Gamma \left (b +\frac {1}{2}, -a \,x^{2}\right )-1\right ) \left (a \,x^{2}+b \right )}{x \left (-c_{1} \Gamma \left (b +\frac {1}{2}\right )+c_{1} \Gamma \left (b +\frac {1}{2}, -a \,x^{2}\right )-1\right )} \]
✓ Solution by Mathematica
Time used: 1.225 (sec). Leaf size: 128
DSolve[x^2*y'[x]==x^2*y[x]^2-a^2*x^4+a*(1-2*b)*x^2-b*(b+1),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x^{2 b+1} \left (a x^2+b\right ) \Gamma \left (b+\frac {1}{2},-a x^2\right )-2 \left (-a x^2\right )^{b+\frac {1}{2}} \left (-e^{a x^2} x^{2 b+1}+c_1 \left (a x^2+b\right )\right )}{x^{2 b+2} \Gamma \left (b+\frac {1}{2},-a x^2\right )-2 c_1 x \left (-a x^2\right )^{b+\frac {1}{2}}} \\ y(x)\to a x+\frac {b}{x} \\ \end{align*}