Internal problem ID [10399]
Internal file name [OUTPUT/9347_Monday_June_06_2022_02_14_40_PM_15338845/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: 70.
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]
\[ \boxed {x^{2} \left (x^{2}+a \right ) \left (y^{\prime }+\lambda y^{2}\right )+x \left (b \,x^{2}+c \right ) y=-s} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {y^{2} \lambda \,x^{4}+y^{2} a \lambda \,x^{2}+b \,x^{3} y +c x y +s}{x^{2} \left (x^{2}+a \right )} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\frac {x^{2} y^{2} \lambda }{x^{2}+a}-\frac {y^{2} a \lambda }{x^{2}+a}-\frac {x b y}{x^{2}+a}-\frac {y c}{x \left (x^{2}+a \right )}-\frac {s}{\left (x^{2}+a \right ) x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {s}{\left (x^{2}+a \right ) x^{2}}\), \(f_1(x)=-\frac {b \,x^{3}+c x}{x^{2} \left (x^{2}+a \right )}\) and \(f_2(x)=-\frac {\lambda \,x^{4}+\lambda \,x^{2} a}{x^{2} \left (x^{2}+a \right )}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {\left (\lambda \,x^{4}+\lambda \,x^{2} a \right ) u}{x^{2} \left (x^{2}+a \right )}} \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 {4 \lambda \,x^{3}+2 a \lambda x}{x^{2} \left (x^{2}+a \right )}+\frac {2 \lambda \,x^{4}+2 \lambda \,x^{2} a}{x^{3} \left (x^{2}+a \right )}+\frac {2 \lambda \,x^{4}+2 \lambda \,x^{2} a}{x \left (x^{2}+a \right )^{2}}\\ f_1 f_2 &=\frac {\left (b \,x^{3}+c x \right ) \left (\lambda \,x^{4}+\lambda \,x^{2} a \right )}{x^{4} \left (x^{2}+a \right )^{2}}\\ f_2^2 f_0 &=-\frac {\left (\lambda \,x^{4}+\lambda \,x^{2} a \right )^{2} s}{x^{6} \left (x^{2}+a \right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -\frac {\left (\lambda \,x^{4}+\lambda \,x^{2} a \right ) u^{\prime \prime }\left (x \right )}{x^{2} \left (x^{2}+a \right )}-\left (-\frac {4 \lambda \,x^{3}+2 a \lambda x}{x^{2} \left (x^{2}+a \right )}+\frac {2 \lambda \,x^{4}+2 \lambda \,x^{2} a}{x^{3} \left (x^{2}+a \right )}+\frac {2 \lambda \,x^{4}+2 \lambda \,x^{2} a}{x \left (x^{2}+a \right )^{2}}+\frac {\left (b \,x^{3}+c x \right ) \left (\lambda \,x^{4}+\lambda \,x^{2} a \right )}{x^{4} \left (x^{2}+a \right )^{2}}\right ) u^{\prime }\left (x \right )-\frac {\left (\lambda \,x^{4}+\lambda \,x^{2} a \right )^{2} s u \left (x \right )}{x^{6} \left (x^{2}+a \right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (x^{2}+a \right )^{\frac {\left (2-b \right ) a +c}{2 a}} \left (c_{2} x^{-\frac {-a +c +\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{2 a}} \operatorname {hypergeom}\left (\left [-\frac {-3 a -c +\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{4 a}, \frac {-2 a b +5 a +c -\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{4 a}\right ], \left [1-\frac {\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{2 a}\right ], -\frac {x^{2}}{a}\right )+c_{1} x^{\frac {a -c +\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{2 a}} \operatorname {hypergeom}\left (\left [\frac {3 a +c +\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{4 a}, \frac {-2 a b +5 a +c +\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{4 a}\right ], \left [1+\frac {\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{2 a}\right ], -\frac {x^{2}}{a}\right )\right ) \] The above shows that \[ \text {Expression too large to display} \] Using the above in (1) gives the solution \[ \text {Expression too large to display} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ \text {Expression too large to display} \]
The solution(s) found are the following \begin{align*} \tag{1} \text {Expression too large to display} \\ \end{align*}
Verification of solutions
\[ \text {Expression too large to display} \] Warning, solution could not be verified
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x^{2}+a \right ) \left (y^{\prime }+\lambda y^{2}\right )+x \left (b \,x^{2}+c \right ) y=-s \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y^{2} \lambda \,x^{4}+y^{2} a \lambda \,x^{2}+b \,x^{3} y+c x y+s}{x^{2} \left (x^{2}+a \right )} \end {array} \]
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 to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -(b*x^2+c)*(diff(y(x), x))/(x*(x^2+a))-lambda*s*y(x)/(x^2*(x^2+a)), y( 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 -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach <- heuristic approach successful <- hypergeometric successful <- special function solution successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 1329
dsolve(x^2*(x^2+a)*(diff(y(x),x)+lambda*y(x)^2)+x*(b*x^2+c)*y(x)+s=0,y(x), singsol=all)
\[ \text {Expression too large to display} \]
✓ Solution by Mathematica
Time used: 7.158 (sec). Leaf size: 1470
DSolve[x^2*(x^2+a)*(y'[x]+\[Lambda]*y[x]^2)+x*(b*x^2+c)*y[x]+s==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\frac {\left (a-c-\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}\right ) c_1 \left (\left (-2 b a+a+c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {-5 a+c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},-\frac {-a (2 b+3)+c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},2-\frac {\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{2 a},-\frac {x^2}{a}\right ) x^2+2 a \left (2 a-\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {-a+c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},-\frac {-2 b a+a+c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},1-\frac {\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{2 a},-\frac {x^2}{a}\right )\right ) a^{\frac {\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{2 a}}}{2 a-\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}+2 x^{\frac {\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{a}} \left (a-c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {-2 b a+a+c-\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},\frac {a-c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},\frac {\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{2 a}+1,-\frac {x^2}{a}\right ) a-\frac {x^{\frac {\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{a}+2} \left (a-c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}\right ) \left (a (2 b-1)-c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}\right ) \operatorname {Hypergeometric2F1}\left (\frac {5 a-c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},\frac {a (2 b+3)-c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},\frac {\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{2 a}+2,-\frac {x^2}{a}\right )}{2 a+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}}{4 a^2 x \lambda \left (c_1 \operatorname {Hypergeometric2F1}\left (-\frac {-a+c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},-\frac {-2 b a+a+c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},1-\frac {\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{2 a},-\frac {x^2}{a}\right ) a^{\frac {\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{2 a}}+x^{\frac {\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {-2 b a+a+c-\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},\frac {a-c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},\frac {\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{2 a}+1,-\frac {x^2}{a}\right )\right )} \\ y(x)\to \frac {x \left (a^3 (-b)+a^2 \left (b \sqrt {a^2-2 a (c+2 \lambda s)+c^2}-4 (b-1) \lambda s+c\right )+a \left (b c \left (\sqrt {a^2-2 a (c+2 \lambda s)+c^2}+c\right )-c \sqrt {a^2-2 a (c+2 \lambda s)+c^2}+2 \lambda s \sqrt {a^2-2 a (c+2 \lambda s)+c^2}+4 c \lambda s\right )-c^2 \left (\sqrt {a^2-2 a (c+2 \lambda s)+c^2}+c\right )\right ) \operatorname {Hypergeometric2F1}\left (-\frac {-5 a+c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},-\frac {-a (2 b+3)+c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},2-\frac {\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{2 a},-\frac {x^2}{a}\right )}{2 a^2 \lambda \left (3 a^2+2 a (c+2 \lambda s)-c^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {-a+c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},-\frac {-2 b a+a+c+\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{4 a},1-\frac {\sqrt {a^2-2 (c+2 s \lambda ) a+c^2}}{2 a},-\frac {x^2}{a}\right )}-\frac {\sqrt {a^2-2 a (c+2 \lambda s)+c^2}-a+c}{2 a \lambda x} \\ \end{align*}