Internal problem ID [10403]
Internal file name [OUTPUT/9351_Monday_June_06_2022_02_15_01_PM_58658321/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: 74.
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]
\[ \boxed {x^{2} \left (x^{n} a -1\right ) \left (y^{\prime }+\lambda y^{2}\right )+\left (p \,x^{n}+q \right ) x y=-r \,x^{n}-s} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {x^{n} y^{2} a \lambda \,x^{2}-y^{2} \lambda \,x^{2}+x^{n} y p x +y q x +r \,x^{n}+s}{x^{2} \left (x^{n} a -1\right )} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\frac {x^{n} y^{2} a \lambda }{x^{n} a -1}+\frac {y^{2} \lambda }{x^{n} a -1}-\frac {x^{n} y p}{x \left (x^{n} a -1\right )}-\frac {y q}{x \left (x^{n} a -1\right )}-\frac {r \,x^{n}}{x^{2} \left (x^{n} a -1\right )}-\frac {s}{x^{2} \left (x^{n} a -1\right )} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {r \,x^{n}+s}{x^{2} \left (x^{n} a -1\right )}\), \(f_1(x)=-\frac {x^{n} p x +q x}{x^{2} \left (x^{n} a -1\right )}\) and \(f_2(x)=-\frac {x^{n} a \lambda \,x^{2}-\lambda \,x^{2}}{x^{2} \left (x^{n} a -1\right )}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {\left (x^{n} a \lambda \,x^{2}-\lambda \,x^{2}\right ) u}{x^{2} \left (x^{n} a -1\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 {x^{n} n x a \lambda +2 x^{n} a \lambda x -2 \lambda x}{x^{2} \left (x^{n} a -1\right )}+\frac {2 x^{n} a \lambda \,x^{2}-2 \lambda \,x^{2}}{x^{3} \left (x^{n} a -1\right )}+\frac {\left (x^{n} a \lambda \,x^{2}-\lambda \,x^{2}\right ) x^{n} n a}{x^{3} \left (x^{n} a -1\right )^{2}}\\ f_1 f_2 &=\frac {\left (x^{n} p x +q x \right ) \left (x^{n} a \lambda \,x^{2}-\lambda \,x^{2}\right )}{x^{4} \left (x^{n} a -1\right )^{2}}\\ f_2^2 f_0 &=-\frac {\left (x^{n} a \lambda \,x^{2}-\lambda \,x^{2}\right )^{2} \left (r \,x^{n}+s \right )}{x^{6} \left (x^{n} a -1\right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -\frac {\left (x^{n} a \lambda \,x^{2}-\lambda \,x^{2}\right ) u^{\prime \prime }\left (x \right )}{x^{2} \left (x^{n} a -1\right )}-\left (-\frac {x^{n} n x a \lambda +2 x^{n} a \lambda x -2 \lambda x}{x^{2} \left (x^{n} a -1\right )}+\frac {2 x^{n} a \lambda \,x^{2}-2 \lambda \,x^{2}}{x^{3} \left (x^{n} a -1\right )}+\frac {\left (x^{n} a \lambda \,x^{2}-\lambda \,x^{2}\right ) x^{n} n a}{x^{3} \left (x^{n} a -1\right )^{2}}+\frac {\left (x^{n} p x +q x \right ) \left (x^{n} a \lambda \,x^{2}-\lambda \,x^{2}\right )}{x^{4} \left (x^{n} a -1\right )^{2}}\right ) u^{\prime }\left (x \right )-\frac {\left (x^{n} a \lambda \,x^{2}-\lambda \,x^{2}\right )^{2} \left (r \,x^{n}+s \right ) u \left (x \right )}{x^{6} \left (x^{n} a -1\right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{\frac {1}{2}+\frac {q}{2}} \left (c_{2} x^{-\frac {\sqrt {4 \lambda s +q^{2}+2 q +1}}{2}} \operatorname {hypergeom}\left (\left [\frac {-a \sqrt {4 \lambda s +q^{2}+2 q +1}+q a +\sqrt {a^{2}+\left (-4 \lambda r -2 p \right ) a +p^{2}}+p}{2 a n}, -\frac {a \sqrt {4 \lambda s +q^{2}+2 q +1}-q a +\sqrt {a^{2}+\left (-4 \lambda r -2 p \right ) a +p^{2}}-p}{2 a n}\right ], \left [1-\frac {\sqrt {4 \lambda s +q^{2}+2 q +1}}{n}\right ], x^{n} a \right )+c_{1} x^{\frac {\sqrt {4 \lambda s +q^{2}+2 q +1}}{2}} \operatorname {hypergeom}\left (\left [\frac {a \sqrt {4 \lambda s +q^{2}+2 q +1}+q a -\sqrt {a^{2}+\left (-4 \lambda r -2 p \right ) a +p^{2}}+p}{2 a n}, \frac {a \sqrt {4 \lambda s +q^{2}+2 q +1}+q a +\sqrt {a^{2}+\left (-4 \lambda r -2 p \right ) a +p^{2}}+p}{2 a n}\right ], \left [1+\frac {\sqrt {4 \lambda s +q^{2}+2 q +1}}{n}\right ], x^{n} a \right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {\left (\left (\left (\frac {a \,q^{2}}{2}+\frac {\left (-a n +a +p \right ) q}{2}+\lambda a s +\lambda r -\frac {p n}{2}+\frac {p}{2}\right ) \sqrt {4 \lambda s +q^{2}+2 q +1}-\frac {a \,q^{3}}{2}+\left (\frac {1}{2} a n -a -\frac {1}{2} p \right ) q^{2}+\left (\frac {\left (a +p \right ) n}{2}-2 \lambda a s -\frac {a}{2}-p \right ) q +\left (\lambda a s +\lambda r +\frac {1}{2} p \right ) n -2 \lambda p s -\frac {p}{2}\right ) c_{2} x^{n -\frac {\sqrt {4 \lambda s +q^{2}+2 q +1}}{2}} \operatorname {hypergeom}\left (\left [\frac {-\sqrt {a^{2}+\left (-4 \lambda r -2 p \right ) a +p^{2}}-a \sqrt {4 \lambda s +q^{2}+2 q +1}+\left (2 n +q \right ) a +p}{2 a n}, \frac {\sqrt {a^{2}+\left (-4 \lambda r -2 p \right ) a +p^{2}}-a \sqrt {4 \lambda s +q^{2}+2 q +1}+\left (2 n +q \right ) a +p}{2 a n}\right ], \left [\frac {2 n -\sqrt {4 \lambda s +q^{2}+2 q +1}}{n}\right ], x^{n} a \right )-\left (\left (\frac {a \,q^{2}}{2}+\frac {\left (-a n +a +p \right ) q}{2}+\lambda a s +\lambda r -\frac {p n}{2}+\frac {p}{2}\right ) \sqrt {4 \lambda s +q^{2}+2 q +1}+\frac {a \,q^{3}}{2}+\left (-\frac {1}{2} a n +a +\frac {1}{2} p \right ) q^{2}+\left (\frac {\left (-a -p \right ) n}{2}+2 \lambda a s +\frac {a}{2}+p \right ) q +\left (-\lambda a s -\lambda r -\frac {1}{2} p \right ) n +2 \lambda p s +\frac {p}{2}\right ) c_{1} x^{n +\frac {\sqrt {4 \lambda s +q^{2}+2 q +1}}{2}} \operatorname {hypergeom}\left (\left [\frac {-\sqrt {a^{2}+\left (-4 \lambda r -2 p \right ) a +p^{2}}+a \sqrt {4 \lambda s +q^{2}+2 q +1}+\left (2 n +q \right ) a +p}{2 a n}, \frac {\sqrt {a^{2}+\left (-4 \lambda r -2 p \right ) a +p^{2}}+a \sqrt {4 \lambda s +q^{2}+2 q +1}+\left (2 n +q \right ) a +p}{2 a n}\right ], \left [\frac {2 n +\sqrt {4 \lambda s +q^{2}+2 q +1}}{n}\right ], x^{n} a \right )+2 \left (x^{-\frac {\sqrt {4 \lambda s +q^{2}+2 q +1}}{2}} c_{2} \left (-1-q +\sqrt {4 \lambda s +q^{2}+2 q +1}\right ) \operatorname {hypergeom}\left (\left [\frac {-a \sqrt {4 \lambda s +q^{2}+2 q +1}+q a +\sqrt {a^{2}+\left (-4 \lambda r -2 p \right ) a +p^{2}}+p}{2 a n}, -\frac {a \sqrt {4 \lambda s +q^{2}+2 q +1}-q a +\sqrt {a^{2}+\left (-4 \lambda r -2 p \right ) a +p^{2}}-p}{2 a n}\right ], \left [\frac {n -\sqrt {4 \lambda s +q^{2}+2 q +1}}{n}\right ], x^{n} a \right )-c_{1} x^{\frac {\sqrt {4 \lambda s +q^{2}+2 q +1}}{2}} \operatorname {hypergeom}\left (\left [\frac {a \sqrt {4 \lambda s +q^{2}+2 q +1}+q a -\sqrt {a^{2}+\left (-4 \lambda r -2 p \right ) a +p^{2}}+p}{2 a n}, \frac {a \sqrt {4 \lambda s +q^{2}+2 q +1}+q a +\sqrt {a^{2}+\left (-4 \lambda r -2 p \right ) a +p^{2}}+p}{2 a n}\right ], \left [\frac {n +\sqrt {4 \lambda s +q^{2}+2 q +1}}{n}\right ], x^{n} a \right ) \left (1+q +\sqrt {4 \lambda s +q^{2}+2 q +1}\right )\right ) \left (\lambda s -\frac {1}{4} n^{2}+\frac {1}{4} q^{2}+\frac {1}{2} q +\frac {1}{4}\right )\right ) x^{-\frac {1}{2}+\frac {q}{2}}}{4 \lambda s -n^{2}+q^{2}+2 q +1} \] 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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x^{n} a -1\right ) \left (y^{\prime }+\lambda y^{2}\right )+\left (p \,x^{n}+q \right ) x y=-r \,x^{n}-s \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {x^{n} y^{2} a \lambda \,x^{2}-y^{2} \lambda \,x^{2}+x^{n} y p x +y q x +r \,x^{n}+s}{x^{2} \left (x^{n} a -1\right )} \end {array} \]
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_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -(x^(n-1)*p*x+q)*(diff(y(x), x))/(x*(x^n*a-1))-(x^(n-2)*r*x^2+s)*lambd Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying an equivalence, under non-integer power transformations, to LODEs admitting Liouvillian solutions. -> 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 <- heuristic approach successful <- hypergeometric successful <- special function solution successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 1222
dsolve(x^2*(a*x^n-1)*(diff(y(x),x)+lambda*y(x)^2)+(p*x^n+q)*x*y(x)+r*x^n+s=0,y(x), singsol=all)
\[ \text {Expression too large to display} \]
✓ Solution by Mathematica
Time used: 7.968 (sec). Leaf size: 2419
DSolve[x^2*(a*x^n-1)*(y'[x]+\[Lambda]*y[x]^2)+(p*x^n+q)*x*y[x]+r*x^n+s==0,y[x],x,IncludeSingularSolutions -> True]
Too large to display