Internal problem ID [10334]
Internal file name [OUTPUT/9282_Monday_June_06_2022_01_45_27_PM_95068594/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: 5.
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]
\[ \boxed {y^{\prime }-y^{2}=a n \,x^{-1+n}-a^{2} x^{2 n}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+a n \,x^{-1+n}-a^{2} x^{2 n} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+\frac {a n \,x^{n}}{x}-a^{2} x^{2 n} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a n \,x^{-1+n}-a^{2} x^{2 n}\), \(f_1(x)=0\) 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 &=a n \,x^{-1+n}-a^{2} x^{2 n} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\left (a n \,x^{-1+n}-a^{2} x^{2 n}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = -\frac {c_{2} x^{-1-\frac {3 n}{2}} \left (2+n \right )^{2} \operatorname {WhittakerM}\left (\frac {2+n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x^{1+n} a}{1+n}\right )}{2}+\left (\left (-\frac {n}{2}-1\right ) x^{-1-\frac {3 n}{2}}+a \,x^{-\frac {n}{2}}\right ) c_{2} \left (1+n \right ) \operatorname {WhittakerM}\left (-\frac {n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x^{1+n} a}{1+n}\right )+c_{1} {\mathrm e}^{-\frac {x^{1+n} a}{1+n}} \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {\left (-\frac {3 c_{2} \left (2+n \right ) \left (\left (\frac {1}{3} n^{2}+n +\frac {2}{3}\right ) x^{-\frac {n}{2}}+a x \,x^{\frac {n}{2}} \left (n +\frac {4}{3}\right )\right ) \operatorname {WhittakerM}\left (\frac {2+n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )}{2}+c_{2} \left (1+n \right ) \left (\left (-\frac {1}{2} n^{2}-\frac {3}{2} n -1\right ) x^{-\frac {n}{2}}+\left (\left (-\frac {n}{2}-1\right ) x^{\frac {n}{2}}+a x \,x^{\frac {3 n}{2}}\right ) x a \right ) \operatorname {WhittakerM}\left (-\frac {n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )+{\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (\frac {3}{2}+n \right ) c_{2} \left (2+n \right )^{2} x^{-\frac {n}{2}} \left (-\frac {2 x \,x^{n} a}{1+n}\right )^{\frac {4+3 n}{2+2 n}}+a c_{1} x^{2} x^{2 n} {\mathrm e}^{-\frac {x \,x^{n} a}{1+n}}\right ) x^{-n}}{x^{2}} \] Using the above in (1) gives the solution \[ y = \frac {\left (-\frac {3 c_{2} \left (2+n \right ) \left (\left (\frac {1}{3} n^{2}+n +\frac {2}{3}\right ) x^{-\frac {n}{2}}+a x \,x^{\frac {n}{2}} \left (n +\frac {4}{3}\right )\right ) \operatorname {WhittakerM}\left (\frac {2+n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )}{2}+c_{2} \left (1+n \right ) \left (\left (-\frac {1}{2} n^{2}-\frac {3}{2} n -1\right ) x^{-\frac {n}{2}}+\left (\left (-\frac {n}{2}-1\right ) x^{\frac {n}{2}}+a x \,x^{\frac {3 n}{2}}\right ) x a \right ) \operatorname {WhittakerM}\left (-\frac {n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )+{\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (\frac {3}{2}+n \right ) c_{2} \left (2+n \right )^{2} x^{-\frac {n}{2}} \left (-\frac {2 x \,x^{n} a}{1+n}\right )^{\frac {4+3 n}{2+2 n}}+a c_{1} x^{2} x^{2 n} {\mathrm e}^{-\frac {x \,x^{n} a}{1+n}}\right ) x^{-n}}{x^{2} \left (-\frac {c_{2} x^{-1-\frac {3 n}{2}} \left (2+n \right )^{2} \operatorname {WhittakerM}\left (\frac {2+n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x^{1+n} a}{1+n}\right )}{2}+\left (\left (-\frac {n}{2}-1\right ) x^{-1-\frac {3 n}{2}}+a \,x^{-\frac {n}{2}}\right ) c_{2} \left (1+n \right ) \operatorname {WhittakerM}\left (-\frac {n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x^{1+n} a}{1+n}\right )+c_{1} {\mathrm e}^{-\frac {x^{1+n} a}{1+n}}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (-\frac {3 \,{\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (2+n \right ) \left (\left (\frac {1}{3} n^{2}+n +\frac {2}{3}\right ) x^{-\frac {n}{2}}+a x \,x^{\frac {n}{2}} \left (n +\frac {4}{3}\right )\right ) \operatorname {WhittakerM}\left (\frac {2+n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )}{2}+{\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (1+n \right ) \left (\left (-\frac {1}{2} n^{2}-\frac {3}{2} n -1\right ) x^{-\frac {n}{2}}+\left (\left (-\frac {n}{2}-1\right ) x^{\frac {n}{2}}+a x \,x^{\frac {3 n}{2}}\right ) x a \right ) \operatorname {WhittakerM}\left (-\frac {n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )+\left (\frac {3}{2}+n \right ) \left (2+n \right )^{2} x^{-\frac {n}{2}} {\mathrm e}^{\frac {2 x \,x^{n} a}{1+n}} \left (-\frac {2 x \,x^{n} a}{1+n}\right )^{\frac {4+3 n}{2+2 n}}+a c_{3} x^{2} x^{2 n}\right ) x^{-n}}{\left (-\frac {x^{-\frac {3 n}{2}} {\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (2+n \right )^{2} \operatorname {WhittakerM}\left (\frac {2+n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )}{2}+{\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (a x \,x^{-\frac {n}{2}}-\frac {\left (2+n \right ) x^{-\frac {3 n}{2}}}{2}\right ) \left (1+n \right ) \operatorname {WhittakerM}\left (-\frac {n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )+c_{3} x \right ) x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-\frac {3 \,{\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (2+n \right ) \left (\left (\frac {1}{3} n^{2}+n +\frac {2}{3}\right ) x^{-\frac {n}{2}}+a x \,x^{\frac {n}{2}} \left (n +\frac {4}{3}\right )\right ) \operatorname {WhittakerM}\left (\frac {2+n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )}{2}+{\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (1+n \right ) \left (\left (-\frac {1}{2} n^{2}-\frac {3}{2} n -1\right ) x^{-\frac {n}{2}}+\left (\left (-\frac {n}{2}-1\right ) x^{\frac {n}{2}}+a x \,x^{\frac {3 n}{2}}\right ) x a \right ) \operatorname {WhittakerM}\left (-\frac {n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )+\left (\frac {3}{2}+n \right ) \left (2+n \right )^{2} x^{-\frac {n}{2}} {\mathrm e}^{\frac {2 x \,x^{n} a}{1+n}} \left (-\frac {2 x \,x^{n} a}{1+n}\right )^{\frac {4+3 n}{2+2 n}}+a c_{3} x^{2} x^{2 n}\right ) x^{-n}}{\left (-\frac {x^{-\frac {3 n}{2}} {\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (2+n \right )^{2} \operatorname {WhittakerM}\left (\frac {2+n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )}{2}+{\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (a x \,x^{-\frac {n}{2}}-\frac {\left (2+n \right ) x^{-\frac {3 n}{2}}}{2}\right ) \left (1+n \right ) \operatorname {WhittakerM}\left (-\frac {n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )+c_{3} x \right ) x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-\frac {3 \,{\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (2+n \right ) \left (\left (\frac {1}{3} n^{2}+n +\frac {2}{3}\right ) x^{-\frac {n}{2}}+a x \,x^{\frac {n}{2}} \left (n +\frac {4}{3}\right )\right ) \operatorname {WhittakerM}\left (\frac {2+n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )}{2}+{\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (1+n \right ) \left (\left (-\frac {1}{2} n^{2}-\frac {3}{2} n -1\right ) x^{-\frac {n}{2}}+\left (\left (-\frac {n}{2}-1\right ) x^{\frac {n}{2}}+a x \,x^{\frac {3 n}{2}}\right ) x a \right ) \operatorname {WhittakerM}\left (-\frac {n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )+\left (\frac {3}{2}+n \right ) \left (2+n \right )^{2} x^{-\frac {n}{2}} {\mathrm e}^{\frac {2 x \,x^{n} a}{1+n}} \left (-\frac {2 x \,x^{n} a}{1+n}\right )^{\frac {4+3 n}{2+2 n}}+a c_{3} x^{2} x^{2 n}\right ) x^{-n}}{\left (-\frac {x^{-\frac {3 n}{2}} {\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (2+n \right )^{2} \operatorname {WhittakerM}\left (\frac {2+n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )}{2}+{\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (a x \,x^{-\frac {n}{2}}-\frac {\left (2+n \right ) x^{-\frac {3 n}{2}}}{2}\right ) \left (1+n \right ) \operatorname {WhittakerM}\left (-\frac {n}{2+2 n}, \frac {3+2 n}{2+2 n}, -\frac {2 x \,x^{n} a}{1+n}\right )+c_{3} x \right ) x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}=a n \,x^{-1+n}-a^{2} x^{2 n} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+a n \,x^{-1+n}-a^{2} x^{2 n} \end {array} \]
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_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (-a*n*x^(n-1)+a^2*x^(2*n))*y(x), y(x)` *** Sublevel 2 *** 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 A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible <- Kovacics algorithm successful <- Equivalence, under non-integer power transformations successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 388
dsolve(diff(y(x),x)=y(x)^2+a*n*x^(n-1)-a^2*x^(2*n),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-3 \left (n +2\right ) c_{1} \left (\left (\frac {1}{3} n^{2}+n +\frac {2}{3}\right ) x^{-\frac {3 n}{2}}+a x \,x^{-\frac {n}{2}} \left (n +\frac {4}{3}\right )\right ) {\mathrm e}^{\frac {a x \,x^{n}}{n +1}} \operatorname {WhittakerM}\left (\frac {n +2}{2 n +2}, \frac {2 n +3}{2 n +2}, -\frac {2 a x \,x^{n}}{n +1}\right )+2 c_{1} {\mathrm e}^{\frac {a x \,x^{n}}{n +1}} \left (n +1\right ) \left (\left (-\frac {1}{2} n^{2}-\frac {3}{2} n -1\right ) x^{-\frac {3 n}{2}}+x a \left (\left (-\frac {n}{2}-1\right ) x^{-\frac {n}{2}}+a x \,x^{\frac {n}{2}}\right )\right ) \operatorname {WhittakerM}\left (-\frac {n}{2 n +2}, \frac {2 n +3}{2 n +2}, -\frac {2 a x \,x^{n}}{n +1}\right )+2 \left (n +2\right )^{2} c_{1} \left (n +\frac {3}{2}\right ) {\mathrm e}^{\frac {2 a x \,x^{n}}{n +1}} x^{-\frac {3 n}{2}} \left (-\frac {2 a x \,x^{n}}{n +1}\right )^{\frac {3 n +4}{2 n +2}}+2 x^{2} a \,x^{n}}{2 \left (-\frac {{\mathrm e}^{\frac {a x \,x^{n}}{n +1}} x^{-\frac {3 n}{2}} c_{1} \left (n +2\right )^{2} \operatorname {WhittakerM}\left (\frac {n +2}{2 n +2}, \frac {2 n +3}{2 n +2}, -\frac {2 a x \,x^{n}}{n +1}\right )}{2}+c_{1} \left (\left (-\frac {n}{2}-1\right ) x^{-\frac {3 n}{2}}+a x \,x^{-\frac {n}{2}}\right ) {\mathrm e}^{\frac {a x \,x^{n}}{n +1}} \left (n +1\right ) \operatorname {WhittakerM}\left (-\frac {n}{2 n +2}, \frac {2 n +3}{2 n +2}, -\frac {2 a x \,x^{n}}{n +1}\right )+x \right ) x} \]
✓ Solution by Mathematica
Time used: 1.61 (sec). Leaf size: 227
DSolve[y'[x]==y[x]^2+a*n*x^(n-1)-a^2*x^(2*n),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {2^{\frac {1}{n+1}} (n+1) \left (-\frac {a x^{n+1}}{n+1}\right )^{\frac {1}{n+1}} \left (a x^n-c_1 e^{\frac {2 a x^{n+1}}{n+1}}\right )-a c_1 x^{n+1} \Gamma \left (\frac {1}{n+1},-\frac {2 a x^{n+1}}{n+1}\right )}{2^{\frac {1}{n+1}} (n+1) \left (-\frac {a x^{n+1}}{n+1}\right )^{\frac {1}{n+1}}-c_1 x \Gamma \left (\frac {1}{n+1},-\frac {2 a x^{n+1}}{n+1}\right )} \\ y(x)\to \frac {2^{\frac {1}{n+1}} (n+1) e^{\frac {2 a x^{n+1}}{n+1}} \left (-\frac {a x^{n+1}}{n+1}\right )^{\frac {1}{n+1}}}{x \Gamma \left (\frac {1}{n+1},-\frac {2 a x^{n+1}}{n+1}\right )}+a x^n \\ \end{align*}