Internal problem ID [10339]
Internal file name [OUTPUT/9287_Monday_June_06_2022_01_46_51_PM_95752007/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: 10.
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]
\[ \boxed {y^{\prime }-a \,x^{n} y^{2}=b m \,x^{m -1}-a \,b^{2} x^{n +2 m}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,x^{n} y^{2}+b m \,x^{m -1}-a \,b^{2} x^{n +2 m} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,x^{n} y^{2}+\frac {b \,x^{m} m}{x}-a \,b^{2} x^{n} x^{2 m} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=b m \,x^{m -1}-a \,b^{2} x^{n +2 m}\), \(f_1(x)=0\) and \(f_2(x)=x^{n} a\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{x^{n} a 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' &=\frac {a n \,x^{n}}{x}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=x^{2 n} a^{2} \left (b m \,x^{m -1}-a \,b^{2} x^{n +2 m}\right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} x^{n} a u^{\prime \prime }\left (x \right )-\frac {a n \,x^{n} u^{\prime }\left (x \right )}{x}+x^{2 n} a^{2} \left (b m \,x^{m -1}-a \,b^{2} x^{n +2 m}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = -\frac {x^{-\frac {3 m}{2}-1-n} c_{2} \left (m +2 n +2\right )^{2} \operatorname {WhittakerM}\left (\frac {m +2 n +2}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )}{2}+\left (\left (-\frac {m}{2}-n -1\right ) x^{-\frac {3 m}{2}-1-n}+x^{-\frac {m}{2}} a b \right ) c_{2} \left (1+m +n \right ) \operatorname {WhittakerM}\left (-\frac {m}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )+c_{1} {\mathrm e}^{-\frac {a b \,x^{1+m +n}}{1+m +n}} \] The above shows that \[ u^{\prime }\left (x \right ) = -\left (-\frac {3 \left (a b \left (m +\frac {4 n}{3}+\frac {4}{3}\right ) x^{1+n +\frac {m}{2}}+\frac {x^{-\frac {m}{2}} \left (m +2 n +2\right ) \left (1+m +n \right )}{3}\right ) \left (m +2 n +2\right ) c_{2} \operatorname {WhittakerM}\left (\frac {m +2 n +2}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )}{2}+\left (a^{2} b^{2} x^{\frac {3 m}{2}+2 n +2}-\frac {\left (m +2 n +2\right ) \left (a b \,x^{1+n +\frac {m}{2}}+x^{-\frac {m}{2}} \left (1+m +n \right )\right )}{2}\right ) c_{2} \left (1+m +n \right ) \operatorname {WhittakerM}\left (-\frac {m}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )+x^{-\frac {m}{2}} \left (m +2 n +2\right )^{2} \left (m +\frac {3 n}{2}+\frac {3}{2}\right ) {\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} c_{2} \left (-\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )^{\frac {3 m +4 n +4}{2+2 m +2 n}}+a b c_{1} x^{2+2 m +2 n} {\mathrm e}^{-\frac {a b \,x^{1+m +n}}{1+m +n}}\right ) x^{-m -2-n} \] Using the above in (1) gives the solution \[ y = \frac {\left (-\frac {3 \left (a b \left (m +\frac {4 n}{3}+\frac {4}{3}\right ) x^{1+n +\frac {m}{2}}+\frac {x^{-\frac {m}{2}} \left (m +2 n +2\right ) \left (1+m +n \right )}{3}\right ) \left (m +2 n +2\right ) c_{2} \operatorname {WhittakerM}\left (\frac {m +2 n +2}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )}{2}+\left (a^{2} b^{2} x^{\frac {3 m}{2}+2 n +2}-\frac {\left (m +2 n +2\right ) \left (a b \,x^{1+n +\frac {m}{2}}+x^{-\frac {m}{2}} \left (1+m +n \right )\right )}{2}\right ) c_{2} \left (1+m +n \right ) \operatorname {WhittakerM}\left (-\frac {m}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )+x^{-\frac {m}{2}} \left (m +2 n +2\right )^{2} \left (m +\frac {3 n}{2}+\frac {3}{2}\right ) {\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} c_{2} \left (-\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )^{\frac {3 m +4 n +4}{2+2 m +2 n}}+a b c_{1} x^{2+2 m +2 n} {\mathrm e}^{-\frac {a b \,x^{1+m +n}}{1+m +n}}\right ) x^{-m -2-n} x^{-n}}{a \left (-\frac {x^{-\frac {3 m}{2}-1-n} c_{2} \left (m +2 n +2\right )^{2} \operatorname {WhittakerM}\left (\frac {m +2 n +2}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )}{2}+\left (\left (-\frac {m}{2}-n -1\right ) x^{-\frac {3 m}{2}-1-n}+x^{-\frac {m}{2}} a b \right ) c_{2} \left (1+m +n \right ) \operatorname {WhittakerM}\left (-\frac {m}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )+c_{1} {\mathrm e}^{-\frac {a b \,x^{1+m +n}}{1+m +n}}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {x^{-1-m -n} \left (-\frac {3 \left (m +2 n +2\right ) {\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} \left (a b \left (m +\frac {4 n}{3}+\frac {4}{3}\right ) x^{1+n +\frac {m}{2}}+\frac {x^{-\frac {m}{2}} \left (m +2 n +2\right ) \left (1+m +n \right )}{3}\right ) \operatorname {WhittakerM}\left (\frac {m +2 n +2}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )}{2}+{\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} \left (a^{2} b^{2} x^{\frac {3 m}{2}+2 n +2}-\frac {\left (m +2 n +2\right ) \left (a b \,x^{1+n +\frac {m}{2}}+x^{-\frac {m}{2}} \left (1+m +n \right )\right )}{2}\right ) \left (1+m +n \right ) \operatorname {WhittakerM}\left (-\frac {m}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )+x^{-\frac {m}{2}} \left (m +2 n +2\right )^{2} \left (m +\frac {3 n}{2}+\frac {3}{2}\right ) {\mathrm e}^{\frac {2 a b \,x^{1+m +n}}{1+m +n}} \left (-\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )^{\frac {3 m +4 n +4}{2+2 m +2 n}}+a b c_{3} x^{2+2 m +2 n}\right )}{\left (-\frac {x^{-\frac {3 m}{2}} {\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} \left (m +2 n +2\right )^{2} \operatorname {WhittakerM}\left (\frac {m +2 n +2}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )}{2}+{\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} \left (x^{1-\frac {m}{2}+n} a b -\frac {x^{-\frac {3 m}{2}} \left (m +2 n +2\right )}{2}\right ) \left (1+m +n \right ) \operatorname {WhittakerM}\left (-\frac {m}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )+x^{1+n} c_{3} \right ) a} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{-1-m -n} \left (-\frac {3 \left (m +2 n +2\right ) {\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} \left (a b \left (m +\frac {4 n}{3}+\frac {4}{3}\right ) x^{1+n +\frac {m}{2}}+\frac {x^{-\frac {m}{2}} \left (m +2 n +2\right ) \left (1+m +n \right )}{3}\right ) \operatorname {WhittakerM}\left (\frac {m +2 n +2}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )}{2}+{\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} \left (a^{2} b^{2} x^{\frac {3 m}{2}+2 n +2}-\frac {\left (m +2 n +2\right ) \left (a b \,x^{1+n +\frac {m}{2}}+x^{-\frac {m}{2}} \left (1+m +n \right )\right )}{2}\right ) \left (1+m +n \right ) \operatorname {WhittakerM}\left (-\frac {m}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )+x^{-\frac {m}{2}} \left (m +2 n +2\right )^{2} \left (m +\frac {3 n}{2}+\frac {3}{2}\right ) {\mathrm e}^{\frac {2 a b \,x^{1+m +n}}{1+m +n}} \left (-\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )^{\frac {3 m +4 n +4}{2+2 m +2 n}}+a b c_{3} x^{2+2 m +2 n}\right )}{\left (-\frac {x^{-\frac {3 m}{2}} {\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} \left (m +2 n +2\right )^{2} \operatorname {WhittakerM}\left (\frac {m +2 n +2}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )}{2}+{\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} \left (x^{1-\frac {m}{2}+n} a b -\frac {x^{-\frac {3 m}{2}} \left (m +2 n +2\right )}{2}\right ) \left (1+m +n \right ) \operatorname {WhittakerM}\left (-\frac {m}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )+x^{1+n} c_{3} \right ) a} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{-1-m -n} \left (-\frac {3 \left (m +2 n +2\right ) {\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} \left (a b \left (m +\frac {4 n}{3}+\frac {4}{3}\right ) x^{1+n +\frac {m}{2}}+\frac {x^{-\frac {m}{2}} \left (m +2 n +2\right ) \left (1+m +n \right )}{3}\right ) \operatorname {WhittakerM}\left (\frac {m +2 n +2}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )}{2}+{\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} \left (a^{2} b^{2} x^{\frac {3 m}{2}+2 n +2}-\frac {\left (m +2 n +2\right ) \left (a b \,x^{1+n +\frac {m}{2}}+x^{-\frac {m}{2}} \left (1+m +n \right )\right )}{2}\right ) \left (1+m +n \right ) \operatorname {WhittakerM}\left (-\frac {m}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )+x^{-\frac {m}{2}} \left (m +2 n +2\right )^{2} \left (m +\frac {3 n}{2}+\frac {3}{2}\right ) {\mathrm e}^{\frac {2 a b \,x^{1+m +n}}{1+m +n}} \left (-\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )^{\frac {3 m +4 n +4}{2+2 m +2 n}}+a b c_{3} x^{2+2 m +2 n}\right )}{\left (-\frac {x^{-\frac {3 m}{2}} {\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} \left (m +2 n +2\right )^{2} \operatorname {WhittakerM}\left (\frac {m +2 n +2}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )}{2}+{\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} \left (x^{1-\frac {m}{2}+n} a b -\frac {x^{-\frac {3 m}{2}} \left (m +2 n +2\right )}{2}\right ) \left (1+m +n \right ) \operatorname {WhittakerM}\left (-\frac {m}{2+2 m +2 n}, \frac {2 m +3 n +3}{2+2 m +2 n}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )+x^{1+n} c_{3} \right ) a} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-a \,x^{n} y^{2}=b m \,x^{m -1}-a \,b^{2} x^{n +2 m} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \,x^{n} y^{2}+b m \,x^{m -1}-a \,b^{2} x^{n +2 m} \end {array} \]
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 sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = n*(diff(y(x), x))/x-a*x^n*b*(-x^(n+2*m)*b*a+x^(m-1)*m)*y(x), y(x)` 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: 522
dsolve(diff(y(x),x)=a*x^n*y(x)^2+b*m*x^(m-1)-a*b^2*x^(n+2*m),y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{-n -1} \left (-\frac {3 \left (a b \left (m +\frac {4 n}{3}+\frac {4}{3}\right ) x^{n +1-\frac {m}{2}}+\frac {x^{-\frac {3 m}{2}} \left (m +2 n +2\right ) \left (1+m +n \right )}{3}\right ) c_{1} \left (m +2 n +2\right ) {\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} \operatorname {WhittakerM}\left (\frac {m +2 n +2}{2 n +2 m +2}, \frac {2 m +3 n +3}{2 n +2 m +2}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )}{2}+\left (a^{2} b^{2} x^{2 n +2+\frac {m}{2}}-\frac {\left (x^{n +1-\frac {m}{2}} a b +x^{-\frac {3 m}{2}} \left (1+m +n \right )\right ) \left (m +2 n +2\right )}{2}\right ) \left (1+m +n \right ) c_{1} {\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} \operatorname {WhittakerM}\left (-\frac {m}{2 n +2 m +2}, \frac {2 m +3 n +3}{2 n +2 m +2}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )+{\mathrm e}^{\frac {2 a b \,x^{1+m +n}}{1+m +n}} \left (m +\frac {3 n}{2}+\frac {3}{2}\right ) c_{1} \left (m +2 n +2\right )^{2} x^{-\frac {3 m}{2}} \left (-\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )^{\frac {3 m +4 n +4}{2 n +2 m +2}}+a b \,x^{m +2 n +2}\right )}{\left (-\frac {{\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} x^{-\frac {3 m}{2}} c_{1} \left (m +2 n +2\right )^{2} \operatorname {WhittakerM}\left (\frac {m +2 n +2}{2 n +2 m +2}, \frac {2 m +3 n +3}{2 n +2 m +2}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )}{2}+\left (1+m +n \right ) c_{1} \left (x^{n +1-\frac {m}{2}} a b -\frac {x^{-\frac {3 m}{2}} \left (m +2 n +2\right )}{2}\right ) {\mathrm e}^{\frac {a b \,x^{1+m +n}}{1+m +n}} \operatorname {WhittakerM}\left (-\frac {m}{2 n +2 m +2}, \frac {2 m +3 n +3}{2 n +2 m +2}, -\frac {2 a b \,x^{1+m +n}}{1+m +n}\right )+x^{n +1}\right ) a} \]
✓ Solution by Mathematica
Time used: 2.322 (sec). Leaf size: 306
DSolve[y'[x]==a*x^n*y[x]^2+b*m*x^(m-1)-a*b^2*x^(n+2*m),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {2^{\frac {n+1}{m+n+1}} (m+n+1) \left (-\frac {a b x^{m+n+1}}{m+n+1}\right )^{\frac {n+1}{m+n+1}} \left (a b x^m-c_1 e^{\frac {2 a b x^{m+n+1}}{m+n+1}}\right )-a b c_1 x^{m+n+1} \Gamma \left (\frac {n+1}{m+n+1},-\frac {2 a b x^{m+n+1}}{m+n+1}\right )}{a \left (2^{\frac {n+1}{m+n+1}} (m+n+1) \left (-\frac {a b x^{m+n+1}}{m+n+1}\right )^{\frac {n+1}{m+n+1}}-c_1 x^{n+1} \Gamma \left (\frac {n+1}{m+n+1},-\frac {2 a b x^{m+n+1}}{m+n+1}\right )\right )} \\ y(x)\to b x^m-\frac {b 2^{\frac {n+1}{m+n+1}} x^m e^{\frac {2 a b x^{m+n+1}}{m+n+1}} \left (-\frac {a b x^{m+n+1}}{m+n+1}\right )^{-\frac {m}{m+n+1}}}{\Gamma \left (\frac {n+1}{m+n+1},-\frac {2 a b x^{m+n+1}}{m+n+1}\right )} \\ \end{align*}