Internal problem ID [10401]
Internal file name [OUTPUT/9349_Monday_June_06_2022_02_14_56_PM_31336711/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: 72.
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]
\[ \boxed {x^{1+n} y^{\prime }-x^{2 n} y^{2} a -y x^{n} b=c \,x^{m}+d} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \left (x^{2 n} a \,y^{2}+x^{n} b y +c \,x^{m}+d \right ) x^{-1-n} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {x^{n} a \,y^{2}}{x}+\frac {y b}{x}+\frac {x^{-n} c \,x^{m}}{x}+\frac {x^{-n} d}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\left (c \,x^{m}+d \right ) x^{-1-n}\), \(f_1(x)=b \,x^{n} x^{-1-n}\) and \(f_2(x)=a \,x^{2 n} x^{-1-n}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{a \,x^{2 n} x^{-1-n} 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 {2 a \,x^{2 n} n \,x^{-1-n}}{x}-\frac {a \,x^{2 n} x^{-1-n} \left (1+n \right )}{x}\\ f_1 f_2 &=b \,x^{n} x^{-2 n -2} a \,x^{2 n}\\ f_2^2 f_0 &=a^{2} x^{4 n} x^{-3-3 n} \left (c \,x^{m}+d \right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a \,x^{2 n} x^{-1-n} u^{\prime \prime }\left (x \right )-\left (\frac {2 a \,x^{2 n} n \,x^{-1-n}}{x}-\frac {a \,x^{2 n} x^{-1-n} \left (1+n \right )}{x}+b \,x^{n} x^{-2 n -2} a \,x^{2 n}\right ) u^{\prime }\left (x \right )+a^{2} x^{4 n} x^{-3-3 n} \left (c \,x^{m}+d \right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{\frac {b}{2}} x^{\frac {n}{2}} \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{2} +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {x^{-1+\frac {b}{2}+\frac {n}{2}} \left (-2 \sqrt {c a}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}+1, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{1} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}+1, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{2} \right ) x^{\frac {m}{2}}+\left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{2} +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{1} \right ) \left (\sqrt {-4 a d +b^{2}+2 b n +n^{2}}+b +n \right )\right )}{2} \] Using the above in (1) gives the solution \[ y = -\frac {x^{-1+\frac {b}{2}+\frac {n}{2}} \left (-2 \sqrt {c a}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}+1, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{1} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}+1, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{2} \right ) x^{\frac {m}{2}}+\left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{2} +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{1} \right ) \left (\sqrt {-4 a d +b^{2}+2 b n +n^{2}}+b +n \right )\right ) x^{-2 n} x^{1+n} x^{-\frac {b}{2}} x^{-\frac {n}{2}}}{2 a \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{2} +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {x^{-n} \left (-2 \sqrt {c a}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}+1, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}+1, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right )\right ) x^{\frac {m}{2}}+\left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right )+\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{3} \right ) \left (\sqrt {-4 a d +b^{2}+2 b n +n^{2}}+b +n \right )\right )}{2 a \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right )+\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {x^{-n} \left (-2 \sqrt {c a}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}+1, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}+1, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right )\right ) x^{\frac {m}{2}}+\left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right )+\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{3} \right ) \left (\sqrt {-4 a d +b^{2}+2 b n +n^{2}}+b +n \right )\right )}{2 a \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right )+\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {x^{-n} \left (-2 \sqrt {c a}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}+1, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}+1, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right )\right ) x^{\frac {m}{2}}+\left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right )+\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{3} \right ) \left (\sqrt {-4 a d +b^{2}+2 b n +n^{2}}+b +n \right )\right )}{2 a \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right )+\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {c a}\, x^{\frac {m}{2}}}{m}\right ) c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{1+n} y^{\prime }-x^{2 n} y^{2} a -y x^{n} b =c \,x^{m}+d \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x^{2 n} y^{2} a +y x^{n} b +c \,x^{m}+d}{x^{1+n}} \end {array} \]
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) = (b+n-1)*(diff(y(x), x))/x-x^(n-1)*a*(x^(-n-1+m)*c+x^(-n-1)*d)*y(x), y( 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 <- Bessel successful <- special function solution successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 284
dsolve(x^(n+1)*diff(y(x),x)=a*x^(2*n)*y(x)^2+b*x^n*y(x)+c*x^m+d,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\sqrt {a c}\, \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}+1, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}+1, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )\right ) x^{\frac {m}{2}}-\frac {\left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )\right ) \left (\sqrt {-4 a d +b^{2}+2 b n +n^{2}}+b +n \right )}{2}\right ) x^{-n}}{a \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b n +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )\right )} \]
✓ Solution by Mathematica
Time used: 3.153 (sec). Leaf size: 2576
DSolve[x^(n+1)*y'[x]==a*x^(2*n)*y[x]^2+b*x^n*y[x]+c*x^m+d,y[x],x,IncludeSingularSolutions -> True]
Too large to display