Internal problem ID [10380]
Internal file name [OUTPUT/9328_Monday_June_06_2022_01_51_52_PM_78719202/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: 51.
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]
\[ \boxed {x^{2} y^{\prime }-a \,x^{2} y^{2}-y b x=c \,x^{n}+s} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {a \,x^{2} y^{2}+b x y +c \,x^{n}+s}{x^{2}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,y^{2}+\frac {y b}{x}+\frac {c \,x^{n}}{x^{2}}+\frac {s}{x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {c \,x^{n}+s}{x^{2}}\), \(f_1(x)=\frac {b}{x}\) and \(f_2(x)=a\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{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' &=0\\ f_1 f_2 &=\frac {a b}{x}\\ f_2^2 f_0 &=\frac {a^{2} \left (c \,x^{n}+s \right )}{x^{2}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a u^{\prime \prime }\left (x \right )-\frac {a b u^{\prime }\left (x \right )}{x}+\frac {a^{2} \left (c \,x^{n}+s \right ) u \left (x \right )}{x^{2}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{\frac {b}{2}} \sqrt {x}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {x^{-\frac {1}{2}+\frac {b}{2}} \left (-2 \sqrt {c a}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}+1, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}+1, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{2} \right ) x^{\frac {n}{2}}+\left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{2} \right ) \left (\sqrt {-4 a s +b^{2}+2 b +1}+b +1\right )\right )}{2} \] Using the above in (1) gives the solution \[ y = -\frac {x^{-\frac {1}{2}+\frac {b}{2}} \left (-2 \sqrt {c a}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}+1, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}+1, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{2} \right ) x^{\frac {n}{2}}+\left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{2} \right ) \left (\sqrt {-4 a s +b^{2}+2 b +1}+b +1\right )\right ) x^{-\frac {b}{2}}}{2 a \sqrt {x}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {2 \sqrt {c a}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}+1, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}+1, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right )\right ) x^{\frac {n}{2}}-\left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right )\right ) \left (\sqrt {-4 a s +b^{2}+2 b +1}+b +1\right )}{2 a x \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {2 \sqrt {c a}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}+1, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}+1, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right )\right ) x^{\frac {n}{2}}-\left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right )\right ) \left (\sqrt {-4 a s +b^{2}+2 b +1}+b +1\right )}{2 a x \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {2 \sqrt {c a}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}+1, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}+1, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right )\right ) x^{\frac {n}{2}}-\left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right )\right ) \left (\sqrt {-4 a s +b^{2}+2 b +1}+b +1\right )}{2 a x \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {c a}\, x^{\frac {n}{2}}}{n}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime }-a \,x^{2} y^{2}-y b x =c \,x^{n}+s \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a \,x^{2} y^{2}+y b x +c \,x^{n}+s}{x^{2}} \end {array} \]
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*(diff(y(x), x))/x-a*(x^(n-2)*c*x^2+s)*y(x)/x^2, y(x)` *** Suble 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: 263
dsolve(x^2*diff(y(x),x)=a*x^2*y(x)^2+b*x*y(x)+c*x^n+s,y(x), singsol=all)
\[ y \left (x \right ) = \frac {2 \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}+1, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}+1, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right )\right ) \sqrt {a c}\, x^{\frac {n}{2}}-\left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right )\right ) \left (\sqrt {-4 a s +b^{2}+2 b +1}+b +1\right )}{2 x a \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a s +b^{2}+2 b +1}}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right )\right )} \]
✓ Solution by Mathematica
Time used: 2.637 (sec). Leaf size: 2281
DSolve[x^2*y'[x]==a*x^2*y[x]^2+b*x*y[x]+c*x^n+s,y[x],x,IncludeSingularSolutions -> True]
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