Internal problem ID [10381]
Internal file name [OUTPUT/9329_Monday_June_06_2022_01_51_53_PM_16676066/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: 52.
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]
\[ \boxed {x^{2} y^{\prime }-a \,x^{2} y^{2}-y b x=c \,x^{2 n}+s \,x^{n}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {a \,x^{2} y^{2}+b x y +c \,x^{2 n}+s \,x^{n}}{x^{2}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,y^{2}+\frac {y b}{x}+\frac {c \,x^{2 n}}{x^{2}}+\frac {s \,x^{n}}{x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {c \,x^{2 n}+s \,x^{n}}{x^{2}}\), \(f_1(x)=\frac {b}{x}\) 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^{2 n}+s \,x^{n}\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^{2 n}+s \,x^{n}\right ) u \left (x \right )}{x^{2}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c_{2} +\operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c_{1} \right ) x^{\frac {b}{2}+\frac {1}{2}-\frac {n}{2}} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {x^{-\frac {1}{2}+\frac {b}{2}-\frac {n}{2}} \left (-\left (i \sqrt {a}\, \sqrt {c}\, s -c \left (1+b +n \right )\right ) c_{1} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s -2 n \sqrt {c}}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right )-2 c n c_{2} \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s -2 n \sqrt {c}}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right )+\left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c_{2} +\operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c_{1} \right ) \left (2 i \sqrt {a}\, c^{\frac {3}{2}} x^{n}+i \sqrt {a}\, \sqrt {c}\, s +c \left (b -n +1\right )\right )\right )}{2 c} \] Using the above in (1) gives the solution \[ y = -\frac {x^{-\frac {1}{2}+\frac {b}{2}-\frac {n}{2}} \left (-\left (i \sqrt {a}\, \sqrt {c}\, s -c \left (1+b +n \right )\right ) c_{1} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s -2 n \sqrt {c}}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right )-2 c n c_{2} \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s -2 n \sqrt {c}}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right )+\left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c_{2} +\operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c_{1} \right ) \left (2 i \sqrt {a}\, c^{\frac {3}{2}} x^{n}+i \sqrt {a}\, \sqrt {c}\, s +c \left (b -n +1\right )\right )\right ) x^{\frac {n}{2}-\frac {1}{2}-\frac {b}{2}}}{2 c a \left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c_{2} +\operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {c_{3} \left (i \sqrt {a}\, \sqrt {c}\, s -c \left (1+b +n \right )\right ) \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}+1, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right )+2 \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}+1, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c n -\left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right )+\operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c_{3} \right ) \left (2 i \sqrt {a}\, c^{\frac {3}{2}} x^{n}+i \sqrt {a}\, \sqrt {c}\, s +c \left (b -n +1\right )\right )}{2 a c x \left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right )+\operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{3} \left (i \sqrt {a}\, \sqrt {c}\, s -c \left (1+b +n \right )\right ) \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}+1, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right )+2 \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}+1, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c n -\left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right )+\operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c_{3} \right ) \left (2 i \sqrt {a}\, c^{\frac {3}{2}} x^{n}+i \sqrt {a}\, \sqrt {c}\, s +c \left (b -n +1\right )\right )}{2 a c x \left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right )+\operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{3} \left (i \sqrt {a}\, \sqrt {c}\, s -c \left (1+b +n \right )\right ) \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}+1, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right )+2 \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}+1, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c n -\left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right )+\operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c_{3} \right ) \left (2 i \sqrt {a}\, c^{\frac {3}{2}} x^{n}+i \sqrt {a}\, \sqrt {c}\, s +c \left (b -n +1\right )\right )}{2 a c x \left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right )+\operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, s}{2 n \sqrt {c}}, \frac {1+b}{2 n}, \frac {2 i \sqrt {c}\, \sqrt {a}\, x^{n}}{n}\right ) c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime }-a \,x^{2} y^{2}-y b x =c \,x^{2 n}+s \,x^{n} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a \,x^{2} y^{2}+y b x +c \,x^{2 n}+s \,x^{n}}{x^{2}} \end {array} \]
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^(2*n-2)*c+x^(n-2)*s)*y(x), y(x)` *** Sub 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 -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Kummer successful <- special function solution successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 373
dsolve(x^2*diff(y(x),x)=a*x^2*y(x)^2+b*x*y(x)+c*x^(2*n)+s*x^n,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\operatorname {KummerM}\left (\frac {\left (b -n +1\right ) \sqrt {c}+i \sqrt {a}\, s}{2 \sqrt {c}\, n}, \frac {b +n +1}{n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right ) \left (\left (-b -n -1\right ) \sqrt {c}+i \sqrt {a}\, s \right )+2 \sqrt {c}\, \operatorname {KummerU}\left (\frac {\left (b -n +1\right ) \sqrt {c}+i \sqrt {a}\, s}{2 \sqrt {c}\, n}, \frac {b +n +1}{n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right ) c_{1} n -2 \left (\operatorname {KummerU}\left (\frac {\left (b +n +1\right ) \sqrt {c}+i \sqrt {a}\, s}{2 \sqrt {c}\, n}, \frac {b +n +1}{n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right ) c_{1} +\operatorname {KummerM}\left (\frac {\left (b +n +1\right ) \sqrt {c}+i \sqrt {a}\, s}{2 \sqrt {c}\, n}, \frac {b +n +1}{n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )\right ) \left (\frac {\left (b -n +1\right ) \sqrt {c}}{2}+i \sqrt {a}\, \left (c \,x^{n}+\frac {s}{2}\right )\right )}{2 \sqrt {c}\, x a \left (\operatorname {KummerU}\left (\frac {\left (b +n +1\right ) \sqrt {c}+i \sqrt {a}\, s}{2 \sqrt {c}\, n}, \frac {b +n +1}{n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right ) c_{1} +\operatorname {KummerM}\left (\frac {\left (b +n +1\right ) \sqrt {c}+i \sqrt {a}\, s}{2 \sqrt {c}\, n}, \frac {b +n +1}{n}, \frac {2 i \sqrt {a}\, \sqrt {c}\, x^{n}}{n}\right )\right )} \]
✓ Solution by Mathematica
Time used: 1.839 (sec). Leaf size: 819
DSolve[x^2*y'[x]==a*x^2*y[x]^2+b*x*y[x]+c*x^(2*n)+s*x^n,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {i \sqrt {a} c_1 x^n \left (\sqrt {c} (b+n+1)-i \sqrt {a} s\right ) \operatorname {HypergeometricU}\left (\frac {b+3 n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n},\frac {b+2 n+1}{n},-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )+c_1 n \left (i \sqrt {a} \sqrt {c} x^n+b+1\right ) \operatorname {HypergeometricU}\left (\frac {b+n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n},\frac {b+n+1}{n},-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )+n \left (2 i \sqrt {a} \sqrt {c} x^n L_{-\frac {b+3 n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n}}^{\frac {b+n+1}{n}}\left (-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )+\left (i \sqrt {a} \sqrt {c} x^n+b+1\right ) L_{-\frac {b+n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n}}^{\frac {b+1}{n}}\left (-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )\right )}{a n x \left (c_1 \operatorname {HypergeometricU}\left (\frac {b+n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n},\frac {b+n+1}{n},-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )+L_{-\frac {b+n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n}}^{\frac {b+1}{n}}\left (-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )\right )} \\ y(x)\to -\frac {\frac {\sqrt {a} x^n \left (\sqrt {a} s+i \sqrt {c} (b+n+1)\right ) \operatorname {HypergeometricU}\left (\frac {b+3 n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n},\frac {b+2 n+1}{n},-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )}{n \operatorname {HypergeometricU}\left (\frac {b+n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n},\frac {b+n+1}{n},-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )}+i \sqrt {a} \sqrt {c} x^n+b+1}{a x} \\ y(x)\to -\frac {\frac {\sqrt {a} x^n \left (\sqrt {a} s+i \sqrt {c} (b+n+1)\right ) \operatorname {HypergeometricU}\left (\frac {b+3 n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n},\frac {b+2 n+1}{n},-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )}{n \operatorname {HypergeometricU}\left (\frac {b+n-\frac {i \sqrt {a} s}{\sqrt {c}}+1}{2 n},\frac {b+n+1}{n},-\frac {2 i \sqrt {a} \sqrt {c} x^n}{n}\right )}+i \sqrt {a} \sqrt {c} x^n+b+1}{a x} \\ \end{align*}