Internal problem ID [10335]
Internal file name [OUTPUT/9283_Monday_June_06_2022_01_46_32_PM_97394906/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: 6.
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]
\[ \boxed {-a y^{2}+y^{\prime }=b \,x^{2 n}+c \,x^{-1+n}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,y^{2}+b \,x^{2 n}+c \,x^{-1+n} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,y^{2}+b \,x^{2 n}+\frac {c \,x^{n}}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=b \,x^{2 n}+c \,x^{-1+n}\), \(f_1(x)=0\) 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 &=0\\ f_2^2 f_0 &=a^{2} \left (b \,x^{2 n}+c \,x^{-1+n}\right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a u^{\prime \prime }\left (x \right )+a^{2} \left (b \,x^{2 n}+c \,x^{-1+n}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{-\frac {n}{2}} \left (c_{1} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )+c_{2} \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (-\left (i \sqrt {a}\, \sqrt {b}\, c -b \left (2+n \right )\right ) c_{1} \operatorname {WhittakerM}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )-2 b c_{2} \left (1+n \right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )+\left (2 i b^{\frac {3}{2}} x \,x^{n} \sqrt {a}+i \sqrt {a}\, \sqrt {b}\, c -b n \right ) \left (c_{1} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )+c_{2} \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )\right )\right ) x^{-\frac {n}{2}}}{2 b x} \] Using the above in (1) gives the solution \[ y = -\frac {-\left (i \sqrt {a}\, \sqrt {b}\, c -b \left (2+n \right )\right ) c_{1} \operatorname {WhittakerM}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )-2 b c_{2} \left (1+n \right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )+\left (2 i b^{\frac {3}{2}} x \,x^{n} \sqrt {a}+i \sqrt {a}\, \sqrt {b}\, c -b n \right ) \left (c_{1} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )+c_{2} \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )\right )}{2 b x a \left (c_{1} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )+c_{2} \operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (i \sqrt {a}\, \sqrt {b}\, c -b \left (2+n \right )\right ) c_{3} \operatorname {WhittakerM}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )+2 b \left (1+n \right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )-\left (2 i b^{\frac {3}{2}} x \,x^{n} \sqrt {a}+i \sqrt {a}\, \sqrt {b}\, c -b n \right ) \left (c_{3} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )+\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )\right )}{2 b x a \left (c_{3} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )+\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (i \sqrt {a}\, \sqrt {b}\, c -b \left (2+n \right )\right ) c_{3} \operatorname {WhittakerM}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )+2 b \left (1+n \right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )-\left (2 i b^{\frac {3}{2}} x \,x^{n} \sqrt {a}+i \sqrt {a}\, \sqrt {b}\, c -b n \right ) \left (c_{3} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )+\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )\right )}{2 b x a \left (c_{3} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )+\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (i \sqrt {a}\, \sqrt {b}\, c -b \left (2+n \right )\right ) c_{3} \operatorname {WhittakerM}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )+2 b \left (1+n \right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )-\left (2 i b^{\frac {3}{2}} x \,x^{n} \sqrt {a}+i \sqrt {a}\, \sqrt {b}\, c -b n \right ) \left (c_{3} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )+\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )\right )}{2 b x a \left (c_{3} \operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )+\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{1+n}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -a y^{2}+y^{\prime }=b \,x^{2 n}+c \,x^{-1+n} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a y^{2}+b \,x^{2 n}+c \,x^{-1+n} \end {array} \]
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 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*(b*x^(2*n)+c*x^(n-1))*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 <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Whittaker successful <- special function solution successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 357
dsolve(diff(y(x),x)=a*y(x)^2+b*x^(2*n)+c*x^(n-1),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\left (\left (\frac {n}{2}+1\right ) \sqrt {b}-\frac {i c \sqrt {a}}{2}\right ) \operatorname {WhittakerM}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i c \sqrt {a}}{\sqrt {b}\, \left (2 n +2\right )}, \frac {1}{2 n +2}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{n +1}\right )-\sqrt {b}\, c_{1} \left (n +1\right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i c \sqrt {a}}{\sqrt {b}\, \left (2 n +2\right )}, \frac {1}{2 n +2}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{n +1}\right )+\left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2 n +2\right )}, \frac {1}{2 n +2}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{n +1}\right ) c_{1} +\operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2 n +2\right )}, \frac {1}{2 n +2}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{n +1}\right )\right ) \left (-\frac {\sqrt {b}\, n}{2}+i \sqrt {a}\, \left (x \,x^{n} b +\frac {c}{2}\right )\right )}{\sqrt {b}\, \left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2 n +2\right )}, \frac {1}{2 n +2}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{n +1}\right ) c_{1} +\operatorname {WhittakerM}\left (-\frac {i \sqrt {a}\, c}{\sqrt {b}\, \left (2 n +2\right )}, \frac {1}{2 n +2}, \frac {2 i \sqrt {a}\, \sqrt {b}\, x \,x^{n}}{n +1}\right )\right ) a x} \]
✓ Solution by Mathematica
Time used: 1.818 (sec). Leaf size: 982
DSolve[y'[x]==a*y[x]^2+b*x^(2*n)+c*x^(n-1),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {x^n \left (\sqrt {b} c_1 (n+1) \sqrt {-(n+1)^2} \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} c}{\sqrt {b} \sqrt {-(n+1)^2}}+\frac {n}{n+1}\right ),\frac {n}{n+1},\frac {2 \sqrt {a} \sqrt {b} x^{n+1}}{\sqrt {-(n+1)^2}}\right )+c_1 \left (\sqrt {a} c (n+1)+\sqrt {b} \sqrt {-(n+1)^2} n\right ) \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} c}{\sqrt {b} \sqrt {-(n+1)^2}}+\frac {3 n+2}{n+1}\right ),\frac {n}{n+1}+1,\frac {2 \sqrt {a} \sqrt {b} x^{n+1}}{\sqrt {-(n+1)^2}}\right )+\sqrt {b} (n+1) \sqrt {-(n+1)^2} \left (L_{-\frac {\sqrt {a} c}{2 \sqrt {b} \sqrt {-(n+1)^2}}-\frac {n}{2 (n+1)}}^{-\frac {1}{n+1}}\left (\frac {2 \sqrt {a} \sqrt {b} x^{n+1}}{\sqrt {-(n+1)^2}}\right )+2 L_{-\frac {\sqrt {a} c}{2 \sqrt {b} \sqrt {-(n+1)^2}}-\frac {3 n+2}{2 n+2}}^{\frac {n}{n+1}}\left (\frac {2 \sqrt {a} \sqrt {b} x^{n+1}}{\sqrt {-(n+1)^2}}\right )\right )\right )}{\sqrt {a} (n+1)^2 \left (L_{-\frac {\sqrt {a} c}{2 \sqrt {b} \sqrt {-(n+1)^2}}-\frac {n}{2 (n+1)}}^{-\frac {1}{n+1}}\left (\frac {2 \sqrt {a} \sqrt {b} x^{n+1}}{\sqrt {-(n+1)^2}}\right )+c_1 \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} c}{\sqrt {b} \sqrt {-(n+1)^2}}+\frac {n}{n+1}\right ),\frac {n}{n+1},\frac {2 \sqrt {a} \sqrt {b} x^{n+1}}{\sqrt {-(n+1)^2}}\right )\right )} \\ y(x)\to \frac {x^n \left (-\frac {\left (\sqrt {a} c (n+1)+\sqrt {b} \sqrt {-(n+1)^2} n\right ) \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} c}{\sqrt {b} \sqrt {-(n+1)^2}}+\frac {n}{n+1}+2\right ),\frac {n}{n+1}+1,\frac {2 \sqrt {a} \sqrt {b} x^{n+1}}{\sqrt {-(n+1)^2}}\right )}{\operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} c}{\sqrt {b} \sqrt {-(n+1)^2}}+\frac {n}{n+1}\right ),\frac {n}{n+1},\frac {2 \sqrt {a} \sqrt {b} x^{n+1}}{\sqrt {-(n+1)^2}}\right )}-\sqrt {b} \sqrt {-(n+1)^2} (n+1)\right )}{\sqrt {a} (n+1)^2} \\ y(x)\to \frac {x^n \left (-\frac {\left (\sqrt {a} c (n+1)+\sqrt {b} \sqrt {-(n+1)^2} n\right ) \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} c}{\sqrt {b} \sqrt {-(n+1)^2}}+\frac {n}{n+1}+2\right ),\frac {n}{n+1}+1,\frac {2 \sqrt {a} \sqrt {b} x^{n+1}}{\sqrt {-(n+1)^2}}\right )}{\operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} c}{\sqrt {b} \sqrt {-(n+1)^2}}+\frac {n}{n+1}\right ),\frac {n}{n+1},\frac {2 \sqrt {a} \sqrt {b} x^{n+1}}{\sqrt {-(n+1)^2}}\right )}-\sqrt {b} \sqrt {-(n+1)^2} (n+1)\right )}{\sqrt {a} (n+1)^2} \\ \end{align*}