Internal problem ID [10354]
Internal file name [OUTPUT/9302_Monday_June_06_2022_01_49_48_PM_63309599/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: 25.
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]
\[ \boxed {y^{\prime }-y^{2}-a \,x^{n} y=a \,x^{-1+n}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+x^{n} a y +a \,x^{-1+n} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+x^{n} a y +\frac {a \,x^{n}}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a \,x^{-1+n}\), \(f_1(x)=x^{n} a\) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{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 &=x^{n} a\\ f_2^2 f_0 &=a \,x^{-1+n} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )-x^{n} a u^{\prime }\left (x \right )+a \,x^{-1+n} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{2} {\mathrm e}^{\frac {a \,x^{1+n}}{2+2 n}} \left (1+n \right ) \left (a \,x^{-\frac {n}{2}}-n \,x^{-1-\frac {3 n}{2}}\right ) \operatorname {WhittakerM}\left (\frac {-n -2}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x^{1+n} a}{1+n}\right )-\operatorname {WhittakerM}\left (\frac {n}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x^{1+n} a}{1+n}\right ) x^{-1-\frac {3 n}{2}} {\mathrm e}^{\frac {a \,x^{1+n}}{2+2 n}} c_{2} n^{2}+c_{1} x \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left ({\mathrm e}^{\frac {x^{n} a x}{2+2 n}} c_{2} \left (1+n \right ) \left (a x \,x^{\frac {n}{2}}+x^{-\frac {n}{2}} n^{2}\right ) \operatorname {WhittakerM}\left (\frac {-n -2}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right )+{\mathrm e}^{\frac {x^{n} a x}{2+2 n}} n c_{2} \left (x^{-\frac {n}{2}} n^{2}+a x \,x^{\frac {n}{2}} \left (1+n \right )\right ) \operatorname {WhittakerM}\left (\frac {n}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right )-2 x^{-\frac {n}{2}} n^{2} {\mathrm e}^{\frac {x \,x^{n} a}{1+n}} c_{2} \left (\frac {1}{2}+n \right ) \left (-\frac {x \,x^{n} a}{1+n}\right )^{\frac {2+3 n}{2+2 n}}+c_{1} x^{2} x^{n}\right ) x^{-n}}{x^{2}} \] Using the above in (1) gives the solution \[ y = -\frac {\left ({\mathrm e}^{\frac {x^{n} a x}{2+2 n}} c_{2} \left (1+n \right ) \left (a x \,x^{\frac {n}{2}}+x^{-\frac {n}{2}} n^{2}\right ) \operatorname {WhittakerM}\left (\frac {-n -2}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right )+{\mathrm e}^{\frac {x^{n} a x}{2+2 n}} n c_{2} \left (x^{-\frac {n}{2}} n^{2}+a x \,x^{\frac {n}{2}} \left (1+n \right )\right ) \operatorname {WhittakerM}\left (\frac {n}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right )-2 x^{-\frac {n}{2}} n^{2} {\mathrm e}^{\frac {x \,x^{n} a}{1+n}} c_{2} \left (\frac {1}{2}+n \right ) \left (-\frac {x \,x^{n} a}{1+n}\right )^{\frac {2+3 n}{2+2 n}}+c_{1} x^{2} x^{n}\right ) x^{-n}}{x^{2} \left (c_{2} {\mathrm e}^{\frac {a \,x^{1+n}}{2+2 n}} \left (1+n \right ) \left (a \,x^{-\frac {n}{2}}-n \,x^{-1-\frac {3 n}{2}}\right ) \operatorname {WhittakerM}\left (\frac {-n -2}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x^{1+n} a}{1+n}\right )-\operatorname {WhittakerM}\left (\frac {n}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x^{1+n} a}{1+n}\right ) x^{-1-\frac {3 n}{2}} {\mathrm e}^{\frac {a \,x^{1+n}}{2+2 n}} c_{2} n^{2}+c_{1} x \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\left ({\mathrm e}^{\frac {x^{n} a x}{2+2 n}} \left (1+n \right ) \left (a x \,x^{\frac {n}{2}}+x^{-\frac {n}{2}} n^{2}\right ) \operatorname {WhittakerM}\left (\frac {-n -2}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right )+{\mathrm e}^{\frac {x^{n} a x}{2+2 n}} n \left (x^{-\frac {n}{2}} n^{2}+a x \,x^{\frac {n}{2}} \left (1+n \right )\right ) \operatorname {WhittakerM}\left (\frac {n}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right )-2 x^{-\frac {n}{2}} n^{2} {\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (\frac {1}{2}+n \right ) \left (-\frac {x \,x^{n} a}{1+n}\right )^{\frac {2+3 n}{2+2 n}}+c_{3} x^{2} x^{n}\right ) x^{-n}}{\left ({\mathrm e}^{\frac {x^{n} a x}{2+2 n}} \left (1+n \right ) \left (x^{-\frac {n}{2}} a x -n \,x^{-\frac {3 n}{2}}\right ) \operatorname {WhittakerM}\left (\frac {-n -2}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right )-\operatorname {WhittakerM}\left (\frac {n}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right ) x^{-\frac {3 n}{2}} {\mathrm e}^{\frac {x^{n} a x}{2+2 n}} n^{2}+c_{3} x^{2}\right ) x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left ({\mathrm e}^{\frac {x^{n} a x}{2+2 n}} \left (1+n \right ) \left (a x \,x^{\frac {n}{2}}+x^{-\frac {n}{2}} n^{2}\right ) \operatorname {WhittakerM}\left (\frac {-n -2}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right )+{\mathrm e}^{\frac {x^{n} a x}{2+2 n}} n \left (x^{-\frac {n}{2}} n^{2}+a x \,x^{\frac {n}{2}} \left (1+n \right )\right ) \operatorname {WhittakerM}\left (\frac {n}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right )-2 x^{-\frac {n}{2}} n^{2} {\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (\frac {1}{2}+n \right ) \left (-\frac {x \,x^{n} a}{1+n}\right )^{\frac {2+3 n}{2+2 n}}+c_{3} x^{2} x^{n}\right ) x^{-n}}{\left ({\mathrm e}^{\frac {x^{n} a x}{2+2 n}} \left (1+n \right ) \left (x^{-\frac {n}{2}} a x -n \,x^{-\frac {3 n}{2}}\right ) \operatorname {WhittakerM}\left (\frac {-n -2}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right )-\operatorname {WhittakerM}\left (\frac {n}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right ) x^{-\frac {3 n}{2}} {\mathrm e}^{\frac {x^{n} a x}{2+2 n}} n^{2}+c_{3} x^{2}\right ) x} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left ({\mathrm e}^{\frac {x^{n} a x}{2+2 n}} \left (1+n \right ) \left (a x \,x^{\frac {n}{2}}+x^{-\frac {n}{2}} n^{2}\right ) \operatorname {WhittakerM}\left (\frac {-n -2}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right )+{\mathrm e}^{\frac {x^{n} a x}{2+2 n}} n \left (x^{-\frac {n}{2}} n^{2}+a x \,x^{\frac {n}{2}} \left (1+n \right )\right ) \operatorname {WhittakerM}\left (\frac {n}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right )-2 x^{-\frac {n}{2}} n^{2} {\mathrm e}^{\frac {x \,x^{n} a}{1+n}} \left (\frac {1}{2}+n \right ) \left (-\frac {x \,x^{n} a}{1+n}\right )^{\frac {2+3 n}{2+2 n}}+c_{3} x^{2} x^{n}\right ) x^{-n}}{\left ({\mathrm e}^{\frac {x^{n} a x}{2+2 n}} \left (1+n \right ) \left (x^{-\frac {n}{2}} a x -n \,x^{-\frac {3 n}{2}}\right ) \operatorname {WhittakerM}\left (\frac {-n -2}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right )-\operatorname {WhittakerM}\left (\frac {n}{2+2 n}, \frac {1+2 n}{2+2 n}, -\frac {x \,x^{n} a}{1+n}\right ) x^{-\frac {3 n}{2}} {\mathrm e}^{\frac {x^{n} a x}{2+2 n}} n^{2}+c_{3} x^{2}\right ) x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}-a \,x^{n} y=a \,x^{-1+n} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+a \,x^{n} y+a \,x^{-1+n} \end {array} \]
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 found: 2 potential symmetries. Proceeding with integration step`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 400
dsolve(diff(y(x),x)=y(x)^2+a*x^n*y(x)+a*x^(n-1),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-{\mathrm e}^{\frac {x^{n} a x}{2 n +2}} \left (-\frac {a x \,x^{n}}{n +1}\right )^{-\frac {n}{2 n +2}} \left (n +1\right )^{2} \left (x^{n} a x -n \right ) \operatorname {WhittakerM}\left (\frac {-n -2}{2 n +2}, \frac {2 n +1}{2 n +2}, -\frac {a x \,x^{n}}{n +1}\right )-2 n \left (-\frac {\left (n +1\right ) n \,{\mathrm e}^{\frac {x^{n} a x}{2 n +2}} \left (-\frac {a x \,x^{n}}{n +1}\right )^{-\frac {n}{2 n +2}} \operatorname {WhittakerM}\left (\frac {n}{2 n +2}, \frac {2 n +1}{2 n +2}, -\frac {a x \,x^{n}}{n +1}\right )}{2}+\left (n +\frac {1}{2}\right ) x \,x^{n} a \left (c_{1} x -{\mathrm e}^{\frac {a x \,x^{n}}{n +1}}\right )\right )}{x \left ({\mathrm e}^{\frac {x^{n} a x}{2 n +2}} \left (-\frac {a x \,x^{n}}{n +1}\right )^{-\frac {n}{2 n +2}} \left (n +1\right )^{2} \left (x^{n} a x -n \right ) \operatorname {WhittakerM}\left (\frac {-n -2}{2 n +2}, \frac {2 n +1}{2 n +2}, -\frac {a x \,x^{n}}{n +1}\right )+2 n \left (-\frac {\left (n +1\right ) n \,{\mathrm e}^{\frac {x^{n} a x}{2 n +2}} \left (-\frac {a x \,x^{n}}{n +1}\right )^{-\frac {n}{2 n +2}} \operatorname {WhittakerM}\left (\frac {n}{2 n +2}, \frac {2 n +1}{2 n +2}, -\frac {a x \,x^{n}}{n +1}\right )}{2}+a \,x^{2} c_{1} x^{n} \left (n +\frac {1}{2}\right )\right )\right )} \]
✓ Solution by Mathematica
Time used: 2.82 (sec). Leaf size: 136
DSolve[y'[x]==y[x]^2+a*x^n*y[x]+a*x^(n-1),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\left (-\frac {a x^{n+1}}{n+1}\right )^{\frac {1}{n+1}} \Gamma \left (-\frac {1}{n+1},-\frac {a x^{n+1}}{n+1}\right )-(n+1) \left (e^{\frac {a x^{n+1}}{n+1}}+c_1 x\right )}{x \left (-\left (-\frac {a x^{n+1}}{n+1}\right )^{\frac {1}{n+1}} \Gamma \left (-\frac {1}{n+1},-\frac {a x^{n+1}}{n+1}\right )+c_1 (n+1) x\right )} \\ y(x)\to -\frac {1}{x} \\ \end{align*}