2.25 problem 25

Internal problem ID [9608]

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.

CAS Maple gives this as type [_Riccati]

\[ \boxed {y^{\prime }-y^{2}-a \,x^{n} y-a \,x^{n -1}=0} \]

Solution by Maple

Time used: 0.016 (sec). Leaf size: 385

dsolve(diff(y(x),x)=y(x)^2+a*x^n*y(x)+a*x^(n-1),y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {{\mathrm e}^{\frac {a \,x^{1+n}}{1+n}}}{x^{2} \left (c_{1} -\frac {\left (\frac {a}{-1-n}\right )^{\frac {1}{1+n}} \left (\frac {\left (-1-n \right )^{2} x^{-1-\frac {1}{1+n}-\frac {n}{1+n}-n} \left (\frac {a}{-1-n}\right )^{-\frac {1}{1+n}} \left (\frac {x^{1+n} a \,n^{2}}{-1-n}+\frac {2 x^{1+n} a n}{-1-n}+n^{2}+\frac {x^{1+n} a}{-1-n}+n \right ) \left (\frac {x^{1+n} a}{-1-n}\right )^{-\frac {n}{2 \left (1+n \right )}} {\mathrm e}^{-\frac {x^{1+n} a}{2 \left (-1-n \right )}} \operatorname {WhittakerM}\left (-\frac {n}{2 \left (1+n \right )}-\frac {1}{1+n}, \frac {n}{2+2 n}+\frac {1}{2}, \frac {x^{1+n} a}{-1-n}\right )}{n \left (2 n +1\right ) a}+\frac {\left (-1-n \right )^{2} x^{-1-\frac {1}{1+n}-\frac {n}{1+n}-n} \left (\frac {a}{-1-n}\right )^{-\frac {1}{1+n}} n \left (\frac {x^{1+n} a}{-1-n}\right )^{-\frac {n}{2 \left (1+n \right )}} {\mathrm e}^{-\frac {x^{1+n} a}{2 \left (-1-n \right )}} \operatorname {WhittakerM}\left (-\frac {n}{2 \left (1+n \right )}-\frac {1}{1+n}+1, \frac {n}{2+2 n}+\frac {1}{2}, \frac {x^{1+n} a}{-1-n}\right )}{\left (2 n +1\right ) a}\right )}{1+n}\right )}-\frac {1}{x} \]

Solution by Mathematica

Time used: 1.868 (sec). Leaf size: 72

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 {1+\frac {(n+1) e^{\frac {a x^{n+1}}{n+1}}}{-\operatorname {ExpIntegralE}\left (1+\frac {1}{n+1},-\frac {a x^{n+1}}{n+1}\right )+c_1 (n+1) x}}{x} \\ y(x)\to -\frac {1}{x} \\ \end{align*}