Internal problem ID [10355]
Internal file name [OUTPUT/9303_Monday_June_06_2022_01_50_21_PM_5228272/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: 26.
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]
\[ \boxed {y^{\prime }-y^{2}-a \,x^{n} y=b \,x^{-1+n}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+x^{n} a y +b \,x^{-1+n} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+x^{n} a y +\frac {x^{n} b}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=b \,x^{-1+n}\), \(f_1(x)=x^{n} a\) 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 &=b \,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 )+b \,x^{-1+n} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x \left (\operatorname {KummerU}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x^{1+n} a}{1+n}\right ) c_{2} +\operatorname {KummerM}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x^{1+n} a}{1+n}\right ) c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {a c_{1} \left (1+n \right ) \left (a -b \right ) \operatorname {KummerM}\left (-\frac {a n +b}{a \left (1+n \right )}+2, \frac {2+n}{1+n}, \frac {x^{1+n} a}{1+n}\right )-\left (c_{2} \left (a -b \right ) \operatorname {KummerU}\left (\frac {\left (2+n \right ) a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x^{1+n} a}{1+n}\right )-a \left (\operatorname {KummerU}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x^{1+n} a}{1+n}\right ) c_{2} +\operatorname {KummerM}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x^{1+n} a}{1+n}\right ) c_{1} \right ) \left (1+n \right )\right ) b}{a^{2} \left (1+n \right )} \] Using the above in (1) gives the solution \[ y = -\frac {a c_{1} \left (1+n \right ) \left (a -b \right ) \operatorname {KummerM}\left (-\frac {a n +b}{a \left (1+n \right )}+2, \frac {2+n}{1+n}, \frac {x^{1+n} a}{1+n}\right )-\left (c_{2} \left (a -b \right ) \operatorname {KummerU}\left (\frac {\left (2+n \right ) a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x^{1+n} a}{1+n}\right )-a \left (\operatorname {KummerU}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x^{1+n} a}{1+n}\right ) c_{2} +\operatorname {KummerM}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x^{1+n} a}{1+n}\right ) c_{1} \right ) \left (1+n \right )\right ) b}{a^{2} \left (1+n \right ) x \left (\operatorname {KummerU}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x^{1+n} a}{1+n}\right ) c_{2} +\operatorname {KummerM}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x^{1+n} a}{1+n}\right ) c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-a c_{3} \left (1+n \right ) \left (a -b \right ) \operatorname {KummerM}\left (\frac {2+n -\frac {b}{a}}{1+n}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right )+\left (\left (a -b \right ) \operatorname {KummerU}\left (\frac {2+n -\frac {b}{a}}{1+n}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right )-a \left (\operatorname {KummerU}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right )+\operatorname {KummerM}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right ) c_{3} \right ) \left (1+n \right )\right ) b}{a^{2} \left (1+n \right ) x \left (\operatorname {KummerU}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right )+\operatorname {KummerM}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right ) c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-a c_{3} \left (1+n \right ) \left (a -b \right ) \operatorname {KummerM}\left (\frac {2+n -\frac {b}{a}}{1+n}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right )+\left (\left (a -b \right ) \operatorname {KummerU}\left (\frac {2+n -\frac {b}{a}}{1+n}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right )-a \left (\operatorname {KummerU}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right )+\operatorname {KummerM}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right ) c_{3} \right ) \left (1+n \right )\right ) b}{a^{2} \left (1+n \right ) x \left (\operatorname {KummerU}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right )+\operatorname {KummerM}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right ) c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {-a c_{3} \left (1+n \right ) \left (a -b \right ) \operatorname {KummerM}\left (\frac {2+n -\frac {b}{a}}{1+n}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right )+\left (\left (a -b \right ) \operatorname {KummerU}\left (\frac {2+n -\frac {b}{a}}{1+n}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right )-a \left (\operatorname {KummerU}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right )+\operatorname {KummerM}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right ) c_{3} \right ) \left (1+n \right )\right ) b}{a^{2} \left (1+n \right ) x \left (\operatorname {KummerU}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right )+\operatorname {KummerM}\left (\frac {a -b}{a \left (1+n \right )}, \frac {2+n}{1+n}, \frac {x \,x^{n} a}{1+n}\right ) c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}-a \,x^{n} y=b \,x^{-1+n} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+a \,x^{n} y+b \,x^{-1+n} \end {array} \]
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 sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (diff(y(x), x))*x^n*a-b*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 -> 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: 267
dsolve(diff(y(x),x)=y(x)^2+a*x^n*y(x)+b*x^(n-1),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-a \left (n +1\right ) \left (a -b \right ) \operatorname {KummerM}\left (\frac {a \left (n +2\right )-b}{a \left (n +1\right )}, \frac {n +2}{n +1}, \frac {a x \,x^{n}}{n +1}\right )+\left (\left (a -b \right ) c_{1} \operatorname {KummerU}\left (\frac {2+n -\frac {b}{a}}{n +1}, \frac {n +2}{n +1}, \frac {a x \,x^{n}}{n +1}\right )-a \left (\operatorname {KummerU}\left (\frac {a -b}{a \left (n +1\right )}, \frac {n +2}{n +1}, \frac {a x \,x^{n}}{n +1}\right ) c_{1} +\operatorname {KummerM}\left (\frac {a -b}{a \left (n +1\right )}, \frac {n +2}{n +1}, \frac {a x \,x^{n}}{n +1}\right )\right ) \left (n +1\right )\right ) b}{a^{2} \left (n +1\right ) x \left (\operatorname {KummerU}\left (\frac {a -b}{a \left (n +1\right )}, \frac {n +2}{n +1}, \frac {a x \,x^{n}}{n +1}\right ) c_{1} +\operatorname {KummerM}\left (\frac {a -b}{a \left (n +1\right )}, \frac {n +2}{n +1}, \frac {a x \,x^{n}}{n +1}\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.956 (sec). Leaf size: 453
DSolve[y'[x]==y[x]^2+a*x^n*y[x]+b*x^(n-1),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\left (x^n\right )^{\frac {1}{n}} \left (-(-1)^{\frac {1}{n+1}} n (n+2) a^{\frac {1}{n+1}} \operatorname {Hypergeometric1F1}\left (\frac {a-b}{n a+a},\frac {n+2}{n+1},\frac {a \left (x^n\right )^{1+\frac {1}{n}}}{n+1}\right )+x^n \left (-(-1)^{\frac {1}{n+1}} n (a-b) a^{\frac {1}{n+1}} \left (x^n\right )^{\frac {1}{n}} \operatorname {Hypergeometric1F1}\left (\frac {n a+2 a-b}{n a+a},\frac {2 n+3}{n+1},\frac {a \left (x^n\right )^{1+\frac {1}{n}}}{n+1}\right )+b c_1 \left (\frac {1}{n}+1\right )^{\frac {1}{n+1}} n^{\frac {1}{n+1}} (n+2) \operatorname {Hypergeometric1F1}\left (\frac {n a+a-b}{n a+a},\frac {2 n+1}{n+1},\frac {a \left (x^n\right )^{1+\frac {1}{n}}}{n+1}\right )\right )\right )}{n (n+2) x \left ((-1)^{\frac {1}{n+1}} a^{\frac {1}{n+1}} \left (x^n\right )^{\frac {1}{n}} \operatorname {Hypergeometric1F1}\left (\frac {a-b}{n a+a},\frac {n+2}{n+1},\frac {a \left (x^n\right )^{1+\frac {1}{n}}}{n+1}\right )+c_1 \left (\frac {1}{n}+1\right )^{\frac {1}{n+1}} n^{\frac {1}{n+1}} \operatorname {Hypergeometric1F1}\left (-\frac {b}{n a+a},\frac {n}{n+1},\frac {a \left (x^n\right )^{1+\frac {1}{n}}}{n+1}\right )\right )} \\ y(x)\to \frac {b x^{n-1} \left (x^n\right )^{\frac {1}{n}} \operatorname {Hypergeometric1F1}\left (\frac {n a+a-b}{n a+a},\frac {2 n+1}{n+1},\frac {a \left (x^n\right )^{1+\frac {1}{n}}}{n+1}\right )}{n \operatorname {Hypergeometric1F1}\left (-\frac {b}{n a+a},\frac {n}{n+1},\frac {a \left (x^n\right )^{1+\frac {1}{n}}}{n+1}\right )} \\ \end{align*}