Internal problem ID [10333]
Internal file name [OUTPUT/9281_Monday_June_06_2022_01_45_26_PM_40145015/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: 4.
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, _special]]
\[ \boxed {-a y^{2}+y^{\prime }=b \,x^{n}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,y^{2}+b \,x^{n} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,y^{2}+b \,x^{n} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=b \,x^{n}\), \(f_1(x)=0\) 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 &=b \,x^{n} a^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a u^{\prime \prime }\left (x \right )+b \,x^{n} a^{2} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (\operatorname {BesselY}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{2} +\operatorname {BesselJ}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{1} \right ) \sqrt {x} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {-\operatorname {BesselJ}\left (\frac {n +3}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) \sqrt {a b}\, x^{1+\frac {n}{2}} c_{1} -\sqrt {a b}\, x^{1+\frac {n}{2}} \operatorname {BesselY}\left (\frac {n +3}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{2} +\operatorname {BesselY}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{2} +\operatorname {BesselJ}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{1}}{\sqrt {x}} \] Using the above in (1) gives the solution \[ y = -\frac {-\operatorname {BesselJ}\left (\frac {n +3}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) \sqrt {a b}\, x^{1+\frac {n}{2}} c_{1} -\sqrt {a b}\, x^{1+\frac {n}{2}} \operatorname {BesselY}\left (\frac {n +3}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{2} +\operatorname {BesselY}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{2} +\operatorname {BesselJ}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{1}}{x a \left (\operatorname {BesselY}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{2} +\operatorname {BesselJ}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\operatorname {BesselJ}\left (\frac {n +3}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) \sqrt {a b}\, x^{1+\frac {n}{2}} c_{3} +\operatorname {BesselY}\left (\frac {n +3}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) \sqrt {a b}\, x^{1+\frac {n}{2}}-\operatorname {BesselY}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right )-\operatorname {BesselJ}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{3}}{x a \left (\operatorname {BesselY}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right )+\operatorname {BesselJ}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\operatorname {BesselJ}\left (\frac {n +3}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) \sqrt {a b}\, x^{1+\frac {n}{2}} c_{3} +\operatorname {BesselY}\left (\frac {n +3}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) \sqrt {a b}\, x^{1+\frac {n}{2}}-\operatorname {BesselY}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right )-\operatorname {BesselJ}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{3}}{x a \left (\operatorname {BesselY}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right )+\operatorname {BesselJ}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\operatorname {BesselJ}\left (\frac {n +3}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) \sqrt {a b}\, x^{1+\frac {n}{2}} c_{3} +\operatorname {BesselY}\left (\frac {n +3}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) \sqrt {a b}\, x^{1+\frac {n}{2}}-\operatorname {BesselY}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right )-\operatorname {BesselJ}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{3}}{x a \left (\operatorname {BesselY}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right )+\operatorname {BesselJ}\left (\frac {1}{2+n}, \frac {2 \sqrt {a b}\, x^{1+\frac {n}{2}}}{2+n}\right ) c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -a y^{2}+y^{\prime }=b \,x^{n} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a y^{2}+b \,x^{n} \end {array} \]
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 <- Riccati Special successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 207
dsolve(diff(y(x),x)=a*y(x)^2+b*x^n,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\operatorname {BesselJ}\left (\frac {3+n}{n +2}, \frac {2 \sqrt {a b}\, x^{\frac {n}{2}+1}}{n +2}\right ) \sqrt {a b}\, x^{\frac {n}{2}+1} c_{1} +\operatorname {BesselY}\left (\frac {3+n}{n +2}, \frac {2 \sqrt {a b}\, x^{\frac {n}{2}+1}}{n +2}\right ) \sqrt {a b}\, x^{\frac {n}{2}+1}-c_{1} \operatorname {BesselJ}\left (\frac {1}{n +2}, \frac {2 \sqrt {a b}\, x^{\frac {n}{2}+1}}{n +2}\right )-\operatorname {BesselY}\left (\frac {1}{n +2}, \frac {2 \sqrt {a b}\, x^{\frac {n}{2}+1}}{n +2}\right )}{x a \left (c_{1} \operatorname {BesselJ}\left (\frac {1}{n +2}, \frac {2 \sqrt {a b}\, x^{\frac {n}{2}+1}}{n +2}\right )+\operatorname {BesselY}\left (\frac {1}{n +2}, \frac {2 \sqrt {a b}\, x^{\frac {n}{2}+1}}{n +2}\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.696 (sec). Leaf size: 605
DSolve[y'[x]==a*y[x]^2+b*x^n,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {\sqrt {a} \sqrt {b} x^{\frac {n}{2}+1} \operatorname {Gamma}\left (1+\frac {1}{n+2}\right ) \operatorname {BesselJ}\left (\frac {1}{n+2}-1,\frac {2 \sqrt {a} \sqrt {b} x^{\frac {n}{2}+1}}{n+2}\right )-\sqrt {a} \sqrt {b} x^{\frac {n}{2}+1} \operatorname {Gamma}\left (1+\frac {1}{n+2}\right ) \operatorname {BesselJ}\left (1+\frac {1}{n+2},\frac {2 \sqrt {a} \sqrt {b} x^{\frac {n}{2}+1}}{n+2}\right )+\operatorname {Gamma}\left (1+\frac {1}{n+2}\right ) \operatorname {BesselJ}\left (\frac {1}{n+2},\frac {2 \sqrt {a} \sqrt {b} x^{\frac {n}{2}+1}}{n+2}\right )-\sqrt {a} \sqrt {b} c_1 x^{\frac {n}{2}+1} \operatorname {Gamma}\left (\frac {n+1}{n+2}\right ) \operatorname {BesselJ}\left (\frac {n+1}{n+2},\frac {2 \sqrt {a} \sqrt {b} x^{\frac {n}{2}+1}}{n+2}\right )+\sqrt {a} \sqrt {b} c_1 x^{\frac {n}{2}+1} \operatorname {Gamma}\left (\frac {n+1}{n+2}\right ) \operatorname {BesselJ}\left (-\frac {n+3}{n+2},\frac {2 \sqrt {a} \sqrt {b} x^{\frac {n}{2}+1}}{n+2}\right )+c_1 \operatorname {Gamma}\left (\frac {n+1}{n+2}\right ) \operatorname {BesselJ}\left (-\frac {1}{n+2},\frac {2 \sqrt {a} \sqrt {b} x^{\frac {n}{2}+1}}{n+2}\right )}{2 a x \left (\operatorname {Gamma}\left (1+\frac {1}{n+2}\right ) \operatorname {BesselJ}\left (\frac {1}{n+2},\frac {2 \sqrt {a} \sqrt {b} x^{\frac {n}{2}+1}}{n+2}\right )+c_1 \operatorname {Gamma}\left (\frac {n+1}{n+2}\right ) \operatorname {BesselJ}\left (-\frac {1}{n+2},\frac {2 \sqrt {a} \sqrt {b} x^{\frac {n}{2}+1}}{n+2}\right )\right )} \\ y(x)\to \frac {\frac {\sqrt {a} \sqrt {b} x^{n/2} \left (\operatorname {BesselJ}\left (\frac {n+1}{n+2},\frac {2 \sqrt {a} \sqrt {b} x^{\frac {n}{2}+1}}{n+2}\right )-\operatorname {BesselJ}\left (-\frac {n+3}{n+2},\frac {2 \sqrt {a} \sqrt {b} x^{\frac {n}{2}+1}}{n+2}\right )\right )}{\operatorname {BesselJ}\left (-\frac {1}{n+2},\frac {2 \sqrt {a} \sqrt {b} x^{\frac {n}{2}+1}}{n+2}\right )}-\frac {1}{x}}{2 a} \\ \end{align*}