Internal problem ID [10363]
Internal file name [OUTPUT/9311_Monday_June_06_2022_01_51_13_PM_28351703/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: 34.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_rational, _Riccati]
\[ \boxed {y^{\prime } x -a y^{2}-y b=c \,x^{2 b}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {a \,y^{2}+y b +c \,x^{2 b}}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {a \,y^{2}}{x}+\frac {c \,x^{2 b}}{x}+\frac {y b}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {c \,x^{2 b}}{x}\), \(f_1(x)=\frac {b}{x}\) and \(f_2(x)=\frac {a}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {a u}{x}} \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' &=-\frac {a}{x^{2}}\\ f_1 f_2 &=\frac {a b}{x^{2}}\\ f_2^2 f_0 &=\frac {a^{2} c \,x^{2 b}}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {a u^{\prime \prime }\left (x \right )}{x}-\left (-\frac {a}{x^{2}}+\frac {a b}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {a^{2} c \,x^{2 b} u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \sin \left (\frac {x^{b} \sqrt {c a}}{b}\right )+c_{2} \cos \left (\frac {x^{b} \sqrt {c a}}{b}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {x^{b} \sqrt {c a}\, \left (c_{1} \cos \left (\frac {x^{b} \sqrt {c a}}{b}\right )-c_{2} \sin \left (\frac {x^{b} \sqrt {c a}}{b}\right )\right )}{x} \] Using the above in (1) gives the solution \[ y = -\frac {x^{b} \sqrt {c a}\, \left (c_{1} \cos \left (\frac {x^{b} \sqrt {c a}}{b}\right )-c_{2} \sin \left (\frac {x^{b} \sqrt {c a}}{b}\right )\right )}{a \left (c_{1} \sin \left (\frac {x^{b} \sqrt {c a}}{b}\right )+c_{2} \cos \left (\frac {x^{b} \sqrt {c a}}{b}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (-c_{3} \cos \left (\frac {x^{b} \sqrt {c a}}{b}\right )+\sin \left (\frac {x^{b} \sqrt {c a}}{b}\right )\right ) x^{b} \sqrt {c a}}{\left (c_{3} \sin \left (\frac {x^{b} \sqrt {c a}}{b}\right )+\cos \left (\frac {x^{b} \sqrt {c a}}{b}\right )\right ) a} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-c_{3} \cos \left (\frac {x^{b} \sqrt {c a}}{b}\right )+\sin \left (\frac {x^{b} \sqrt {c a}}{b}\right )\right ) x^{b} \sqrt {c a}}{\left (c_{3} \sin \left (\frac {x^{b} \sqrt {c a}}{b}\right )+\cos \left (\frac {x^{b} \sqrt {c a}}{b}\right )\right ) a} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-c_{3} \cos \left (\frac {x^{b} \sqrt {c a}}{b}\right )+\sin \left (\frac {x^{b} \sqrt {c a}}{b}\right )\right ) x^{b} \sqrt {c a}}{\left (c_{3} \sin \left (\frac {x^{b} \sqrt {c a}}{b}\right )+\cos \left (\frac {x^{b} \sqrt {c a}}{b}\right )\right ) a} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -a y^{2}-y b =c \,x^{2 b} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a y^{2}+y b +c \,x^{2 b}}{x} \end {array} \]
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 <- Chini successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 34
dsolve(x*diff(y(x),x)=a*y(x)^2+b*y(x)+c*x^(2*b),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\tan \left (\frac {x^{b} \sqrt {a}\, \sqrt {c}-c_{1} b}{b}\right ) \sqrt {c}\, x^{b}}{\sqrt {a}} \]
✓ Solution by Mathematica
Time used: 0.544 (sec). Leaf size: 139
DSolve[x*y'[x]==a*y[x]^2+b*y[x]+c*x^(2*b),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\sqrt {c} x^b \left (-\cos \left (\frac {\sqrt {a} \sqrt {c} x^b}{b}\right )+c_1 \sin \left (\frac {\sqrt {a} \sqrt {c} x^b}{b}\right )\right )}{\sqrt {a} \left (\sin \left (\frac {\sqrt {a} \sqrt {c} x^b}{b}\right )+c_1 \cos \left (\frac {\sqrt {a} \sqrt {c} x^b}{b}\right )\right )} \\ y(x)\to \frac {\sqrt {c} x^b \tan \left (\frac {\sqrt {a} \sqrt {c} x^b}{b}\right )}{\sqrt {a}} \\ \end{align*}