Internal problem ID [10370]
Internal file name [OUTPUT/9318_Monday_June_06_2022_01_51_21_PM_22191330/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: 41.
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]
\[ \boxed {y^{\prime } x -x^{2 n} y^{2}-\left (m -n \right ) y=x^{2 m}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {x^{2 n} y^{2}+m y -n y +x^{2 m}}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {x^{2 n} y^{2}}{x}+\frac {m y}{x}-\frac {n y}{x}+\frac {x^{2 m}}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {x^{2 m}}{x}\), \(f_1(x)=\frac {m -n}{x}\) and \(f_2(x)=\frac {x^{2 n}}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {x^{2 n} 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 {x^{2 n}}{x^{2}}+\frac {2 x^{2 n} n}{x^{2}}\\ f_1 f_2 &=\frac {\left (m -n \right ) x^{2 n}}{x^{2}}\\ f_2^2 f_0 &=\frac {x^{4 n} x^{2 m}}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {x^{2 n} u^{\prime \prime }\left (x \right )}{x}-\left (-\frac {x^{2 n}}{x^{2}}+\frac {2 x^{2 n} n}{x^{2}}+\frac {\left (m -n \right ) x^{2 n}}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {x^{4 n} x^{2 m} u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \sin \left (\frac {x^{m +n}}{m +n}\right )+c_{2} \cos \left (\frac {x^{m +n}}{m +n}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = x^{m +n -1} \left (c_{1} \cos \left (\frac {x^{m +n}}{m +n}\right )-c_{2} \sin \left (\frac {x^{m +n}}{m +n}\right )\right ) \] Using the above in (1) gives the solution \[ y = -\frac {x^{m +n -1} \left (c_{1} \cos \left (\frac {x^{m +n}}{m +n}\right )-c_{2} \sin \left (\frac {x^{m +n}}{m +n}\right )\right ) x \,x^{-2 n}}{c_{1} \sin \left (\frac {x^{m +n}}{m +n}\right )+c_{2} \cos \left (\frac {x^{m +n}}{m +n}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {x^{m -n} \left (-c_{3} \cos \left (\frac {x^{m +n}}{m +n}\right )+\sin \left (\frac {x^{m +n}}{m +n}\right )\right )}{c_{3} \sin \left (\frac {x^{m +n}}{m +n}\right )+\cos \left (\frac {x^{m +n}}{m +n}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{m -n} \left (-c_{3} \cos \left (\frac {x^{m +n}}{m +n}\right )+\sin \left (\frac {x^{m +n}}{m +n}\right )\right )}{c_{3} \sin \left (\frac {x^{m +n}}{m +n}\right )+\cos \left (\frac {x^{m +n}}{m +n}\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{m -n} \left (-c_{3} \cos \left (\frac {x^{m +n}}{m +n}\right )+\sin \left (\frac {x^{m +n}}{m +n}\right )\right )}{c_{3} \sin \left (\frac {x^{m +n}}{m +n}\right )+\cos \left (\frac {x^{m +n}}{m +n}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -x^{2 n} y^{2}-\left (m -n \right ) y=x^{2 m} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x^{2 n} y^{2}+\left (m -n \right ) y+x^{2 m}}{x} \end {array} \]
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: 0.0 (sec). Leaf size: 34
dsolve(x*diff(y(x),x)=x^(2*n)*y(x)^2+(m-n)*y(x)+x^(2*m),y(x), singsol=all)
\[ y \left (x \right ) = \tan \left (\frac {x^{n +m}+\left (-n -m \right ) c_{1}}{n +m}\right ) x^{-n +m} \]
✓ Solution by Mathematica
Time used: 0.727 (sec). Leaf size: 28
DSolve[x*y'[x]==x^(2*n)*y[x]^2+(m-n)*y[x]+x^(2*m),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x^{m-n} \tan \left (\frac {x^{m+n}}{m+n}+c_1\right ) \]