Internal problem ID [10373]
Internal file name [OUTPUT/9321_Monday_June_06_2022_01_51_24_PM_98601438/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: 44.
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]
\[ \boxed {y^{\prime } x -a \,x^{m +2 n} y^{2}-\left (b \,x^{m +n}-n \right ) y=c \,x^{m}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {a \,x^{m +2 n} y^{2}+x^{m +n} b y +c \,x^{m}-n y}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {x^{m} a \,x^{2 n} y^{2}}{x}+\frac {x^{m} x^{n} b y}{x}+\frac {c \,x^{m}}{x}-\frac {n y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {c \,x^{m}}{x}\), \(f_1(x)=\frac {b \,x^{m +n}-n}{x}\) and \(f_2(x)=\frac {a \,x^{m +2 n}}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {a \,x^{m +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 {a \,x^{m +2 n}}{x^{2}}+\frac {a \,x^{m +2 n} \left (m +2 n \right )}{x^{2}}\\ f_1 f_2 &=\frac {\left (b \,x^{m +n}-n \right ) a \,x^{m +2 n}}{x^{2}}\\ f_2^2 f_0 &=\frac {a^{2} x^{2 m +4 n} c \,x^{m}}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {a \,x^{m +2 n} u^{\prime \prime }\left (x \right )}{x}-\left (-\frac {a \,x^{m +2 n}}{x^{2}}+\frac {a \,x^{m +2 n} \left (m +2 n \right )}{x^{2}}+\frac {\left (b \,x^{m +n}-n \right ) a \,x^{m +2 n}}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {a^{2} x^{2 m +4 n} c \,x^{m} u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\frac {b \,x^{m +n}}{2 m +2 n}} \left (c_{1} \sinh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )+c_{2} \cosh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (\left (c_{1} \left (m +n \right ) \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}+c_{2} b \right ) \cosh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )+\left (c_{2} \left (m +n \right ) \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}+c_{1} b \right ) \sinh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )\right ) {\mathrm e}^{\frac {b \,x^{m +n}}{2 m +2 n}} x^{m +n -1}}{2} \] Using the above in (1) gives the solution \[ y = -\frac {\left (\left (c_{1} \left (m +n \right ) \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}+c_{2} b \right ) \cosh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )+\left (c_{2} \left (m +n \right ) \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}+c_{1} b \right ) \sinh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )\right ) x^{m +n -1} x \,x^{-m -2 n}}{2 a \left (c_{1} \sinh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )+c_{2} \cosh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {x^{-n} \left (\left (c_{3} \left (m +n \right ) \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}+b \right ) \cosh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )+\left (\left (m +n \right ) \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}+b c_{3} \right ) \sinh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )\right )}{2 a \left (c_{3} \sinh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )+\cosh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {x^{-n} \left (\left (c_{3} \left (m +n \right ) \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}+b \right ) \cosh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )+\left (\left (m +n \right ) \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}+b c_{3} \right ) \sinh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )\right )}{2 a \left (c_{3} \sinh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )+\cosh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {x^{-n} \left (\left (c_{3} \left (m +n \right ) \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}+b \right ) \cosh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )+\left (\left (m +n \right ) \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}+b c_{3} \right ) \sinh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )\right )}{2 a \left (c_{3} \sinh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )+\cosh \left (\frac {x^{m +n} \sqrt {\frac {-4 c a +b^{2}}{\left (m +n \right )^{2}}}}{2}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -a \,x^{m +2 n} y^{2}-\left (b \,x^{m +n}-n \right ) y=c \,x^{m} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a \,x^{m +2 n} y^{2}+\left (b \,x^{m +n}-n \right ) y+c \,x^{m}}{x} \end {array} \]
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: 0.031 (sec). Leaf size: 78
dsolve(x*diff(y(x),x)=a*x^(2*n+m)*y(x)^2+(b*x^(n+m)-n)*y(x)+c*x^m,y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{-n} \left (\sqrt {4 a \,b^{2} c -b^{4}}\, \tan \left (\frac {\left (x^{n +m} b +c_{1} \left (n +m \right )\right ) \sqrt {4 a \,b^{2} c -b^{4}}}{2 b^{2} \left (n +m \right )}\right )-b^{2}\right )}{2 a b} \]
✓ Solution by Mathematica
Time used: 1.566 (sec). Leaf size: 126
DSolve[x*y'[x]==a*x^(2*n+m)*y[x]^2+(b*x^(n+m)-n)*y[x]+c*x^m,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x^{-n} \left (-b+\frac {\sqrt {b^2-4 a c} \left (-e^{\frac {\sqrt {b^2-4 a c} x^{m+n}}{m+n}}+c_1\right )}{e^{\frac {\sqrt {b^2-4 a c} x^{m+n}}{m+n}}+c_1}\right )}{2 a} \\ y(x)\to \frac {x^{-n} \left (\sqrt {b^2-4 a c}-b\right )}{2 a} \\ \end{align*}