Internal problem ID [10365]
Internal file name [OUTPUT/9313_Monday_June_06_2022_01_51_14_PM_27493872/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: 36.
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]
\[ \boxed {y^{\prime } x -a y^{2}-\left (n +b \,x^{n}\right ) y=c \,x^{2 n}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {x^{n} b y +a \,y^{2}+c \,x^{2 n}+n y}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {x^{n} b y}{x}+\frac {a \,y^{2}}{x}+\frac {c \,x^{2 n}}{x}+\frac {n y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {c \,x^{2 n}}{x}\), \(f_1(x)=\frac {n +b \,x^{n}}{x}\) 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 {\left (n +b \,x^{n}\right ) a}{x^{2}}\\ f_2^2 f_0 &=\frac {a^{2} c \,x^{2 n}}{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 {\left (n +b \,x^{n}\right ) a}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {a^{2} c \,x^{2 n} u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\frac {x^{n} b}{2 n}} \left (c_{1} \sinh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right )+c_{2} \cosh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (\left (\sqrt {\frac {-4 c a +b^{2}}{n^{2}}}\, n c_{1} +c_{2} b \right ) \cosh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right )+\sinh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right ) \left (\sqrt {\frac {-4 c a +b^{2}}{n^{2}}}\, n c_{2} +c_{1} b \right )\right ) {\mathrm e}^{\frac {x^{n} b}{2 n}} x^{-1+n}}{2} \] Using the above in (1) gives the solution \[ y = -\frac {\left (\left (\sqrt {\frac {-4 c a +b^{2}}{n^{2}}}\, n c_{1} +c_{2} b \right ) \cosh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right )+\sinh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right ) \left (\sqrt {\frac {-4 c a +b^{2}}{n^{2}}}\, n c_{2} +c_{1} b \right )\right ) x^{-1+n} x}{2 a \left (c_{1} \sinh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right )+c_{2} \cosh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\left (\left (\sqrt {\frac {-4 c a +b^{2}}{n^{2}}}\, n c_{3} +b \right ) \cosh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right )+\sinh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right ) \left (\sqrt {\frac {-4 c a +b^{2}}{n^{2}}}\, n +b c_{3} \right )\right ) x^{n}}{2 a \left (c_{3} \sinh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right )+\cosh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (\left (\sqrt {\frac {-4 c a +b^{2}}{n^{2}}}\, n c_{3} +b \right ) \cosh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right )+\sinh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right ) \left (\sqrt {\frac {-4 c a +b^{2}}{n^{2}}}\, n +b c_{3} \right )\right ) x^{n}}{2 a \left (c_{3} \sinh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right )+\cosh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left (\left (\sqrt {\frac {-4 c a +b^{2}}{n^{2}}}\, n c_{3} +b \right ) \cosh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right )+\sinh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right ) \left (\sqrt {\frac {-4 c a +b^{2}}{n^{2}}}\, n +b c_{3} \right )\right ) x^{n}}{2 a \left (c_{3} \sinh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right )+\cosh \left (\frac {x^{n} \sqrt {\frac {-4 c a +b^{2}}{n^{2}}}}{2}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -a y^{2}-\left (n +b \,x^{n}\right ) y=c \,x^{2 n} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a y^{2}+\left (n +b \,x^{n}\right ) y+c \,x^{2 n}}{x} \end {array} \]
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: 0.015 (sec). Leaf size: 69
dsolve(x*diff(y(x),x)=a*y(x)^2+(n+b*x^n)*y(x)+c*x^(2*n),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {x^{n} \left (b^{2}-\sqrt {4 a \,b^{2} c -b^{4}}\, \tan \left (\frac {\sqrt {4 a \,b^{2} c -b^{4}}\, \left (b \,x^{n}+c_{1} n \right )}{2 b^{2} n}\right )\right )}{2 a b} \]
✓ Solution by Mathematica
Time used: 1.07 (sec). Leaf size: 114
DSolve[x*y'[x]==a*y[x]^2+(n+b*x^n)*y[x]+c*x^(2*n),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 {x^n \sqrt {b^2-4 a c}}{n}}+c_1\right )}{e^{\frac {x^n \sqrt {b^2-4 a c}}{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*}