Internal problem ID [10443]
Internal file name [OUTPUT/9391_Monday_June_06_2022_02_21_22_PM_4099908/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. subsection 1.2.3-2. Equations with power and
exponential 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
[_Riccati]
\[ \boxed {y^{\prime } x -a \,x^{2 n} {\mathrm e}^{\lambda x} y^{2}-\left (b \,x^{n} {\mathrm e}^{\lambda x}-n \right ) y={\mathrm e}^{\lambda x} c} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {a \,x^{2 n} {\mathrm e}^{\lambda x} y^{2}+x^{n} {\mathrm e}^{\lambda x} b y +{\mathrm e}^{\lambda x} c -n y}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {a \,x^{2 n} {\mathrm e}^{\lambda x} y^{2}}{x}+\frac {{\mathrm e}^{\lambda x} x^{n} b y}{x}+\frac {{\mathrm e}^{\lambda x} c}{x}-\frac {n y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {{\mathrm e}^{\lambda x} c}{x}\), \(f_1(x)=\frac {b \,x^{n} {\mathrm e}^{\lambda x}-n}{x}\) and \(f_2(x)=\frac {{\mathrm e}^{\lambda x} a \,x^{2 n}}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {{\mathrm e}^{\lambda x} a \,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 {\lambda \,{\mathrm e}^{\lambda x} a \,x^{2 n}}{x}-\frac {{\mathrm e}^{\lambda x} a \,x^{2 n}}{x^{2}}+\frac {2 \,{\mathrm e}^{\lambda x} a \,x^{2 n} n}{x^{2}}\\ f_1 f_2 &=\frac {\left (b \,x^{n} {\mathrm e}^{\lambda x}-n \right ) {\mathrm e}^{\lambda x} a \,x^{2 n}}{x^{2}}\\ f_2^2 f_0 &=\frac {{\mathrm e}^{3 \lambda x} a^{2} x^{4 n} c}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {{\mathrm e}^{\lambda x} a \,x^{2 n} u^{\prime \prime }\left (x \right )}{x}-\left (\frac {\lambda \,{\mathrm e}^{\lambda x} a \,x^{2 n}}{x}-\frac {{\mathrm e}^{\lambda x} a \,x^{2 n}}{x^{2}}+\frac {2 \,{\mathrm e}^{\lambda x} a \,x^{2 n} n}{x^{2}}+\frac {\left (b \,x^{n} {\mathrm e}^{\lambda x}-n \right ) {\mathrm e}^{\lambda x} a \,x^{2 n}}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {{\mathrm e}^{3 \lambda x} a^{2} x^{4 n} c u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (c_{1} \operatorname {BesselJ}\left (\frac {\sqrt {3}\, \sqrt {-c a}}{8 b}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {c a \,x^{2 n}}\, x^{-n}}{8 b}\right )+c_{2} \operatorname {BesselY}\left (\frac {\sqrt {3}\, \sqrt {-c a}}{8 b}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {c a \,x^{2 n}}\, x^{-n}}{8 b}\right )\right ) {\mathrm e}^{\frac {\left (\int \left (\frac {b \,x^{n} {\mathrm e}^{\lambda x}}{x}+\lambda +\frac {3 n}{x}\right )d x \right )}{2}} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {{\mathrm e}^{\frac {\left (\int \frac {b \,x^{n} {\mathrm e}^{\lambda x}+\lambda x +3 n}{x}d x \right )}{2}} \left (c_{1} \operatorname {BesselJ}\left (\frac {\sqrt {3}\, \sqrt {-c a}}{8 b}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {c a \,x^{2 n}}\, x^{-n}}{8 b}\right )+c_{2} \operatorname {BesselY}\left (\frac {\sqrt {3}\, \sqrt {-c a}}{8 b}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {c a \,x^{2 n}}\, x^{-n}}{8 b}\right )\right ) \left (b \,x^{n} {\mathrm e}^{\lambda x}+\lambda x +3 n \right )}{2 x} \] Using the above in (1) gives the solution \[ y = -\frac {{\mathrm e}^{\frac {\left (\int \frac {b \,x^{n} {\mathrm e}^{\lambda x}+\lambda x +3 n}{x}d x \right )}{2}} \left (b \,x^{n} {\mathrm e}^{\lambda x}+\lambda x +3 n \right ) {\mathrm e}^{-\lambda x} x^{-2 n} {\mathrm e}^{\int \left (-\frac {b \,x^{-1+n} {\mathrm e}^{\lambda x}}{2}-\frac {\lambda }{2}-\frac {3 n}{2 x}\right )d x}}{2 a} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {x^{-2 n} \left (\left (\lambda x +3 n \right ) {\mathrm e}^{-\lambda x}+b \,x^{n}\right )}{2 a} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {x^{-2 n} \left (\left (\lambda x +3 n \right ) {\mathrm e}^{-\lambda x}+b \,x^{n}\right )}{2 a} \\ \end{align*}
Verification of solutions
\[ y = -\frac {x^{-2 n} \left (\left (\lambda x +3 n \right ) {\mathrm e}^{-\lambda x}+b \,x^{n}\right )}{2 a} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -a \,x^{2 n} {\mathrm e}^{\lambda x} y^{2}-\left (b \,x^{n} {\mathrm e}^{\lambda x}-n \right ) y={\mathrm e}^{\lambda x} c \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a \,x^{2 n} {\mathrm e}^{\lambda x} y^{2}+\left (b \,x^{n} {\mathrm e}^{\lambda x}-n \right ) y+{\mathrm e}^{\lambda x} c}{x} \end {array} \]
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: 86
dsolve(x*diff(y(x),x)=a*x^(2*n)*exp(lambda*x)*y(x)^2+(b*x^n*exp(lambda*x)-n)*y(x)+c*exp(lambda*x),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\left (\tan \left (\frac {\left (x^{n} b \left (\Gamma \left (n , -x \lambda \right )-\Gamma \left (n \right )\right ) \left (-x \lambda \right )^{-n}-c_{1} \right ) \sqrt {4 a \,b^{2} c -b^{4}}}{2 b^{2}}\right ) \sqrt {4 a \,b^{2} c -b^{4}}+b^{2}\right ) x^{-n}}{2 a b} \]
✓ Solution by Mathematica
Time used: 3.62 (sec). Leaf size: 87
DSolve[x*y'[x]==a*x^(2*n)*Exp[\[Lambda]*x]*y[x]^2+(b*x^n*Exp[\[Lambda]*x]-n)*y[x]+c*Exp[\[Lambda]*x],y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{\sqrt {\frac {a x^{2 n}}{c}} y(x)}\frac {1}{K[1]^2-\sqrt {\frac {b^2}{a c}} K[1]+1}dK[1]=-c (\lambda (-x))^{-n} \sqrt {\frac {a x^{2 n}}{c}} \Gamma (n,-x \lambda )+c_1,y(x)\right ] \]