2.4.15 Problem 36

Internal problem ID [13315]
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
Date solved : Wednesday, December 31, 2025 at 01:14:04 PM
CAS classification : [_Riccati]

2.4.15.1 Solved using first_order_ode_riccati

17.483 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime } x&=a \,x^{2 n} {\mathrm e}^{\lambda x} y^{2}+\left ({\mathrm e}^{\lambda x} x^{n} b -n \right ) y+{\mathrm e}^{\lambda x} c \\ \end{align*}

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

\[ y' = \textit {the\_rhs} \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=\frac {{\mathrm e}^{\lambda x} c}{x}\), \(f_1(x)=\frac {b \,x^{n} {\mathrm e}^{\lambda x}}{x}-\frac {n}{x}\) and \(f_2(x)=\frac {x^{2 n} {\mathrm e}^{\lambda x} a}{x}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u a \,x^{2 n} {\mathrm e}^{\lambda x}}{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 {{\mathrm e}^{\lambda x} a \,x^{2 n}}{x^{2}}+\frac {2 a \,x^{2 n} {\mathrm e}^{\lambda x} n}{x^{2}}+\frac {a \,x^{2 n} \lambda \,{\mathrm e}^{\lambda x}}{x}\\ f_1 f_2 &=\frac {\left (\frac {b \,x^{n} {\mathrm e}^{\lambda x}}{x}-\frac {n}{x}\right ) a \,x^{2 n} {\mathrm e}^{\lambda x}}{x}\\ f_2^2 f_0 &=\frac {a^{2} x^{4 n} {\mathrm e}^{3 \lambda x} c}{x^{3}} \end{align*}

Substituting the above terms back in equation (2) gives

\[ \frac {a \,x^{2 n} {\mathrm e}^{\lambda x} u^{\prime \prime }\left (x \right )}{x}-\left (-\frac {{\mathrm e}^{\lambda x} a \,x^{2 n}}{x^{2}}+\frac {2 a \,x^{2 n} {\mathrm e}^{\lambda x} n}{x^{2}}+\frac {a \,x^{2 n} \lambda \,{\mathrm e}^{\lambda x}}{x}+\frac {\left (\frac {b \,x^{n} {\mathrm e}^{\lambda x}}{x}-\frac {n}{x}\right ) a \,x^{2 n} {\mathrm e}^{\lambda x}}{x}\right ) u^{\prime }\left (x \right )+\frac {a^{2} x^{4 n} {\mathrm e}^{3 \lambda x} c u \left (x \right )}{x^{3}} = 0 \]

Unable to solve. Will ask Maple to solve this ode now.

The solution for \(u \left (x \right )\) is

\begin{equation} \tag{3} u \left (x \right ) = {\mathrm e}^{\int \frac {\left (b^{2}+\tan \left (\frac {\sqrt {4 a \,b^{2} c -b^{4}}\, \left (\Gamma \left (n , -\lambda x \right ) x^{n} b \left (-\lambda x \right )^{-n}-\Gamma \left (n \right ) x^{n} b \left (-\lambda x \right )^{-n}-c_1 \right )}{2 b^{2}}\right ) \sqrt {4 a \,b^{2} c -b^{4}}\right ) x^{n -1} {\mathrm e}^{\lambda x}}{2 b}d x} c_2 \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = \frac {\left (b^{2}+\tan \left (\frac {\sqrt {4 a \,b^{2} c -b^{4}}\, \left (\Gamma \left (n , -\lambda x \right ) x^{n} b \left (-\lambda x \right )^{-n}-\Gamma \left (n \right ) x^{n} b \left (-\lambda x \right )^{-n}-c_1 \right )}{2 b^{2}}\right ) \sqrt {4 a \,b^{2} c -b^{4}}\right ) x^{n -1} {\mathrm e}^{\lambda x} {\mathrm e}^{\int \frac {\left (b^{2}+\tan \left (\frac {\sqrt {4 a \,b^{2} c -b^{4}}\, \left (\Gamma \left (n , -\lambda x \right ) x^{n} b \left (-\lambda x \right )^{-n}-\Gamma \left (n \right ) x^{n} b \left (-\lambda x \right )^{-n}-c_1 \right )}{2 b^{2}}\right ) \sqrt {4 a \,b^{2} c -b^{4}}\right ) x^{n -1} {\mathrm e}^{\lambda x}}{2 b}d x} c_2}{2 b} \end{equation}

Substituting equations (3,4) into (1) results in

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{\frac {u a \,x^{2 n} {\mathrm e}^{\lambda x}}{x}} \\ y &= -\frac {\left (b^{2}+\tan \left (\frac {\sqrt {4 a \,b^{2} c -b^{4}}\, \left (\Gamma \left (n , -\lambda x \right ) x^{n} b \left (-\lambda x \right )^{-n}-\Gamma \left (n \right ) x^{n} b \left (-\lambda x \right )^{-n}-c_1 \right )}{2 b^{2}}\right ) \sqrt {4 a \,b^{2} c -b^{4}}\right ) x^{n -1} x \,x^{-2 n}}{2 b a} \\ \end{align*}

Simplifying the above gives

\begin{align*} y &= -\frac {x^{-n} \left (\tan \left (\frac {\sqrt {4 a \,b^{2} c -b^{4}}\, \left (b \,x^{n} \left (\Gamma \left (n , -\lambda x \right )-\Gamma \left (n \right )\right ) \left (-\lambda x \right )^{-n}-c_1 \right )}{2 b^{2}}\right ) \sqrt {4 a \,b^{2} c -b^{4}}+b^{2}\right )}{2 a b} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= -\frac {x^{-n} \left (\tan \left (\frac {\sqrt {4 a \,b^{2} c -b^{4}}\, \left (b \,x^{n} \left (\Gamma \left (n , -\lambda x \right )-\Gamma \left (n \right )\right ) \left (-\lambda x \right )^{-n}-c_1 \right )}{2 b^{2}}\right ) \sqrt {4 a \,b^{2} c -b^{4}}+b^{2}\right )}{2 a b} \\ \end{align*}

2.4.15.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

1.728 (sec)

Entering first order ode riccati guess solver

\begin{align*} y^{\prime } x&=a \,x^{2 n} {\mathrm e}^{\lambda x} y^{2}+\left ({\mathrm e}^{\lambda x} x^{n} b -n \right ) y+{\mathrm e}^{\lambda x} c \\ \end{align*}

This is a Riccati ODE. Comparing the above ODE to solve with the Riccati standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that

\begin{align*} f_0(x) & =\frac {{\mathrm e}^{\lambda x} c}{x}\\ f_1(x) & =\frac {b \,x^{n} {\mathrm e}^{\lambda x}}{x}-\frac {n}{x}\\ f_2(x) &=\frac {x^{2 n} {\mathrm e}^{\lambda x} a}{x} \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = \frac {\left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{-n}}{2 a} \]

Since a particular solution is known, then the general solution is given by

\begin{align*} y &= y_p + \frac { \phi (x) }{ c_1 - \int { \phi (x) f_2 \,dx} } \end{align*}

Where

\begin{align*} \phi (x) &= e^{ \int 2 f_2 y_p + f_1 \,dx } \end{align*}

Evaluating the above gives the general solution as

\[ y = \frac {\left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{-n}}{2 a}+\frac {{\mathrm e}^{\int \left (\frac {\left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{-n} x^{2 n} {\mathrm e}^{\lambda x}}{x}+\frac {b \,x^{n} {\mathrm e}^{\lambda x}}{x}-\frac {n}{x}\right )d x}}{c_1 -\int \frac {{\mathrm e}^{\int \left (\frac {\left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{-n} x^{2 n} {\mathrm e}^{\lambda x}}{x}+\frac {b \,x^{n} {\mathrm e}^{\lambda x}}{x}-\frac {n}{x}\right )d x} a \,x^{2 n} {\mathrm e}^{\lambda x}}{x}d x} \]

Summary of solutions found

\begin{align*} y &= \frac {\left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{-n}}{2 a}+\frac {{\mathrm e}^{\int \left (\frac {\left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{-n} x^{2 n} {\mathrm e}^{\lambda x}}{x}+\frac {b \,x^{n} {\mathrm e}^{\lambda x}}{x}-\frac {n}{x}\right )d x}}{c_1 -\int \frac {{\mathrm e}^{\int \left (\frac {\left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{-n} x^{2 n} {\mathrm e}^{\lambda x}}{x}+\frac {b \,x^{n} {\mathrm e}^{\lambda x}}{x}-\frac {n}{x}\right )d x} a \,x^{2 n} {\mathrm e}^{\lambda x}}{x}d x} \\ \end{align*}

2.4.15.3 ✓ Maple. Time used: 0.009 (sec). Leaf size: 86

ode:=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); 
dsolve(ode,y(x), singsol=all);
 

\[ y = -\frac {\left (\tan \left (\frac {\sqrt {4 b^{2} a c -b^{4}}\, \left (b \,x^{n} \left (\Gamma \left (n , -\lambda x \right )-\Gamma \left (n \right )\right ) \left (-\lambda x \right )^{-n}-c_1 \right )}{2 b^{2}}\right ) \sqrt {4 b^{2} a c -b^{4}}+b^{2}\right ) x^{-n}}{2 a b} \]

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
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y \left (x \right )\right )=a \,x^{26630} {\mathrm e}^{\lambda x} y \left (x \right )^{2}+\left (b \,x^{13315} {\mathrm e}^{\lambda x}-13315\right ) y \left (x \right )+c \,{\mathrm e}^{\lambda x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {a \,x^{26630} {\mathrm e}^{\lambda x} y \left (x \right )^{2}+\left (b \,x^{13315} {\mathrm e}^{\lambda x}-13315\right ) y \left (x \right )+c \,{\mathrm e}^{\lambda x}}{x} \end {array} \]

2.4.15.4 ✓ Mathematica. Time used: 1.315 (sec). Leaf size: 87

ode=x*D[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]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\int _1^{\sqrt {\frac {a x^{2 n}}{c}} y(x)}\frac {1}{K[1]^2-\sqrt {\frac {b^2}{a c}} K[1]+1}dK[1]=-c (\lambda (-x))^{-n} \sqrt {\frac {a x^{2 n}}{c}} \Gamma (n,-x \lambda )+c_1,y(x)\right ] \]

2.4.15.5 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
c = symbols("c") 
lambda_ = symbols("lambda_") 
n = symbols("n") 
y = Function("y") 
ode = Eq(-a*x**(2*n)*y(x)**2*exp(lambda_*x) - c*exp(lambda_*x) + x*Derivative(y(x), x) - (b*x**n*exp(lambda_*x) - n)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE Derivative(y(x), x) - (a*x**(2*n)*y(x)**2*exp(lambda_*x) + b*x**n*y(x)*exp(lambda_*x) + c*exp(lambda_*x) - n*y(x))/x cannot be solved by the factorable group method