2.3.20 Problem 21

Internal problem ID [13300]
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. Equations Containing Exponential Functions
Problem number : 21
Date solved : Wednesday, December 31, 2025 at 01:04:04 PM
CAS classification : [_Riccati]

2.3.20.1 Solved using first_order_ode_riccati

7.763 (sec)

Entering first order ode riccati solver

\begin{align*} \left ({\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c \right ) \left (y^{\prime }-y^{2}\right )+a \,\lambda ^{2} {\mathrm e}^{\lambda x}+\mu ^{2} {\mathrm e}^{\mu x} b&=0 \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {a \,{\mathrm e}^{\lambda x} y^{2}-a \,\lambda ^{2} {\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x} y^{2}-\mu ^{2} {\mathrm e}^{\mu x} b +c y^{2}}{{\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c} \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 \(f_0(x)=-\frac {a \,\lambda ^{2} {\mathrm e}^{\lambda x}}{{\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c}-\frac {\mu ^{2} {\mathrm e}^{\mu x} b}{{\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c}\), \(f_1(x)=0\) and \(f_2(x)=1\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u} \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' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=-\frac {a \,\lambda ^{2} {\mathrm e}^{\lambda x}}{{\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c}-\frac {\mu ^{2} {\mathrm e}^{\mu x} b}{{\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c} \end{align*}

Substituting the above terms back in equation (2) gives

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

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 ) = \left (\int \frac {1}{\left ({\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c \right )^{2}}d x c_1 +c_2 \right ) \left ({\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c \right ) \end{equation}

Taking derivative gives

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

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{u} \\ y &= -\frac {\frac {c_1}{{\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c}+\left (\int \frac {1}{\left ({\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c \right )^{2}}d x c_1 +c_2 \right ) \left (a \lambda \,{\mathrm e}^{\lambda x}+\mu \,{\mathrm e}^{\mu x} b \right )}{\left (\int \frac {1}{\left ({\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c \right )^{2}}d x c_1 +c_2 \right ) \left ({\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c \right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\frac {1}{{\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c}+\left (\int \frac {1}{\left ({\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c \right )^{2}}d x +c_3 \right ) \left (a \lambda \,{\mathrm e}^{\lambda x}+\mu \,{\mathrm e}^{\mu x} b \right )}{\left (\int \frac {1}{\left ({\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c \right )^{2}}d x +c_3 \right ) \left ({\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c \right )} \]

Summary of solutions found

\begin{align*} y &= -\frac {\frac {1}{{\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c}+\left (\int \frac {1}{\left ({\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c \right )^{2}}d x +c_3 \right ) \left (a \lambda \,{\mathrm e}^{\lambda x}+\mu \,{\mathrm e}^{\mu x} b \right )}{\left (\int \frac {1}{\left ({\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c \right )^{2}}d x +c_3 \right ) \left ({\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c \right )} \\ \end{align*}

2.3.20.2 ✓ Maple. Time used: 0.005 (sec). Leaf size: 176

ode:=(a*exp(lambda*x)+b*exp(mu*x)+c)*(diff(y(x),x)-y(x)^2)+a*lambda^2*exp(lambda*x)+b*mu^2*exp(mu*x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = \frac {\left (-a b \left (\lambda +\mu \right ) {\mathrm e}^{x \left (\lambda +\mu \right )}-a^{2} \lambda \,{\mathrm e}^{2 \lambda x}-b^{2} \mu \,{\mathrm e}^{2 \mu x}-c \left (a \lambda \,{\mathrm e}^{\lambda x}+b \mu \,{\mathrm e}^{\mu x}\right )\right ) \int \frac {1}{\left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right )^{2}}d x -a b c_1 \left (\lambda +\mu \right ) {\mathrm e}^{x \left (\lambda +\mu \right )}-a^{2} \lambda \,{\mathrm e}^{2 \lambda x} c_1 -{\mathrm e}^{\lambda x} c_1 a c \lambda -b^{2} \mu \,{\mathrm e}^{2 \mu x} c_1 -{\mathrm e}^{\mu x} c_1 b c \mu -1}{\left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right )^{2} \left (c_1 +\int \frac {1}{\left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right )^{2}}d x \right )} \]

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 
differential order: 1; looking for linear symmetries 
trying exact 
Looking for potential symmetries 
trying Riccati 
trying Riccati sub-methods: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = (a*lambda^2*exp( 
lambda*x)+b*mu^2*exp(mu*x))/(a*exp(lambda*x)+b*exp(mu*x)+c)*y(x), y(x) 
      *** Sublevel 2 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying a symmetry of the form [xi=0, eta=F(x)] 
      <- linear_1 successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right ) \left (\frac {d}{d x}y \left (x \right )-y \left (x \right )^{2}\right )+a \,\lambda ^{2} {\mathrm e}^{\lambda x}+b \,\mu ^{2} {\mathrm e}^{\mu x}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {{\mathrm e}^{\lambda x} y \left (x \right )^{2} a -a \,\lambda ^{2} {\mathrm e}^{\lambda x}+{\mathrm e}^{\mu x} y \left (x \right )^{2} b -b \,\mu ^{2} {\mathrm e}^{\mu x}+y \left (x \right )^{2} c}{a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c} \end {array} \]

2.3.20.3 ✓ Mathematica. Time used: 5.528 (sec). Leaf size: 393

ode=(a*Exp[\[Lambda]*x]+b*Exp[\[Mu]*x]+c)*(D[y[x],x]-y[x]^2)+a*\[Lambda]^2*Exp[\[Lambda]*x]+b*\[Mu]^2*Exp[\[Mu]*x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\int _1^x-\frac {-a e^{\lambda K[1]} \lambda ^2-b e^{\mu K[1]} \mu ^2+a e^{\lambda K[1]} y(x)^2+b e^{\mu K[1]} y(x)^2+c y(x)^2}{\left (e^{\lambda K[1]} a+b e^{\mu K[1]}+c\right ) \left (a e^{\lambda K[1]} \lambda +b e^{\mu K[1]} \mu +a e^{\lambda K[1]} y(x)+b e^{\mu K[1]} y(x)+c y(x)\right )^2}dK[1]+\int _1^{y(x)}\left (\frac {1}{\left (a e^{x \lambda } \lambda +b e^{x \mu } \mu +a e^{x \lambda } K[2]+b e^{x \mu } K[2]+c K[2]\right )^2}-\int _1^x\left (\frac {2 \left (-a e^{\lambda K[1]} \lambda ^2-b e^{\mu K[1]} \mu ^2+a e^{\lambda K[1]} K[2]^2+b e^{\mu K[1]} K[2]^2+c K[2]^2\right )}{\left (a e^{\lambda K[1]} \lambda +b e^{\mu K[1]} \mu +a e^{\lambda K[1]} K[2]+b e^{\mu K[1]} K[2]+c K[2]\right )^3}-\frac {2 a e^{\lambda K[1]} K[2]+2 b e^{\mu K[1]} K[2]+2 c K[2]}{\left (e^{\lambda K[1]} a+b e^{\mu K[1]}+c\right ) \left (a e^{\lambda K[1]} \lambda +b e^{\mu K[1]} \mu +a e^{\lambda K[1]} K[2]+b e^{\mu K[1]} K[2]+c K[2]\right )^2}\right )dK[1]\right )dK[2]=c_1,y(x)\right ] \]

2.3.20.4 ✗ Sympy

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

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