2.3.2 Problem 2

Internal problem ID [13282]
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 : 2
Date solved : Wednesday, December 31, 2025 at 12:54:36 PM
CAS classification : [_Riccati]

2.3.2.1 Solved using first_order_ode_riccati

9.808 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime }&=y^{2}+a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x} \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= y^{2}+a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda 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 \(f_0(x)=a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x}\), \(f_1(x)=0\) 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 &=a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x} \end{align*}

Substituting the above terms back in equation (2) gives

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

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 ) = c_1 \,{\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }}+c_2 \,{\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }} \operatorname {Ei}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = -c_1 \,{\mathrm e}^{\lambda x} a \,{\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }}-c_2 \,{\mathrm e}^{\lambda x} a \,{\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }} \operatorname {Ei}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right )-c_2 \,{\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }} {\mathrm e}^{\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }} \lambda \end{equation}

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{u} \\ y &= -\frac {-c_1 \,{\mathrm e}^{\lambda x} a \,{\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }}-c_2 \,{\mathrm e}^{\lambda x} a \,{\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }} \operatorname {Ei}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right )-c_2 \,{\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }} {\mathrm e}^{\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }} \lambda }{c_1 \,{\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }}+c_2 \,{\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }} \operatorname {Ei}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {-{\mathrm e}^{\lambda x} a \,{\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }}-c_3 \,{\mathrm e}^{\lambda x} a \,{\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }} \operatorname {Ei}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right )-c_3 \,{\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }} {\mathrm e}^{\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }} \lambda }{{\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }}+c_3 \,{\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }} \operatorname {Ei}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right )} \]

Simplifying the above gives

\begin{align*} y &= \frac {c_3 \,{\mathrm e}^{\lambda x} a \,\operatorname {Ei}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right )+c_3 \lambda \,{\mathrm e}^{\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }}+{\mathrm e}^{\lambda x} a}{c_3 \,\operatorname {Ei}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right )+1} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {c_3 \,{\mathrm e}^{\lambda x} a \,\operatorname {Ei}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right )+c_3 \lambda \,{\mathrm e}^{\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }}+{\mathrm e}^{\lambda x} a}{c_3 \,\operatorname {Ei}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right )+1} \\ \end{align*}

2.3.2.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

0.152 (sec)

Entering first order ode riccati guess solver

\begin{align*} y^{\prime }&=y^{2}+a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x} \\ \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) & =a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x}\\ f_1(x) & =0\\ f_2(x) &=1 \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = {\mathrm e}^{\lambda x} a \]

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

\[ y = {\mathrm e}^{\lambda x} a +\frac {{\mathrm e}^{\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }}}{c_1 +\frac {\operatorname {Ei}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right )}{\lambda }} \]

Summary of solutions found

\begin{align*} y &= {\mathrm e}^{\lambda x} a +\frac {{\mathrm e}^{\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }}}{c_1 +\frac {\operatorname {Ei}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right )}{\lambda }} \\ \end{align*}

2.3.2.3 ✓ Maple. Time used: 0.002 (sec). Leaf size: 63

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

\[ y = \frac {{\mathrm e}^{\lambda x} \operatorname {Ei}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) c_1 a +{\mathrm e}^{\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }} c_1 \lambda +{\mathrm e}^{\lambda x} a}{\operatorname {Ei}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) c_1 +1} \]

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 Special 
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*exp( 
lambda*x)+a^2*exp(2*lambda*x))*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)] 
      checking if the LODE is missing y 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a \ 
power @ Moebius 
      -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int\ 
(r(x), dx)) * 2F1([a1, a2], [b1], f) 
      -> Trying changes of variables to rationalize or make the ODE simpler 
         trying a quadrature 
         checking if the LODE has constant coefficients 
         checking if the LODE is of Euler type 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> Trying a Liouvillian solution using Kovacics algorithm 
            A Liouvillian solution exists 
            Reducible group (found an exponential solution) 
            Group is reducible, not completely reducible 
         <- Kovacics algorithm successful 
         Change of variables used: 
            [x = ln(t)/lambda] 
         Linear ODE actually solved: 
            (-a^2*t+a*lambda)*u(t)+lambda^2*diff(u(t),t)+lambda^2*t*diff(diff(u\ 
(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

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

2.3.2.4 ✓ Mathematica. Time used: 1.176 (sec). Leaf size: 107

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

\begin{align*} y(x)&\to \frac {a e^{\lambda x} \int _1^{e^{x \lambda }}\frac {e^{\frac {2 a K[1]}{\lambda }}}{K[1]}dK[1]+\lambda \left (-e^{\frac {2 a e^{\lambda x}}{\lambda }}\right )+a c_1 e^{\lambda x}}{\int _1^{e^{x \lambda }}\frac {e^{\frac {2 a K[1]}{\lambda }}}{K[1]}dK[1]+c_1}\\ y(x)&\to a e^{\lambda x} \end{align*}

2.3.2.5 ✗ Sympy

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

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