2.3.5 Problem 5

Internal problem ID [13285]
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 : 5
Date solved : Wednesday, December 31, 2025 at 12:56:02 PM
CAS classification : [_Riccati]

2.3.5.1 Solved using first_order_ode_riccati

10.525 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime }&=y^{2}+y b +a \left (\lambda -b \right ) {\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)\\ &= -a \,{\mathrm e}^{\lambda x} b +a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x}+y b +y^{2} \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 \(f_0(x)=-a \,{\mathrm e}^{\lambda x} b +a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x}\), \(f_1(x)=b\) 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 &=b\\ f_2^2 f_0 &=-a \,{\mathrm e}^{\lambda x} b +a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x} \end{align*}

Substituting the above terms back in equation (2) gives

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

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

Taking derivative gives

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

Doing change of constants, the above solution becomes

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

Summary of solutions found

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

2.3.5.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

0.053 (sec)

Entering first order ode riccati guess solver

\begin{align*} y^{\prime }&=y^{2}+y b +a \left (\lambda -b \right ) {\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 \,{\mathrm e}^{\lambda x} b +a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x}\\ f_1(x) & =b\\ 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

\[ y = {\mathrm e}^{\lambda x} a +\frac {{\mathrm e}^{b x +\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }}}{c_1 -\int {\mathrm e}^{b x +\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }}d x} \]

Summary of solutions found

\begin{align*} y &= {\mathrm e}^{\lambda x} a +\frac {{\mathrm e}^{b x +\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }}}{c_1 -\int {\mathrm e}^{b x +\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }}d x} \\ \end{align*}

2.3.5.3 ✓ Maple. Time used: 0.002 (sec). Leaf size: 53

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

\[ y = a \,{\mathrm e}^{\lambda x}-\frac {{\mathrm e}^{\frac {x b \lambda +2 a \,{\mathrm e}^{\lambda x}}{\lambda }}}{\int {\mathrm e}^{\frac {x b \lambda +2 a \,{\mathrm e}^{\lambda x}}{\lambda }}d x +c_1} \]

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 sub-methods: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = b*diff(y(x),x)+(a* 
exp(lambda*x)*b-a*exp(lambda*x)*lambda+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)] 
      <- linear_1 successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

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

2.3.5.4 ✓ Mathematica. Time used: 1.1 (sec). Leaf size: 191

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

\begin{align*} y(x)&\to \frac {-2^{b/\lambda } \left (b-a e^{\lambda x}\right ) \left (\frac {a e^{\lambda x}}{\lambda }\right )^{b/\lambda } L_{-\frac {b}{\lambda }}^{\frac {b}{\lambda }}\left (\frac {2 a e^{x \lambda }}{\lambda }\right )+a e^{\lambda x} \left (2^{\frac {b+\lambda }{\lambda }} \left (\frac {a e^{\lambda x}}{\lambda }\right )^{b/\lambda } L_{-\frac {b+\lambda }{\lambda }}^{\frac {b+\lambda }{\lambda }}\left (\frac {2 a e^{x \lambda }}{\lambda }\right )+c_1\right )}{2^{b/\lambda } \left (\frac {a e^{\lambda x}}{\lambda }\right )^{b/\lambda } L_{-\frac {b}{\lambda }}^{\frac {b}{\lambda }}\left (\frac {2 a e^{x \lambda }}{\lambda }\right )+c_1}\\ y(x)&\to a e^{\lambda x} \end{align*}

2.3.5.5 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
lambda_ = symbols("lambda_") 
y = Function("y") 
ode = Eq(a**2*exp(2*lambda_*x) - a*(-b + lambda_)*exp(lambda_*x) - b*y(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*b*exp(lambda_*x) - a*lambda_*exp(lambda_*x) - b*y(x) - y(x)**2 + Derivative(y(x), x) cannot be solved by the factorable group method