2.19.12 Problem 12

Internal problem ID [13460]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.8-1. Equations containing arbitrary functions (but not containing their derivatives).
Problem number : 12
Date solved : Wednesday, December 31, 2025 at 09:41:26 PM
CAS classification : [_Riccati]

2.19.12.1 Solved using first_order_ode_riccati

11.598 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= a \,{\mathrm e}^{\lambda x} y^{2}+a \,{\mathrm e}^{\lambda x} f \left (x \right ) y+\lambda f \left (x \right ) \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 \(f_0(x)=\lambda f \left (x \right )\), \(f_1(x)=a \,{\mathrm e}^{\lambda x} f \left (x \right )\) and \(f_2(x)={\mathrm e}^{\lambda x} a\). Let

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

Substituting the above terms back in equation (2) gives

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

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}^{\lambda x}+c_2 \int \frac {{\mathrm e}^{-\lambda x +a \int f \left (x \right ) {\mathrm e}^{\lambda x}d x}}{\lambda }d x {\mathrm e}^{\lambda x} \end{equation}

Taking derivative gives

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

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

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

Doing change of constants, the above solution becomes

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

Summary of solutions found

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

2.19.12.2 ✓ Maple. Time used: 0.003 (sec). Leaf size: 86

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

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

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*exp(lambda*x)*f(x 
)+lambda)*diff(y(x),x)-a*exp(lambda*x)*lambda*f(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 symmetry of the form [xi=0, eta=F(x)] 
         <- linear_1 successful 
         Change of variables used: 
            [x = ln(t)/lambda] 
         Linear ODE actually solved: 
            a*f(ln(t)/lambda)*u(t)-a*t*f(ln(t)/lambda)*diff(u(t),t)+t*lambda*di\ 
ff(diff(u(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

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

2.19.12.3 ✓ Mathematica. Time used: 1.496 (sec). Leaf size: 166

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

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

2.19.12.4 ✗ Sympy

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

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