2.3.18 Problem 19

Internal problem ID [13298]
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 : 19
Date solved : Wednesday, December 31, 2025 at 01:03:00 PM
CAS classification : [[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]

2.3.18.1 Solved using first_order_ode_riccati

3.970 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

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

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

Substituting the above terms back in equation (2) gives

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

Entering second order change of variable on \(y\) method 1 solverIn normal form the given ode is written as

\begin{align*} \frac {d^{2}u}{d x^{2}}+p \left (x \right ) \left (\frac {d u}{d x}\right )+q \left (x \right ) u&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=\left (2 b \,{\mathrm e}^{x \left (\lambda +2 \mu \right )}-\mu \,{\mathrm e}^{\mu x}\right ) {\mathrm e}^{-\mu x}\\ q \left (x \right )&=b \left ({\mathrm e}^{x \left (2 \lambda +3 \mu \right )} b +{\mathrm e}^{x \left (\lambda +2 \mu \right )} \lambda \right ) {\mathrm e}^{-\mu x} \end{align*}

Calculating the Liouville ode invariant \(Q\) given by

\begin{align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= b \left ({\mathrm e}^{x \left (2 \lambda +3 \mu \right )} b +{\mathrm e}^{x \left (\lambda +2 \mu \right )} \lambda \right ) {\mathrm e}^{-\mu x} - \frac {\left (\left (2 b \,{\mathrm e}^{x \left (\lambda +2 \mu \right )}-\mu \,{\mathrm e}^{\mu x}\right ) {\mathrm e}^{-\mu x}\right )'}{2}- \frac {\left (\left (2 b \,{\mathrm e}^{x \left (\lambda +2 \mu \right )}-\mu \,{\mathrm e}^{\mu x}\right ) {\mathrm e}^{-\mu x}\right )^2}{4} \\ &= b \left ({\mathrm e}^{x \left (2 \lambda +3 \mu \right )} b +{\mathrm e}^{x \left (\lambda +2 \mu \right )} \lambda \right ) {\mathrm e}^{-\mu x} - \frac {\left (\left (2 b \left (\lambda +2 \mu \right ) {\mathrm e}^{x \left (\lambda +2 \mu \right )}-\mu ^{2} {\mathrm e}^{\mu x}\right ) {\mathrm e}^{-\mu x}-\left (2 b \,{\mathrm e}^{x \left (\lambda +2 \mu \right )}-\mu \,{\mathrm e}^{\mu x}\right ) {\mathrm e}^{-\mu x} \mu \right )}{2}- \frac {\left (\left (2 b \,{\mathrm e}^{x \left (\lambda +2 \mu \right )}-\mu \,{\mathrm e}^{\mu x}\right )^{2} {\mathrm e}^{-2 \mu x}\right )}{4} \\ &= b \left ({\mathrm e}^{x \left (2 \lambda +3 \mu \right )} b +{\mathrm e}^{x \left (\lambda +2 \mu \right )} \lambda \right ) {\mathrm e}^{-\mu x} - \left (\frac {\left (2 b \left (\lambda +2 \mu \right ) {\mathrm e}^{x \left (\lambda +2 \mu \right )}-\mu ^{2} {\mathrm e}^{\mu x}\right ) {\mathrm e}^{-\mu x}}{2}-\frac {\left (2 b \,{\mathrm e}^{x \left (\lambda +2 \mu \right )}-\mu \,{\mathrm e}^{\mu x}\right ) {\mathrm e}^{-\mu x} \mu }{2}\right )-\frac {\left (2 b \,{\mathrm e}^{x \left (\lambda +2 \mu \right )}-\mu \,{\mathrm e}^{\mu x}\right )^{2} {\mathrm e}^{-2 \mu x}}{4}\\ &= -\frac {\mu ^{2}}{4} \end{align*}

Since the Liouville ode invariant does not depend on the independent variable \(x\) then the transformation

\begin{align*} u = v \left (x \right ) z \left (x \right )\tag {3} \end{align*}

is used to change the original ode to a constant coefficients ode in \(v\). In (3) the term \(z \left (x \right )\) is given by

\begin{align*} z \left (x \right )&={\mathrm e}^{-\int \frac {p \left (x \right )}{2}d x}\\ &= e^{-\int \frac {\left (2 b \,{\mathrm e}^{x \left (\lambda +2 \mu \right )}-\mu \,{\mathrm e}^{\mu x}\right ) {\mathrm e}^{-\mu x}}{2} }\\ &= {\mathrm e}^{\frac {\mu x}{2}-\frac {b \,{\mathrm e}^{x \left (\lambda +\mu \right )}}{\lambda +\mu }}\tag {5} \end{align*}

Hence (3) becomes

\begin{align*} u = v \left (x \right ) {\mathrm e}^{\frac {\mu x}{2}-\frac {b \,{\mathrm e}^{x \left (\lambda +\mu \right )}}{\lambda +\mu }}\tag {4} \end{align*}

Applying this change of variable to the original ode results in

\begin{align*} -\frac {\left (v \left (x \right ) \mu ^{2}-4 \frac {d^{2}}{d x^{2}}v \left (x \right )\right ) {\mathrm e}^{\frac {-2 b \,{\mathrm e}^{x \left (\lambda +\mu \right )}+3 \mu x \left (\lambda +\mu \right )}{2 \lambda +2 \mu }}}{4} = 0 \end{align*}

Which is now solved for \(v \left (x \right )\).

The above ode simplifies to

\begin{align*} \frac {d^{2}}{d x^{2}}v \left (x \right )-\frac {v \left (x \right ) \mu ^{2}}{4} = 0 \end{align*}

Entering second order linear constant coefficient ode solver

This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

\[ A v''(x) + B v'(x) + C v(x) = 0 \]

Where in the above \(A=1, B=0, C=-\frac {\mu ^{2}}{4}\). Let the solution be \(v \left (x \right )=e^{\lambda x}\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{x \lambda }-\frac {\mu ^{2} {\mathrm e}^{x \lambda }}{4} = 0 \tag {1} \]

Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda x}\) gives

\[ \lambda ^{2}-\frac {\mu ^{2}}{4} = 0 \tag {2} \]

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]

Substituting \(A=1, B=0, C=-\frac {\mu ^{2}}{4}\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (-\frac {\mu ^{2}}{4}\right )}\\ &= \pm \frac {\sqrt {\mu ^{2}}}{2} \end{align*}

Hence

\begin{align*} \lambda _1 &= + \frac {\sqrt {\mu ^{2}}}{2} \\ \lambda _2 &= - \frac {\sqrt {\mu ^{2}}}{2} \\ \end{align*}

Which simplifies to

\begin{align*} \lambda _1 &= \frac {\sqrt {\mu ^{2}}}{2} \\ \lambda _2 &= -\frac {\sqrt {\mu ^{2}}}{2} \\ \end{align*}

Since roots are distinct, then the solution is

\begin{align*} v \left (x \right ) &= c_1 e^{\lambda _1 x} + c_2 e^{\lambda _2 x} \\ v \left (x \right ) &= c_1 e^{\left (\frac {\sqrt {\mu ^{2}}}{2}\right )x} +c_2 e^{\left (-\frac {\sqrt {\mu ^{2}}}{2}\right )x} \\ \end{align*}

Or

\[ v \left (x \right ) =c_1 \,{\mathrm e}^{\frac {\sqrt {\mu ^{2}}\, x}{2}}+c_2 \,{\mathrm e}^{-\frac {\sqrt {\mu ^{2}}\, x}{2}} \]

Now that \(v \left (x \right )\) is known, then

\begin{align*} u&= v \left (x \right ) z \left (x \right )\\ &= \left (c_1 \,{\mathrm e}^{\frac {\sqrt {\mu ^{2}}\, x}{2}}+c_2 \,{\mathrm e}^{-\frac {\sqrt {\mu ^{2}}\, x}{2}}\right ) \left (z \left (x \right )\right )\tag {7} \end{align*}

But from (5)

\begin{align*} z \left (x \right )&= {\mathrm e}^{\frac {\mu x}{2}-\frac {b \,{\mathrm e}^{x \left (\lambda +\mu \right )}}{\lambda +\mu }} \end{align*}

Hence (7) becomes

\begin{align*} u = \left (c_1 \,{\mathrm e}^{\frac {\sqrt {\mu ^{2}}\, x}{2}}+c_2 \,{\mathrm e}^{-\frac {\sqrt {\mu ^{2}}\, x}{2}}\right ) {\mathrm e}^{\frac {\mu x}{2}-\frac {b \,{\mathrm e}^{x \left (\lambda +\mu \right )}}{\lambda +\mu }} \end{align*}

Taking derivative gives

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

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

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

Doing change of constants, the above solution becomes

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

Summary of solutions found

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

2.3.18.2 ✓ Maple. Time used: 0.003 (sec). Leaf size: 38

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

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

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: 
   <- Riccati particular polynomial solution successful
 

Maple step by step

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

2.3.18.3 ✓ Mathematica. Time used: 0.597 (sec). Leaf size: 40

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

\begin{align*} y(x)&\to b e^{\lambda x}+\frac {\mu }{-e^{\mu x}+c_1 \mu }\\ y(x)&\to b e^{\lambda x} \end{align*}

2.3.18.4 ✗ Sympy

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

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