2.4.13 Problem 34

Internal problem ID [13313]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.3-2. Equations with power and exponential functions
Problem number : 34
Date solved : Sunday, January 18, 2026 at 07:13:25 PM
CAS classification : [[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]

2.4.13.1 Solved using first_order_ode_riccati

0.750 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime }&=a \,{\mathrm e}^{\lambda x} \left (y-b \,x^{n}-c \right )^{2}+b n \,x^{n -1} \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= a \,b^{2} {\mathrm e}^{\lambda x} x^{2 n}+2 \,{\mathrm e}^{\lambda x} x^{n} a b c -2 a b \,x^{n} {\mathrm e}^{\lambda x} y+{\mathrm e}^{\lambda x} a \,c^{2}-2 \,{\mathrm e}^{\lambda x} y a c +a \,{\mathrm e}^{\lambda x} y^{2}+b n \,x^{n -1} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = a \,b^{2} {\mathrm e}^{\lambda x} x^{2 n}+2 \,{\mathrm e}^{\lambda x} x^{n} a b c -2 a b \,x^{n} {\mathrm e}^{\lambda x} y+{\mathrm e}^{\lambda x} a \,c^{2}-2 \,{\mathrm e}^{\lambda x} y a c +a \,{\mathrm e}^{\lambda x} y^{2}+\frac {b \,x^{n} n}{x} \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=a \,b^{2} {\mathrm e}^{\lambda x} x^{2 n}+2 \,{\mathrm e}^{\lambda x} x^{n} a b c +{\mathrm e}^{\lambda x} a \,c^{2}+\frac {b \,x^{n} n}{x}\), \(f_1(x)=-2 x^{n} {\mathrm e}^{\lambda x} a b -2 \,{\mathrm e}^{\lambda x} a c\) 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 &=\left (-2 x^{n} {\mathrm e}^{\lambda x} a b -2 \,{\mathrm e}^{\lambda x} a c \right ) {\mathrm e}^{\lambda x} a\\ f_2^2 f_0 &={\mathrm e}^{2 \lambda x} a^{2} \left (a \,b^{2} {\mathrm e}^{\lambda x} x^{2 n}+2 \,{\mathrm e}^{\lambda x} x^{n} a b c +{\mathrm e}^{\lambda x} a \,c^{2}+\frac {b \,x^{n} n}{x}\right ) \end{align*}

Substituting the above terms back in equation (2) gives

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

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 )&=2 a \left (b \,x^{n}+c \right ) {\mathrm e}^{\lambda x}-\lambda \\ q \left (x \right )&=\left (b n \,x^{n -1}+a \,{\mathrm e}^{\lambda x} \left (b^{2} x^{2 n}+2 c \,x^{n} b +c^{2}\right )\right ) {\mathrm e}^{2 \lambda x} a \,{\mathrm e}^{-\lambda x} \end{align*}

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

\begin{align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= \left (b n \,x^{n -1}+a \,{\mathrm e}^{\lambda x} \left (b^{2} x^{2 n}+2 c \,x^{n} b +c^{2}\right )\right ) {\mathrm e}^{2 \lambda x} a \,{\mathrm e}^{-\lambda x} - \frac {\left (2 a \left (b \,x^{n}+c \right ) {\mathrm e}^{\lambda x}-\lambda \right )'}{2}- \frac {\left (2 a \left (b \,x^{n}+c \right ) {\mathrm e}^{\lambda x}-\lambda \right )^2}{4} \\ &= \left (b n \,x^{n -1}+a \,{\mathrm e}^{\lambda x} \left (b^{2} x^{2 n}+2 c \,x^{n} b +c^{2}\right )\right ) {\mathrm e}^{2 \lambda x} a \,{\mathrm e}^{-\lambda x} - \frac {\left (\frac {2 \,{\mathrm e}^{\lambda x} a b \,x^{n} n}{x}+2 a \left (b \,x^{n}+c \right ) \lambda \,{\mathrm e}^{\lambda x}\right )}{2}- \frac {\left ({\left (2 a \left (b \,x^{n}+c \right ) {\mathrm e}^{\lambda x}-\lambda \right )}^{2}\right )}{4} \\ &= \left (b n \,x^{n -1}+a \,{\mathrm e}^{\lambda x} \left (b^{2} x^{2 n}+2 c \,x^{n} b +c^{2}\right )\right ) {\mathrm e}^{2 \lambda x} a \,{\mathrm e}^{-\lambda x} - \left (\frac {{\mathrm e}^{\lambda x} a b \,x^{n} n}{x}+a \left (b \,x^{n}+c \right ) \lambda \,{\mathrm e}^{\lambda x}\right )-\frac {{\left (2 a \left (b \,x^{n}+c \right ) {\mathrm e}^{\lambda x}-\lambda \right )}^{2}}{4}\\ &= -\frac {\lambda ^{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 {2 a \left (b \,x^{n}+c \right ) {\mathrm e}^{\lambda x}-\lambda }{2} }\\ &= {\mathrm e}^{-\frac {\int \left (2 a \left (b \,x^{n}+c \right ) {\mathrm e}^{\lambda x}-\lambda \right )d x}{2}}\tag {5} \end{align*}

Hence (3) becomes

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

Applying this change of variable to the original ode results in

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

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

The above ode simplifies to

\begin{align*} \frac {a \left (-v \left (x \right ) \lambda ^{2}+4 \frac {d^{2}}{d x^{2}}v \left (x \right )\right )}{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 \(A=a, B=0, C=-\frac {a \,\lambda ^{2}}{4}\). Let the solution be \(v \left (x \right )=e^{\lambda x}\). Substituting this into the ODE gives

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

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

\[ a \,\lambda ^{2}-\frac {1}{4} a \,\lambda ^{2} = 0 \tag {2} \]

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

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

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

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

Hence

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

Which simplifies to

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

Since the roots are distinct, 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 {a^{2} \lambda ^{2}}}{2 a}\right )x} +c_2 e^{\left (-\frac {\sqrt {a^{2} \lambda ^{2}}}{2 a}\right )x} \\ \end{align*}

Or

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

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 {a^{2} \lambda ^{2}}\, x}{2 a}}+c_2 \,{\mathrm e}^{-\frac {\sqrt {a^{2} \lambda ^{2}}\, x}{2 a}}\right ) \left (z \left (x \right )\right )\tag {7} \end{align*}

But from (5)

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

Hence (7) becomes

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

Taking derivative gives

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

Doing change of constants, the above solution becomes

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

Summary of solutions found

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

2.4.13.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

0.125 (sec)

Entering first order ode riccati guess solver

\begin{align*} y^{\prime }&=a \,{\mathrm e}^{\lambda x} \left (y-b \,x^{n}-c \right )^{2}+b n \,x^{n -1} \\ \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 \,b^{2} {\mathrm e}^{\lambda x} x^{2 n}+2 \,{\mathrm e}^{\lambda x} x^{n} a b c +{\mathrm e}^{\lambda x} a \,c^{2}+\frac {b \,x^{n} n}{x}\\ f_1(x) & =-2 x^{n} {\mathrm e}^{\lambda x} a b -2 \,{\mathrm e}^{\lambda x} a c\\ f_2(x) &={\mathrm e}^{\lambda x} a \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = \left (b \,{\mathrm e}^{n \ln \left (x \right )}+c \right ) {\mathrm e}^{-\frac {\ln \left (x \right ) b n \,{\mathrm e}^{n \ln \left (x \right )}}{b \,{\mathrm e}^{n \ln \left (x \right )}+c}} x^{\frac {b n \,{\mathrm e}^{n \ln \left (x \right )}}{b \,{\mathrm e}^{n \ln \left (x \right )}+c}} \]

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

\[ y = b \,x^{n}+c -\frac {\lambda }{{\mathrm e}^{\lambda x} a -c_1 \lambda } \]

Summary of solutions found

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

2.4.13.3 ✓ Maple. Time used: 0.002 (sec). Leaf size: 38

ode:=diff(y(x),x) = a*exp(lambda*x)*(y(x)-b*x^n-c)^2+b*n*x^(n-1); 
dsolve(ode,y(x), singsol=all);
 

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

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

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=a \,{\mathrm e}^{\lambda x} \left (y \left (x \right )-b \,x^{13313}-c \right )^{2}+13313 b \,x^{13312} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=a \,{\mathrm e}^{\lambda x} \left (y \left (x \right )-b \,x^{13313}-c \right )^{2}+13313 b \,x^{13312} \end {array} \]

2.4.13.4 ✓ Mathematica. Time used: 1.262 (sec). Leaf size: 40

ode=D[y[x],x]==a*Exp[\[Lambda]*x]*(y[x]-b*x^n-c)^2+b*n*x^(n-1); 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

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

2.4.13.5 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
c = symbols("c") 
lambda_ = symbols("lambda_") 
n = symbols("n") 
y = Function("y") 
ode = Eq(-a*(-b*x**n - c + y(x))**2*exp(lambda_*x) - b*n*x**(n - 1) + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

Timed Out
 

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0