2.35.37 Problem 37

Internal problem ID [13961]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.3-1. Equations with exponential functions
Problem number : 37
Date solved : Sunday, January 18, 2026 at 09:37:44 PM
CAS classification : [[_2nd_order, _exact, _linear, _homogeneous]]

2.35.37.1 second order linear exact ode

0.317 (sec)

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

Entering second order linear exact ode solverAn ode of the form

\begin{align*} p \left (x \right ) y^{\prime \prime }+q \left (x \right ) y^{\prime }+r \left (x \right ) y&=s \left (x \right ) \end{align*}

is exact if

\begin{align*} p''(x) - q'(x) + r(x) &= 0 \tag {1} \end{align*}

For the given ode the above values are

\begin{align*} p(x) &= {\mathrm e}^{\lambda x} a +b\\ q(x) &= 0\\ r(x) &= -a \,\lambda ^{2} {\mathrm e}^{\lambda x}\\ s(x) &= 0 \end{align*}

Hence

\begin{align*} p''(x) &= a \,\lambda ^{2} {\mathrm e}^{\lambda x}\\ q'(x) &= 0 \end{align*}

Therefore (1) becomes

\begin{align*} a \,\lambda ^{2} {\mathrm e}^{\lambda x}- \left (0\right ) + \left (-a \,\lambda ^{2} {\mathrm e}^{\lambda x}\right )&=0 \end{align*}

This shows the ode is exact. Since the ode is exact, it can be written as

\begin{align*} \left (p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y\right )' &= s(x) \end{align*}

Integrating the above gives

\begin{align*} p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y&=\int {s \left (x \right )\, dx} \end{align*}

Substituting the values of \(p,q,r,s\) into the above results in

\begin{align*} \left ({\mathrm e}^{\lambda x} a +b \right ) y^{\prime }-{\mathrm e}^{\lambda x} y a \lambda &=c_1 \end{align*}

Entering first order ode linear solverIn canonical form a linear first order is

\begin{align*} y^{\prime } + q(x)y &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-\frac {a \lambda }{{\mathrm e}^{-\lambda x} b +a}\\ p(x) &=\frac {c_1}{{\mathrm e}^{\lambda x} a +b} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {a \lambda }{{\mathrm e}^{-\lambda x} b +a}d x}\\ &= \frac {1}{{\mathrm e}^{\lambda x} a +b} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_1}{{\mathrm e}^{\lambda x} a +b}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{{\mathrm e}^{\lambda x} a +b}\right ) &= \left (\frac {1}{{\mathrm e}^{\lambda x} a +b}\right ) \left (\frac {c_1}{{\mathrm e}^{\lambda x} a +b}\right ) \\ \mathrm {d} \left (\frac {y}{{\mathrm e}^{\lambda x} a +b}\right ) &= \left (\frac {c_1}{\left ({\mathrm e}^{\lambda x} a +b \right )^{2}}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} \frac {y}{{\mathrm e}^{\lambda x} a +b}&= \int {\frac {c_1}{\left ({\mathrm e}^{\lambda x} a +b \right )^{2}} \,dx} \\ &=\frac {c_1 \left (-\left ({\mathrm e}^{\lambda x} a +b \right ) \ln \left ({\mathrm e}^{\lambda x} a +b \right )+\left ({\mathrm e}^{\lambda x} a +b \right ) \ln \left ({\mathrm e}^{\lambda x}\right )+b \right )}{\lambda \,b^{2} \left ({\mathrm e}^{\lambda x} a +b \right )} + c_2 \end{align*}

Dividing throughout by the integrating factor \(\frac {1}{{\mathrm e}^{\lambda x} a +b}\) gives the final solution

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

Simplifying the above gives

\begin{align*} y &= \frac {-c_1 \left ({\mathrm e}^{\lambda x} a +b \right ) \ln \left ({\mathrm e}^{\lambda x} a +b \right )+c_1 \left ({\mathrm e}^{\lambda x} a +b \right ) \ln \left ({\mathrm e}^{\lambda x}\right )+b \left (a b c_2 \lambda \,{\mathrm e}^{\lambda x}+c_2 \lambda \,b^{2}+c_1 \right )}{\lambda \,b^{2}} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {-c_1 \left ({\mathrm e}^{\lambda x} a +b \right ) \ln \left ({\mathrm e}^{\lambda x} a +b \right )+c_1 \left ({\mathrm e}^{\lambda x} a +b \right ) \ln \left ({\mathrm e}^{\lambda x}\right )+b \left (a b c_2 \lambda \,{\mathrm e}^{\lambda x}+c_2 \lambda \,b^{2}+c_1 \right )}{\lambda \,b^{2}} \\ \end{align*}

2.35.37.2 second order integrable as is

0.143 (sec)

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

Entering second order integrable as is solverIntegrating both sides of the ODE w.r.t \(x\) gives

\begin{align*} \int \left (\left ({\mathrm e}^{\lambda x} a +b \right ) y^{\prime \prime }-a \,\lambda ^{2} {\mathrm e}^{\lambda x} y\right )d x &= 0 \\ -{\mathrm e}^{\lambda x} y a \lambda -\left (-{\mathrm e}^{\lambda x} a -b \right ) y^{\prime } = c_1 \end{align*}

Which is now solved for \(y\). Entering first order ode linear solverIn canonical form a linear first order is

\begin{align*} y^{\prime } + q(x)y &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-\frac {a \lambda }{{\mathrm e}^{-\lambda x} b +a}\\ p(x) &=\frac {c_1}{{\mathrm e}^{\lambda x} a +b} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {a \lambda }{{\mathrm e}^{-\lambda x} b +a}d x}\\ &= \frac {1}{{\mathrm e}^{\lambda x} a +b} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_1}{{\mathrm e}^{\lambda x} a +b}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{{\mathrm e}^{\lambda x} a +b}\right ) &= \left (\frac {1}{{\mathrm e}^{\lambda x} a +b}\right ) \left (\frac {c_1}{{\mathrm e}^{\lambda x} a +b}\right ) \\ \mathrm {d} \left (\frac {y}{{\mathrm e}^{\lambda x} a +b}\right ) &= \left (\frac {c_1}{\left ({\mathrm e}^{\lambda x} a +b \right )^{2}}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} \frac {y}{{\mathrm e}^{\lambda x} a +b}&= \int {\frac {c_1}{\left ({\mathrm e}^{\lambda x} a +b \right )^{2}} \,dx} \\ &=\frac {c_1 \left (-\left ({\mathrm e}^{\lambda x} a +b \right ) \ln \left ({\mathrm e}^{\lambda x} a +b \right )+\left ({\mathrm e}^{\lambda x} a +b \right ) \ln \left ({\mathrm e}^{\lambda x}\right )+b \right )}{\lambda \,b^{2} \left ({\mathrm e}^{\lambda x} a +b \right )} + c_2 \end{align*}

Dividing throughout by the integrating factor \(\frac {1}{{\mathrm e}^{\lambda x} a +b}\) gives the final solution

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

Simplifying the above gives

\begin{align*} y &= \frac {-c_1 \left ({\mathrm e}^{\lambda x} a +b \right ) \ln \left ({\mathrm e}^{\lambda x} a +b \right )+c_1 \left ({\mathrm e}^{\lambda x} a +b \right ) \ln \left ({\mathrm e}^{\lambda x}\right )+b \left (a b c_2 \lambda \,{\mathrm e}^{\lambda x}+c_2 \lambda \,b^{2}+c_1 \right )}{\lambda \,b^{2}} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {-c_1 \left ({\mathrm e}^{\lambda x} a +b \right ) \ln \left ({\mathrm e}^{\lambda x} a +b \right )+c_1 \left ({\mathrm e}^{\lambda x} a +b \right ) \ln \left ({\mathrm e}^{\lambda x}\right )+b \left (a b c_2 \lambda \,{\mathrm e}^{\lambda x}+c_2 \lambda \,b^{2}+c_1 \right )}{\lambda \,b^{2}} \\ \end{align*}

2.35.37.3 ✓ Maple. Time used: 0.002 (sec). Leaf size: 52

ode:=(a*exp(lambda*x)+b)*diff(diff(y(x),x),x)-a*lambda^2*exp(lambda*x)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = -c_1 \left (a \,{\mathrm e}^{\lambda x}+b \right ) \ln \left (a \,{\mathrm e}^{\lambda x}+b \right )+c_1 \left (a \,{\mathrm e}^{\lambda x}+b \right ) \ln \left ({\mathrm e}^{\lambda x}\right )+{\mathrm e}^{\lambda x} c_2 a +b \left (c_1 +c_2 \right ) \]

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
<- linear_1 successful
 

2.35.37.4 ✗ Mathematica

ode=(a*Exp[\[Lambda]*x]+b)*D[y[x],{x,2}]-a*\[Lambda]^2*Exp(\[Lambda]*x)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

Not solved

2.35.37.5 ✗ Sympy

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

False
 

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

classify_ode(ode,func=y(x)) 
 
('2nd_power_series_ordinary',)