2.35.32 Problem 32

Internal problem ID [13956]
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 : 32
Date solved : Thursday, January 01, 2026 at 04:06:18 AM
CAS classification : [[_2nd_order, _exact, _linear, _homogeneous]]

2.35.32.1 second order linear exact ode

0.948 (sec)

\begin{align*} y^{\prime \prime }+\left ({\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c \right ) y^{\prime }+\left (a \lambda \,{\mathrm e}^{\lambda x}+\mu \,{\mathrm e}^{\mu x} b \right ) 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 we have

\begin{align*} p(x) &= 1\\ q(x) &= {\mathrm e}^{\lambda x} a +{\mathrm e}^{\mu x} b +c\\ r(x) &= a \lambda \,{\mathrm e}^{\lambda x}+\mu \,{\mathrm e}^{\mu x} b\\ s(x) &= 0 \end{align*}

Hence

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

Therefore (1) becomes

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

Hence the ode is exact. Since we now know 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 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 above values for \(p,q,r,s\) gives

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

We now have a first order ode to solve which is

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

The integrating factor \(\mu \) is

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

Integrating gives

\begin{align*} y \,{\mathrm e}^{c x +\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }+\frac {b \,{\mathrm e}^{\mu x}}{\mu }}&= \int {c_1 \,{\mathrm e}^{c x +\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }+\frac {b \,{\mathrm e}^{\mu x}}{\mu }} \,dx} \\ &=\int c_1 \,{\mathrm e}^{c x +\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }+\frac {b \,{\mathrm e}^{\mu x}}{\mu }}d x + c_2 \end{align*}

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

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

Summary of solutions found

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

2.35.32.2 second order integrable as is

0.574 (sec)

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

The integrating factor \(\mu \) is

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

Integrating gives

\begin{align*} y \,{\mathrm e}^{c x +\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }+\frac {b \,{\mathrm e}^{\mu x}}{\mu }}&= \int {c_1 \,{\mathrm e}^{c x +\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }+\frac {b \,{\mathrm e}^{\mu x}}{\mu }} \,dx} \\ &=\int c_1 \,{\mathrm e}^{c x +\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }+\frac {b \,{\mathrm e}^{\mu x}}{\mu }}d x + c_2 \end{align*}

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

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

Summary of solutions found

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

2.35.32.3 ✓ Maple. Time used: 0.003 (sec). Leaf size: 70

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

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

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
   One independent solution has integrals. Trying a hypergeometric solution fre\ 
e of integrals... 
   -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
No hypergeometric solution was found. 
<- linear_1 successful
 

2.35.32.4 ✓ Mathematica. Time used: 29.203 (sec). Leaf size: 113

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

\begin{align*} y(x)&\to e^{-\frac {a e^{\lambda x}}{\lambda }-\frac {b e^{\mu x}}{\mu }-c x} \left (\int _1^xe^{\frac {e^{\lambda K[1]} a}{\lambda }+c K[1]+\frac {b e^{\mu K[1]}}{\mu }} c_1dK[1]+c_2\right )\\ y(x)&\to c_2 e^{-\frac {a e^{\lambda x}}{\lambda }-\frac {b e^{\mu x}}{\mu }-c x} \end{align*}

2.35.32.5 ✗ Sympy

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

False