2.35.17 Problem 17

Internal problem ID [13941]
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 : 17
Date solved : Sunday, January 18, 2026 at 09:36:04 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

2.35.17.1 second order change of variable on x method 2

0.350 (sec)

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

Entering second order change of variable on \(x\) method 2 solverIn normal form the ode

\begin{align*} y^{\prime \prime }-\left (a +2 b \,{\mathrm e}^{a x}\right ) y^{\prime }+b^{2} {\mathrm e}^{2 a x} y = 0\tag {1} \end{align*}

Becomes

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

Where

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

Applying change of variables \(\tau = g \left (x \right )\) to (2) gives

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end{align*}

Where \(\tau \) is the new independent variable, and

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end{align*}

Let \(p_{1} = 0\). Eq (4) simplifies to

\begin{align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end{align*}

This ode is solved resulting in

\begin{align*} \tau &= \int {\mathrm e}^{-\int p \left (x \right )d x}d x\\ &= \int {\mathrm e}^{-\int \left (-2 b \,{\mathrm e}^{a x}-a \right )d x}d x\\ &= \int e^{a x +\frac {2 b \,{\mathrm e}^{a x}}{a}} \,dx\\ &= \int {\mathrm e}^{a x +\frac {2 b \,{\mathrm e}^{a x}}{a}}d x\\ &= \frac {{\mathrm e}^{\frac {2 b \,{\mathrm e}^{a x}}{a}}}{2 b}\tag {6} \end{align*}

Using (6) to evaluate \(q_{1}\) from (5) gives

\begin{align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &= \frac {b^{2} {\mathrm e}^{2 a x}}{{\mathrm e}^{2 a x +\frac {4 b \,{\mathrm e}^{a x}}{a}}}\\ &= b^{2} {\mathrm e}^{-\frac {4 b \,{\mathrm e}^{a x}}{a}}\tag {7} \end{align*}

Substituting the above in (3) and noting that now \(p_{1} = 0\) results in

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+b^{2} {\mathrm e}^{-\frac {4 b \,{\mathrm e}^{a x}}{a}} y \left (\tau \right )&=0 \\ \end{align*}

But in terms of \(\tau \)

\begin{align*} b^{2} {\mathrm e}^{-\frac {4 b \,{\mathrm e}^{a x}}{a}}&=\frac {1}{4 \tau ^{2}} \end{align*}

Hence the above ode becomes

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {y \left (\tau \right )}{4 \tau ^{2}}&=0 \end{align*}

The above ode is now solved for \(y \left (\tau \right )\). Entering second order euler ode solverThis is Euler second order ODE. Let the solution be \(y \left (\tau \right ) = \tau ^r\), then \(y'=r \tau ^{r-1}\) and \(y''=r(r-1) \tau ^{r-2}\). Substituting these back into the given ODE gives

\[ 4 \tau ^{2}(r(r-1))\tau ^{r-2}+0 r \tau ^{r-1}+\tau ^{r} = 0 \]

Simplifying gives

\[ 4 r \left (r -1\right )\tau ^{r}+0\,\tau ^{r}+\tau ^{r} = 0 \]

Since \(\tau ^{r}\neq 0\) then dividing throughout by \(\tau ^{r}\) gives

\[ 4 r \left (r -1\right )+0+1 = 0 \]

Or

\[ 4 r^{2}-4 r +1 = 0 \tag {1} \]

Equation (1) is the characteristic equation. Its roots determine the form of the general solution. Using the quadratic equation the roots are

\begin{align*} r_1 &= {\frac {1}{2}}\\ r_2 &= {\frac {1}{2}} \end{align*}

Since the roots are equal, then the general solution is

\[ y \left (\tau \right )= c_1 y_1 + c_2 y_2 \]

Where \(y_1 = \tau ^{r}\) and \(y_2 = \tau ^{r} \ln \left (\tau \right )\). Hence

\[ y \left (\tau \right ) = c_1 \sqrt {\tau }+c_2 \sqrt {\tau }\, \ln \left (\tau \right ) \]

The above solution is now transformed back to \(y\) using (6) which results in

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

Summary of solutions found

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

2.35.17.2 second order change of variable on y method 1

0.273 (sec)

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

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

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

Where

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

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

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

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

\begin{align*} y = 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 b \,{\mathrm e}^{a x}-a}{2} }\\ &= {\mathrm e}^{\frac {a x}{2}+\frac {b \,{\mathrm e}^{a x}}{a}}\tag {5} \end{align*}

Hence (3) becomes

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

Applying this change of variable to the original ode results in

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

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

The above ode simplifies to

\begin{align*} v^{\prime \prime }\left (x \right )-\frac {v \left (x \right ) a^{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 \(A=1, B=0, C=-\frac {a^{2}}{4}\). Let the solution be \(v \left (x \right )=e^{\lambda x}\). Substituting this into the ODE gives

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

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

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

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

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

Substituting \(A=1, B=0, C=-\frac {a^{2}}{4}\) 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 {a^{2}}{4}\right )}\\ &= \pm \frac {\sqrt {a^{2}}}{2} \end{align*}

Hence

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

Which simplifies to

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

Or

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

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

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

Hence (7) becomes

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

Summary of solutions found

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

2.35.17.3 ✓ Maple. Time used: 0.003 (sec). Leaf size: 39

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

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

Maple trace

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 
-> Trying a Liouvillian solution using Kovacics algorithm 
   A Liouvillian solution exists 
   Reducible group (found an exponential solution) 
   Reducible group (found another exponential solution) 
<- Kovacics algorithm successful
 

2.35.17.4 ✓ Mathematica. Time used: 0.026 (sec). Leaf size: 35

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

\begin{align*} y(x)&\to \frac {e^{\frac {b e^{a x}}{a}} \left (b c_2 e^{a x}+a c_1\right )}{a} \end{align*}

2.35.17.5 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
y = Function("y") 
ode = Eq(b**2*y(x)*exp(2*a*x) - (a + 2*b*exp(a*x))*Derivative(y(x), x) + 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)) 
 
('factorable', '2nd_power_series_ordinary')