2.28.36 Problem 46

Internal problem ID [13707]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.2-2
Problem number : 46
Date solved : Sunday, January 18, 2026 at 09:08:53 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

2.28.36.1 second order ode solved by an integrating factor

0.105 (sec)

\begin{align*} y^{\prime \prime }+2 a \,x^{n} y^{\prime }+a \left (a \,x^{2 n}+n \,x^{n -1}\right ) y&=0 \\ \end{align*}

Entering second order ode solved by an integrating factor solverThe ode satisfies this form

\[ y^{\prime \prime }+p \left (x \right ) y^{\prime }+\frac {\left (p \left (x \right )^{2}+p^{\prime }\left (x \right )\right ) y}{2} = f \left (x \right ) \]

Where \( p(x) = 2 x^{n} a\). Therefore, there is an integrating factor given by

\begin{align*} M(x) &= e^{\frac {1}{2} \int p \, dx} \\ &= e^{ \int 2 x^{n} a \, dx} \\ &= {\mathrm e}^{\frac {a \,x^{n +1}}{n +1}} \end{align*}

Multiplying both sides of the ODE by the integrating factor \(M(x)\) makes the left side of the ODE a complete differential

\begin{align*} \left ( M(x) y \right )'' &= 0 \\ \left ( {\mathrm e}^{\frac {a \,x^{n +1}}{n +1}} y \right )'' &= 0 \\ \end{align*}

Integrating once gives

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

Integrating again gives

\[ \left ( {\mathrm e}^{\frac {a \,x^{n +1}}{n +1}} y \right ) = c_1 x +c_2 \]

Hence the solution is

\begin{align*} y &= \frac {c_1 x +c_2}{{\mathrm e}^{\frac {a \,x^{n +1}}{n +1}}} \\ \end{align*}

Or

\[ y = c_1 x \,{\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}}+c_2 \,{\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}} \]

Summary of solutions found

\begin{align*} y &= c_1 x \,{\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}}+c_2 \,{\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}} \\ \end{align*}

2.28.36.2 second order change of variable on y method 1

0.177 (sec)

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

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

\begin{align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= a^{2} x^{2 n}+a n \,x^{n -1} - \frac {\left (2 x^{n} a\right )'}{2}- \frac {\left (2 x^{n} a\right )^2}{4} \\ &= a^{2} x^{2 n}+a n \,x^{n -1} - \frac {\left (\frac {2 a n \,x^{n}}{x}\right )}{2}- \frac {\left (4 a^{2} x^{2 n}\right )}{4} \\ &= a^{2} x^{2 n}+a n \,x^{n -1} - \left (\frac {a n \,x^{n}}{x}\right )-a^{2} x^{2 n}\\ &= 0 \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 x^{n} a}{2} }\\ &= {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}}\tag {5} \end{align*}

Hence (3) becomes

\begin{align*} y = v \left (x \right ) {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}}\tag {4} \end{align*}

Applying this change of variable to the original ode results in

\begin{align*} v^{\prime \prime }\left (x \right ) {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}} = 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 ) = 0 \end{align*}

Entering second order ode quadrature solverIntegrating twice gives the solution

\[ v \left (x \right )= c_1 x + c_2 \]

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

\begin{align*} y&= v \left (x \right ) z \left (x \right )\\ &= \left (c_1 x +c_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^{n +1}}{n +1}} \end{align*}

Hence (7) becomes

\begin{align*} y = \left (c_1 x +c_2 \right ) {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}} \end{align*}

Summary of solutions found

\begin{align*} y &= \left (c_1 x +c_2 \right ) {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}} \\ \end{align*}

2.28.36.3 ✓ Maple. Time used: 0.003 (sec). Leaf size: 24

ode:=diff(diff(y(x),x),x)+2*a*x^n*diff(y(x),x)+a*(a*x^(2*n)+n*x^(n-1))*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}} \left (c_2 x +c_1 \right ) \]

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) 
<- Kovacics algorithm successful
 

2.28.36.4 ✓ Mathematica. Time used: 0.058 (sec). Leaf size: 28

ode=D[y[x],{x,2}]+2*a*x^n*D[y[x],x]+a*(a*x^(2*n)+n*x^(n-1))*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to (c_2 x+c_1) e^{-\frac {a x^{n+1}}{n+1}} \end{align*}

2.28.36.5 ✗ Sympy

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

RecursionError : maximum recursion depth exceeded
 

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')