2.34.26 Problem 264

Internal problem ID [13924]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.2-8. Other equations.
Problem number : 264
Date solved : Thursday, January 01, 2026 at 04:01:56 AM
CAS classification : [[_2nd_order, _with_linear_symmetries], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]

2.34.26.1 Solved as second order ode adjoint method

5.635 (sec)

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

Entering second order ode lagrange adjoint equation method solverIn normal form the ode

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

Becomes

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

Where

\begin{align*} p \left (x \right )&=\left (x^{n} a +b \right )^{-m}\\ q \left (x \right )&=-a n m \,x^{n -1} \left (x^{n} a +b \right )^{-1-m}\\ r \left (x \right )&=0 \end{align*}

The Lagrange adjoint ode is given by

\begin{align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (\left (x^{n} a +b \right )^{-m} \xi \left (x \right )\right )' + \left (-a n m \,x^{n -1} \left (x^{n} a +b \right )^{-1-m} \xi \left (x \right )\right ) &= 0\\ \xi ^{\prime \prime }\left (x \right )-\left (x^{n} a +b \right )^{-m} \xi ^{\prime }\left (x \right )&= 0 \end{align*}

Which is solved for \(\xi (x)\). Entering second order ode missing \(y\) solverThis is second order ode with missing dependent variable \(\xi \). Let

\begin{align*} u(x) &= \xi ^{\prime } \end{align*}

Then

\begin{align*} u'(x) &= \xi ^{\prime \prime } \end{align*}

Hence the ode becomes

\begin{align*} u^{\prime }\left (x \right )-\left (x^{n} a +b \right )^{-m} u \left (x \right ) = 0 \end{align*}

Which is now solved for \(u(x)\) as first order ode.

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

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

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-\left (x^{n} a +b \right )^{-m}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\left (x^{n} a +b \right )^{-m}d x}\\ &= {\mathrm e}^{\int -\left (x^{n} a +b \right )^{-m}d x} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu u &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (u \,{\mathrm e}^{\int -\left (x^{n} a +b \right )^{-m}d x}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \,{\mathrm e}^{\int -\left (x^{n} a +b \right )^{-m}d x}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\int -\left (x^{n} a +b \right )^{-m}d x}\) gives the final solution

\[ u \left (x \right ) = {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x} c_1 \]

In summary, these are the solution found for \(\xi \)

\begin{align*} u \left (x \right ) &= {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x} c_1 \\ \end{align*}

For solution \(u \left (x \right ) = {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x} c_1\), since \(u=\xi ^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} \xi ^{\prime } = {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x} c_1 \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(\xi ^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {d\xi } &= \int {{\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1\, dx}\\ \xi &= \int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x + c_2 \end{align*}

\begin{align*} \xi &= \int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x +c_2 \end{align*}

In summary, these are the solution found for \((\xi )\)

\begin{align*} \xi &= \int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x +c_2 \\ \end{align*}

The original ode now reduces to first order ode

\begin{align*} \xi \left (x \right ) y^{\prime }-y \xi ^{\prime }\left (x \right )+\xi \left (x \right ) p \left (x \right ) y&=\int \xi \left (x \right ) r \left (x \right )d x\\ y^{\prime }+y \left (p \left (x \right )-\frac {\xi ^{\prime }\left (x \right )}{\xi \left (x \right )}\right )&=\frac {\int \xi \left (x \right ) r \left (x \right )d x}{\xi \left (x \right )} \end{align*}

Or

\begin{align*} y^{\prime }+y \left (\left (x^{n} a +b \right )^{-m}-\frac {{\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1}{\int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x +c_2}\right )&=0 \end{align*}

Which is now a first order ode. This is now solved for \(y\). Entering first order ode separable solverThe ode

\begin{equation} y^{\prime } = -\frac {y \left (\left (x^{n} a +b \right )^{-m} \int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x +\left (x^{n} a +b \right )^{-m} c_2 -{\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 \right )}{\int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x +c_2} \end{equation}

is separable as it can be written as

\begin{align*} y^{\prime }&= -\frac {y \left (\left (x^{n} a +b \right )^{-m} \int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x +\left (x^{n} a +b \right )^{-m} c_2 -{\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 \right )}{\int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x +c_2}\\ &= f(x) g(y) \end{align*}

Where

\begin{align*} f(x) &= -\frac {\left (x^{n} a +b \right )^{-m} \int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x +\left (x^{n} a +b \right )^{-m} c_2 -{\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1}{\int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x +c_2}\\ g(y) &= y \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx} \\ \int { \frac {1}{y}\,dy} &= \int { -\frac {\left (x^{n} a +b \right )^{-m} \int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x +\left (x^{n} a +b \right )^{-m} c_2 -{\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1}{\int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x +c_2} \,dx} \\ \end{align*}

\[ \ln \left (y\right )=\int -\frac {\left (x^{n} a +b \right )^{-m} \int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x +\left (x^{n} a +b \right )^{-m} c_2 -{\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1}{\int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x +c_2}d x +c_3 \]

We now need to find the singular solutions, these are found by finding for what values \(g(y)\) is zero, since we had to divide by this above. Solving \(g(y)=0\) or

\[ y=0 \]

for \(y\) gives

\begin{align*} y&=0 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \ln \left (y\right ) &= \int -\frac {\left (x^{n} a +b \right )^{-m} \int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x +\left (x^{n} a +b \right )^{-m} c_2 -{\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1}{\int {\mathrm e}^{-\int -\left (x^{n} a +b \right )^{-m}d x} c_1 d x +c_2}d x +c_3 \\ y &= 0 \\ \end{align*}

Solving for \(y\) gives

\begin{align*} y &= 0 \\ y &= {\mathrm e}^{-c_1 \int \frac {\left (x^{n} a +b \right )^{-m} \int {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x}d x}{c_1 \int {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x}d x +c_2}d x +c_1 \int \frac {{\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x}}{c_1 \int {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x}d x +c_2}d x -c_2 \int \frac {\left (x^{n} a +b \right )^{-m}}{c_1 \int {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x}d x +c_2}d x +c_3} \\ \end{align*}

Hence, the solution found using Lagrange adjoint equation method is

\begin{align*} y &= 0 \\ y &= {\mathrm e}^{-c_1 \int \frac {\left (x^{n} a +b \right )^{-m} \int {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x}d x}{c_1 \int {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x}d x +c_2}d x +c_1 \int \frac {{\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x}}{c_1 \int {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x}d x +c_2}d x -c_2 \int \frac {\left (x^{n} a +b \right )^{-m}}{c_1 \int {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x}d x +c_2}d x +c_3} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= 0 \\ y &= {\mathrm e}^{-c_1 \int \frac {\left (x^{n} a +b \right )^{-m} \int {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x}d x}{c_1 \int {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x}d x +c_2}d x +c_1 \int \frac {{\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x}}{c_1 \int {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x}d x +c_2}d x -c_2 \int \frac {\left (x^{n} a +b \right )^{-m}}{c_1 \int {\mathrm e}^{\int \left (x^{n} a +b \right )^{-m}d x}d x +c_2}d x +c_3} \\ \end{align*}

2.34.26.2 ✓ Maple. Time used: 0.027 (sec). Leaf size: 41

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

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

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 solution in terms of special functions: 
   -> Bessel 
   -> elliptic 
   -> Legendre 
   -> Whittaker 
      -> hyper3: Equivalence to 1F1 under a power @ Moebius 
   -> hypergeometric 
      -> heuristic approach 
      -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
   -> Mathieu 
      -> Equivalence to the rational form of Mathieu ODE under a power @ Moebi\ 
us 
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power \ 
@ Moebius 
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power \ 
@ Moebius 
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x),\ 
 dx)) * 2F1([a1, a2], [b1], f) 
-> Trying changes of variables to rationalize or make the ODE simpler 
<- unable to find a useful change of variables 
   trying a symmetry of the form [xi=0, eta=F(x)] 
   trying differential order: 2; exact nonlinear 
   trying symmetries linear in x and y(x) 
   trying to convert to a linear ODE with constant coefficients 
   trying 2nd order, integrating factor of the form mu(x,y) 
   trying a symmetry of the form [xi=0, eta=F(x)] 
   checking if the LODE is missing y 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      -> elliptic 
      -> Legendre 
      -> Whittaker 
         -> hyper3: Equivalence to 1F1 under a power @ Moebius 
      -> hypergeometric 
         -> heuristic approach 
         -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
      -> Mathieu 
         -> Equivalence to the rational form of Mathieu ODE under a power @ Mo\ 
ebius 
   -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a pow\ 
er @ Moebius 
   -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a pow\ 
er @ Moebius 
   -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(\ 
x), dx)) * 2F1([a1, a2], [b1], f) 
   -> Trying changes of variables to rationalize or make the ODE simpler 
   <- unable to find a useful change of variables 
      trying a symmetry of the form [xi=0, eta=F(x)] 
   trying to convert to an ODE of Bessel type 
   -> trying reduction of order to Riccati 
      trying Riccati sub-methods: 
         trying Riccati_symmetries 
         -> trying a symmetry pattern of the form [F(x)*G(y), 0] 
         -> trying a symmetry pattern of the form [0, F(x)*G(y)] 
         -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] 
--- Trying Lie symmetry methods, 2nd order --- 
   -> Computing symmetries using: way = 3 
[0, y] 
   <- successful computation of symmetries. 
   -> Computing symmetries using: way = 5 
[0, y], [y, -(a*x^n+b)^(-m)*y^2] 
   <- successful computation of symmetries.
 

2.34.26.3 ✓ Mathematica. Time used: 60.096 (sec). Leaf size: 116

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

\begin{align*} y(x)&\to \exp \left (-x \left (a x^n+b\right )^{-m} \left (\frac {a x^n}{b}+1\right )^m \operatorname {Hypergeometric2F1}\left (m,\frac {1}{n},1+\frac {1}{n},-\frac {a x^n}{b}\right )\right ) \left (\int _1^x\exp \left (\operatorname {Hypergeometric2F1}\left (m,\frac {1}{n},1+\frac {1}{n},-\frac {a K[1]^n}{b}\right ) K[1] \left (a K[1]^n+b\right )^{-m} \left (\frac {a K[1]^n}{b}+1\right )^m\right ) c_1dK[1]+c_2\right ) \end{align*}

2.34.26.4 ✗ Sympy

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

TypeError : Add object cannot be interpreted as an integer