2.28.45 Problem 55

Internal problem ID [13716]
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 : 55
Date solved : Sunday, January 18, 2026 at 09:09:37 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

2.28.45.1 second order change of variable on y method 2

0.774 (sec)

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

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

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

Applying change of variables on the depndent variable \(y = v \left (x \right ) x^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (x \right )\) and not \(y\).

\begin{align*} v^{\prime \prime }\left (x \right )+\left (\frac {2 n}{x}+p \right ) v^{\prime }\left (x \right )+\left (\frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q \right ) v \left (x \right )&=0 \tag {3} \end{align*}

Let the coefficient of \(v \left (x \right )\) above be zero. Hence

\begin{align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q&=0 \tag {4} \end{align*}

Substituting the earlier values found for \(p \left (x \right )\) and \(q \left (x \right )\) into (4) gives

\begin{align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n \left (x^{n} a +b \,x^{m}\right )}{x}+\frac {-x^{n} a -b \,x^{m}}{x}&=0 \tag {5} \end{align*}

Solving (5) for \(n\) gives

\begin{align*} n&=1 \tag {6} \end{align*}

Substituting this value in (3) gives

\begin{align*} v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}+x^{n} a +b \,x^{m}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}+x^{n} a +b \,x^{m}\right ) v^{\prime }\left (x \right )&=0 \tag {7} \\ \end{align*}

Using the substitution

\begin{align*} u \left (x \right ) = v^{\prime }\left (x \right ) \end{align*}

Then (7) becomes

\begin{align*} u^{\prime }\left (x \right )+\left (\frac {2}{x}+x^{n} a +b \,x^{m}\right ) u \left (x \right ) = 0 \tag {8} \\ \end{align*}

The above is now solved for \(u \left (x \right )\). 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) &=-\frac {-2-x^{n +1} a -b \,x^{1+m}}{x}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {-2-x^{n +1} a -b \,x^{1+m}}{x}d x}\\ &= x^{2} {\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}} \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 \,x^{2} {\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \,x^{2} {\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \(x^{2} {\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}\) gives the final solution

\[ u \left (x \right ) = \frac {{\mathrm e}^{-\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}} c_1}{x^{2}} \]

Simplifying the above gives

\begin{align*} u \left (x \right ) &= \frac {{\mathrm e}^{-\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}} c_1}{x^{2}} \\ \end{align*}

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

\begin{align*} v^{\prime }\left (x \right )&= u \left (x \right )\\ v \left (x \right )&= \int u \left (x \right )d x +c_2\\ &= \int \frac {{\mathrm e}^{-\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}} c_1}{x^{2}}d x +c_2 \end{align*}

Hence

\begin{align*} y&= v \left (x \right ) x^{n}\\ &= \left (\int \frac {{\mathrm e}^{-\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}} c_1}{x^{2}}d x +c_2 \right ) x\\ &= \left (c_1 \int \frac {{\mathrm e}^{-\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}}{x^{2}}d x +c_2 \right ) x\\ \end{align*}

Summary of solutions found

\begin{align*} y &= \left (\int \frac {{\mathrm e}^{-\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}} c_1}{x^{2}}d x +c_2 \right ) x \\ \end{align*}

2.28.45.2 ✓ Maple. Time used: 0.042 (sec). Leaf size: 47

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

\[ y = x \left (\int \frac {{\mathrm e}^{-\frac {x \left (b \left (1+n \right ) x^{m}+a \,x^{n} \left (1+m \right )\right )}{\left (1+n \right ) \left (1+m \right )}}}{x^{2}}d x c_2 +c_1 \right ) \]

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 an equivalence, under non-integer power transformations, 
   to LODEs admitting Liouvillian solutions. 
-> Trying a solution in terms of special functions: 
   -> Bessel 
   -> elliptic 
   -> Legendre 
   -> Kummer 
      -> 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 2nd order exact linear 
   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)] 
      One independent solution has integrals. Trying a hypergeometric solution \ 
free of integrals... 
      -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
   No hypergeometric solution was found. 
   <- linear_1 successful 
   <- 2nd order, integrating factors of the form mu(x,y) successful
 

2.28.45.3 ✓ Mathematica. Time used: 0.655 (sec). Leaf size: 55

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

\begin{align*} y(x)&\to x \left (c_2 \int _1^x\frac {\exp \left (K[1] \left (-\frac {b K[1]^m}{m+1}-\frac {a K[1]^n}{n+1}\right )\right )}{K[1]^2}dK[1]+c_1\right ) \end{align*}

2.28.45.4 ✗ 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*x**n + b*x**m)*Derivative(y(x), x) - (a*x**(n - 1) + b*x**(m - 1))*y(x) + Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

TypeError : Add object cannot be interpreted as an integer
 

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')