2.29.35 Problem 95

Internal problem ID [13756]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.2-3
Problem number : 95
Date solved : Sunday, January 18, 2026 at 09:13:21 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

2.29.35.1 second order change of variable on y method 2

0.629 (sec)

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

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

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

Where

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

Solving (5) for \(n\) gives

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

Substituting this value in (3) gives

\begin{align*} v^{\prime \prime }\left (x \right )+\left (\frac {-2 b +2}{x}+\frac {x^{n} a +b}{x}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\frac {\left (-b +2+x^{n} a \right ) v^{\prime }\left (x \right )}{x}&=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 )+\frac {\left (-b +2+x^{n} a \right ) u \left (x \right )}{x} = 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 {b -2-x^{n} a}{x}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

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

Integrating gives

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

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

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

Simplifying the above gives

\begin{align*} u \left (x \right ) &= {\mathrm e}^{-\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{\frac {b -2}{n}} c_1 \\ \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\\ &= \frac {\left (x^{n}\right )^{\frac {b -2}{n}} x^{-b +2} \left (\frac {a}{n}\right )^{-\frac {b}{n}+\frac {1}{n}} c_1 \left (\frac {n^{3} x^{-n +b -1} \left (\frac {a}{n}\right )^{\frac {b}{n}-\frac {1}{n}} \left (x^{n} a +b +n -1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n -1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {b -1}{n}-\frac {b +n -1}{2 n}, \frac {b +n -1}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (b -1\right ) \left (b +n -1\right ) \left (b +2 n -1\right ) a}+\frac {n^{2} x^{-n +b -1} \left (\frac {a}{n}\right )^{\frac {b}{n}-\frac {1}{n}} \left (b +n -1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n -1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {b -1}{n}-\frac {b +n -1}{2 n}+1, \frac {b +n -1}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (b -1\right ) a \left (b +2 n -1\right )}\right )}{n}+c_2 \end{align*}

Hence

\begin{align*} y&= v \left (x \right ) x^{n}\\ &= \left (\frac {\left (x^{n}\right )^{\frac {b -2}{n}} x^{-b +2} \left (\frac {a}{n}\right )^{-\frac {b}{n}+\frac {1}{n}} c_1 \left (\frac {n^{3} x^{-n +b -1} \left (\frac {a}{n}\right )^{\frac {b}{n}-\frac {1}{n}} \left (x^{n} a +b +n -1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n -1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {b -1}{n}-\frac {b +n -1}{2 n}, \frac {b +n -1}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (b -1\right ) \left (b +n -1\right ) \left (b +2 n -1\right ) a}+\frac {n^{2} x^{-n +b -1} \left (\frac {a}{n}\right )^{\frac {b}{n}-\frac {1}{n}} \left (b +n -1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n -1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {b -1}{n}-\frac {b +n -1}{2 n}+1, \frac {b +n -1}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (b -1\right ) a \left (b +2 n -1\right )}\right )}{n}+c_2 \right ) x^{-b +1}\\ &= \left (\frac {x^{1-n} c_1 \left (x^{n}\right )^{\frac {b -2}{n}} \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n -1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \left (\frac {\operatorname {WhittakerM}\left (\frac {-n +b -1}{2 n}, \frac {b +2 n -1}{2 n}, \frac {a \,x^{n}}{n}\right ) n^{3} \left (x^{n} a +b +n -1\right )}{b +n -1}+\operatorname {WhittakerM}\left (\frac {b +n -1}{2 n}, \frac {b +2 n -1}{2 n}, \frac {a \,x^{n}}{n}\right ) n^{2} \left (b +n -1\right )\right )}{a n \left (b -1\right ) \left (b +2 n -1\right )}+c_2 \right ) x^{-b +1}\\ \end{align*}

Summary of solutions found

\begin{align*} y &= \left (\frac {\left (x^{n}\right )^{\frac {b -2}{n}} x^{-b +2} \left (\frac {a}{n}\right )^{-\frac {b}{n}+\frac {1}{n}} c_1 \left (\frac {n^{3} x^{-n +b -1} \left (\frac {a}{n}\right )^{\frac {b}{n}-\frac {1}{n}} \left (x^{n} a +b +n -1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n -1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {b -1}{n}-\frac {b +n -1}{2 n}, \frac {b +n -1}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (b -1\right ) \left (b +n -1\right ) \left (b +2 n -1\right ) a}+\frac {n^{2} x^{-n +b -1} \left (\frac {a}{n}\right )^{\frac {b}{n}-\frac {1}{n}} \left (b +n -1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n -1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {b -1}{n}-\frac {b +n -1}{2 n}+1, \frac {b +n -1}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (b -1\right ) a \left (b +2 n -1\right )}\right )}{n}+c_2 \right ) x^{-b +1} \\ \end{align*}

2.29.35.2 ✓ Maple. Time used: 0.016 (sec). Leaf size: 115

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

\[ y = c_1 \,x^{-b +1}+c_2 \,x^{-\frac {3 n}{2}+\frac {1}{2}-\frac {b}{2}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \left (n \left (a \,x^{n}+b +n -1\right ) \operatorname {WhittakerM}\left (\frac {b -1-n}{2 n}, \frac {b +2 n -1}{2 n}, \frac {a \,x^{n}}{n}\right )+\operatorname {WhittakerM}\left (\frac {b +n -1}{2 n}, \frac {b +2 n -1}{2 n}, \frac {a \,x^{n}}{n}\right ) \left (b +n -1\right )^{2}\right ) \]

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 Liouvillian solution using Kovacics algorithm 
      A Liouvillian solution exists 
      Reducible group (found an exponential solution) 
      Group is reducible, not completely reducible 
   <- Kovacics algorithm successful 
<- Equivalence, under non-integer power transformations successful
 

2.29.35.3 ✓ Mathematica. Time used: 0.055 (sec). Leaf size: 90

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

\begin{align*} y(x)&\to (-1)^{-\frac {b}{n}} n^{\frac {b-n-1}{n}} a^{\frac {1-b}{n}} \left (x^n\right )^{\frac {1-b}{n}} \left ((b-1) c_1 (-1)^{b/n} \Gamma \left (\frac {b-1}{n},0,\frac {a x^n}{n}\right )+c_2 (-1)^{\frac {1}{n}} n\right ) \end{align*}

2.29.35.4 ✗ Sympy

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

ValueError : Expected Expr or iterable but got None
 

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_regular')