2.29.32 Problem 92

Internal problem ID [13753]
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 : 92
Date solved : Thursday, January 01, 2026 at 02:31:48 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

2.29.32.1 second order change of variable on y method 2

1.184 (sec)

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

The integrating factor \(\mu \) is

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

Integrating gives

\begin{align*} u \,{\mathrm e}^{\frac {a \,x^{n}}{n}}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\frac {a \,x^{n}}{n}}\) gives the final solution

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

Simplifying the above gives

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

Hence

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

Summary of solutions found

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

2.29.32.2 ✓ Maple. Time used: 0.026 (sec). Leaf size: 92

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

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

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
 

2.29.32.3 ✓ Mathematica. Time used: 0.036 (sec). Leaf size: 62

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

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

2.29.32.4 ✗ Sympy

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

False