2.28.34 Problem 44

Internal problem ID [13705]
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 : 44
Date solved : Sunday, January 18, 2026 at 09:08:43 PM
CAS classification : [[_2nd_order, _missing_y]]

2.28.34.1 second order ode missing y

0.271 (sec)

\begin{align*} y^{\prime \prime }+a \,x^{n} y^{\prime }&=0 \\ \end{align*}

Entering second order ode missing \(y\) solverThis is second order ode with missing dependent variable \(y\). Let

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

Then

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

Hence the ode becomes

\begin{align*} u^{\prime }\left (x \right )+a \,x^{n} 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) &=x^{n} a\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

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

Integrating gives

\begin{align*} u \,{\mathrm e}^{\frac {a \,x^{n +1}}{n +1}}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\frac {a \,x^{n +1}}{n +1}}\) gives the final solution

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

Simplifying the above gives

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

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

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

For solution \(u \left (x \right ) = c_1 \,{\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}}\), since \(u=y^{\prime }\) then the new first order ode to solve is

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

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

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

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

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

Summary of solutions found

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

2.28.34.2 ✓ Maple. Time used: 0.001 (sec). Leaf size: 159

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

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

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 
<- LODE missing y successful
 

2.28.34.3 ✓ Mathematica. Time used: 0.027 (sec). Leaf size: 56

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

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

2.28.34.4 ✓ Sympy. Time used: 1.118 (sec). Leaf size: 258

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

\[ y{\left (x \right )} = \begin {cases} - \frac {C_{1} n \Gamma \left (\frac {n}{n + 1} + \frac {2}{n + 1}\right ) \gamma \left (\frac {1}{n + 1}, \frac {a x^{n + 1}}{\operatorname {polar\_lift}{\left (n + 1 \right )}}\right ) \operatorname {polar\_lift}^{\frac {1}{n + 1}}{\left (n + 1 \right )}}{a^{\frac {1}{n + 1}} n^{2} \Gamma \left (\frac {2 n}{n + 1} + \frac {3}{n + 1}\right ) + 2 a^{\frac {1}{n + 1}} n \Gamma \left (\frac {2 n}{n + 1} + \frac {3}{n + 1}\right ) + a^{\frac {1}{n + 1}} \Gamma \left (\frac {2 n}{n + 1} + \frac {3}{n + 1}\right )} - \frac {2 C_{1} \Gamma \left (\frac {n}{n + 1} + \frac {2}{n + 1}\right ) \gamma \left (\frac {1}{n + 1}, \frac {a x^{n + 1}}{\operatorname {polar\_lift}{\left (n + 1 \right )}}\right ) \operatorname {polar\_lift}^{\frac {1}{n + 1}}{\left (n + 1 \right )}}{a^{\frac {1}{n + 1}} n^{2} \Gamma \left (\frac {2 n}{n + 1} + \frac {3}{n + 1}\right ) + 2 a^{\frac {1}{n + 1}} n \Gamma \left (\frac {2 n}{n + 1} + \frac {3}{n + 1}\right ) + a^{\frac {1}{n + 1}} \Gamma \left (\frac {2 n}{n + 1} + \frac {3}{n + 1}\right )} - C_{2} & \text {for}\: n > -\infty \wedge n < \infty \wedge n \neq -1 \\\frac {C_{1} x}{a e^{a \log {\left (x \right )}} - e^{a \log {\left (x \right )}}} - C_{2} & \text {for}\: a \neq 1 \\- C_{1} \log {\left (x \right )} - C_{2} & \text {otherwise} \end {cases} \]

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', 'Liouville', 'nth_order_reducible', '2nd_power_series_ordinary', 'Liouville_Integral')