2.2.6 Problem 6

Internal problem ID [10417]
Book : Second order enumerated odes
Section : section 2
Problem number : 6
Date solved : Wednesday, April 29, 2026 at 09:15:29 AM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

2.2.6.1 second order ode missing x

16.986 (sec)

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

Entering second order ode missing \(x\) solverThis is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using

\begin{align*} y' &= p \end{align*}

Then

\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= p \frac {dp}{dy} \end{align*}

Hence the ode becomes

\begin{align*} p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )+y^{n} = 0 \end{align*}

Which is now solved as first order ode for \(p(y)\).

Entering first order ode separable solverThe ode

\begin{equation} p^{\prime } = -\frac {y^{n}}{p^{2}} \end{equation}

is separable as it can be written as

\begin{align*} p^{\prime }&= -\frac {y^{n}}{p^{2}}\\ &= f(y) g(p) \end{align*}

Where

\begin{align*} f(y) &= -y^{n}\\ g(p) &= \frac {1}{p^{2}} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy} \\ \int { p^{2}\,dp} &= \int { -y^{n} \,dy} \\ \end{align*}

\[ \frac {p^{3}}{3}=-\frac {y^{n +1}}{n +1}+c_1 \]

Solving for \(p\) gives

\begin{align*} p &= \frac {\left (-3 \left (n +1\right )^{2} \left (-c_1 n +y^{n +1}-c_1 \right )\right )^{{1}/{3}}}{n +1} \\ p &= -\frac {\left (-3 \left (n +1\right )^{2} \left (-c_1 n +y^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 \left (n +1\right )}-\frac {i \sqrt {3}\, \left (-3 \left (n +1\right )^{2} \left (-c_1 n +y^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 \left (n +1\right )} \\ p &= -\frac {\left (-3 \left (n +1\right )^{2} \left (-c_1 n +y^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 \left (n +1\right )}+\frac {i \sqrt {3}\, \left (-3 \left (n +1\right )^{2} \left (-c_1 n +y^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 n +2} \\ \end{align*}

For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is

\begin{align*} y^{\prime } = \frac {\left (-3 \left (n +1\right )^{2} \left (-c_1 n +y^{n +1}-c_1 \right )\right )^{{1}/{3}}}{n +1} \end{align*}

Entering first order ode autonomous solverIntegration could not be done or too complicated. Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as

\begin{align*} \int _{}^{y}-\frac {n +1}{\left (-3 \left (n +1\right )^{2} \left (-c_1 n +\tau ^{n +1}-c_1 \right )\right )^{{1}/{3}}}d \tau +x +c_2 = 0\end{align*}

For solution (2) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is

\begin{align*} y^{\prime } = -\frac {\left (-3 \left (n +1\right )^{2} \left (-c_1 n +y^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 \left (n +1\right )}-\frac {i \sqrt {3}\, \left (-3 \left (n +1\right )^{2} \left (-c_1 n +y^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 \left (n +1\right )} \end{align*}

Entering first order ode autonomous solverIntegration could not be done or too complicated. Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as

\begin{align*} \int _{}^{y}-\frac {1}{-\frac {\left (-3 \left (n +1\right )^{2} \left (-c_1 n +\tau ^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 \left (n +1\right )}-\frac {i \sqrt {3}\, \left (-3 \left (n +1\right )^{2} \left (-c_1 n +\tau ^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 \left (n +1\right )}}d \tau +x +c_3 = 0\end{align*}

For solution (3) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is

\begin{align*} y^{\prime } = -\frac {\left (-3 \left (n +1\right )^{2} \left (-c_1 n +y^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 \left (n +1\right )}+\frac {i \sqrt {3}\, \left (-3 \left (n +1\right )^{2} \left (-c_1 n +y^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 n +2} \end{align*}

Entering first order ode autonomous solverIntegration could not be done or too complicated. Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as

\begin{align*} \int _{}^{y}-\frac {1}{-\frac {\left (-3 \left (n +1\right )^{2} \left (-c_1 n +\tau ^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 \left (n +1\right )}+\frac {i \sqrt {3}\, \left (-3 \left (n +1\right )^{2} \left (-c_1 n +\tau ^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 n +2}}d \tau +x +c_4 = 0\end{align*}

Summary of solutions found

\begin{align*} \int _{}^{y}-\frac {1}{-\frac {\left (-3 \left (n +1\right )^{2} \left (-c_1 n +\tau ^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 \left (n +1\right )}-\frac {i \sqrt {3}\, \left (-3 \left (n +1\right )^{2} \left (-c_1 n +\tau ^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 \left (n +1\right )}}d \tau +x +c_3 &= 0 \\ \int _{}^{y}-\frac {1}{-\frac {\left (-3 \left (n +1\right )^{2} \left (-c_1 n +\tau ^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 \left (n +1\right )}+\frac {i \sqrt {3}\, \left (-3 \left (n +1\right )^{2} \left (-c_1 n +\tau ^{n +1}-c_1 \right )\right )^{{1}/{3}}}{2 n +2}}d \tau +x +c_4 &= 0 \\ \int _{}^{y}-\frac {n +1}{\left (-3 \left (n +1\right )^{2} \left (-c_1 n +\tau ^{n +1}-c_1 \right )\right )^{{1}/{3}}}d \tau +x +c_2 &= 0 \\ \end{align*}

2.2.6.2 ✓ Maple. Time used: 0.011 (sec). Leaf size: 172

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

\begin{align*} \frac {2 \left (1+n \right ) \int _{}^{y}\frac {1}{{\left (-\left (3 \textit {\_a}^{1+n}-c_1 \right ) \left (1+n \right )^{2}\right )}^{{1}/{3}}}d \textit {\_a} +\left (x +c_2 \right ) \left (1+i \sqrt {3}\right )}{-1-i \sqrt {3}} &= 0 \\ -\frac {2 i \left (1+n \right ) \int _{}^{y}\frac {1}{{\left (-\left (3 \textit {\_a}^{1+n}-c_1 \right ) \left (1+n \right )^{2}\right )}^{{1}/{3}}}d \textit {\_a} +\left (x +c_2 \right ) \left (\sqrt {3}+i\right )}{\sqrt {3}+i} &= 0 \\ \int _{}^{y}\frac {1}{{\left (-\left (3 \textit {\_a}^{1+n}-c_1 \right ) \left (1+n \right )^{2}\right )}^{{1}/{3}}}d \textit {\_a} n +\int _{}^{y}\frac {1}{{\left (-\left (3 \textit {\_a}^{1+n}-c_1 \right ) \left (1+n \right )^{2}\right )}^{{1}/{3}}}d \textit {\_a} -c_2 -x &= 0 \\ \end{align*}

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
trying 2nd order WeierstrassP 
trying 2nd order JacobiSN 
differential order: 2; trying a linearization to 3rd order 
trying 2nd order ODE linearizable_by_differentiation 
trying 2nd order, 2 integrating factors of the form mu(x,y) 
trying differential order: 2; missing variables 
   -> Computing symmetries using: way = 3 
-> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)+_a^n/_b(_a) = 0, _b(_a) 
, HINT = [[-3/(n-2)*_a, -_b*(1+n)/(n-2)]] 
   *** Sublevel 2 *** 
   symmetry methods on request 
   1st order, trying reduction of order with given symmetries: 
[-3/(n-2)*_a, -_b*(1+n)/(n-2)] 
   1st order, trying the canonical coordinates of the invariance group 
      -> Calling odsolve with the ODE, diff(y(x),x) = 1/3*y(x)*(1+n)/x, y(x) 
         *** Sublevel 3 *** 
         Methods for first order ODEs: 
         --- Trying classification methods --- 
         trying a quadrature 
         trying 1st order linear 
         <- 1st order linear successful 
   <- 1st order, canonical coordinates successful 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful
 

2.2.6.3 ✓ Mathematica. Time used: 1.291 (sec). Leaf size: 910

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

\begin{align*} \text {Solution too large to show}\end{align*}

2.2.6.4 ✗ Sympy

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

NotImplementedError : The given ODE y(x)**n/Derivative(y(x), (x, 2)) + Derivative(y(x), x) cannot be
 

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',)