2.2.6 Problem 6

Internal problem ID [10417]
Book : Second order enumerated odes
Section : section 2
Problem number : 6
Date solved : Monday, January 26, 2026 at 10:19:12 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

2.2.6.1 Solved as second order ode by reversing roles of dependent and independent variables

7.996 (sec)

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

Entering second order ode flip role solverreversing the roles of the dependent and independent variables, the ode becomes

\begin{align*} \frac {d^{2}}{d y^{2}}x \left (y \right ) = y^{n} \left (\frac {d}{d y}x \left (y \right )\right )^{4} \end{align*}

Which is now solved for \(x \left (y \right )\) instead for \(y\) Entering second order ode missing \(y\) solverThis is second order ode with missing dependent variable \(x \left (y \right )\). Let

\begin{align*} u(y) &= \frac {d}{d y}x \left (y \right ) \end{align*}

Then

\begin{align*} u'(y) &= \frac {d^{2}}{d y^{2}}x \left (y \right ) \end{align*}

Hence the ode becomes

\begin{align*} \frac {d}{d y}u \left (y \right )-y^{n} u \left (y \right )^{4} = 0 \end{align*}

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

Entering first order ode separable solverThe ode

\begin{equation} \frac {d}{d y}u \left (y \right ) = y^{n} u \left (y \right )^{4} \end{equation}

is separable as it can be written as

\begin{align*} \frac {d}{d y}u \left (y \right )&= y^{n} u \left (y \right )^{4}\\ &= f(y) g(u) \end{align*}

Where

\begin{align*} f(y) &= y^{n}\\ g(u) &= u^{4} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(y) \,dy} \\ \int { \frac {1}{u^{4}}\,du} &= \int { y^{n} \,dy} \\ \end{align*}

\[ -\frac {1}{3 u \left (y \right )^{3}}=\frac {y^{n +1}}{n +1}+c_3 \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ u^{4}=0 \]

for \(u \left (y \right )\) gives

\begin{align*} u \left (y \right )&=0 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} -\frac {1}{3 u \left (y \right )^{3}} &= \frac {y^{n +1}}{n +1}+c_3 \\ u \left (y \right ) &= 0 \\ \end{align*}

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

\begin{align*} -\frac {1}{3 u \left (y \right )^{3}} &= \frac {y^{n +1}}{n +1}+c_3 \\ u \left (y \right ) &= 0 \\ \end{align*}

For solution \(-\frac {1}{3 u \left (y \right )^{3}} = \frac {y^{n +1}}{n +1}+c_3\), since \(u=\frac {d}{d y}x \left (y \right )\) then the new first order ode to solve is

\begin{align*} -\frac {1}{3 \left (\frac {d}{d y}x \left (y \right )\right )^{3}} = \frac {y^{n +1}}{n +1}+c_3 \end{align*}

Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} \frac {d}{d y}x \left (y \right )&=\frac {{\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}}}{3 c_3 n +3 c_3 +3 y^{n +1}} \\ \tag{2} \frac {d}{d y}x \left (y \right )&=-\frac {{\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}}}{6 \left (c_3 n +y^{n +1}+c_3 \right )}-\frac {i \sqrt {3}\, {\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}}}{6 \left (c_3 n +y^{n +1}+c_3 \right )} \\ \tag{3} \frac {d}{d y}x \left (y \right )&=-\frac {{\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}}}{6 \left (c_3 n +y^{n +1}+c_3 \right )}+\frac {i \sqrt {3}\, {\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}}}{6 c_3 n +6 y^{n +1}+6 c_3} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

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

\begin{align*} \int {dx} &= \int {\frac {{\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}}}{3 c_3 n +3 c_3 +3 y^{n +1}}\, dy}\\ x \left (y \right ) &= \int \frac {{\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}}}{3 c_3 n +3 c_3 +3 y^{n +1}}d y + c_5 \end{align*}

\begin{align*} x \left (y \right )&= \int \frac {{\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}}}{3 c_3 n +3 c_3 +3 y^{n +1}}d y +c_5 \end{align*}

Solving Eq. (2)

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

\begin{align*} \int {dx} &= \int {-\frac {{\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}{6 \left (c_3 n +y^{n +1}+c_3 \right )}\, dy}\\ x \left (y \right ) &= \int -\frac {{\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}{6 \left (c_3 n +y^{n +1}+c_3 \right )}d y + c_6 \end{align*}

\begin{align*} x \left (y \right )&= \int -\frac {{\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}{6 \left (c_3 n +y^{n +1}+c_3 \right )}d y +c_6 \end{align*}

Solving Eq. (3)

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

\begin{align*} \int {dx} &= \int {\frac {{\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}} \left (-1+i \sqrt {3}\right )}{6 c_3 n +6 y^{n +1}+6 c_3}\, dy}\\ x \left (y \right ) &= \int \frac {{\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}} \left (-1+i \sqrt {3}\right )}{6 c_3 n +6 y^{n +1}+6 c_3}d y + c_7 \end{align*}

\begin{align*} x \left (y \right )&= \int \frac {{\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}} \left (-1+i \sqrt {3}\right )}{6 c_3 n +6 y^{n +1}+6 c_3}d y +c_7 \end{align*}

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

\begin{align*} x \left (y \right ) &= \int \frac {{\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}}}{3 c_3 n +3 c_3 +3 y^{n +1}}d y +c_5 \\ x \left (y \right ) &= \int -\frac {{\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}{6 \left (c_3 n +y^{n +1}+c_3 \right )}d y +c_6 \\ x \left (y \right ) &= \int \frac {{\left (\left (-9 n -9\right ) \left (c_3 n +y^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}} \left (-1+i \sqrt {3}\right )}{6 c_3 n +6 y^{n +1}+6 c_3}d y +c_7 \\ \end{align*}

For solution \(u \left (y \right ) = 0\), since \(u=\frac {d}{d y}x \left (y \right )\) then the new first order ode to solve is

\begin{align*} \frac {d}{d y}x \left (y \right ) = 0 \end{align*}

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

\begin{align*} \int {dx} &= \int {0\, dy} + c_8 \\ x \left (y \right ) &= c_8 \end{align*}

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

\begin{align*} x \left (y \right ) &= c_8 \\ \end{align*}

Now that the reversed roles ode was solved, we will change back to the original roles. This results in the above solution becoming the following.

\begin{align*} x &= \int _{}^{y}\frac {{\left (\left (-9 n -9\right ) \left (c_3 n +\textit {\_a}^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}}}{3 c_3 n +3 \textit {\_a}^{n +1}+3 c_3}d \textit {\_a} +c_5 \\ x &= \int _{}^{y}\frac {{\left (\left (-9 n -9\right ) \left (c_3 n +\textit {\_a}^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}} \left (-1+i \sqrt {3}\right )}{6 c_3 n +6 \textit {\_a}^{n +1}+6 c_3}d \textit {\_a} +c_7 \\ x &= \int _{}^{y}-\frac {{\left (\left (-9 n -9\right ) \left (c_3 n +\textit {\_a}^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}{6 \left (c_3 n +\textit {\_a}^{n +1}+c_3 \right )}d \textit {\_a} +c_6 \\ \end{align*}

Summary of solutions found

\begin{align*} x &= \int _{}^{y}\frac {{\left (\left (-9 n -9\right ) \left (c_3 n +\textit {\_a}^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}}}{3 c_3 n +3 \textit {\_a}^{n +1}+3 c_3}d \textit {\_a} +c_5 \\ x &= \int _{}^{y}\frac {{\left (\left (-9 n -9\right ) \left (c_3 n +\textit {\_a}^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}} \left (-1+i \sqrt {3}\right )}{6 c_3 n +6 \textit {\_a}^{n +1}+6 c_3}d \textit {\_a} +c_7 \\ x &= \int _{}^{y}-\frac {{\left (\left (-9 n -9\right ) \left (c_3 n +\textit {\_a}^{n +1}+c_3 \right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}{6 \left (c_3 n +\textit {\_a}^{n +1}+c_3 \right )}d \textit {\_a} +c_6 \\ \end{align*}

2.2.6.2 ✓ Maple. Time used: 0.011 (sec). Leaf size: 174

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

\begin{align*} \frac {\left (-2-2 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',)