2.2.5 Problem 5

Internal problem ID [10416]
Book : Second order enumerated odes
Section : section 2
Problem number : 5
Date solved : Monday, January 26, 2026 at 10:19:01 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

2.2.5.1 second order ode missing x

5.914 (sec)

\begin{align*} y^{\prime } y^{\prime \prime }+y^{2}&=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^{2} = 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^{2}}{p^{2}} \end{equation}

is separable as it can be written as

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

Where

\begin{align*} f(y) &= -y^{2}\\ 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^{2} \,dy} \\ \end{align*}

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

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

\begin{align*} \frac {{y^{\prime }}^{3}}{3} = -\frac {y^{3}}{3}+c_1 \end{align*}

Entering first order ode dAlembert solverLet \(p=y^{\prime }\) the ode becomes

\begin{align*} \frac {p^{3}}{3} = -\frac {y^{3}}{3}+c_1 \end{align*}

Solving for \(y\) from the above results in

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

This has the form

\begin{align*} y=x f(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=y'(x)\). Each of the above ode’s is dAlembert ode which is now solved.

Solving ode 1A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= 0\\ g &= \left (-p^{3}+3 c_1 \right )^{{1}/{3}} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = -\frac {p^{2} p^{\prime }\left (x \right )}{\left (-p^{3}+3 c_1 \right )^{{2}/{3}}} \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = 3^{{1}/{3}} c_1^{{1}/{3}} \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = -\frac {\left (-p \left (x \right )^{3}+3 c_1 \right )^{{2}/{3}}}{p \left (x \right )} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{p \left (x \right )}-\frac {\tau }{\left (-\tau ^{3}+3 c_1 \right )^{{2}/{3}}}d \tau = x +c_2 \]

Singular solutions are found by solving

\begin{align*} -\frac {\left (-p^{3}+3 c_1 \right )^{{2}/{3}}}{p}&= 0 \end{align*}

for \(p \left (x \right )\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 3^{{1}/{3}} c_1^{{1}/{3}}\\ p \left (x \right ) = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}-\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2} \end{align*}

Substituing the above solution for \(p\) in (2A) gives

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

Solving ode 2A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= 0\\ g &= \frac {\left (-p^{3}+3 c_1 \right )^{{1}/{3}} \left (-1+i \sqrt {3}\right )}{2} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \left (\frac {p^{2}}{2 \left (-p^{3}+3 c_1 \right )^{{2}/{3}}}-\frac {i p^{2} \sqrt {3}}{2 \left (-p^{3}+3 c_1 \right )^{{2}/{3}}}\right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = -\frac {c_1^{{1}/{3}} \left (-i 3^{{5}/{6}}+3^{{1}/{3}}\right )}{2} \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )}{\frac {p \left (x \right )^{2}}{2 \left (-p \left (x \right )^{3}+3 c_1 \right )^{{2}/{3}}}-\frac {i p \left (x \right )^{2} \sqrt {3}}{2 \left (-p \left (x \right )^{3}+3 c_1 \right )^{{2}/{3}}}} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{p \left (x \right )}-\frac {\tau \left (-1+i \sqrt {3}\right )}{2 \left (-\tau ^{3}+3 c_1 \right )^{{2}/{3}}}d \tau = x +c_3 \]

Singular solutions are found by solving

\begin{align*} -\frac {2 \left (-p^{3}+3 c_1 \right )^{{2}/{3}}}{p \left (-1+i \sqrt {3}\right )}&= 0 \end{align*}

for \(p \left (x \right )\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 3^{{1}/{3}} c_1^{{1}/{3}}\\ p \left (x \right ) = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}-\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2} \end{align*}

Substituing the above solution for \(p\) in (2A) gives

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

Solving ode 3A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= 0\\ g &= -\frac {\left (-p^{3}+3 c_1 \right )^{{1}/{3}} \left (1+i \sqrt {3}\right )}{2} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \left (\frac {p^{2}}{2 \left (-p^{3}+3 c_1 \right )^{{2}/{3}}}+\frac {i p^{2} \sqrt {3}}{2 \left (-p^{3}+3 c_1 \right )^{{2}/{3}}}\right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = -\frac {c_1^{{1}/{3}} \left (i 3^{{5}/{6}}+3^{{1}/{3}}\right )}{2} \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )}{\frac {p \left (x \right )^{2}}{2 \left (-p \left (x \right )^{3}+3 c_1 \right )^{{2}/{3}}}+\frac {i p \left (x \right )^{2} \sqrt {3}}{2 \left (-p \left (x \right )^{3}+3 c_1 \right )^{{2}/{3}}}} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{p \left (x \right )}\frac {\tau \left (1+i \sqrt {3}\right )}{2 \left (-\tau ^{3}+3 c_1 \right )^{{2}/{3}}}d \tau = x +c_4 \]

Singular solutions are found by solving

\begin{align*} \frac {2 \left (-p^{3}+3 c_1 \right )^{{2}/{3}}}{p \left (1+i \sqrt {3}\right )}&= 0 \end{align*}

for \(p \left (x \right )\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 3^{{1}/{3}} c_1^{{1}/{3}}\\ p \left (x \right ) = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}-\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2} \end{align*}

Substituing the above solution for \(p\) in (2A) gives

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

The solution \(y = {\left (-\operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}-\frac {\tau }{\left (-\tau ^{3}+3 c_1 \right )^{{2}/{3}}}d \tau +x +c_2 \right )^{3}+3 c_1 \right )}^{{1}/{3}}\) simplifies to

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

The solution \(y = \frac {{\left (-\operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}-\frac {\tau \left (-1+i \sqrt {3}\right )}{2 \left (-\tau ^{3}+3 c_1 \right )^{{2}/{3}}}d \tau +x +c_3 \right )^{3}+3 c_1 \right )}^{{1}/{3}} \left (-1+i \sqrt {3}\right )}{2}\) simplifies to

\begin{align*} y &= \frac {{\left (-\operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}-\frac {\tau \left (-1+i \sqrt {3}\right )}{2 \left (-\tau ^{3}+3 c_1 \right )^{{2}/{3}}}d \tau +x +c_3 \right )^{3}+3 c_1 \right )}^{{1}/{3}} \left (-1+i \sqrt {3}\right )}{2} \\ \end{align*}

The solution \(y = -\frac {{\left (-\operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}\frac {\tau \left (1+i \sqrt {3}\right )}{2 \left (-\tau ^{3}+3 c_1 \right )^{{2}/{3}}}d \tau +x +c_4 \right )^{3}+3 c_1 \right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}{2}\) simplifies to

\begin{align*} y &= -\frac {{\left (-\operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}\frac {\tau \left (1+i \sqrt {3}\right )}{2 \left (-\tau ^{3}+3 c_1 \right )^{{2}/{3}}}d \tau +x +c_4 \right )^{3}+3 c_1 \right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}{2} \\ \end{align*}

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

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

2.2.5.2 ✓ Maple. Time used: 0.025 (sec). Leaf size: 61

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

\begin{align*} y &= 0 \\ y &= {\mathrm e}^{\frac {\sqrt {3}\, \int \tan \left (\operatorname {RootOf}\left (-\sqrt {3}\, \ln \left (\cos \left (\textit {\_Z} \right )^{2}\right )-2 \sqrt {3}\, \ln \left (\tan \left (\textit {\_Z} \right )+\sqrt {3}\right )+6 \sqrt {3}\, c_1 +6 \sqrt {3}\, x +6 \textit {\_Z} \right )\right )d x}{2}+c_2 +\frac {x}{2}} \\ \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 
Try integration with the canonical coordinates of the symmetry [0, y] 
-> Calling odsolve with the ODE, diff(_b(_a),_a) = -(_b(_a)^3+1)/_b(_a), _b(_a) 
, explicit, HINT = [[1, 0]] 
   *** Sublevel 2 *** 
   symmetry methods on request 
   1st order, trying reduction of order with given symmetries: 
[1, 0] 
   1st order, trying the canonical coordinates of the invariance group 
   <- 1st order, canonical coordinates successful 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful
                                                                                   
                                                                                   
 

2.2.5.3 ✓ Mathematica. Time used: 0.284 (sec). Leaf size: 55

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

\begin{align*} y(x)&\to c_2 \exp \left (\int _1^x\text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {K[1]}{(K[1]+1) \left (K[1]^2-K[1]+1\right )}dK[1]\&\right ][c_1-K[2]]dK[2]\right ) \end{align*}

2.2.5.4 ✗ Sympy

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

NotImplementedError : The given ODE y(x)**2/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',)