2.5.1 Problem 1
Internal
problem
ID
[10236]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
5.0
Problem
number
:
1
Date
solved
:
Monday, January 26, 2026 at 09:25:13 PM
CAS
classification
:
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
2.5.1.1 Solved as second_order_ode_missing_x algorithm
31.160 (sec)
\begin{align*}
y^{\prime \prime }&=A y^{{2}/{3}} \\
\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 ) \left (\frac {d}{d y}p \left (y \right )\right ) = A \,y^{{2}/{3}} \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 {A \,y^{{2}/{3}}}{p}
\end{equation}
is separable as it can be written as
\begin{align*} p^{\prime }&= \frac {A \,y^{{2}/{3}}}{p}\\ &= f(y) g(p) \end{align*}
Where
\begin{align*} f(y) &= A \,y^{{2}/{3}}\\ g(p) &= \frac {1}{p} \end{align*}
Integrating gives
\begin{align*}
\int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy} \\
\int { p\,dp} &= \int { A \,y^{{2}/{3}} \,dy} \\
\end{align*}
\[
\frac {p^{2}}{2}=\frac {3 y^{{5}/{3}} A}{5}+c_1
\]
Solving for \(p\) gives \begin{align*}
p &= -\frac {\sqrt {30 y^{{5}/{3}} A +50 c_1}}{5} \\
p &= \frac {\sqrt {30 y^{{5}/{3}} A +50 c_1}}{5} \\
\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 {\sqrt {30 y^{{5}/{3}} A +50 c_1}}{5} \end{align*}
Entering first order ode autonomous solverUnable to integrate (or intergal too complicated), and
since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}-\frac {5}{\sqrt {30 \tau ^{{5}/{3}} A +50 c_1}}d \tau = x +c_2 \]
Singular solutions are
found by solving \begin{align*} -\frac {\sqrt {30 y^{{5}/{3}} A +50 c_1}}{5}&= 0 \end{align*}
for \(y\). 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*} y = \frac {5^{{3}/{5}} 3^{{2}/{5}} \left (-c_1 \,A^{4}\right )^{{3}/{5}}}{3 A^{3}} \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 {\sqrt {30 y^{{5}/{3}} A +50 c_1}}{5} \end{align*}
Entering first order ode autonomous solverUnable to integrate (or intergal too complicated), and
since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {5}{\sqrt {30 \tau ^{{5}/{3}} A +50 c_1}}d \tau = x +c_3 \]
Singular solutions are
found by solving \begin{align*} \frac {\sqrt {30 y^{{5}/{3}} A +50 c_1}}{5}&= 0 \end{align*}
for \(y\). 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*} y = \frac {5^{{3}/{5}} 3^{{2}/{5}} \left (-c_1 \,A^{4}\right )^{{3}/{5}}}{3 A^{3}} \end{align*}
The above solution was found not to satisfy the ode or the IC. Hence it is removed.
2.5.1.2 ✓ Maple. Time used: 0.026 (sec). Leaf size: 61
ode:=diff(diff(y(x),x),x) = A*y(x)^(2/3);
dsolve(ode,y(x), singsol=all);
\begin{align*}
y &= 0 \\
-5 \int _{}^{y}\frac {1}{\sqrt {30 \textit {\_a}^{{5}/{3}} A -5 c_1}}d \textit {\_a} -x -c_2 &= 0 \\
5 \int _{}^{y}\frac {1}{\sqrt {30 \textit {\_a}^{{5}/{3}} A -5 c_1}}d \textit {\_a} -x -c_2 &= 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*_a^(2/3) = 0, _b(_a),
HINT = [[_a, 5/6*_b]]
*** Sublevel 2 ***
symmetry methods on request
1st order, trying reduction of order with given symmetries:
[_a, 5/6*_b]
1st order, trying the canonical coordinates of the invariance group
-> Calling odsolve with the ODE, diff(y(x),x) = 5/6*y(x)/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
2.5.1.3 ✓ Mathematica. Time used: 0.056 (sec). Leaf size: 75
ode=D[y[x],{x,2}]==A*y[x]^(2/3);
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\[
\text {Solve}\left [\frac {y(x)^2 \left (1+\frac {6 A y(x)^{5/3}}{5 c_1}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{5},\frac {8}{5},-\frac {6 A y(x)^{5/3}}{5 c_1}\right ){}^2}{\frac {6}{5} A y(x)^{5/3}+c_1}=(x+c_2){}^2,y(x)\right ]
\]
2.5.1.4 ✓ Sympy. Time used: 11.672 (sec). Leaf size: 99
from sympy import *
x = symbols("x")
A = symbols("A")
y = Function("y")
ode = Eq(-A*y(x)**(2/3) + Derivative(y(x), (x, 2)),0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
\[
\left [ \frac {3 y{\left (x \right )} \Gamma \left (\frac {3}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{5} \\ \frac {8}{5} \end {matrix}\middle | {\frac {6 A e^{i \pi } y^{\frac {5}{3}}{\left (x \right )}}{5 C_{1}}} \right )}}{5 \sqrt {C_{1}} \Gamma \left (\frac {8}{5}\right )} = C_{2} + x, \ \frac {3 y{\left (x \right )} \Gamma \left (\frac {3}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{5} \\ \frac {8}{5} \end {matrix}\middle | {\frac {6 A e^{i \pi } y^{\frac {5}{3}}{\left (x \right )}}{5 C_{1}}} \right )}}{5 \sqrt {C_{1}} \Gamma \left (\frac {8}{5}\right )} = C_{2} - x\right ]
\]
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))
('2nd_nonlinear_autonomous_conserved', '2nd_nonlinear_autonomous_conserved_Integral')