29.14.26 problem 407
Internal
problem
ID
[5005]
Book
:
Ordinary
differential
equations
and
their
solutions.
By
George
Moseley
Murphy.
1960
Section
:
Various
14
Problem
number
:
407
Date
solved
:
Sunday, March 30, 2025 at 04:30:09 AM
CAS
classification
:
[_separable]
\begin{align*} y^{\prime } \left (x^{3}+1\right )^{{2}/{3}}+\left (1+y^{3}\right )^{{2}/{3}}&=0 \end{align*}
✓ Maple. Time used: 0.014 (sec). Leaf size: 119
ode:=diff(y(x),x)*(x^3+1)^(2/3)+(1+y(x)^3)^(2/3) = 0;
dsolve(ode,y(x), singsol=all);
\[
c_1 +\frac {2 \pi \sqrt {3}\, \left (\left (-y^{3}\right )^{{1}/{6}} \operatorname {LegendreP}\left (-\frac {1}{3}, -\frac {1}{3}, \frac {-x^{3}+1}{x^{3}+1}\right ) \left (1+y^{3}\right )^{{1}/{3}} x +\operatorname {LegendreP}\left (-\frac {1}{3}, -\frac {1}{3}, \frac {-y^{3}+1}{1+y^{3}}\right ) \left (-x^{3}\right )^{{1}/{6}} y \left (x^{3}+1\right )^{{1}/{3}}\right )}{9 \left (-x^{3}\right )^{{1}/{6}} \left (x^{3}+1\right )^{{1}/{3}} \left (-y^{3}\right )^{{1}/{6}} \left (1+y^{3}\right )^{{1}/{3}} \Gamma \left (\frac {2}{3}\right )} = 0
\]
✓ Mathematica. Time used: 0.464 (sec). Leaf size: 72
ode=D[y[x],x](1+x^3)^(2/3)+(1+y[x]^3)^(2/3)==0;
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to \text {InverseFunction}\left [\text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\text {$\#$1}^3\right )\&\right ]\left [-x \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-x^3\right )+c_1\right ] \\
y(x)\to -1 \\
y(x)\to \sqrt [3]{-1} \\
y(x)\to -(-1)^{2/3} \\
\end{align*}
✓ Sympy. Time used: 0.735 (sec). Leaf size: 61
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq((x**3 + 1)**(2/3)*Derivative(y(x), x) + (y(x)**3 + 1)**(2/3),0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
\[
\frac {y{\left (x \right )} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {4}{3} \end {matrix}\middle | {e^{i \pi } y^{3}{\left (x \right )}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} = C_{1} - \frac {x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {4}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {4}{3}\right )}
\]