52.1.52 problem 52

Internal problem ID [10423]
Book : Second order enumerated odes
Section : section 1
Problem number : 52
Date solved : Tuesday, September 30, 2025 at 07:23:27 PM
CAS classification : [[_2nd_order, _missing_x]]

\begin{align*} y {y^{\prime \prime }}^{3}+y^{3} {y^{\prime }}^{5}&=0 \end{align*}

✓ Maple. Time used: 0.275 (sec). Leaf size: 208

ode:=y(x)*diff(diff(y(x),x),x)^3+y(x)^3*diff(y(x),x)^5 = 0; 
dsolve(ode,y(x), singsol=all);
 

\begin{align*} y &= 0 \\ y &= c_1 \\ \int _{}^{y}\frac {1}{\operatorname {RootOf}\left (5 \int _{\textit {\_g}}^{\textit {\_Z}}\frac {1}{\textit {\_a} \left (-\textit {\_a}^{2} \textit {\_f}^{2}\right )^{{1}/{3}}-5 \textit {\_f}}d \textit {\_f} -\ln \left (\textit {\_a}^{5}+125\right )+5 c_1 \right )}d \textit {\_a} -x -c_2 &= 0 \\ \int _{}^{y}\frac {1}{\operatorname {RootOf}\left (-i \ln \left (\textit {\_a}^{5}+125\right )+\ln \left (\textit {\_a}^{5}+125\right ) \sqrt {3}+20 \int _{\textit {\_g}}^{\textit {\_Z}}\frac {1}{2 i \textit {\_a} \left (-\textit {\_a}^{2} \textit {\_f}^{2}\right )^{{1}/{3}}+5 i \textit {\_f} +5 \sqrt {3}\, \textit {\_f}}d \textit {\_f} -20 c_1 \right )}d \textit {\_a} -x -c_2 &= 0 \\ \int _{}^{y}\frac {1}{\operatorname {RootOf}\left (\ln \left (\textit {\_a}^{5}+125\right ) \sqrt {3}+i \ln \left (\textit {\_a}^{5}+125\right )+20 \int _{\textit {\_g}}^{\textit {\_Z}}\frac {1}{-2 i \textit {\_a} \left (-\textit {\_a}^{2} \textit {\_f}^{2}\right )^{{1}/{3}}+5 \sqrt {3}\, \textit {\_f} -5 i \textit {\_f}}d \textit {\_f} -20 c_1 \right )}d \textit {\_a} -x -c_2 &= 0 \\ \end{align*}

✓ Mathematica. Time used: 22.559 (sec). Leaf size: 449

ode=y[x]*D[y[x],{x,2}]^3+y[x]^3*D[y[x],x]^5==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to 0\\ y(x)&\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right )}{c_1{}^3}\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},-\frac {3 i \left (-i+\sqrt {3}\right ) \text {$\#$1}^{5/3}}{10 c_1}\right )}{c_1{}^3}\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},\frac {3 i \left (i+\sqrt {3}\right ) \text {$\#$1}^{5/3}}{10 c_1}\right )}{c_1{}^3}\&\right ][x+c_2]\\ y(x)&\to 0\\ y(x)&\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},\frac {3 \text {$\#$1}^{5/3}}{5 (-c_1)}\right )}{(-c_1){}^3}\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},-\frac {3 i \left (-i+\sqrt {3}\right ) \text {$\#$1}^{5/3}}{10 (-c_1)}\right )}{(-c_1){}^3}\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},\frac {3 i \left (i+\sqrt {3}\right ) \text {$\#$1}^{5/3}}{10 (-c_1)}\right )}{(-c_1){}^3}\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right )}{c_1{}^3}\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},-\frac {3 i \left (-i+\sqrt {3}\right ) \text {$\#$1}^{5/3}}{10 c_1}\right )}{c_1{}^3}\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},\frac {3 i \left (i+\sqrt {3}\right ) \text {$\#$1}^{5/3}}{10 c_1}\right )}{c_1{}^3}\&\right ][x+c_2] \end{align*}

✓ Sympy. Time used: 7.213 (sec). Leaf size: 3

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(y(x)**3*Derivative(y(x), x)**5 + y(x)*Derivative(y(x), (x, 2))**3,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

\[ y{\left (x \right )} = 0 \]