2.1.52 Problem 52
Internal
problem
ID
[9123]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
52
Date
solved
:
Friday, February 21, 2025 at 09:17:40 PM
CAS
classification
:
[[_2nd_order, _missing_x]]
Solve
\begin{align*} y {y^{\prime \prime }}^{3}+y^{3} {y^{\prime }}^{5}&=0 \end{align*}
Factoring the ode gives these factors
\begin{align*}
\tag{1} y &= 0 \\
\tag{2} y^{2} {y^{\prime }}^{5}+{y^{\prime \prime }}^{3} &= 0 \\
\end{align*}
Now each of the above equations is solved in turn.
Solving equation (1)
Solving for \(y\) from
\begin{align*} y = 0 \end{align*}
Solving gives \(y = 0\)
Solving equation (2)
Solved as second order missing x ode
Time used: 5.569 (sec)
This 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*} y p \left (y \right )^{3} \left (\frac {d}{d y}p \left (y \right )\right )^{3}+y^{3} p \left (y \right )^{5} = 0 \end{align*}
Which is now solved as first order ode for \(p(y)\).
Factoring the ode gives these factors
\begin{align*}
\tag{1} p^{3} &= 0 \\
\tag{2} p^{2} y^{2}+{p^{\prime }}^{3} &= 0 \\
\end{align*}
Now each of the above equations is solved in turn.
Solving equation (1)
Solving for \(p\) from
\begin{align*} p^{3} = 0 \end{align*}
Solving gives \(p = 0\)
Solving equation (2)
The ode has the form
\begin{align*} (p')^{\frac {n}{m}} &= f(y) g(p)\tag {1} \end{align*}
Where \(n=3, m=1, f=-y^{2} , g=p^{2}\). Hence the ode is
\begin{align*} (p')^{3} &= -p^{2} y^{2} \end{align*}
Solving for \(p^{\prime }\) from (1) gives
\begin{align*} p^{\prime } &=\left (f g \right )^{{1}/{3}}\\ p^{\prime } &=-\frac {\left (f g \right )^{{1}/{3}}}{2}+\frac {i \sqrt {3}\, \left (f g \right )^{{1}/{3}}}{2}\\ p^{\prime } &=-\frac {\left (f g \right )^{{1}/{3}}}{2}-\frac {i \sqrt {3}\, \left (f g \right )^{{1}/{3}}}{2} \end{align*}
To be able to solve as separable ode, we have to now assume that \(f>0,g>0\).
\begin{align*} -y^{2} &> 0\\ p^{2} &> 0 \end{align*}
Under the above assumption the differential equations become separable and can be written as
\begin{align*} p^{\prime } &=f^{{1}/{3}} g^{{1}/{3}}\\ p^{\prime } &=\frac {f^{{1}/{3}} g^{{1}/{3}} \left (-1+i \sqrt {3}\right )}{2}\\ p^{\prime } &=-\frac {f^{{1}/{3}} g^{{1}/{3}} \left (1+i \sqrt {3}\right )}{2} \end{align*}
Therefore
\begin{align*} \frac {1}{g^{{1}/{3}}} \, dp &= \left (f^{{1}/{3}}\right )\,dy\\ \frac {2}{g^{{1}/{3}} \left (-1+i \sqrt {3}\right )} \, dp &= \left (f^{{1}/{3}}\right )\,dy\\ -\frac {2}{g^{{1}/{3}} \left (1+i \sqrt {3}\right )} \, dp &= \left (f^{{1}/{3}}\right )\,dy \end{align*}
Replacing \(f(y),g(p)\) by their values gives
\begin{align*} \frac {1}{\left (p^{2}\right )^{{1}/{3}}} \, dp &= \left (\left (-y^{2}\right )^{{1}/{3}}\right )\,dy\\ \frac {2}{\left (p^{2}\right )^{{1}/{3}} \left (-1+i \sqrt {3}\right )} \, dp &= \left (\left (-y^{2}\right )^{{1}/{3}}\right )\,dy\\ -\frac {2}{\left (p^{2}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right )} \, dp &= \left (\left (-y^{2}\right )^{{1}/{3}}\right )\,dy \end{align*}
Integrating now gives the following solutions
\begin{align*} \int \frac {1}{\left (p^{2}\right )^{{1}/{3}}}d p &= \int \left (-y^{2}\right )^{{1}/{3}}d y +c_1\\ \frac {3 \left (p^{2}\right )^{{2}/{3}}}{p} &= \frac {3 y \left (-y^{2}\right )^{{1}/{3}}}{5}\\ \int \frac {2}{\left (p^{2}\right )^{{1}/{3}} \left (-1+i \sqrt {3}\right )}d p &= \int \left (-y^{2}\right )^{{1}/{3}}d y +c_1\\ -\frac {3 \left (p^{2}\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )}{2 p} &= \frac {3 y \left (-y^{2}\right )^{{1}/{3}}}{5}\\ \int -\frac {2}{\left (p^{2}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right )}d p &= \int \left (-y^{2}\right )^{{1}/{3}}d y +c_1\\ \frac {3 \left (p^{2}\right )^{{2}/{3}} \left (-1+i \sqrt {3}\right )}{2 p} &= \frac {3 y \left (-y^{2}\right )^{{1}/{3}}}{5} \end{align*}
Therefore
\begin{align*}
\frac {3 \left (p^{2}\right )^{{2}/{3}}}{p} &= \frac {3 y \left (-y^{2}\right )^{{1}/{3}}}{5}+c_1 \\
p &= -\frac {y^{5}}{125}+\frac {\left (-y^{2}\right )^{{2}/{3}} c_1 \,y^{2}}{25}+\frac {\left (-y^{2}\right )^{{1}/{3}} c_1^{2} y}{15}+\frac {c_1^{3}}{27} \\
p &= -\frac {y^{5}}{125}+\frac {\left (-y^{2}\right )^{{2}/{3}} c_1 \,y^{2}}{25}+\frac {\left (-y^{2}\right )^{{1}/{3}} c_1^{2} y}{15}+\frac {c_1^{3}}{27} \\
\end{align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} \frac {3 \left ({y^{\prime }}^{2}\right )^{{2}/{3}}}{y^{\prime }} = \frac {3 y \left (-y^{2}\right )^{{1}/{3}}}{5}+c_1 \end{align*}
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} y^{\prime }&=0 \\
\tag{2} y^{\prime }&=\frac {\left (\frac {\left (-y^{2}\right )^{{2}/{3}} y^{2} \sqrt {15}}{25}+\frac {2 c_1 \left (-y^{2}\right )^{{1}/{3}} y \sqrt {15}}{15}+\frac {\sqrt {15}\, c_1^{2}}{9}\right ) \sqrt {15}\, \left (3 y \left (-y^{2}\right )^{{1}/{3}}+5 c_1 \right )}{225} \\
\end{align*}
Now each of the above is solved separately.
Solving Eq. (1)
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {0\, dx} + c_2 \\ y &= c_2 \end{align*}
Solving Eq. (2)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {3375}{\left (9 \tau ^{2} \left (-\tau ^{2}\right )^{{2}/{3}}+30 c_1 \left (-\tau ^{2}\right )^{{1}/{3}} \tau +25 c_1^{2}\right ) \left (3 \tau \left (-\tau ^{2}\right )^{{1}/{3}}+5 c_1 \right )}d \tau = x +c_3 \]
For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = 0 \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {0\, dx} + c_4 \\ y &= c_4 \end{align*}
For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = -\frac {y^{5}}{125}+\frac {y^{2} \left (-y^{2}\right )^{{2}/{3}} c_1}{25}+\frac {y \left (-y^{2}\right )^{{1}/{3}} c_1^{2}}{15}+\frac {c_1^{3}}{27} \end{align*}
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {1}{-\frac {\tau ^{5}}{125}+\frac {\left (-\tau ^{2}\right )^{{2}/{3}} c_1 \,\tau ^{2}}{25}+\frac {\left (-\tau ^{2}\right )^{{1}/{3}} c_1^{2} \tau }{15}+\frac {c_1^{3}}{27}}d \tau = x +c_5 \]
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
\int _{}^{y}\frac {1}{-\frac {\tau ^{5}}{125}+\frac {\left (-\tau ^{2}\right )^{{2}/{3}} c_1 \,\tau ^{2}}{25}+\frac {\left (-\tau ^{2}\right )^{{1}/{3}} c_1^{2} \tau }{15}+\frac {c_1^{3}}{27}}d \tau &= x +c_5 \\
\int _{}^{y}\frac {3375}{\left (9 \tau ^{2} \left (-\tau ^{2}\right )^{{2}/{3}}+30 c_1 \left (-\tau ^{2}\right )^{{1}/{3}} \tau +25 c_1^{2}\right ) \left (3 \tau \left (-\tau ^{2}\right )^{{1}/{3}}+5 c_1 \right )}d \tau &= x +c_3 \\
y &= c_2 \\
y &= c_4 \\
\end{align*}
Maple step by step solution
Maple trace
`Methods for second order ODEs:
*** Sublevel 2 ***
Methods for second order ODEs:
Successful isolation of d^2y/dx^2: 3 solutions were found. Trying to solve each resulting ODE.
*** Sublevel 3 ***
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^2*_b(_a)^2)^(1/3)*_b(_a) = 0, _b(_a), HINT = [[_a, 5*_b]]`
symmetry methods on request
`, `1st order, trying reduction of order with given symmetries:`[_a, 5*_b]
Maple dsolve solution
Solving time : 0.214 (sec)
Leaf size : 208
dsolve(y(x)*diff(diff(y(x),x),x)^3+y(x)^3*diff(y(x),x)^5 = 0,y(x),singsol=all)
\begin{align*}
y &= 0 \\
y &= c_{1} \\
\int _{}^{y}\frac {1}{\operatorname {RootOf}\left (5 \left (\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} \right )-\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 (\sqrt {3}\, \ln \left (\textit {\_a}^{5}+125\right )-i \ln \left (\textit {\_a}^{5}+125\right )+20 \left (\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} \right )-20 c_{1} \right )}d \textit {\_a} -x -c_{2} &= 0 \\
\int _{}^{y}\frac {1}{\operatorname {RootOf}\left (\sqrt {3}\, \ln \left (\textit {\_a}^{5}+125\right )+i \ln \left (\textit {\_a}^{5}+125\right )+20 \left (\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} \right )-20 c_{1} \right )}d \textit {\_a} -x -c_{2} &= 0 \\
\end{align*}
✓Mathematica DSolve solution
Solving time : 24.151 (sec)
Leaf size : 449
DSolve[{y[x]*D[y[x],{x,2}]^3+y[x]^3*D[y[x],x]^5==0,{}},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 solution
Solving time : 13.171 (sec)
Leaf size : 3
Python version: 3.13.1 (main, Dec 4 2024, 18:05:56) [GCC 14.2.1 20240910]
Sympy version 1.13.3
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)
Eq(y(x), 0)
\[
y{\left (x \right )} = 0
\]