1.52 problem 52
Internal
problem
ID
[8089]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
52
Date
solved
:
Monday, October 21, 2024 at 04:49:19 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^{\prime }}^{5} y^{2} &= 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)
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} y^{\prime }&=0 \\
\tag{2} y^{\prime }&=0 \\
\tag{3} y^{\prime }&=0 \\
\tag{4} y^{\prime }&=0 \\
\tag{5} y^{\prime }&=0 \\
\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_1 \\ y &= c_1 \end{align*}
Solving Eq. (2)
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. (3)
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_3 \\ y &= c_3 \end{align*}
Solving Eq. (4)
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*}
Solving Eq. (5)
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_5 \\ y &= c_5 \end{align*}
1.52.1 Maple step by step solution
1.52.2 Maple trace
Methods for second order ODEs:
1.52.3 Maple dsolve solution
Solving time : 0.112 (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 (-i \ln \left (\textit {\_a}^{5}+125\right )+\sqrt {3}\, \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*}
1.52.4 Mathematica DSolve solution
Solving time : 26.332 (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*}