80.8.12 problem 16
Internal
problem
ID
[21369]
Book
:
A
Textbook
on
Ordinary
Differential
Equations
by
Shair
Ahmad
and
Antonio
Ambrosetti.
Second
edition.
ISBN
978-3-319-16407-6.
Springer
2015
Section
:
Chapter
8.
Qualitative
analysis
of
2
by
2
systems
and
nonlinear
second
order
equations.
Excercise
8.5
at
page
184
Problem
number
:
16
Date
solved
:
Sunday, October 12, 2025 at 05:51:29 AM
CAS
classification
:
system_of_ODEs
\begin{align*} \frac {d}{d t}x \left (t \right )&=x \left (t \right ) \left (3-y \left (t \right )\right )\\ \frac {d}{d t}y \left (t \right )&=y \left (t \right ) \left (x \left (t \right )-5\right ) \end{align*}
✓ Maple. Time used: 0.574 (sec). Leaf size: 109
ode:=[diff(x(t),t) = x(t)*(3-y(t)), diff(y(t),t) = y(t)*(x(t)-5)];
dsolve(ode);
\begin{align*}
[\{x \left (t \right ) = 0\}, \{y \left (t \right ) &= c_1 \,{\mathrm e}^{-5 t}\}] \\
\left [\left \{x \left (t \right ) &= \operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}\frac {\textit {\_a}^{{2}/{3}} {\mathrm e}^{\operatorname {LambertW}\left (\frac {{\mathrm e}^{\frac {c_1}{3}} {\mathrm e}^{\frac {\textit {\_a}}{3}} {\mathrm e}^{-1}}{3 \textit {\_a}^{{5}/{3}}}\right )}}{3 \textit {\_a}^{{5}/{3}} {\mathrm e}^{\operatorname {LambertW}\left (\frac {{\mathrm e}^{\frac {c_1}{3}} {\mathrm e}^{\frac {\textit {\_a}}{3}} {\mathrm e}^{-1}}{3 \textit {\_a}^{{5}/{3}}}\right )}+{\mathrm e}^{\frac {c_1}{3}} {\mathrm e}^{\frac {\textit {\_a}}{3}} {\mathrm e}^{-1}}d \textit {\_a} +t +c_2 \right )\right \}, \left \{y \left (t \right ) = \frac {3 x \left (t \right )-\frac {d}{d t}x \left (t \right )}{x \left (t \right )}\right \}\right ] \\
\end{align*}
✓ Mathematica. Time used: 0.163 (sec). Leaf size: 569
ode={D[x[t],t]==x[t]*(3-y[t]),D[y[t],t]==y[t]*(x[t]-5) };
ic={};
DSolve[{ode,ic},{x[t],y[t]},t,IncludeSingularSolutions->True]
\begin{align*} y(t)&\to -3 W\left (\frac {1}{3} \sqrt [3]{-\frac {\exp \left (\text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[1] \left (W\left (\frac {1}{3} \sqrt [3]{-\frac {e^{K[1]-c_1}}{K[1]^5}}\right )+1\right )}dK[1]\&\right ][3 t+c_2]-c_1\right )}{\text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[1] \left (W\left (\frac {1}{3} \sqrt [3]{-\frac {e^{K[1]-c_1}}{K[1]^5}}\right )+1\right )}dK[1]\&\right ][3 t+c_2]{}^5}}\right )\\ x(t)&\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[1] \left (W\left (\frac {1}{3} \sqrt [3]{-\frac {e^{K[1]-c_1}}{K[1]^5}}\right )+1\right )}dK[1]\&\right ][3 t+c_2]\\ y(t)&\to -3 W\left (-\frac {1}{3} \sqrt [3]{-1} \sqrt [3]{-\frac {\exp \left (\text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[2] \left (W\left (-\frac {1}{3} \sqrt [3]{-1} \sqrt [3]{-\frac {e^{K[2]-c_1}}{K[2]^5}}\right )+1\right )}dK[2]\&\right ][3 t+c_2]-c_1\right )}{\text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[2] \left (W\left (-\frac {1}{3} \sqrt [3]{-1} \sqrt [3]{-\frac {e^{K[2]-c_1}}{K[2]^5}}\right )+1\right )}dK[2]\&\right ][3 t+c_2]{}^5}}\right )\\ x(t)&\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[2] \left (W\left (-\frac {1}{3} \sqrt [3]{-1} \sqrt [3]{-\frac {e^{K[2]-c_1}}{K[2]^5}}\right )+1\right )}dK[2]\&\right ][3 t+c_2]\\ y(t)&\to -3 W\left (\frac {1}{3} (-1)^{2/3} \sqrt [3]{-\frac {\exp \left (\text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[3] \left (W\left (\frac {1}{3} (-1)^{2/3} \sqrt [3]{-\frac {e^{K[3]-c_1}}{K[3]^5}}\right )+1\right )}dK[3]\&\right ][3 t+c_2]-c_1\right )}{\text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[3] \left (W\left (\frac {1}{3} (-1)^{2/3} \sqrt [3]{-\frac {e^{K[3]-c_1}}{K[3]^5}}\right )+1\right )}dK[3]\&\right ][3 t+c_2]{}^5}}\right )\\ x(t)&\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[3] \left (W\left (\frac {1}{3} (-1)^{2/3} \sqrt [3]{-\frac {e^{K[3]-c_1}}{K[3]^5}}\right )+1\right )}dK[3]\&\right ][3 t+c_2] \end{align*}
✗ Sympy
from sympy import *
t = symbols("t")
x = Function("x")
y = Function("y")
ode=[Eq((y(t) - 3)*x(t) + Derivative(x(t), t),0),Eq((5 - x(t))*y(t) + Derivative(y(t), t),0)]
ics = {}
dsolve(ode,func=[x(t),y(t)],ics=ics)