64.25.8 problem 8

Internal problem ID [13703]
Book : Differential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi. 2004.
Section : Chapter 13, Limit cycles and periodic solutions. Section 13.4, Exercises page 706
Problem number : 8
Date solved : Tuesday, January 28, 2025 at 08:24:56 PM
CAS classification : system_of_ODEs

\begin{align*} x^{\prime }&=x-x^{2}\\ y^{\prime }\left (t \right )&=2 y \left (t \right )-y \left (t \right )^{2} \end{align*}

✓ Solution by Maple

Time used: 0.138 (sec). Leaf size: 32

dsolve([diff(x(t),t)=x(t)-x(t)^2,diff(y(t),t)=2*y(t)-y(t)^2],singsol=all)
 

\begin{align*} \left \{x \left (t \right ) &= \frac {1}{1+c_{2} {\mathrm e}^{-t}}\right \} \\ \left \{y \left (t \right ) &= \frac {2}{1+2 c_{1} {\mathrm e}^{-2 t}}\right \} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.448 (sec). Leaf size: 238

DSolve[{D[x[t],t]==x[t]-x[t]^2,D[y[t],t]==2*y[t]-y[t]^2},{x[t],y[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} x(t)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{(K[1]-1) K[1]}dK[1]\&\right ][-t+c_1] \\ y(t)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{(K[2]-2) K[2]}dK[2]\&\right ][-t+c_2] \\ x(t)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{(K[1]-1) K[1]}dK[1]\&\right ][-t+c_1] \\ y(t)\to 0 \\ x(t)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{(K[1]-1) K[1]}dK[1]\&\right ][-t+c_1] \\ y(t)\to 2 \\ x(t)\to 0 \\ y(t)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{(K[2]-2) K[2]}dK[2]\&\right ][-t+c_2] \\ x(t)\to 0 \\ y(t)\to 0 \\ x(t)\to 0 \\ y(t)\to 2 \\ x(t)\to 1 \\ y(t)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{(K[2]-2) K[2]}dK[2]\&\right ][-t+c_2] \\ x(t)\to 1 \\ y(t)\to 0 \\ x(t)\to 1 \\ y(t)\to 2 \\ \end{align*}