Pencil-and-paper exercise set 5

Systems of differential equations: Linear systems of differential equations

Pencil-and-paper exercise set 5

$\beta=\mathrm{sp}=a+d$ and $\gamma=\det=ad-bc$ for the system $\left\{\begin{aligned} \frac{\dd x}{\dd t} &= a\, x+ b\, y\\[0.25cm] \frac{\dd y}{\dd t} &= c\, x + d\, y\end{aligned}\right.$

Use the above stability scheme to determine the nature of the equilibrium $(0,0)$ of the below system of differential equations: $\left\{\begin{aligned} \frac{\dd x}{\dd t} &= -4x+ 2y\\ \\ \frac{\dd y}{\dd t} &= x -2y\end{aligned}\right.$

In this case: $a=-4,\quad b=2,\quad c=1,\quad d=-2$ Then $\beta\,(=a+d)=-6,\quad \gamma\,(=ad-bc)=6$ Because $\beta<0$ , $\gamma>0$ and $\beta^2-4\gamma=12>0$ , there is, according to the stability scheme, an attracting equilibrium $(0,0)$ with solution curves that approach it, but not in a spiral shape.

Use the above stability scheme to determine the nature of the equilibrium $(0,0)$ of the below system of differential equations: $\left\{\begin{aligned} \frac{\dd x}{\dd t} &= x-y\\ \\ \frac{\dd y}{\dd t} &= x-2y\end{aligned}\right.$

In this case: $a=1,\quad b=-1,\quad c=1,\quad d=-2$ Then $\beta\,(=a+d)=-1,\quad \gamma\,(=ad-bc)=-1$ Because $\beta<0$ and $\gamma<0$ there is, according to the stability scheme, a saddle point in $(0,0)$ .

Use the above stability scheme to determine the nature of the equilibrium $(0,0)$ of the below system of differential equations: $\left\{\begin{aligned} \frac{\dd x}{\dd t} &= x-y\\ \\ \frac{\dd y}{\dd t} &= 2x-y\end{aligned}\right.$

In this case: $a=1,\quad b=-1,\quad c=2,\quad d=-1$ Then $\beta\,(=a+d)=0,\quad \gamma\,(=ad-bc)=1$ Because $\beta=0$ and $\gamma>0$ there is, according to the stability scheme, an equilibrium $(0,0)$ around which the solution curves move around in elliptical orbits.

Use the above stability scheme to determine the nature of the equilibrium $(0,0)$ of the below system of differential equations: $\left\{\begin{aligned} \frac{\dd x}{\dd t} &= x-y\\ \\ \frac{\dd y}{\dd t} &= 2x+y\end{aligned}\right.$

In this case: $a=1,\quad b=-1,\quad c=2,\quad d=1$ Then $\beta\,(=a+d)=2,\quad \gamma\,(=ad-bc)=3$ Because $\beta<0$ , $\gamma>0$ and $\beta^2-4\gamma=-5<0$ there is, according to the stability scheme, a repelling equilibrium $(0,0)$ with solution curves that move away from it in spiral form.

Use the above stability scheme to determine the nature of the equilibrium $(0,0)$ of the below system of differential equations: $\left\{\begin{aligned} \frac{\dd x}{\dd t} &= -x-y\\ \\ \frac{\dd y}{\dd t} &= 2x-y\end{aligned}\right.$

In this case $a=-1,\quad b=-1,\quad c=2,\quad d=-1$ Then $\beta\,(=a+d)=-2,\quad \gamma\,(=ad-bc)=3$ Because $\beta<0$ , $\gamma>0$ and $\beta^2-4\gamma=-8<0$ there is, according to the stability diagram, an attracting equilibrium $(0,0)$ with solution curves that approach it in a spiral form.