Find the critical points and solve the related phase plane differential equation for the system $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{,}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{y}\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{1}\mathbf{)}$.Describe (without using computer software) the asymptotic behavior of trajectories (as role="math" localid="1664302603010" $\mathbf{t}\mathbf{\to }\mathbf{\infty }$) that start at

1. (3, 2)
2. (2, 1/2)
3. ( -2, 1/2)
4. (3, -2)

Here the equation is:

$$\begin{gathered} \frac{dx}{dt}=(x-1)(y-1) \\ \frac{dy}{dt}=y(y-1) \end{gathered}$$

If y=0 then x=1.

And in (2) ($y\ne 0$) the y=1,

This is satisfied for all numbers x. So, all points are of the form (x, 1). So, the critical point is (1, 0),(x,1).

Here,

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}-1}$

Solve this by variable separating.

Y=C(x-1).

Now implied that $x-1\ne 0$so that could have it in the denominator. The solution x=1 makes the system.

$$\begin{gathered} \frac{dx}{dt}=0 \\ \frac{dy}{dx}=y(y-1) \end{gathered}$$

This curve x=1 is also a solution of the system because when we differentiate it by t so,$\frac{dx}{dt}=0$.so the value of y is y=C(x-1) or x=1.

The solution that starts at (3, 2) is above the lines of critical points (x, 1) and it is also right of the x=1. that means that the corresponding trajectory is of the form y=C(x-1). So put x=3, y=2 then C=1.

So, the trajectory line is y=x-1.

So, the system is for y=1.

$$\begin{gathered} \frac{\mathrm{dx}}{\mathrm{dt}}>0 \\ \frac{\mathrm{dy}}{\mathrm{dx}}>0 \end{gathered}$$

So, the solution along the trajectory starting at (3,2) is increasing and therefore going to infinity as$\mathrm{t}\to \infty$.

Put x=1 and y=1/2 then C=1/2.

The trajectory of this solution is $y=\frac{x-1}{2}$.

Since one is below the line y=1 line and above y=0 line that y-1<0 and y>0. So,

$$\begin{gathered} \frac{\mathrm{dx}}{\mathrm{dt}}<0 \\ \frac{\mathrm{dy}}{\mathrm{dx}}<0 \end{gathered}$$

That means the trajectory decreases until it hits the critical point (1,0).

Put x=-2, y=1/2 then C=-1/6.

The trajectory of this solution is $y=-\frac{x-1}{6}$.

Since one is below the line y=1 line and above y=0 line that y-1<0 and y>0. So,

$$\begin{gathered} \frac{\mathrm{dx}}{\mathrm{dt}}>0 \\ \frac{\mathrm{dy}}{\mathrm{dx}}<0 \end{gathered}$$

That means the trajectory increases along the x-axis and decreases along the y-axis until it hits the critical point (1,0).

Put x = 3, y = -2 then C = -1.

The trajectory of this solution is $y=-(x-1)$.

Thus, one is below the line y=1 line and above y = 0 line that y -1 < 0 and y > 0. So,

$$\begin{gathered} \frac{\mathrm{dx}}{\mathrm{dt}}<0 \\ \frac{\mathrm{dy}}{\mathrm{dx}}>0 \end{gathered}$$

That means the trajectory decreases along the x-axis and increases along the y-axis until it hits the critical point (1,0).