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Q5.4-29E

Expert-verifiedFound in: Page 272

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**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**

**(3, 2)****(2, 1/2)****( -2, 1/2)****(3, -2)**

**a. The solution approaches infinity.**

**b. The solution approaches point (1,0).**

**c. The solution approaches point (1,0).**

**d. The solution approaches point (1,0).**

Here the equation is:

$\frac{dx}{dt}=(x-1)(y-1)\phantom{\rule{0ex}{0ex}}\frac{dy}{dt}=y(y-1)$

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.

$\frac{dx}{dt}=0\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=y(y-1)$

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.

$\frac{\mathrm{dx}}{\mathrm{dt}}>0\phantom{\rule{0ex}{0ex}}\frac{\mathrm{dy}}{\mathrm{dx}}>0$

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,

$\frac{\mathrm{dx}}{\mathrm{dt}}<0\phantom{\rule{0ex}{0ex}}\frac{\mathrm{dy}}{\mathrm{dx}}<0$

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,

$\frac{\mathrm{dx}}{\mathrm{dt}}>0\phantom{\rule{0ex}{0ex}}\frac{\mathrm{dy}}{\mathrm{dx}}<0$

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,

$\frac{\mathrm{dx}}{\mathrm{dt}}<0\phantom{\rule{0ex}{0ex}}\frac{\mathrm{dy}}{\mathrm{dx}}>0$

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

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