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Q5.4-29E

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Found in: Page 272

Fundamentals Of Differential Equations And Boundary Value Problems

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 $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{\left(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbf{y}\mathbf{-}\mathbf{1}\mathbf{\right)}{\mathbf{,}}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{y}\mathbf{\left(}\mathbf{y}\mathbf{-}\mathbf{1}\mathbf{\right)}$.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).

See the step by step solution

Find the critical point

Here the equation is:

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

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).

Find the value of y

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.

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

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.

Solve for points (3,2)

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 $\mathrm{t}\to \infty$.

Evaluate for point (2,12)

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).

Solve for point (-2,12).

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).

Determine the value for point (3,-2).

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

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

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).