Question

Consider the system of equations

Dx/dt = x(4 – x – 5y)
Dy/dt = y(1 – 4x),
taking (x, y) > 0.

Recall that a nullcline of this system is a line on which dx/dt = dy/dt = 0. Likewise, a vertical nullcline of this system is a line on wh dx/dt = 0, and a horizontal nullcline of this system is a line on which dy/dt = 0
(a) Write an equation for the (non-zero) vertical (x-)nullcline of this system:
_____ (Enter your equation, e.g., y=x.)
And for the (non-zero) horizontal (y-)nullcline:_____

Answer 1

The non-zero vertical (x-) nullcline of the system is y = ⅔(4 - x), and the non-zero horizontal (y-) nullcline is x = ¼.

Step-by-step explanation:

To find the equations for the non-zero vertical (x-) nullcline and horizontal (y-) nullcline of the given system of differential equations, we need to set Dx/dt and Dy/dt to zero separately, and solve for y and x, respectively.

Vertical (x-) Nullcline

The equation for Dx/dt is given by:

Dx/dt = x(4 - x - 5y)

For Dx/dt to be zero, the product of the terms must be zero. Assuming we're interested in the non-zero nullcline (where x > 0), we can set the inner expression to zero:

4 - x - 5y = 0

Rearranging terms gives us the equation of the vertical nullcline:

y = ⅔(4 - x)

Horizontal (y-) Nullcline

The equation for Dy/dt is given by:

Dy/dt = y(1 - 4x)

Similarly, for Dy/dt to be zero and with y > 0, we must have:

1 - 4x = 0

Solving for x gives us the equation of the horizontal nullcline:

x = ¼