Find an autonomous differential equation with all of the following properties: equilibrium solutions at y=0 and y=3,
y' > 0 for 0 y' < 0 for -inf < y < 0 and 3 < y < inf
dy/dx =

dy/dt = y ( 3 - y )

Step-by-step explanation:

Given data:

Determine an autonomous differential equation with the following properties

y = 0 and Y = 3

y' > 0 for  0 < y < 3

y' < 0 for  -∞ < y < 0  and  3 < y < ∞

considering an autonomous differential equation

dy/dt = y ( 3 - y )

y = 0 and 3 represents equilibrium solutions

if 0 < y < 3  then y ( 3 - y ) > 0  for   0 < y < 3

hence : dy / dt = y' > 0  for 0 < y < 3

y ( 3 - y ) <  0  for -∞ < y < 0 and  3 < y < ∞

hence : dy / dt = y' < 0  for  -∞ < y < 0  and  3 < y < ∞

this shows that the autonomous differential equation satisfies every condition hence the autonomous differential equation is :

dy/dt = y ( 3 - y )

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!