Answer:
dy = 2(c-3y)dt
Step-by-step explanation:
Given the Equilibrium solutions at y=0 and y=3, y' > 0 for 0 < y < 3; and y' < 0 for -inf < y < 0 and 3 < y < inf.
from the boundary conditions given,
- since our task is to find the differential of y wrt t i.e dy/dt,
- from the first condition, it implies that if we are to assume from the range of values of y = {0,1,2}, assume when y =0, t = 0
from y = 3, y-3 = 0
integrate wrt dy i.e Integral (y-3) dy = 0
y2/2 - 3y + c = 0, where c is the constant of integration
hence, y2-6y+2c = 0
from the equation, above for the differential of y (dy/dx) to be greater than zero, for the boundary conditions, 0 < y < 3, the constant of integrative must be negative or zero.
hence the equation y2-6y+2c = 0 can be written as
dy/dx =2c-6y, dy = 2(c-3y)dt
in terms of t
