Answer:
a.
b. 1.5
c. 1.5
d. No
Step-by-step explanation:
a. First, let's solve the differential equation:

Divide both sides by
and multiply both sides by dt:

Integrate both sides:

Evaluate the integrals and simplify:

Where C1 is an arbitrary constant
I sketched the direction field using a computer software. You can see it in the picture that I attached you.
b. First let's find the constant C1 for the initial condition given:

Solving for C1:

Now, let's evaluate the limit:

The expression
tends to zero as x approaches ∞ . Hence:

c. As we did before, let's find the constant C1 for the initial condition given:

Solving for C1:

Now, let's evaluate the limit:

The expression
tends to zero as x approaches ∞ . Hence:

d. To figure out that, we need to do the same procedure as we did before. So, let's find the constant C1 for the initial condition given:

Solving for C1:

Can a population of 2000 ever decline to 800? well, let's find the limit of the function when it approaches to ∞:

The expression
tends to zero as x approaches ∞ . Hence:

Therefore, a population of 2000 never will decline to 800.