Answer:
a.
b. 1.5
c. 1.5
d. No
Step-by-step explanation:
a. First, let's solve the differential equation:
[tex]\frac{dp}{dt} =3p-2p^2[/tex]
Divide both sides by [tex]3p-2p^2[/tex] and multiply both sides by dt:
[tex]\frac{dp}{3p-2p^2}=dt[/tex]
Integrate both sides:
[tex]\int\ \frac{1}{3p-2p^2} dp =\int\ dt[/tex]
Evaluate the integrals and simplify:
[tex]p(t)=\frac{3e^{3t} }{C_1+2e^{3t}}[/tex]
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:
[tex]p(0)=3=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}[/tex]
Solving for C1:
[tex]C_1=-1[/tex]
Now, let's evaluate the limit:
[tex]\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-e^{-3x} }[/tex]
The expression [tex]-e^{-3x}[/tex] tends to zero as x approaches ∞ . Hence:
[tex]\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } =\frac{3}{2} =1.5[/tex]
c. As we did before, let's find the constant C1 for the initial condition given:
[tex]p(0)=0.8=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}[/tex]
Solving for C1:
[tex]C_1=1.75[/tex]
Now, let's evaluate the limit:
[tex]\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2+1.75e^{-3x} }[/tex]
The expression [tex]-e^{-3x}[/tex] tends to zero as x approaches ∞ . Hence:
[tex]\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } =\frac{3}{2} =1.5[/tex]
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:
[tex]p(0)=2=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}[/tex]
Solving for C1:
[tex]C_1=-\frac{1}{2} =-0.5[/tex]
Can a population of 2000 ever decline to 800? well, let's find the limit of the function when it approaches to ∞:
[tex]\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-0.5e^{-3x} }[/tex]
The expression [tex]-e^{-3x}[/tex] tends to zero as x approaches ∞ . Hence:
[tex]\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } =\frac{3}{2} =1.5[/tex]
Therefore, a population of 2000 never will decline to 800.
