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alexdok [17]
3 years ago
5

The logistic equation for the population​ (in thousands) of a certain species is given by:

Mathematics
1 answer:
Eva8 [605]3 years ago
6 0

Answer:

a.

b. 1.5

c. 1.5

d. No

Step-by-step explanation:

a. First, let's solve the differential equation:

\frac{dp}{dt} =3p-2p^2

Divide both sides by 3p-2p^2  and multiply both sides by dt:

\frac{dp}{3p-2p^2}=dt

Integrate both sides:

\int\ \frac{1}{3p-2p^2}  dp =\int\ dt

Evaluate the integrals and simplify:

p(t)=\frac{3e^{3t} }{C_1+2e^{3t}}

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:

p(0)=3=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}

Solving for C1:

C_1=-1

Now, let's evaluate the limit:

\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}  }

The expression -e^{-3x} tends to zero as x approaches ∞ . Hence:

\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } =\frac{3}{2} =1.5

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

p(0)=0.8=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}

Solving for C1:

C_1=1.75

Now, let's evaluate the limit:

\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}  }

The expression -e^{-3x} tends to zero as x approaches ∞ . Hence:

\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } =\frac{3}{2} =1.5

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:

p(0)=2=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}

Solving for C1:

C_1=-\frac{1}{2} =-0.5

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

\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}  }

The expression -e^{-3x} tends to zero as x approaches ∞ . Hence:

\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } =\frac{3}{2} =1.5

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

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