Question

A one-parameter family of solutions of the de p' = p(1 − p) is given below.

Answer 1

Answer:

A solution curve pass through the point (0,4) when $c_{1} = -\frac{4}{3}$.

There is not a solution curve passing through the point(0,1).

Step-by-step explanation:

We have the following solution:

$P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}$

Does any solution curve pass through the point (0, 4)?

We have to see if P = 4 when t = 0.

$P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}$

$4 = \frac{c_{1}}{1 + c_{1}}$

$4 + 4c_{1} = c_{1}$

$c_{1} = -\frac{4}{3}$

A solution curve pass through the point (0,4) when $c_{1} = -\frac{4}{3}$.

Through the point (0, 1)?

Same thing as above

$P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}$

$1 = \frac{c_{1}}{1 + c_{1}}$

$1 + c_{1} = c_{1}$

$0c_{1} = 1$

No solution.

So there is not a solution curve passing through the point(0,1).