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Answer 1

The general solution of the logistic equation is $y(t) = \frac{14}{1 - C\cdot e^{-\frac{14\cdot t}{3} }}$.

The particular solution of the logistic equation is $y(t) = \frac{14}{1 + 0.4 \cdot e^{-\frac{14\cdot t}{3} }}$.

What are the general and particular solutions of the logistic equation?

In this question we are before a type of ordinary differential equation known as equation with separable variables, that is to say, that variables t and y can be separated at each side of the expression in order to find a solution:

dy / dt = 3 · y · (1 - y /14)

dy / [(- 3 / 14 ) · y · (y - 14)] = dt

Then, we simplify the expression by partial fractions and integrate the resulting expression:

- (1 / 14) ∫ dy / y + (1 / 14)∫ dy / (y - 14) = - (14 / 3)∫ dt

- (1 / 14) · ㏑ |y| + (1 / 14) · ㏑ |y - 14| = - (14 / 3) · t + C, where C is the integration constant.

㏑ |(y - 14) / y| = - (14 / 3) · t + C

$1 - \frac{14}{y} = C\cdot e^{-\frac{14\cdot t}{3} }$

$\frac{14}{y} = 1 - C \cdot e^{-\frac{14\cdot t}{3} }$

$y(t) = \frac{14}{1 - C\cdot e^{-\frac{14\cdot t}{3} }}$

The general solution of the logistic equation is $y(t) = \frac{14}{1 - C\cdot e^{-\frac{14\cdot t}{3} }}$.

If y(0) = 10, then the particular solution of the differential equation is:

10 = 14 /(1 - C)

1 - C = 1.4

C = - 0.4

The particular solution of the logistic equation is $y(t) = \frac{14}{1 + 0.4 \cdot e^{-\frac{14\cdot t}{3} }}$.

To learn more on differential equations: https://brainly.com/question/14620493

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