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Suppose a population follows the equation $\frac{dy}{dt} = ry[1-\frac{y}{K}]$

(let $y_i =$ initial value of $y$)

a.) $y(0) = \frac{K}{3}$ find the time $T$ at which the initial population has doubled fish the value of $T$ corresponding to $r = 0.025$

b.) if $\dfrac{y_i}{K} = a$, find the time $t$ at which $\dfrac{y(t)}{K} = b$, where $0<a, b<1$ find the value of $t$ for $r=0.025, a =0.1$ and $b =0.9$

I have began to solve the equation, but I am stuck at $\dfrac{y}{\left[1-\dfrac{y}{K}\right]}=ce^{rt}$

solving for y my equation is $y= k\dfrac{y_i e^{rt} +K}{1-y_i}$. I'm not sure that this is correct, and would like to see how the algebra would be done more gracefully. I'm not sure where to proceed from there.

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I take it $y_i = y(0)$? And, in b), what is o? –  Ron Gordon Jan 31 '13 at 19:31
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2 Answers 2

Starting with

$$\frac{y}{1-\frac{y}{K}} = c e^{r t}$$

Multiply both sides by $1-\frac{y}{K}$:

$$y = c e^{r t} \left (1-\frac{y}{K} \right ) = c e^{r t} - c e^{r t}\frac{y}{K}$$

Add $c e^{r t}\frac{y}{K}$ to both sides and factor out the $y$:

$$y \left ( 1 + \frac{c}{K} e^{r t} \right ) = c e^{r t} $$

(It should be noted that the above is a multiplication and not an evaluation of $y$ at the time in parentheses.) And you get an expression for $y(t)$:

$$y(t) = \frac{c e^{r t}}{1 + \frac{c}{K} e^{r t}} $$

EDIT

It appears you want this expression in terms of $y_i = y(0)$. To do this, notice from the top equation that

$$c = \frac{y_i}{1-\frac{y_i}{K}}$$

Putting this into the previous equation, we get

$$\begin{align}\frac{c e^{r t}}{1 + \frac{c}{K} e^{r t}} &= \frac{\frac{y_i}{1-\frac{y_i}{K}} e^{r t}}{1 + \frac{1}{K}\frac{y_i}{1-\frac{y_i}{K}} e^{r t}} \\ &= \frac{K \frac{y_i}{K-y_i} e^{r t}}{1+ \frac{y_i}{K-y_i} e^{r t}} \\ &= \frac{K y_i e^{r t}}{K + y_i (e^{r t} - 1)} \\ \end{align} $$

I do not think this agrees with your solution. Check the algebra here and make sure you understand it.

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Hint: take logarithm of both sides of $y/(1-y/K) = c e^{rt}$

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