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# is there a difference?

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I have the equation:

$$P(t)=\frac{K}{9e^{-rt}+1}$$

this equation was derived from solving the logistic growth equation:

$$\frac{dP}{dt}=rP\big(1-\frac{P}{K}\big)$$

where,

r is the rate of growth = 1.13

K is the maximum population = 10,000

I've been asked to change the growth rate to:

$$r = 0.83 + \frac{0.3}{1+0.3t}$$

My question is,

Do I have to solve the entire logistic growth equation again, that is,

or can I just substituted r into the derived equation and rearrage, that is,

I wanted to get some clarification as I've tried doing both process with a simpler equation, however the answer still get quite complex and I'm not sure If I've done it correctly.

Thank you.

vest4R  Mar 29, 2018
edited by vest4R  Mar 29, 2018
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In my opinion I don't think it matters if $$r = 0.83 + \frac{0.3}{1+0.3t}$$ is substituted staight into p(t)

but I'm not overly confident about that.

vest4R  Mar 29, 2018

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