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Rudiy27
2 years ago
10

The differential equation in Example 3 of Section 2.1 is a well-known population model. Suppose the DE is changed to dP dt = P(a

P − b), where a and b are positive constants. Discuss what happens to the population P as time t increases.
Mathematics
1 answer:
LuckyWell [14K]2 years ago
6 0

Answer:

Decreases

Step-by-step explanation:

We need to determine the integral of the DE;

dP/dt=P(aP-b)

dP=P(aP-b)dt

1/(dP^2-bP)dP=dt

We can solve this by integration by parts on the left side. We expand the fraction 1/P²:

1/(d-b/P)\cdot{P^2} dP

let

u=d-b/P

du/dP=b/P^2

dP=\int\limits {P^2/b} \, du

P=lnu/b

Substitute u in:

P=ln(d-b/P)/b

Therefore the equation is:

ln(d-b/P)/b=t

We simplify:

d-b/P=e^b^t

P=b/(d-e^b^t)

As t increases to infinity P will decrease

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I need help with this.
likoan [24]

a quick clarification and then some.

Profit is what's leftover after the cost is subtracted from the revenue, or namely, if you sell a product for some amount, say 10, and you sold 100 of those, so you made 10*100 or 1000, that's the revenue or the income coming in, however, in making the product you had to cover some expenses, like if it's clothing, well, you have to buy the garment and saw it, have premises and machines to make the clothing and so on, all that's expenses, and since it's out of pocket money, is Cost, if you subtract that Cost from the 1000 in Revenue, what's leftover, that surplus is Profit.

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\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{15\% of 206970}}{\left( \cfrac{15}{100} \right)206970}\implies 31045.5

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3 years ago
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