Write the solution of the initial-value problem 0.05P (1-p/5000) , where P(0)=200. Use it to find the population sizes P(100) and P(150). At what time does the population reach 4000?

`P(t) = (5000 e^(0.05t))/(24+e^(0.05t))`

`P(100) approx. 4304, P(150) approx. 4934, P=4000` at ` t approx. 91.3.`

Expert Answers

An illustration of the letter 'A' in a speech bubbles

I suppose that the lowercase p is the same as the uppercase. Then the following equation has sense: `P' = 0.05 P ( 1 - P / 5000 ) . ` Here `P = P ( t ) , ` a function of the independent variable `t ` (time).

This differential equation is separable because we can divide both sides by the right side and obtain an equivalent equation

`(P') / ( 0.05 P ( 1 - P / 5000 ) ) = 1 .`

Now we can integrate the right side and obtain `t + C , ` but we can also integrate the left side:

`int (P') / ( 0.05 P ( 1 - P / 5000 ) ) dt = int ( dP ) / ( 0.05 P ( 1 - P / 5000 ) ) . `

It is an integral of a rational function; to solve it, use partial fraction decomposition:

`1 / ( 0.05 P ( 1 - P / 5000 ) ) = A / ( 0.05 P ) + B ( 1 - P / 5000 ) .`

Multiply by the denominator to determine `A , B :`

`1 = A ( 1 - P / 5000 ) + B ( 0.05 P ) ; ` in other words, `1 = P ( 0.05B - A/5000) + A. ` This gives `A = 1 , 0.05B = 1/5000, ` so `B = 1 / 250 ` and the function is equal to `20/P - 20/(P-5000), ` which integrates easily:

`20 ln | P | - 20 ln|P-5000|=t+C, ` or `ln|P/(P-5000)|=0.05t+C,`
or `P/(P-5000)=1+5000/(P-5000) = Ce^(0.05t). ` At `t=0 ` this becomes `-200/4800=C, ` so `C=-1/24.`

From here we finally find `P=-5000/(1+1/24 e^(0.05t))+5000 = (5000 e^(0.05t))/(24+e^(0.05t)). `

This tends to 5000 from below as `t ` tends to (plus) infinity. Now you can easily answer all specific questions using a graphing calculator (see the attached picture).

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
Last Updated by eNotes Editorial on
Soaring plane image

We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

  • 30,000+ book summaries
  • 20% study tools discount
  • Ad-free content
  • PDF downloads
  • 300,000+ answers
  • 5-star customer support
Start your 48-Hour Free Trial