# 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.`

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).

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