differential equation

xref228

A reservoir initially contains 2000m3 pure water. At time t=0 water contaminated with a liquid pollutant begins flowing into the reservoir at the rate of 200 m3 per month. The well-mixed water in the reservoir flows out at the same rate.

a) Assume the pollutant concentration c (t) =20 [1 + cos (t)](L/m3), find the amount P (t) of pollutant in the reservoir after t months;

b) Use Euler’s method with h=0.5, h=0.25 to graph the solution on the interval [0, 10]. Estimate the amount of pollutant in the reservoir after 10 months and tabulate the results with appropriate headings in each step.

thanks all

xref228

part a
The initial condition is dc/dt = input rate - output rate
rate flow in = 200(m3/moth) x 20 [1+cos (t)] L/m3
         = 4000 [1+cos (t)] L/month
The concentration of the pollutant in to reservoir is c (t) / 2000 L/m3
So the rate of flow out = 200 x c / 2000 = c / 10 L/month
The differential equation is dc/dt = 4000 [1+cos (t)] – c/10
⇒ dc/dt + c/10 = 4000 [1+cos (t)]

true ?

Naozumi

唔多o岩,你input果個rate又係用 c(t), 係水塘果個rate又用 c(t), 到底 c(t)係入黎果d pollutant的 concentration定係水塘的concentration?

Naozumi

條題目有d唔太清楚, 係果d pollutant係t = 0果一刹那入左水塘,跟住就冇,定係不停有pollutant入,定點呢?

[ 本帖最後由 Naozumi 於 11-4-24 12:33 PM 編輯 ]