D.E 兩條20分超易急

2009-09-25 6:22 am

THE POPULATION P OF A TOWN AFTER T YEARS IS GROWING IN SUCH A WAY THAT IT SATISFIES THE D.E

dP/dT=0.0468P+10000

solve the equation
give that initial population is 200000

find the population after 2years

2. a can of water is being heated at a rate which is decreasing steadily with time . the temperature A after t minutes satisfies the D.E

dA/dt=8-(4/3)t

FIND A WHEN T=4 IF A=32 INITIALLY

DO YOU THINK THE MODEL WILL STILL BE VALID AT TIME 10?

回答 (1)

用戶9112

2009-09-25 6:45 am

✔ 最佳答案

1. dP/dT = 0.0468P + 10000
dP/(0.0468P + 10000) = dT
[ln(0.0468P + 10000)]/0.0468 = T + C where C is a constant
ln(0.0468P + 10000) = 0.0468T + C1 where C1 is a constant
0.0468P + 10000 = exp(0.0468T + C1) = Aexp(0.0468T) where A is a constant
0.0468P = Aexp(0.0468T) - 10000
P = A1exp(0.0468T) - 213675 where A1 is a constant
At T = 0, P = 200000
200000 = A1exp(0) - 213675.21
A = 413675
P = 413675.21exp(0.0468T) - 213675.21

After 2 years, P = 413675.21exp(0.0938) - 213675.21 = 240590

2. dA/dt = 8 - (4/3)t
dA = (8 - 4t/3)dt
A = 8t - 2t^2/3 + C where C is a constant
At t = 0, A = C = 32
A = 8t - 2t^2/3 + 32
When t = 4, A = 32 - 32/3 + 32 = 160/3

dA/dt = 8 - (4/3)t
At t = 6, dA/dt = 0. Therefore after time t = 6, the heater stops heating the water. The heating equation is no longer true at t = 10 which is after t = 6.

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D.E 兩條20分 超易 急

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D.E 兩條20分超易急