first oder linear II

Let the amount of substance in the tank at time t is x(t). Then
dx/dt=the rate of substance flow in - the rate of substance flow out

At time t, there is 4*3=12 g per litre of substance flow into the tank and there is (x/40)*4=x/10 g per litre of substance flow out the tank.

So dx/dt=12-x/10
10(dx/dt)=120-x
dx/(120-x)=dt/10
-ln(120-x)=t/10+C (where C is a constant)
x=120-Ce^(-t/10)

Sub. x(0)=80=>C=40 and x(t)=120-40e^(-t/10)
The amount of substance after 15 minutes is x(15)=111.1 g

Let x be the amount of the substance in the tank measured in g, then we have x = 2 when t = 0

Flow in rate of the substance = 4 x 3 = 12 g/min

Flow out rate of the substance = x/10 g/min (since 4L is 1/10 of 40L)

So we have"

dx/dt = 12 - x/10

dx/dt + x/10 = 12

Integrating factor = exp (dt/10) = e^(t/10)

Thus:

e^(t/10) dx/dt + xe^(t/10)/10 = 12e^(t/10)

d [x e^(t/10)]/dt = 12e^(t/10)

x e^(t/10) = 12∫e^(t/10) dt

= 120e^(t/10) + C

x = 120 + Ce^(-t/10)

With x = 2 when t = 0, we have 2 = 120 + C

So C = -118 and hence:

x = 120 - 118e^(-t/10)

So when t = 15 mins:

x = 120 - 118e^(-3/2)

= 93.7 g

2010-11-11 17:26:09 補充：
Correction:

It should be x = 80 when t = 0, thus we have:

80 = 120 + C

C = -40

Thus

x = 120 - 40e^(-t/10)

When t = 15:

x = 120 - 40e^(-3/2)

= 111.1 g