# Homework Help: Mixture DE

1. Oct 9, 2009

### KillerZ

1. The problem statement, all variables and given/known data

A tank contains 200 liters of fluid which 30 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 4 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

2. Relevant equations

initial amount of liquid = 200L
A(t=0) = 30g
rate in = 4L/min
rate out = 4L/min
concentration of the inflow liquid = 1 g/L

3. The attempt at a solution

$$\frac{dA}{dt} = (1g/L)(4L/min) - \frac{(4L/min)(A(t))g}{(200L) + (4L/min - 4L/min)t}$$

$$\frac{dA}{dt} = (4) - \frac{4A}{200}$$

$$\frac{dA}{dt} + \frac{4A}{200} = (4)$$

$$\frac{dA}{dt} + \frac{A}{50} = (4)$$

$$I(t) = e^{\int\frac{A}{50}dt}$$

$$I(t) = e^{\frac{At}{50}}$$

$$f(t) = \frac{1}{e^{\frac{At}{50}}}[\int (e^{\frac{At}{50}})(4) dt + c]$$

$$f(t) = \frac{1}{e^{\frac{At}{50}}}[\frac{200e^{\frac{At}{50}}}{A} + c]$$

$$f(t) = \frac{200e^{\frac{At}{50}}}{Ae^{\frac{At}{50}}} + \frac{c}{e^{\frac{At}{50}}}]$$

2. Oct 9, 2009

### HallsofIvy

No, no, no! "A" is your dependent variable, not a constant. You find the integrating factor by integrating the coefficient of A. Your integrating factor is
$$e^{\int\frac{dt}{50}}= e^{\frac{t}{50}}$$

And now you've completely lost me! Where did "f(t)" come from? There was no "f" before this line. You are supposed to be finding A(t).

Since the integrating factor is
$$e^{\int\frac{dt}{50}}= e^{\frac{t}{50}}$$
you have
$$e^{\frac{t}{50}}\frac{dA}{dt}-e^{\frac{t}{50}}\frac{A}{50}= 4e^{\frac{t}{50}}$$
$$\frac{d}{dt}\left(e^{\frac{t}{50}}A\right)= 4e^{\frac{t}{50}}[/itex] Integrate that to find A. The left side is, of course, just [tex]e^{\frac{t}{50}}A(t)$$

Frankly, I think it would be easier not to use the "integrating factor" method at all. This is a separable equation.

$$\frac{dA}{dt}= 4-\frac{A}{50}= \frac{200- A}{50}$$
$$\frac{dA}{200- A}= \frac{dt}{50}$$
Integrate both sides of that.

3. Oct 9, 2009

### KillerZ

opps I made some typos and I don't know what I was thinking doing the integrating factor .

Here is my next attempt:

$$\frac{dA}{dt} = (1g/L)(4L/min) - \frac{(4L/min)(A(t))g}{(200L) + (4L/min - 4L/min)t}$$

$$\frac{dA}{dt} = (4) - \frac{4A}{200}$$

$$\frac{dA}{dt} + \frac{4A}{200} = (4)$$

$$\frac{dA}{dt} + \frac{A}{50} = (4)$$

$$I(t) = e^{\int\frac{1}{50}dt}$$

$$I(t) = e^{\frac{t}{50}}$$

$$A(t) = \frac{1}{e^{\frac{t}{50}}}[\int (e^{\frac{t}{50}})(4) dt + c]$$

$$A(t) = \frac{1}{e^{\frac{t}{50}}}[200e^{\frac{t}{50}} + c]$$

$$A(t) = \frac{200e^{\frac{t}{50}}}{e^{\frac{t}{50}}} + \frac{c}{e^{\frac{t}{50}}}$$

4. Oct 9, 2009

### HallsofIvy

I hope you understand that this can be simplified!

And, of course, use the fact that A(0)= 30 to find c.

5. Oct 9, 2009

### KillerZ

$$A(t) = \frac{200e^{\frac{t}{50}}}{e^{\frac{t}{50}}} + \frac{c}{e^{\frac{t}{50}}}$$

$$A(t) = 200 + \frac{c}{e^{\frac{t}{50}}}$$

$$A(0) = 30 = 200 + \frac{c}{e^{\frac{0}{50}}}$$

$$c = 30 - 200 = -170$$

$$A(t) = 200 - \frac{170}{e^{\frac{t}{50}}}$$