# Homework Help: Diff Equations, Mixing

1. Apr 19, 2006

### verd

I'm having a bit of difficulty with these. I'm sort of new to this so I'm somewhat confused, maybe someone could lighten a bit of my confusion.

Here's the first problem:
http://www.synthdriven.com/images/deletable/img01.jpg [Broken]

This is what I've done, please let me know if I'm doing this correctly...

$$\frac{dy}{dt}=$$rate in - rate out
rate in =$$(0kg/G)(5G/hr)=0$$
rate out =$$(\frac{y(t)}{10G})(5G/hr)=\frac{5y(t)}{10}=\frac{1}{2}y(t)$$
$$\frac{dy}{dt}=-\frac{1}{2}y(t)$$
$$\int \frac{1}{y}=-\int \frac{1}{2}dt$$
$$\ln{y}=-\frac{1}{2}t$$

This is where I've ended up... Am I to find y and then solve for y=(1/2)10=5?? If so, that'd give me the following:

$$\ln{5}=-\frac{1}{2}t$$
$$t=-2\ln{5}$$

Is this correct??

The second problem is the following:
http://www.synthdriven.com/images/deletable/img02.jpg [Broken]

For this, I got the following:
$$\frac{dy}{dt}=$$rate in - rate out
rate in=$$(0.15\frac{ft^3}{min})(0.06\frac{C0}{ft^3})$$
rate out=$$(\frac{y(t)CO}{1800ft^3})(\frac{0.15ft^3}{min})=\frac{0.15y}{1800}CO/min$$
$$\frac{dy}{dt}=0.009-\frac{0.15y}{1800}$$
$$\frac{dy}{dt}=-(\frac{0.15}{1800}-0.009)$$
$$\int{\frac{1800}{0.15y}-\frac{1}{0.009}dy}=-\int{dt}$$
$$\frac{1800}{0.15}\ln{y}-\frac{y}{0.009}=-t$$

I would then plug in 0.00018 for y, right?
When I do that, I get 51735.34 min... Is this correct...?

Any pointers?

Thanks

Last edited by a moderator: May 2, 2017
2. Apr 19, 2006

### AKG

For number 1, you're forgetting your constant of integration, so you should have:

ln(y) = C-t/2
y = ce-t/2 where c = eC

You need to solve for c, using the fact that y(0) = 10. This is easy, c=10. Now you want to find when y(t) = (1/2)10 = 5, so you set:

5 = 10e-t/2
-2ln(0.5) = t
t = ln4

For your second problem, its correct where you say dy/dt = 0.009 - 0.15y/1800. After that, it appears you've made some mistakes. Observe that you have dy/dt = a + by for some number as and b. So:

dy/(a+by) = dt
ln(a+by)/b = t + C

You should be able to get the rest from here. Note that in the end, you want to find the smallest t such that y(t)/1800 > 0.00018, i.e. you want to solve y(t)/1800 = 0.00018, not y(t) = 0.00018. In case you get stuck, the solution is below, but only use it if you get stuck.

ln(a+by) = tb + D
a+by = cetb
y = (cetb-a)/b
y = detb - a/b

where a = 0.009, b = -0.15/1800, and d has to be determined. Again, you know y(0) = 0, so d = a/b, giving:

y = (a/b)(etb-1)

When is y/1800 > 0.00018?

y/1800 > 0.00018
y > 0.324
etb-1[/sup] < -0.003
etb < 0.997
tb < ln(0.997)
t > 36.054108243584661759010672195441...

3. Apr 20, 2006

### e(ho0n3

I don't get it. What is y(t) represent? Gallons of water in the first problem and cubic feet of CO in the second problem at time t?

4. Apr 20, 2006

### AKG

Yes, that's what it represents.