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.