# Mixture Problem ODE

1. Jun 21, 2009

### dipset24

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

A 500 gallon tank is filled with pure water. A solution containing 4lbs of salt per gallon is added at a rate of 2 gallons per minute. The well-mixed solution is drained at a rate of 3 gallons per minute.

A) How long does it take for the container to achieve a concentration of 2lbs/gal?

B) What is the domain of x(t)?

2. Relevant equations
ODE

3. The attempt at a solution
I found x(t) to be (500-t)^2*(4000-t^2)

then I tried using x(t)/v(t)=2 lbs/gal and this is where I am stuck

2. Jun 21, 2009

### Dick

If x(t)=(500-t)^2*(4000-t^2), then if t=0 then x(0)=500^2*4000. What does that mean? What is x(t) supposed to be? What ODE did you solve? How did you get that strange answer?

3. Jun 21, 2009

### dipset24

x'+(3/(500-t))x=8 was the DE I used

4. Jun 21, 2009

### Dick

I'm going to guess x(t) is the amount of salt. The ODE looks right. The solution doesn't. Can you show us more of how you solved it?

Last edited: Jun 21, 2009
5. Jun 22, 2009

### dipset24

Integrating Factor

e^$$\int(3/(500-t))$$ dt = (500-t)^(-3)

Dt$$\int [(500-t)^(-3)*x]dt$$=8$$\int(500-t)^(-3)$$

(500-t)^(-3) X=4/(t-500)^2 + C

6. Jun 22, 2009

### Dick

That seems ok again. Now what's C? What is x(0) supposed to be?

Last edited: Jun 22, 2009
7. Jun 22, 2009

### HallsofIvy

Staff Emeritus
Surely this is not the exact statement of the problem! There is NO "x(t)" in the original statement. What does x(t) mean here?

8. Jun 22, 2009

### dipset24

The initial conditions are x(0)=0 since it is all pure water at time 0

After solving using the integrating factor my x(t)(pounds of salt)= -4(t-500)+c(t-500)^3

9. Jun 22, 2009

### Dick

Ok, but now you have to find C using x(0)=0. Then find where the concentration x(t)/(500-t)=2lbs/gallon, right?

10. Jun 22, 2009

### dipset24

Yes that is where I am caught up I tried several ways to get c and then plug it into x(t)/(500-t)=2lbs/gallon

The c I come up with is .000016

11. Jun 22, 2009

### Dick

That's gives you x(0)=0 doesn't it? What's wrong with that?

12. Jun 22, 2009

### dipset24

Yes but when I plug that into my X(t)/v(t)

(-4(t-500)+.000016(t-500)^3)/((500-t)=2

which I can reduce to 4-.000016(t-500)^2=2

but the answer t=853.55 is not correct I just need some help verifying my answer

13. Jun 22, 2009

### Dick

It's a quadratic equation. It has two solutions, not just one. Can you find the other one?

14. Jun 22, 2009

### dipset24

Yea i am sorry t=146.44 t= 853.553
However doesnt t have to be less then 500 going all the way back to the integrating factor.

e^(-3*ln(500-t)) t<500???

15. Jun 22, 2009

### Dick

Sure, t has to be less than 500. At t=500 the tank is drained. I think that's the sense to the question about what is the domain of x(t). The ODE has a solution outside of that range, but it doesn't make physical sense.

Last edited: Jun 22, 2009
16. Jun 22, 2009

### dipset24

Yes thank you so the Domain is [0,500] and when t=146.446609 min our concentration is 2lbs/gal