Diff. Eq. Mixture problem. Did most of the work, Need to solve for C

In summary, the conversation discusses finding the correct concentration for a solution and the net change in volume over time. The final solution is A*(t-50)^(-5/2)=-8(50-t)^(-3/2)+c, and a link to Wolfram Alpha is provided to check the integration.
  • #1
Jeff12341234
179
0
I need to solve for C. I know it's probably simple but i don't remember how to. This is what I have so far:
FgCaDxf.jpg
 
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  • #2


Use the initial conditions given.

Also, your solution's incorrect.

You only have 100L of solution at one point in time, so at only one point in time will the concentration leaving the tank be A/100. There's a net change in volume as time passes.
 
  • #3


Mangoes said:
Use the initial conditions given.

Also, your solution's incorrect.

You only have 100L of solution at one point in time, so at only one point in time will the concentration leaving the tank be A/100. There's a net change in volume as time passes.

so what should it be instead of A/100?? A/35?
 
  • #4


The rate out will be the the concentration times the rate at which the fluid flows out per minute.

You already know the fluid flows out at 5L/min, but your concentration's wrong.

It's correct that the concentration will depend on the amount A of chemical X, but the concentration also depends on the total volume of the solution. Five grams in 1 L is much more concentrated than five grams in 100L.

You just need to write the concentration in a way that's dependent on time since your volume also depends on time.
 
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  • #5


so it will be A/(100+5t)?

or maybe A/(100-2t) <--- -2t came from Rate in rate out
 
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  • #6


No, but you're getting closer.

V(t) = 5t + 100 would imply that the volume is increasing with time. It's not.

Look at the net change in volume.
 
  • #7


Close, but you need to consider the rate into the tank as well.
 
  • #8


A/(100-2t)? <--- -2t came from Rate in - rate out
 
  • #9


A/(100 - 2t) is correct, but don't forget to multiply it by 5 representing the liters/min to get the total rate out.
 
  • #10


Right. Thanks
 
  • #11


I get a non-real result now when I try to solve for c. Where did I go wrong?

OVhz36N.jpg
 
  • #12


That shouldn't be giving you a non-real solution. It looks fine to me.
 
  • #13


no, it doesn't work out.. :/
 
  • #14


Oh, I read it wrong. It should be 50-t, not t-50.
 
  • #15


are you saying it should be A*(t-50)^(-5/2)=-8(50-t)^(-3/2)+c ?
How do you figure??
 
  • #16


The integral of 5/(100-2t) is (-5/2)ln(50-t).
 
  • #17


I get a different answer:

GesMlaO.jpg


Is there another source we can check?
 
  • #18


That's strange that it's giving you that answer. I checked by doing it by hand, as well as wolfram alpha and I get (-5/2)ln(50-t).
 
  • #19


do you have a wolfram alpha link? I want to try it.
 
  • #21


wolfram says: (-5*Log[-2*(-50 + x)])/2
 
  • #22


tkumM93.png


That's not the answer I got, nor is it the answer you got...
 
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1. What is a differential equation mixture problem?

A differential equation mixture problem is a type of mathematical problem that involves finding the rate at which a substance is being added to a mixture, while taking into account the changing concentrations of the mixture over time.

2. How is a differential equation mixture problem solved?

To solve a differential equation mixture problem, you need to first write out the differential equation that represents the problem. Then, you can use various mathematical techniques, such as separation of variables or substitution, to find the solution. Finally, you can solve for any unknown constants, such as the concentration of the substance being added to the mixture.

3. What is the role of the constant (C) in a differential equation mixture problem?

The constant (C) in a differential equation mixture problem represents the initial concentration of the mixture. It is often referred to as the "constant of integration" and is necessary to find the specific solution to the problem.

4. Can a differential equation mixture problem have multiple solutions?

Yes, a differential equation mixture problem can have multiple solutions. This is because there are often multiple ways to solve a differential equation, and each solution may be valid. However, the initial conditions of the problem can help determine which solution is the most appropriate.

5. How are differential equation mixture problems used in science?

Differential equation mixture problems are commonly used in science, particularly in fields such as chemistry, biology, and physics. They are used to model and understand real-world processes, such as chemical reactions, population growth, and fluid dynamics. By solving these problems, scientists can make predictions and analyze the behavior of complex systems.

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