Mixing Problem with Constant Coefficient Differential Equation

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The discussion revolves around solving a differential equation related to a tank containing a sugar solution. The initial setup involves a tank with 1160 liters of pure water, where a solution with 0.03 kg of sugar per liter enters at 7 L/min. The differential equation is formulated as ds/dt = 0.21 - (7s/1160), representing the rate of sugar entering and leaving the tank. Participants confirm that the equation is a constant coefficient linear differential equation, suggesting methods such as integrating factors or separation of variables for solving it. The conversation emphasizes the importance of understanding the equation's structure to find the amount of sugar over time.
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Homework Statement



A tank contains 1160 L of pure water. A solution that contains 0.03 kg of sugar per liter enters a tank at the rate 7 L/min The solution is mixed and drains from the tank at the same rate.

a.) How much sugar is in the tank initially?
b.) Find the amount of sugar in the tank after t minutes.
c.) Find the concentration of sugar in the solution in the tank after 72 minutes.

Homework Equations


The Attempt at a Solution



I made s(t) be the amount of salt at time 't'

ds/dt = rate in - rate out

Well the rate in would be the concentration times the rate... so

ds/dt = (.03kg)(7L/min) - (s/1160)(7L/min)

ds/dt = .21 - (7s/1160)

Kind of get confused from there...I figure I'll have to get the 's' on one side right? separable?Am I even on the right track here? Just starting out in the class.
 
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CollegeStudent said:

Homework Statement



A tank contains 1160 L of pure water. A solution that contains 0.03 kg of sugar per liter enters a tank at the rate 7 L/min The solution is mixed and drains from the tank at the same rate.

a.) How much sugar is in the tank initially?
b.) Find the amount of sugar in the tank after t minutes.
c.) Find the concentration of sugar in the solution in the tank after 72 minutes.

Homework Equations





The Attempt at a Solution



I made s(t) be the amount of salt at time 't'

ds/dt = rate in - rate out

Well the rate in would be the concentration times the rate... so

ds/dt = (.03kg)(7L/min) - (s/1160)(7L/min)

ds/dt = .21 - (7s/1160)

Kind of get confused from there...I figure I'll have to get the 's' on one side right? separable?


Am I even on the right track here? Just starting out in the class.

Yes, you are on the right track. So far so good.

If you call ##a = \frac 7 {1160}## your equation can be rewritten$$
\frac{ds}{dt} +as = .21$$It is constant coefficient, linear, and separable, so any of these methods would work. I would suggest the constant coefficient method first, linear (integrating factor) second, and separation of variables last in order of preference.
 
Constant coefficient? I wouldn't just integrate here? I'm looking all over the web for an explanation of "constant coefficient" but not seeing too much
 
CollegeStudent said:
Constant coefficient? I wouldn't just integrate here? I'm looking all over the web for an explanation of "constant coefficient" but not seeing too much

That's "constant coefficient differential equation". If you haven't studied those yet use the integrating factor method for first order linear DE's.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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