# Setting up a Linear ODE

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• Jan 24th 2011, 03:46 PM
lilaziz1
Setting up a Linear ODE
Hey. My book is being a really bad teacher right now so I went on google for some help and I got this: http://college.cengage.com/mathemati...lc7eap1502.pdf

The linear ODE stuff makes sense, but I'm stuck on setting up a differential equation (the one on page 5). It says:

A tank contains 50 gallons of a solution composed of 90% water and 10% alcohol. A second solution containing 50% water and 50% alcohol is added to the tank at the rate of 4 gallons per minute. As the second solution is being added, the tank is being drained at the rate of 5 gallons per minute. Assuming the solution in the tank is stirred constantly, how much alcohol is in the tank after 10 minutes?

Solution: Let y be the number of gallons of alcohol in the tank at any time t. You know that y = 5 when t = 0. Because the number of gallons of solution in the tank at any time is 50 - t and the tank loses 5 gallons of solution per minute, it must lose
$\frac{5}{50-t}y$
gallons of alcohol per minute.

Ok. I'm having a problem understanding the 50 - t. Shouldn't it be 50 - 5t??

Thanks in advance!
• Jan 24th 2011, 04:59 PM
adkinsjr
Quote:

Originally Posted by lilaziz1
Hey. My book is being a really bad teacher right now so I went on google for some help and I got this: http://college.cengage.com/mathemati...lc7eap1502.pdf

The linear ODE stuff makes sense, but I'm stuck on setting up a differential equation (the one on page 5). It says:

A tank contains 50 gallons of a solution composed of 90% water and 10% alcohol. A second solution containing 50% water and 50% alcohol is added to the tank at the rate of 4 gallons per minute. As the second solution is being added, the tank is being drained at the rate of 5 gallons per minute. Assuming the solution in the tank is stirred constantly, how much alcohol is in the tank after 10 minutes?

Solution: Let y be the number of gallons of alcohol in the tank at any time t. You know that y = 5 when t = 0. Because the number of gallons of solution in the tank at any time is 50 - t and the tank loses 5 gallons of solution per minute, it must lose
$\frac{5}{50-t}y$
gallons of alcohol per minute.

Ok. I'm having a problem understanding the 50 - t. Shouldn't it be 50 - 5t??

Thanks in advance!

Fluid is flowing out faster than it's flowing in. So the amount of fluid in the tank at any time t is:

50 gallons fluid + (4gallon Soln/min - 5 gallons of fluid)*t

Remember, 4 gallons are being pumped IN every time 5 gallons is removed. So the difference is -1. So the amount of fluid in the tank at any time t is given by the function 50-t.

Remember that's gallons of FLUID = water +alcohol. So to get the rate out for alcohol alone you just have to pay attention to the units.

$\frac{5(gallons of fluid per minute)}{(50-t)gallons of flud}*(y) gallons of Alcohol$

So,

$\frac{5}{50-t}y$ has the units: gallons of Alcohol / min.