# Separable Equations please help

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• August 11th 2008, 01:21 PM
eawolbert
I've been trying to solve this problem for quite a while now but can't seem to get the right answer. please any pointer will be greatly appreciated.

A tank contains 50 kg of salt and 1000 L of water. A solution of a concentration 0.025 kg of salt per liter enters a tank at the rate 7 L/min. The solution is mixed and drains from the tank at the same rate.

Find the amount of salt in the tank after 2.5 hours.

Find the concentration of salt in the solution in the tank as time approaches infinity.

my dy/dt equation looks like this dy/dt = .175-7y/1000

after doing the steps my book shows I end up with a really long equation that doesn't help at all.
I know that y(0)=80 kg has to help me find c but the answer is totally wrong. I don't know where did I messed up. Please help.
• August 11th 2008, 01:26 PM
Matt Westwood
Easiest way to address stuff like this is replace the numbers with letters and put the numbers back in afterwards. So you get dy/dt = a - by/d which indeed is separable and should give you a solution involving exponentials. Yes it will be a bit messy (these revolting things are). Take it slowly, do it neatly, works for me every time.
• August 11th 2008, 02:04 PM
eawolbert
I tried but i made a mess and didn't come up with the right equation. :b
what really messes me up it's that extra 7y.

can someone help me figure out the equation with respect to y. please I really need to see what steps I messed up. Thanks.
• August 11th 2008, 04:28 PM
mr fantastic
Quote:

Originally Posted by eawolbert
I've been trying to solve this problem for quite a while now but can't seem to get the right answer. please any pointer will be greatly appreciated.

A tank contains 50 kg of salt and 1000 L of water. A solution of a concentration 0.025 kg of salt per liter enters a tank at the rate 7 L/min. The solution is mixed and drains from the tank at the same rate.

Find the amount of salt in the tank after 2.5 hours.

Find the concentration of salt in the solution in the tank as time approaches infinity.

my dy/dt equation looks like this dy/dt = .175-7y/1000

after doing the steps my book shows I end up with a really long equation that doesn't help at all.
I know that y(0)=80 kg has to help me find c but the answer is totally wrong. I don't know where did I messed up. Please help.

The DE can be re-written as

$\frac{dy}{dt} = \frac{175 - 7y}{1000}$

$\Rightarrow \frac{dt}{dy} = \frac{1000}{175-7y} = - \frac{1000}{7y - 175}$ subject to the boundary condition y(0) = 50 (or is it 80??).

If you're solving differential equations then it's expected that you can integrate things like $\frac{1000}{7y - 175}$ with respect to y.
• August 11th 2008, 05:59 PM
eawolbert
I know how to get there but after I try to find c(which is y(0)=50). I end up with this equation:

-1/7 ln(175-7y)= t/1000 + c

my c= -1/7ln(-175)

leaving

-1/7 ln(175-7y)= t/1000 -1/7ln(-175)

from here is where I mess something up and cannot find the answer.
• August 11th 2008, 07:52 PM
mr fantastic
Quote:

Originally Posted by eawolbert
I know how to get there but after I try to find c(which is y(0)=50). I end up with this equation:

-1/7 ln(175-7y)= t/1000 + c

my c= -1/7ln(-175)

leaving

-1/7 ln(175-7y)= t/1000 -1/7ln(-175)

from here is where I mess something up and cannot find the answer.

My advice is to get y in terms of t before using the boundary condition:

$\ln | 175 - 7y| = -\frac{7t}{1000} - 7C$

note the use of absolute value | | NOT () ......

$\Rightarrow 175 - 7y = e^{-\frac{7t}{1000} - 7C} = e^{-\frac{7t}{1000}} e^{- 7C} = A e^{-\frac{7t}{1000}}$

where $A = e^{- 7C}$ is just as arbitrary as $e^{- 7C}$

$\Rightarrow y = \left(175 - A e^{-\frac{7t}{1000}}\right)/7$.

Now substitute y = 50 (80??) when t = 0 and solve for A.
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