# mixture problem solution

suppose we are given following problem.

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? i guessed that it is related with differential equation,also i know that at time t=0,in tank there is 45 gallons water and 5 gallons alcohol,because per minute 4 gallons is added ,it means that 2 gallons water and 2 gallons alcohol is added so it one minute 7 gallons alcohol is in tank yes?and also because 5 gallons drained per minute ,we must substract 2.5 -2.5 each water and alcohol yes?please help me how to define how much alcohol will be in 10 minutes.

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This problem with solution can be found here , example 4 of section 15.2. –  Américo Tavares Jan 7 '12 at 14:37

V(t)=total volume of water g(t)=volume of alcohol

$V=(4-5)t+50$

$\frac{dg}{dt}=2-5\frac{g}{v}$

use integrating factor method:

$I=e^{\int\frac{5}{-t+50}}=(50-t)^{-5}$

$Ig(t)=\int2(50-t)^{-5}=)=(\frac{1}{2}(t-50)^{-4}+C$

$g(t)=(\frac{1}{2}(t-50)+C(50-t)^{5}$

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thanks @Angela Richardson and one more question,if instead of alcohol,we are interested in water then? –  dato datuashvili Jan 6 '12 at 10:51
Amount of water=total volume - amount of alcohol. Total volume=50-t. –  Angela Richardson Jan 6 '12 at 12:34
I think that $V(t)=50-t$ is the total volume of the water and alcohol solution. –  Américo Tavares Jan 6 '12 at 14:15
+1, Typos: $$I=e^{\int\frac{5}{-t+50}dt}=(50-t)^{-5}$$ $$Ig(t)=\int2(50-t)^{-5}dt=\frac{1}{2}(t-50)^{-4}+C$$ $$g(t)=\frac{1}{2}(50-t)+C(50-t)^{5}.$$ –  Américo Tavares Jan 6 '12 at 17:14