?Differential Equation Applications: A tank contains 30 gallons of a solution co
ID: 2853227 • Letter: #
Question
?Differential Equation Applications:
A tank contains 30 gallons of a solution composed of 80% water and 20% alcohol. A second solution containing half water and half alcohol is added to the tank at the rate of 6 gallons per minute. At the same time, the well-stirred solution is withdrawn at the same rate, as shown in the figure. Find the amount y of alcohol in the solution as a function of t by solving the differential equation dy/dt = -6(y/30) + 3. Find the amount of alcohol in the tank after 10 minutes.Explanation / Answer
a) To find the amount y of alcohal in the solution as a
function of t by solving the differential equation
dy/dt = -6 (y/30) + 3
dy/dt = (-y/5) + 3
dy/dt + (y/5) = 3
This is the linear differential equation of the form
dy/dt = Py + Q
and , Integrating factor (IF ) = e^[Integral Pdt ]
Here , I F = e^[Integral dt /5 ]
I F = e^ [t / 5]
Now, its Complete Solution will be
ye^(t/5) = Integral [ 3 e^ [t / 5] dt ]
ye^(t/5) = 3 Integral [ e^ [t / 5] dt ]
ye^(t/5) = 15 [ e^ [t / 5] ] + C
y = 15 + C e^(-t/5)
b) To find the Amount of Alcohal in the tank after 10 minutes
Initially, the tank contains 30 gallon (20% alcohol) = 60 gallon of alcohol,
y(0) = 30
Substituting y = 30 and t = 0
y = 15 + C e^(-t/5)
30 = 15 + Ce^(0/5)
15 = C
On Putting the value , we get
y = 15 + 15 e^(-t/5)
Now, to find the amount of alcohal in the tank after
10 minutes
y = 15 + 15 e^(-10/5)
y = 15 + 15 e^(-2)
y = 15 ( 1 + e^(-2) )
y = 15 ( 1 + 1/7.389)
y = 15 ( 1 + 0.1353)
y = 17.0300 gallon
Hence , Amount of alcohal in the tank after 10 minutes
is 17. 0300 gallon
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