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a tank contains 200 liters of fluid in which 30 grams of salt is dissolved. brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 4 l/min; the well-mixed solution is pumped out at the same rate. find the number a(t) of grams of salt in the tank at time t



Answer :

The number a(t) of grams of salt in the tank at time t is 200-170e^-1/50t

Let the amount of salt be A(t). Then,

A'(t)=Rate of salt coming in-Rate of salt going out

      = 4 - A(t)/50

,Then

A(t)/50 + A'(t) =4

We know that the linear differential equation: y+ay = g(x) has solution in the form of

e^dx y= ∫e^ax g(x)dx + C.

Here, a= 1/50 ,g(x)=4.

So, the solution is,

e^1/50t A(t) = ∫e^1/50t (4) dt +C

                   =4∫e^1/50t dt +C

By solving,

                  = 200+Ce^1/50t

Plug in the initial condition: A(0)=30 in the above differential equation.

200+Ce^1/50(0) = 30 =-170

200+C-30

C=30-200

  =-170

Thus, the number A(t) of grams of salt in the tank at time t is,

A(t)=200-170e^-1/50t

To learn more about differential equation click here:

brainly.com/question/14620493

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