Atank initially holds 200 gallons of fresh water (no salt). a brine solution containing 2 pounds of salt per gallon is poured in at a rate of 3 gallons per minute. the tank is kept well-mixed and drained at the rate of 3 gallons per minute. write and solve an initial value differential equation to model the amount of salt, a, in the tank at time, t. 2.

1.99 pounds per gallon of salt in t=2 in the tank.

Step-by-step explanation:

First we consider the matter balance equation that contemplates the input and output; the generation and consumption equal to the acomulation.  

In this case we have no Generation neither do Consumption so, if we consider Acomulation = A(t), the rate of change of A(t) in time is given by:

---(1)

where C is th concentration, with the initial value statement that A(t=0) = 0 because there is no salt in the time cero in the tank, only water.

Given the integral factor -> and multipying the entire (1) by it, we have:

Solving this integrals we obtain:

So given the initial value condition A(t=0)=0 we have:

,

and the solution is,

.

If we give the actual values we obtain then,

.

So in t= 2 we have .

step-by-step explanation:

0.6, 3/5, and 6% are the 3 same portion.

but, when they're all changed to a decimal they would equal to 0.6.

hope this , good luck!