Answer:
A) A(t) = 10(100 - t) + c(100 - t)²
B) Tank will be empty after 100 minutes.
Explanation:
A) The differential equation of this problem is;
dA/dt = R_in - R_out
Where;
R_in is the rate at which salt enters
R_out is the rate at which salt exits
R_in = (concentration of salt in inflow) × (input rate of brine)
We are given;
Concentration of salt in inflow = 2 lb/gal
Input rate of brine = 5 gal/min
Thus;
R_in = 2 × 5 = 10 lb/min
Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min
So, after t minutes, there will be (500 - 5t) gallons in the tank
Therefore;
R_out = (concentration of salt in outflow) × (output rate of brine)
R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min
R_out = 10A(t)/(500 - 5t) lb/min
So, we substitute the values of R_in and R_out into the Differential equation to get;
dA/dt = 10 - 10A(t)/(500 - 5t)
This simplifies to;
dA/dt = 10 - 2A(t)/(100 - t)
Rearranging, we have;
dA/dt + 2A(t)/(100 - t) = 10
This is a linear differential equation in standard form.
Thus, the integrating factor is;
e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²
Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².
We have;
So, we ;
(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²
Integrating this, we now have;
A(t)/(100 - t)² = ∫10/(100 - t)²
This gives;
A(t)/(100 - t)² = (10/(100 - t)) + c
Multiplying through by (100 - t)²,we have;
A(t) = 10(100 - t) + c(100 - t)²
B) At initial condition, A(0) = 0.
So,0 = 10(100 - 0) + c(100 - 0)²
1000 + 10000c = 0
10000c = -1000
c = -1000/10000
c = -0.1
Thus;
A(t) = 10(100 - t) + -0.1(100 - t)²
A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)
A(t) = 1000 - 10t - 1000 + 20t - 0.1t²
A(t) = 10t - 0.1t²
Tank will be empty when A(t) = 0
So, 0 = 10t - 0.1t²
0.1t² = 10t
Divide both sides by 0.1t to give;
t = 10/0.1
t = 100 minutes
