Answer:
dS/dt = 8 - (3S(t)/(90 + t))
Step-by-step explanation:
The differential equation of this problem is;
dS/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 = 4 gal/min
Thus;
R_in = 2 × 4 = 8 lb/min
Due to the fact that mixture is removed out at a slower rate, thus it is accumulating at the rate of (4 - 3)gal/min = 1 gal/min
Now, R_out = 3(concentration of solution outflow)
Now, concentration of solution outflow is given by the formula;
c_out = S(t)/v(t)
Where v is volume of fluid in tank at time t.
Since initial volume is 90 and we have a net total of 1 gallon added every minute. Thus, v(t) = 90 + t
So, c_out = S(t)/(90 + t)
R_out = 3S(t)/(90 + t)
So, final final differential equation is now;
dS/dt = R_in - R_out = 8 - 3S(t)/(90 + t)
