Two tanks A and B each holding 40 L of the tears of differen
Solution
Let start with generalize the differential equation
Consider a tank, which initially hold V0 L of brine that contains a kg of salt. Another brine solution, containing salt b kg /L, is poured into the tank at the rate of e L/min, the well stirred solution leaves the tank at the rate of f L/min.
Let Q denote the amount ( in kg) of salt in the tank at any time. The time rate of change of Q, dQ/dt, equal the rate at which enters the tank minus the rate at which salt leaves the tank.
Salt enters the tank at the rate of be kg/min.
Volume of brine in the tank at any time t = V0 + et + ft
The concentration of salt in the tank at any time = Q/( V0 + et + ft)
Salt leaves the tank at the rate of f Q/( V0 + et + ft) kg/min
dQ/dt = be - f Q/( V0 + et + ft)
dQ/dt + f Q/( V0 + et + ft) = be , which is required generalized differential equation.
Now V0 = 40L , f = 3L/min, e = 4L/min, b=0
Thus dQ/dt + 3Q/(40 + t) = 0
On solving Q(40 + t)3 = c
Now at t= 0 , Q=a = 3.5 c = 224000
Q(40 + t)3 = 224000

