Answer:
a. dy/dt = - 2y/(200 + 2t) where y(0) = 100 g
b. y = 20000/(200 + t)
Explanation:
a. Write an IVP to determine y, the number of grams of salt in the tank at time, t.
Let y be the mass of salt.
The net flow into the tank dy/dt = mass flow in - mass flow out
Since only water flows into the tank, the mass flow in = 0 g/min
Let m be the mass of salt in the tank at time t. Since the volume of the tank is 200 L and water flows in at a rate of 4 L/min and out at a rate of 2 L/min, the net rate of increase of the volume of the tank is rate in - rate out = 4 L/min - 2 L/min = 2L/min. So, in time, t, the volume of the water in the tank increases by 2t. So, the volume of the tank in time, t is V = 200 + 2t.
So, the concentration of salt in the tank at time t is mass/volume = m/(200 + 2t).
Since the well mixed solution leaves at a rate of 2 L/min, the mass flow out is concentration × volume flow out = y/(200 + 2t) × 2 = 2y/(200 + 2t)
The net flow into the tank dy/dt = mass flow in - mass flow out
dy/dt = 0 - 2y/(200 + 2t)
dy/dt = - 2y/(200 + 2t)
Since the initial mass of salt in the tank is 100 g, y(0) = 100 g
So, the initial value problem IVP is
dy/dt = - 2y/(200 + 2t) where y(0) = 100 g
b. Solve an IVP to determine, the number of grams of salt in the tank at time, t.
Solving the IVP, we have
dy/dt = - 2y/(200 + 2t) where y(0) = 100 g
Separating the variables, we have
dy/y = - 2dt/(200 + 2t)
Integrating both sides, we have
∫dy/y = - ∫2dt/(200 + 2t)
㏑y = - ㏑(200 + t) + ㏑C
㏑y + ㏑(200 + t) = ㏑C
㏑[y(200 + t)] = ㏑C
y(200 + t)] = C
y = C/(200 + t) since y(0) = 100, we have
100 = C/(200 + 0)
100 = C/200
C = 100 × 200
C = 20000
So, y = C/(200 + t)
y = 20000/(200 + t)