A tank contains 200 L of salt solution which contains 100 grams of salt. Pure water enters the tank ata rate of 4 L/min, but the thoroughly mixed solution leaves the tank at a rate of 2 L/min. Write andsolve an IVP to determiney, the number of grams of salt in the tank at timet.

a. dy/dt = - 2y/(200 + 2t)  where y(0) = 100 g

b. y = 20000/(200 + t)

Step-by-step 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)

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step-by-step explanation:

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