Answer:
dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0
Step-by-step explanation:
The mass flow rate dQ(t)/dt = mass flowing in - mass flowing out
Since 5 g/L of salt is pumped in at a rate of 4 L/min, the mass flow in is thus 5 g/L × 4 L/min = 20 g/min.
Let Q(t) be the mass present at any time, t. The concentration at any time ,t is thus Q(t)/volume = Q(t)/10. Since water drains at a rate of 4 L/min, the mass flow out is thus, Q(t)/10 g/L × 4 L/min = 2Q(t)/5 g/min.
So, dQ(t)/dt = mass flowing in - mass flowing out
dQ(t)/dt = 20 g/min - 2Q(t)/5 g/min
Since the salt just begins to be pumped in, the initial mass of salt in the tank is zero. So Q(0) = 0
So, the initial value problem is thus
dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0
