Question

A tank contains 80 kg of salt and 2000 L of water. Pure water enters a tank at the rate 12 L/min. The solution is mixed and drains from the tank at the rate 6 L/min. Let y be the number of kg of salt in the tank after t minutes. The differential equation for this situation would be ?

Answer 1

Answer:

$y'(t) = \frac{-y(t)}{2000+6t}$

Step-by-step explanation:

Let y be the number of kg of salt in the tank after t minutes

y(0) = 80 kg

Water enters at 12l per minutes and drains at 6 l /minute

At time t water content = 2000+(12-6) t= 2000+6t

Rate of change of quantify of salt

= $y'(t) = 0*12 -\frac{y(t)}{2000+6t}$

This is because we assume that pure water entering is mixed uniformly and drained at the rate of 6 litres per minute

So rate of change of salt content = Salt in the container incoming - salt drained out

Incoming is pure water with salt content =0

Outgoing is Salt in the water at that time divided by 2000+6t (total water)

So differential equation would be

$y'(t) = \frac{-y(t)}{2000+6t}$