The salt that is left after t minutes is y = 16e^[-t/100]
We have to mix problems we use:
dy / dt = (rate in) - (rate out)
The water entering the tank at a speed of 10 L / min but it has no salt, therefore
rate in = 0
The tank has 1000 liters of brine with 16 kg of salt initially, therefore the concentration of salt at time "t" is:
y (t) / 1000
then rate out would be:
rate out = y (t) / 1000 * 10 = y (t) / 100
The difference equation would then be:
dy / dt = 0 - y (t) / 100
1 / y (t) * dy = 1/100 * dt
We integrate from both sides and we have:
ln y = - (1/100) * t + C1
y = e ^ [- (1/100) * t + C1]
y = e ^ [- (1/100) * t] * e ^ [C1]
We assume that C = e ^ [C1]
Thus:
y = C * e ^ [- (1/100) * t]
now y = 16 to t = 0, replacing:
16 = C * e ^ [- (1/100) * 0]
16 = C, therefore we would have:
y = 16*e ^ [- t / 100]
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