Let <em>x(t)</em> denote the amount of salt (in kg) in the tank at time <em>t</em>. The tank starts with 18 kg of salt, so <em>x</em> (0) = 18.
The solution is drained from the tank at a rate of 90 L/min, so that the amount of salt in the tank changes according to the differential equation
d<em>x(t)</em>/d<em>t</em> = - (<em>x(t)</em> kg)/(9000 L) × (90 L/min) = -1/100 <em>x(t)</em> kg/min
or, more succintly,
<em>x'</em> = -1/100 <em>x</em>
This equation is separable as
d<em>x</em>/<em>x</em> = -1/100 d<em>t</em>
Integrating both sides gives
∫ d<em>x</em>/<em>x</em> = -1/100 ∫ d<em>t</em>
ln|<em>x</em>| = -1/100 <em>t</em> + <em>C</em>
<em>x</em> = exp(-1/100 <em>t</em> + <em>C </em>)
<em>x</em> = <em>C</em> exp(-<em>t</em>/100)
(a) Using the initial condition <em>x</em> (0) = 18, we find
18 = <em>C</em> exp(0) ==> <em>C</em> = 18
so that
<em>x(t)</em> = 18 exp(-<em>t</em>/100)
(b) After 20 minutes, we have
<em>x</em> (20) = 18 exp(-20/100) = 18 exp(-1/5) ≈ 14.74
so the tank contains approximately 14.74 kg of salt after this time.
