Let <em>s(t)</em> be the amount of salt in the tank at time <em>t</em>. Then <em>s(0)</em> = 50 lb.
Salt flows into the tank at a rate of
(2 gal/min) (6 lb/gal) = 12 lb/min
and flows out at a rate of
(2 gal/min) (<em>s(t)</em>/300 lb/gal) = <em>s(t)</em>/150 lb/min
so that the net rate at which salt is exchanged through the tank is
d<em>s(t)</em>/d<em>t</em> = 12 - <em>s(t)</em>/150 … (lb/min)
Solve for <em>s(t)</em>. The DE is separable, so we have
d<em>s</em>/d<em>t</em> = 12 - <em>s</em>/150
150 d<em>s</em>/d<em>t</em> = 1800 - <em>s</em>
150/(1800 - <em>s</em>) d<em>s</em> = d<em>t</em>
Integrate both sides to get
-150 ln|1800 - <em>s</em>| = <em>t</em> + <em>C</em>
Solve for <em>s</em> :
ln|1800 - <em>s</em>| = -<em>t</em>/150 + <em>C</em>
1800 - <em>s</em> = exp(-<em>t</em>/150 + <em>C </em>)
1800 - <em>s</em> = <em>C</em> exp(-<em>t</em>/150)
<em>s</em> = 1800 - <em>C</em> exp(-<em>t</em>/150)
Now given that <em>s(0)</em> = 50, we solve for <em>C</em> :
50 = 1800 - <em>C</em> exp(-0/150)
50 = 1800 - <em>C</em>
<em>C</em> = 1750
Then the amount of salt in the tank at any time <em>t</em> ≥ 0 is
<em>s(t)</em> = 1800 - 1750 exp(-<em>t</em>/150)
To find the time it takes for the tank to hold 100 lb of salt, solve for <em>t</em> in
100 = 1800 - 1750 exp(-<em>t</em>/150)
1700 = 1750 exp(-<em>t</em>/150)
34/35 = exp(-<em>t</em>/150)
ln(34/35) = -<em>t</em>/150
<em>t</em> = -150 ln(34/35) ≈ 4.348
So it would take approximately 4.348 minutes.
By the way, we didn't have to solve for <em>s(t)</em>, we could have instead stopped with
-150 ln|1800 - <em>s</em>| = <em>t</em> + <em>C</em>
Solve for <em>C</em> - this <em>C</em> <u>is not</u> the same as the one we found using the other method. <em>s(0)</em> = 50, so
-150 ln|1800 - 50| = 0 + <em>C</em>
<em>C</em> = -150 ln|1750|
==> <em>t</em> = 150 ln(1750) - 150 ln|1800 - <em>s</em>|
Then <em>s(t)</em> = 100 lb when
<em>t</em> = 150 ln(1750) - 150 ln(1700)
<em>t</em> = 150 ln(1750/1700)
<em>t</em> = 150 ln(35/34)
<em>t</em> ≈ 4.348
