problem in differential equation?

Question

1. into a 100 gal tank initially filled with brine containing 50 lb of salt flow 3 gal/min of brine containing 2 lb/gal of salt and the solution,kept uniform by stirring,flows out at the rate of 2 gal/min.How much salt is in the tank at the end of 100 minutes? ans. 350 lb

2.A tank initially holds 80 gal of brine containing 1/18 lb of salt per gallon.Another brine solution,containing 1 lb/gal of salt, is poured into the tank at the rate of 4 gal/min and the well-stirred mixture leaves at the rate of 8 gal/min.Find the amount of salt in the tank a.) at any time. b.) after 3 mins c.) the time when the tank will hold 40 gallons of solution?

thanks for the help

Answer 1

Ill show you one and let you learn so you can do the other or you wont learn anything (I don't think you can bring yahoo into exams).

Let t be the time in minutes from the start of the experiment, s(t) be the amount of salt in lbs and v(t) be the volume of the tank in gallons.

we are filling at a rate of 3 gallons / 6 lbs of salt a minute
and emptying at a rate of 2 gallons a minute
a net increase of 1 gallon per minute
v(t) = 100 + t

s(0) = 50
we are adding 6 lbs/minute (3 *2)
the 2 g/m leakage includes 2s/v of salt (perfectly mixed)
s' = +6 - 2s/(100+t)

s' + [2/(100+t)]s = 6
let p(t) = 2/(100+t)
integral p(t) = 2 ln(100+t) + constant
let mu = exp(integral( p(t))) = k exp( 2ln(100+t))
= k(100+t)^2
mu is defined so that mu' = p(t) mu

mu s' + s mu' = 6 mu
(mu s)' = 6 mu = 6k(100+t)^2

mu s = 2k (100+t)^3 + c where c is a constant
s = { [2k( 100+t)^3] +c}/ [k(100+t)^2]
s = 2(100+t) + (c/k)/(100+t)^2
s(0) = 200 + (c/k)/10000 = 50
so c/k = -1500000
s(100)= 2 * 200 -1500000/40000
= 400 - 150/4 = 375

presumably I have made a small slip somewhere, and it will help you to find it.