Thread: Firs-Order linear diff story problem

1. Firs-Order linear diff story problem

A tank initially contains 200 gallons of brine, with 50 pounds of salt in solution. Brine containing 2 pounds of salt per gallon is entering the tank at the rate of 4 gallons per minute and is flowing out at the same rate. If the mixture in the tank is kept uniform by constant stirring, find the amount of salt in the tank at the end of 40 minutes.

Amount of salt after 40 minutes: ___ pound(s).

the formula is dy/dt = rate in - rate out

here what i have so for:
dy/dt = 8 - (1/100y)

dy/dt + (1/100y) = 8

d/dt (y exp(1/100*t)) = 8 exp(1/100*t) dt

y exp(1/100*t) = int(8*exp(1/100*t), t)

i'm stuck from this point on.

2. Originally Posted by viet
A tank initially contains 200 gallons of brine, with 50 pounds of salt in solution. Brine containing 2 pounds of salt per gallon is entering the tank at the rate of 4 gallons per minute and is flowing out at the same rate. If the mixture in the tank is kept uniform by constant stirring, find the amount of salt in the tank at the end of 40 minutes.

Amount of salt after 40 minutes: ___ pound(s).

the formula is dy/dt = rate in - rate out

here what i have so for:
dy/dt = 8 - (1/100y)
The inflow rate is 8 lb/min, but 1/50 of the tank content flows out per min
(that is 4 gal/min out of a total of 200 gal.), so if the tank contains y lb the
mass outflow rate is y/50 lb/min, so:

$\displaystyle dy/dt=8-y/50$.

This is of variables seperable type so:

$\displaystyle \int \frac{1}{8-y/50} dy=\int dt$

or:

$\displaystyle -50 \ln(8-y/50) = t+C$

or exponentiating:

$\displaystyle 8-y/50 = A\,e^{-t/50}.$

Now when $\displaystyle t=0, \ y=50$, so: $\displaystyle A=7$ so:

$\displaystyle y=400-350 \,e^{-t/50}$ pounds.

RonL