Salt Mixture Problem (First-Order Linear ODE Problem)

**Aryth** · February 29th 2008, 07:49 PM

I just want to verify my work... Hopefully I was right:

A tank initially contains 100 gallons of brine in which 50 lbs. of salt are dissolved. A brine containing 2 lbs/gal of salt runs into the tank at the rate of 5 gal/min. The mixture is kept uniform by stirring and flows out of the tank at 4 gal/min.

Initial Information:

IN: 2 lbs/gal AT: 5 gal/min
OUT: 4 gal/min

$V(t) = Volume$

$y(t) = Concentration$

$\frac{dy}{dt} = \text{RATE IN} - \text{RATE OUT}$

a) At what rate (lbs/min) does the salt enter the container at time t?

$\text{RATE IN} = 2\frac{lbs}{gal}*5\frac{gal}{min} = 10\frac{lbs}{min}$

b) What is the volume of brine in the tank at time t?

$V(t) = 100 + t$

c) At what rate (lbs/min) does salt leave the tank at time t?

$\text{RATE OUT} = 4*\left(\frac{y(t)}{100 + t}\right)$

d) Write and solve initial value problem.

$\frac{dy}{dt} = 10 - \left(\frac{4}{100 + t}\right)y$

$\frac{dy}{dt} + \left(\frac{4}{100 + t}\right)y = 10$

$IF = (100 + t)^4$

$[(100 + t)^4y]' = 10(100 + t)^4$

$\int [(100 + t)^4y]' = \int 10(100 + t)^4 dt$

$(100 + t)^4y = \frac{(100 + t)^5}{2} + C$

$y = \frac{100 + t}{2} + C$

$y(0) = 50 + C = 0$

$C = -50$

$y = \frac{100 + t}{2} - 50$

$y = \frac{t}{2}$

e) Find the concentration of salt in the tank at 25 minutes after process starts.

$y(25) = \frac{25}{2}$

$y = 12.5 lbs.$

**mr fantastic** · February 29th 2008, 09:19 PM

Originally Posted by Aryth

I just want to verify my work... Hopefully I was right:

A tank initially contains 100 gallons of brine in which 50 lbs. of salt are dissolved. A brine containing 2 lbs/gal of salt runs into the tank at the rate of 5 gal/min. The mixture is kept uniform by stirring and flows out of the tank at 4 gal/min.

Initial Information:

IN: 2 lbs/gal AT: 5 gal/min
OUT: 4 gal/min

$V(t) = Volume$

$y(t) = Concentration$

$\frac{dy}{dt} = \text{RATE IN} - \text{RATE OUT}$

a) At what rate (lbs/min) does the salt enter the container at time t?

$\text{RATE IN} = 2\frac{lbs}{gal}*5\frac{gal}{min} = 10\frac{lbs}{min}$

b) What is the volume of brine in the tank at time t?

$V(t) = 100 + t$

c) At what rate (lbs/min) does salt leave the tank at time t?

$\text{RATE OUT} = 4*\left(\frac{y(t)}{100 + t}\right)$

d) Write and solve initial value problem.

$\frac{dy}{dt} = 10 - \left(\frac{4}{100 + t}\right)y$

$\frac{dy}{dt} + \left(\frac{4}{100 + t}\right)y = 10$

$IF = (100 + t)^4$

$[(100 + t)^4y]' = 10(100 + t)^4$

$\int [(100 + t)^4y]' = \int 10(100 + t)^4 dt$

$(100 + t)^4y = \frac{(100 + t)^5}{2} + C$ *

Mr F says: This line should be $(100 + t)^4y = {\color{blue}2} (100 + t)^5 + C$ . But that's the least of your worries ......

$y = \frac{100 + t}{2} + C$ Mr F says: No. It's $y = \frac{100 + t}{2} + \frac{C}{(100 + t)^4}$ . So I'm afraid that everything that follows is no good.

$y(0) = 50 + C = 0$

$C = -50$

$y = \frac{100 + t}{2} - 50$

$y = \frac{t}{2}$

e) Find the concentration of salt in the tank at 25 minutes after process starts.

$y(25) = \frac{25}{2}$

$y = 12.5 lbs.$

Two things:

1. y is the amount (in lb) of salt, not concentration.

2. I'm afraid your integration goes South after the line marked *.

You could have seen that your solution was no good by substituting it back into the original DE ...... It doesn't work.

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