1. ## Solution Mixing Problem

Each of two tanks contains 400 gal of water, with Tank 1 containing 100 lbs of dissolved fertilizer and Tank 2 containing 40 lbs of dissolved fertilizer. Tank 1 has inflow of 48 gal/min of pure water and 16 gal/min from Tank 2, with an outflow of 64 gal/min to Tank 2. Additionally, Tank 2 has an outflow of 48 gal/min. The mixture is kept uniform by stirring. Find the fertilizer contents y1(t) in Tank 1 and y2(t) in Tank 2.

Start of solution:

y is going to equal the inflow/min - outflow/min. Since the pure water coming into Tank 1 does not add to the fertilizer content in Tank 1, the equations become:
y1' = (16/400)y2 - (64/400)y1
y2' = (64/400)y1 - (16/400)y2 - (48/400)y2

or simplified:

y1' = .04y2 - .16y1
y2' = .16y1 - .16y2

At this point, I honestly don't know what to do next. A clue would be much appreciated.

2. Originally Posted by dsprice
Each of two tanks contains 400 gal of water, with Tank 1 containing 100 lbs of dissolved fertilizer and Tank 2 containing 40 lbs of dissolved fertilizer. Tank 1 has inflow of 48 gal/min of pure water and 16 gal/min from Tank 2, with an outflow of 64 gal/min to Tank 2. Additionally, Tank 2 has an outflow of 48 gal/min. The mixture is kept uniform by stirring. Find the fertilizer contents y1(t) in Tank 1 and y2(t) in Tank 2.

Start of solution:

y is going to equal the inflow/min - outflow/min. Since the pure water coming into Tank 1 does not add to the fertilizer content in Tank 1, the equations become:
y1' = (16/400)y2 - (64/400)y1
y2' = (64/400)y1 - (16/400)y2 - (48/400)y2

or simplified:

y1' = .04y2 - .16y1
y2' = .16y1 - .16y2

At this point, I honestly don't know what to do next. A clue would be much appreciated.
Differentiate one of them and then substitute the derivative from the other into this to give a second order ODE in one variable.

CB