Differential Equation mixing problem - read details?

Question

Two large tanks, each holding 100 L of brine. The tanks are interconnected by pipes
that allows fluid to flow between the tanks. The liquid flows from tank A to tank B
at a rate of 3 L/min, and from tank B to tank A at a rate of 1 L/min. A brine
solution flows into tank A at a rate of 6 L/min with concentration of 0.3 kg/L.
Solution leaves tank A at a rate of 4 L/min and from tank B at 2 L/min. If there was
initially 10 kg of salt in tank A and 20 kg of salt in tank B, find an equation for the
amount of salt in both tanks at time t.

Answer 1

Let A(t) denote the amount of salt in tank A at time t, and B(t) be similarly for tank B. Also notice that each minute 6+1 L of water flows into the tank A and 4+3L leaves it. Similarly 3L water enters tank B each minute and 2+1 L leaves it. So both tanks contain 100L of water anytime. Now notice also that each minute 6(0.3) kg plus (1/100)B(t) kg of salt enters tank A and (3/100 + 4/100)A(t) kg salt leaves it. Similarly (3/100)A(t) kg salt enters tank B and (1/100+2/100) B(t) kg salt leaves it. This gives you the following system of linear differential equations:

dA/dt = 1.8 + (3/100) B(t)-(7/100) A(t)
dB/dt = (3/100)A(t) -(3/100) B(t)

Now you write the corresponding matrix, find eigenvalues and use the initial conditions to solve the system.

Answer 2

EDIT: I found several errors, so new files have been posted below. Please check.

EDIT AGAIN: Another correction on page 2: coeff of 1st derivative is 10^3, not 10^5

FINAL EDIT: I have completed the solution for tank A. There are five pages, new files are posted below.

I have made a stab at this, but because of complex formulas I did it in Word. Find it here (five pages)

EDITED FILES

http://img400.imageshack.us/img400/1651/twotanks1li4.png
http://img400.imageshack.us/img400/8776/twotanks2ew2.png
http://img248.imageshack.us/img248/7577/twotanks3oy2.png
http://img505.imageshack.us/img505/6245/twotanks4km7.png
http://img248.imageshack.us/img248/971/twotanks5bs5.png

To complete the solution, use the same process to derive a differential eq for tank B, or use the value of cA(t) from the above in the flow eq for tank B, where it becomes the forcing function of the inhomogeneous differential eq.

Answer 3

the two are " replace in salt = incoming salt - outgoing salt " permit S = salt in tank at time t #2. dS / dt = [3][a million] - [2][ S / { 2 hundred + (a million)t} ] , S(0) = one hundred..{gal * concentration consistent with gallon} discover S(t) and S(3 hundred) 1st is comparable yet extra handy