Let A(t), B(t), and C(t) denote the amounts (in hg) of dye in tanks A, B, and C, respectively.
Both A and B start with 20 hg of dye, so A(0) = B(0) = 20. C starts off containing only pure water, so C(0) = 0.
The amount of dye in a given tank changes at a net rate equal to the difference between how much dye flows in and how much flows out.
• Tank A:
•• Dye flows in from an independent source at a rate of
(3 hg/L) (50 L/hr) = 3/50 hg/hr
That is,
(concentration of dye) (flow rate) = rate of change in amount of dye
and the concentration is equal to the amount of dye per unit volume.
Dye also flows in from tank B at a rate of
(B(t)/100 hg/L) (10 L/hr) = B(t)/10 hg/hr
•• Dye flows out of A into B at a rate of
(A(t)/200 hg/L) (40 L/hr) = A(t)/5 hg/hr
and out the exhaust spout at a rate of
(A(t)/200 hg/L) (20 L/hr) = A(t)/10 hg/hr
•• Then the net rate dA/dt is
• Tank B:
•• Dye flows in from A at a rate of
(A(t)/200 hg/L) (40 L/hr) = A(t)/5 hg/hr
and from C at a rate of
(C(t)/100 hg/L) (10 L/hr) = C(t)/10 hg/hr
•• Dye flows out into A at a rate of
(B(t)/100 hg/L) (10 L/hr) = B(t)/10 hg/hr,
out into C at a rate of
(B(t)/100 hg/L) (30 L/hr) = 3B(t)/10 hg/hr,
and out the exhaust spout at a rate of
(B(t)/100 hg/L) (10 L/hr) = B(t)/10 hg/hr
•• The net rate dB/dt is then
• Tank C:
•• Dye flows from B at a rate of
(B(t)/100 hg/L) (30 L/hr) = 3B(t)/10 hg/hr
•• Dye flows out into B at a rate of
(C(t)/100 hg/L) (10 L/hr) = C(t)/10 hg/hr
and out the exhaust spout at a rate of
(C(t)/100 hg/L) (20 L/hr) = C(t)/5 hg/hr
•• The net rate dC/dt is then
Condensed into a matrix equation, the system of ODEs can be expressed as
where
and with initial condition
Just for fun, here's a plot of the solution to this system for the first 50 hours after the pumps are started: