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
[tex]\dfrac{\mathrm dA}{\mathrm dt} = \dfrac3{50}+\dfrac{B(t)-3A(t)}{10}[/tex]
• 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
[tex]\dfrac{\mathrm dB}{\mathrm dt} = \dfrac{2A(t)+C(t)-5B(t)}{10}[/tex]
• 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
[tex]\dfrac{\mathrm dC}{\mathrm dt} = \dfrac{3B(t)-3C(t)}{10}[/tex]
Condensed into a matrix equation, the system of ODEs can be expressed as
[tex]\dfrac{\mathrm d\mathbf X}{\mathrm dt} = \dfrac1{10}\begin{pmatrix}-3&1&0\\2&-5&1\\0&3&-3\end{pmatrix}\mathbf X + \dfrac1{50}\begin{pmatrix}3\\0\\0\end{pmatrix}[/tex]
where
[tex]\mathbf X = \begin{pmatrix}A(t)\\B(t)\\C(t)\end{pmatrix}[/tex]
and with initial condition
[tex]\mathbf X(0) = \begin{pmatrix}20\\20\\0\end{pmatrix}[/tex]
Just for fun, here's a plot of the solution to this system for the first 50 hours after the pumps are started:
