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Solving two independent ordinary differential equations. Concentrations of solute in flowing from one tank to another.

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Two tanks A and B, each of volume V, are filled with water at time t=0. For t > 0, volume v of solution containing mass m of solute flows into tank A per second; mixture flows from tank A to tank B at the same rate and mixture flows away from tank B at the same rate. The differential equations used to model this system are given by:

d(SA)/dt+v/V*SA= m/V
d(SB)/dt+v/V*SB= v/V*SA

Where S stands for sigma (concentration of solute) in tank A or B.

Show that the mass of solute in tank B is given by:

mV/v *{1-e^(-vt/V)} - mt*e^(-vt/V)

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Solution Summary

The solution thoroughly explains the process to solve a linear ordinary differential equation (ODE). Both the homogeneous and non-homogeneous are studied.

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The fact that there are two differential equations might be daunting, but if we look closer we can make sense of it.
There are actually two wholly separate differential equations. This means that we can solve for the concentration of A first, then use this solution to solve for the concentration of B and ultimately the mass. Let us begin.

The first thing we must note is that the question does not supply us with the needed initial conditions. This means we must come up with them from physical or logical sense.
What do we know about the tanks at t=0?
We know that they do not contain any of the solute. Thus, we can write
sigma_A (0)= 0 (1)

This states exactly what we have said in words; the tank holds no solute at time t = 0. We can write the same thing for tank B.
With the initial conditions set, we can now solve the problem. First, let us begin by finding the concentration of tank A. The governing differential equation is

(dsigma_A)/dt+v/V sigma_A=m/V (2)

To solve any differential equation, we need to solve it twice: once for the homogeneous solution, and another for the so called particular solution.
We ...

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