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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)

https://brainmass.com/math/ordinary-differential-equations/solving-two-independent-ordinary-differential-equations-450209

## SOLUTION This solution is FREE courtesy of BrainMass!

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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 begin by solving for the homogeneous solution. To do this, we set the right hand side (the forcing function, which here is constant) to zero.

(dsigma_A)/dt+v/V sigma_A=0 (3)

The solution of this particular Ordinary Differential Equation (ODE) is attained by making the guess

sigma_(A,h)=C_1 e^Lt (4)

Here, C1 is a constant coefficient and the subscript h denotes the homogeneous solution. Differentiating equation (4) with respect to t gives:

(dsigma_(A,h))/dt=C_1 Le^Lt

Plugging this and (4) into equation (3) then results in:

C_1 Le^Lt+v/V C_1 e^Lt=0

Factoring,

C_1 e^Lt (L+v/V)=0

Because we want a non trivial solution, the coefficient must be nonzero. Thus, we must have the following to be true.

L+v/V=0
Or
L=-v/V

Thus, plugging this into (4) we get the homogeneous solution.

sigma_(A,h)=C_1 e^Lt=C_1 e^(-vt/V) (5)

For the particular solution, we must make a guess similar to the forcing function (the right hand term) in our original ODE (equation 2). We see that this is a constant (independent of t). Our guess then needs to only be a constant.

sigma_(A,p)=C_2 (6)

Where C_2 is a constant and p stands for particular. The derivative of a constant is zero, so plugging (6) into (2) gives us

(vC_2)/V=m/V

Thus,
C_2=m/v

The solution is then the sum of the particular (5) and homogeneous (6).

sigma_A= sigma_(A,p)+sigma_(A,h)=m/v+ C_1 e^(-vt/V) (7)

Finally, we can apply the initial condition (1) to (7) to solve for the unknown constant. Doing this:

0=m/v+ C_1 e^(-v*0/V)=m/v+C_1 => C_1= -m/v

After rewriting (7),

sigma_A=m/v (1- e^(-vt/V) ) (8)

Now we can solve the second ODE.

(dsigma_B)/dt+v/V sigma_B=v/V sigma_A (9)

We follow the same procedure as above to solve for the homogeneous solution. Once this is done, we get

sigma_(B,h)=C_1 e^(-vt/V) (10)

In order to find the particular solution to the ODE given in (9), we must use our solution for the concentration in tank A (equation (8)). Equation (9) becomes

(dsigma_(B,p))/dt+v/V sigma_(B,p)=v/V (m/v (1- e^(-vt/V) ) )

Expanding the right hand side to make things more clear.

(dsigma_(B,p))/dt+v/V sigma_(B,p)=m/V-m/V e^((-vt)/V) (11)

There are now two terms on the right hand side of our ODE. One is a constant term for which we will again guess a constant. The other is of the form of our homogeneous solution and we can no longer guess this form. This is because any solution to the homogeneous problem cannot be a solution to the particular problem. In these cases, we multiply by t or by higher powers in t when necessary. Thus the form of the particular solution is:

sigma_(B,p)=C_2+C_3 t e^(-vt/V) (12)

Upon differentiating (12), we obtain:

(dsigma_(B,p))/dt=C_3 e^(-vt/V) (1-vt/V)

Once we plug this and (12) into (11) we get

C_3 e^(-vt/V) * (1-vt/V)+v/V * {C_2+C_3 t e^(-vt/V) }=m/V-m/V e^(-vt/V)

It can easily be shown by expanding the left hand side that this reduces to

C_3 e^(-vt/V)+v/V C_2=m/V-m/V e^(-vt/V)

And it is therefore evident that the coefficients must:

C_2=m/v ; C_3=-m/V

Substituting these into our particular solution (12) then results in:

sigma_(B,p)=m/v-m/V t e^(-vt/V) (13)

The complete solution again being the sum of the particular and homogeneous.

sigma_B=C_1 e^(-vt/V)+m/v-m/V t e^(-vt/V) (14)

The final coefficient is found from the initial condition and it equal to -m/v. Finally we write the final form for the concentration of solute in tank B.

sigma_B=m/v * (1-e^(-vt/V) )-m/V t e^(-vt/V) (15)

To find the mass of solute in tank B, we use the simple fact that the mass is the concentration multiplied by the volume. In other words,
mass= sigma_B*V.

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