2- Type the solution please ( No hand writting ) Consider a closed system of thr

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2- Type the solution please ( No hand writting )

Consider a closed system of three well-mixed brine tanks. Tank 1 has volume 20 gallons, tank 2 has volume 15 gallons, and tank 3 has volume 4 gallons. Mixed brine flows from tank 1 to tank 2, from tank 2 to tank 3, and from tank 3 back to tank 1. The flow rate between each pair of tanks is 60 gallons per minute. At time zero, tank 1 contains 28 lb of salt, tank 2 contains 11 lb of salt, and tank 3 contain no salt. Solve for the amount (lb) of salt in each tank at time t (minutes). Also determine the limiting amount (as t rightarrow infinity) of salt in each tank. (Solve the problem by using Eigenvalues and Laplace Transform)

Explanation / Answer

Let x1(t), x2(t), x3(t) represent the amount of salt in tanks 1, 2, and 3, respectively, at time t. The amounts are governed by the following system of differential equations:

x 1 = 60 /20 *x1 + 60 /4 *x3

x 2 = 60 /15 * x2 + 60 /20 * x1

x 3 = 60 /4 * x3 + 60/ 15 * x2

with inital conditions x1(0) = 28, x2(0) = 11, x3(0) = 0. In matrix form:

x ‘ =[3 0 15

          3 4 0                          x1

            0 4 15 ]

x (0) =[28

          11                       

            0 ]

We find eigenvalues

| 3 0 15

3 4 0

0 4 15 –           | =0

Expanding along the first row we get (3 )(4 )(15 ) + (15)(3)(4) = 0. Simplifying and factoring we get ( + 9)( + 13) = 0.

So the eigenvalues are 1 = 0, 2 = 9, 3 = 13. Next we solve for the corresponding eigenvalues

[3 0 15]                     [0 ]                                                         [20 ]

[ 3 4 0]                     [0]                                                         [15]

[ 0 4 15 ] v1     =     [0 ] solved by v1                                 [4 ]

[6 0 15]                     [0 ]                                                         [-5]

[ 3 9 0]                     [0]                                                         [3]

[ 0 4 2 ] v2     =     [0 ] solved by v2                                [2 ]

[10 0 15]                     [0 ]                                                         [-3 ]

[ 3 9 0]                     [0]                                                         [15]

[ 0 4 2 ] v3     =     [0 ] solved by v3                                 [2 ]

So the general solution is

x = c1 [ 20]                               [ -5]                              [ -3]

             [ 15 ]       +            c2 [3 ] e^-9t +            c3 [1 ] e^-1.3t

               [4 ]                             [ 2 ]                                    [ 2 ]

Plugging in the initial conditions leads to the following system for the unknown constants

[20 -5 -3]                     [c1 ]       =          [28]

[ 15 3 1]                     [c2]       =           [11]

[ 4 2 2 ]                      [c3 ]      =            [0 ]

which has solution c1 = 1, c2 = 1, c3 = 1. The solution is

x1(t) = 20 + 5e 9t + 3e 13t

x2(t) = 15 3e 9t e 13t

x3(t) = 4 2e 9t 2e 13t

As t , the exponential terms in the above functions go to zero, and

x1(t) 20, x2(t) 15, x3(t) 4.

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