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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 306

Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

For the interconnected tanks problem of Section 5.1, page 241 , suppose that instead of pure water being fed into the tank A, a brine solution with concentration $0.2 kg / L$ is used; all other data remain the same. Determine the mass of salt in each tank at time $t$ if the initial masses are and $y_{0} = 0.3 kg$ .

The change of mass of salt n tanks A and B are:

$x (t) = - 1 . 225 e^{- t / 2} - 3.475 e^{- t / 6} + 4.8 y (t) = 2 . 45 e^{- t / 2} - 6.95 e^{- t / 6} + 4.8$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Finding the change of concentration of the tank A,B

Let’s first derive a system of differential equations that describes the change of salt in each tank at a time t. One knows that $dx / dt = inputrate - outputrate$ and in the tank t, two pipes bring salt in it, the left one at the rate role="math" localid="1664171528030" $0.2 kg / L \times 6 L / \min$ , and the right lower pipe brings salt at the rate $2 L / \min \times y / (24 L)$ . The upper pipe carries salt out of the tank A at the rate of $2 L / \min \times x / (24 L)$ . So, one has that the change of the concentration of salt in the tank A is $\frac{dx}{dt} = 0.2 \times 6 + 2 \frac{y}{24} - 8 \frac{x}{24}$ .

One has that the upper pipe brings salt into the tank B by the rate of $8 L / \min \times x / (24 L)$ , the lower pipe carries salt out of tank B by the rate $2 L / \min \times y / (24 L)$ and the right pipe carries salt out by the rate of $6 L / \min \times y / (24 L)$ so the change of concentration of salt in the tank B is $\frac{dy}{dt} = 8 \frac{x}{24} - 2 \frac{y}{24} - 6 \frac{y}{24}$

Step 2: Substituting the values

So, to determine the mass of salt in each rank one has to solve the following system:

$\frac{dx}{dt} = - \frac{x}{3} + \frac{y}{12} + 1.2 \frac{dy}{dt} = - \frac{y}{3} + \frac{x}{3}$

The second equation gives us that

$\frac{x}{3} = \frac{dy}{dt} + \frac{y}{3} \frac{dx}{dt} = 3 \frac{d^{2} y}{{dt}^{2}} + \frac{dy}{dt}$

Substituting this into the first equation of the system one will get

$3 \frac{d^{2} y}{{dt}^{2}} + \frac{dy}{dt} = - \frac{dy}{dt} - \frac{y}{3} + \frac{y}{12} + 1.2 \Rightarrow 3 \frac{d^{2} y}{t^{2}} + 2 \frac{dy}{dt} + \frac{y}{4} = 1.2 \Rightarrow 12 \frac{d^{2} y}{{dt}^{2}} + 8 \frac{dy}{dt} + y = 4.8$

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