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Found in: Page 306

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# 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 ${\mathbf{0}}{\mathbf{.}}{\mathbf{2}}{\mathbf{}}{\mathbf{kg}}{\mathbf{/}}{\mathbf{L}}$ is used; all other data remain the same. Determine the mass of salt in each tank at time ${\mathbit{t}}$ if the initial masses are and ${{\mathbf{y}}}_{{\mathbf{0}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{3}}{\mathbf{}}{\mathbf{kg}}$.

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

$\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{225}{\mathbf{e}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{475}{\mathbf{e}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{6}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{8}\phantom{\rule{0ex}{0ex}}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{45}{\mathbf{e}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{2}}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{95}{\mathbf{e}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{6}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{8}$

See the step by step solution

## 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 $\mathbf{dx}\mathbf{/}\mathbf{dt}\mathbf{=}\mathbf{inputrate}\mathbf{}\mathbf{-}\mathbf{outputrate}$ and in the tank t, two pipes bring salt in it, the left one at the rate role="math" localid="1664171528030" $\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{kg}\mathbf{/}\mathbf{L}\mathbf{×}\mathbf{6}\mathbf{}\mathbf{L}\mathbf{/}\mathbf{min}$ , and the right lower pipe brings salt at the rate $\mathbf{2}\mathbf{}\mathbf{L}\mathbf{/}\mathbf{min}\mathbf{×}\mathbf{y}\mathbf{/}\left(\mathbf{24}\mathbf{L}\right)$. The upper pipe carries salt out of the tank A at the rate of $2\mathbf{L}\mathbf{/}\mathbf{min}\mathbf{×}\mathbf{x}\mathbf{/}\mathbf{\left(}\mathbf{24}\mathbf{}\mathbf{L}\mathbf{\right)}$. So, one has that the change of the concentration of salt in the tank A is $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{×}\mathbf{6}\mathbf{+}\mathbf{2}\frac{\mathbf{y}}{\mathbf{24}}\mathbf{-}\mathbf{8}\frac{\mathbf{x}}{\mathbf{24}}$.

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

## Step 2: Substituting the values

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

$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{\mathbf{x}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{12}}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{2}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{\mathbf{y}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{x}}{\mathbf{3}}$

The second equation gives us that

$\frac{\mathbf{x}}{\mathbf{3}}\mathbf{=}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{3}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{3}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dt}}$

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

$\mathbf{3}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{-}\frac{\mathbf{y}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{12}}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{2}\phantom{\rule{0ex}{0ex}}⇒\mathbf{3}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\mathbf{2}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\phantom{\rule{0ex}{0ex}}⇒\mathbf{12}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{+}\mathbf{8}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{8}$

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