For the interconnected tanks problem of Section5.1, page241, suppose that instead of pure water being fed into the tankA, 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 $\mathit{t}$if the initial masses are and ${\mathbf{y}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{kg}$.

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" $\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{kg}\mathbf{/}\mathbf{L}\mathbf{\times }\mathbf{6}\mathbf{}\mathbf{L}\mathbf{/}\mathbf{min}$, and the right lower pipe brings salt at the rate $\mathbf{2}\mathbf{}\mathbf{L}\mathbf{/}\mathbf{min}\mathbf{\times }\mathbf{y}\mathbf{/}\left(\mathbf{24}\mathbf{L}\right)$. The upper pipe carries salt out of the tank A at the rate of $2\mathbf{L}\mathbf{/}\mathbf{min}\mathbf{\times }\mathbf{x}\mathbf{/}\mathbf{(}\mathbf{24}\mathbf{}\mathbf{L}\mathbf{)}$. So, one has that the change of the concentration of salt in the tank A is $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{\times }\mathbf{6}\mathbf{+}\mathbf{2}\frac{\mathbf{y}}{\mathbf{24}}\mathbf{-}\mathbf{8}\frac{\mathbf{x}}{\mathbf{24}}$.

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

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

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{\mathbf{x}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{12}}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{2} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{\mathbf{y}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{x}}{\mathbf{3}} \end{gathered}$$

The second equation gives us that

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

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

$$\begin{gathered} \mathbf{3}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{-}\frac{\mathbf{y}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{12}}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{2} \\ \Rightarrow \mathbf{3}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\mathbf{2}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2} \\ \Rightarrow \mathbf{12}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{+}\mathbf{8}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{8} \end{gathered}$$