If \(x(t)\) and \(y(t)\) are the amounts of salt in Tank \(A\) and Tank \(B\text{,}\) respectively, then our problem can be modeled with a linear system of two equations,
\begin{align*}
\frac{dx}{dt} \amp = \text{rate in} - \text{rate out} = - r_A \frac{x}{V} + r_B \frac{y}{V}\\
\frac{dy}{dt} \amp = \text{rate in} - \text{rate out} = r_A \frac{x}{V} - r_B \frac{y}{V} - r_{\text{out}} \frac{y}{V}.
\end{align*}
Furthermore, \(r_A = r_B + r_{\text{out}}\text{,}\) since the volume in Tank \(B\) is constant. Consequently, our system now becomes
\begin{align*}
\frac{dx}{dt} \amp = - r_A \frac{x}{V} + r_B \frac{y}{V}\\
\frac{dy}{dt} \amp = r_A \frac{x}{V} - r_A \frac{y}{V}.
\end{align*}
If we have initial conditions \(x(0) = x_0\) and \(y(0) = y_0\text{,}\) it is not too difficult to deduce that the amount of salt in each tank will approach zero as \(t \to \infty\text{,}\) and we will have a stable equilibrium solution at \((0, 0)\text{.}\) Determining the nature of the equilibrium solution is a more difficult question. For example, is it ever possible that the equilibrium solution is a spiral sink? One solution is provided by studying the trace-determinant plane.