Mixtures ( mixture in volume A is circulated with water in volume B at what point of time both the tanks contain equal percentage of salt) Two tanks are connected to each other through two pipes. Tank A contains 200 liters of water in which 60 grams of salt is dissolved (mixture) and Tank B contains 200 liters of pure water. The mixture and the pure water is circulated between the two tank A and tank B such that outflow of mixture from tank A to tank B is 20 Liters/min and inflow of mixture from tank B to tank A is 5Liters/ min. pure water is added to tank A at 15 Liters/min through inlet pipe and mixture from tank B is removed through outlet pipe at 15 liters/min. How much time does it take to reach such that both the tanks contain equal percentage of salt dissolved in both the tanks?
 A: We model the situation by a system of two homogeneous linear differential equations with constant coefficients.   
Let $x=x(t)$ be the quantity, in grams, of salt in tank A at time $t$, and $y=y(t)$ the quantity of salt in tank B at time $t$.
We are given the initial conditions $x(0)=60$ and $y(0)=0$. 
We now examine the flow of salt in and out of tank A.  Brine is flowing out to B at the rate of $20$ liters per minute.  At any time $t$, the amount of salt in a liter of water is $x/200$, so salt is flowing out at the rate $(20/200)x$.  Also, salty (after a while) water is flowing in from B at rate $5$ liters per minute. Thus salt is flowing into A at the rate $(5/200)y$.  The above out-in analysis for the salt can be written as the differential equation
$$\frac{dx}{dt}=-\frac{20}{200}x +\frac{5}{200}y \qquad\qquad\text{(Equation $1$)}$$
Now we do a similar analysis for tank B.  Salt is flowing in from A at the rate of $(20/100)x$.  Salt is flowing out to A at the rate $(5/200)y$, and to the outlet pipe at the rate $(15/200)y$, for a combined out rate of $(20/200)y$. This yields the differential equation
$$\frac{dy}{dt}=\frac{20}{200}x -\frac{20}{200}y \qquad\qquad\text{(Equation $2$)}$$
We run through the solution process. Let $M$ be the matrix 
$$\begin{pmatrix} -\frac{20}{200}& \frac{5}{200}\\ \frac{20}{200}& -\frac{20}{200}
\end{pmatrix}$$
We first find the eigenvalues and associated eigenvectors of the matrix $M$.  Standard calculation shows that the eigenvalues are $-1/20$ and $-3/20$.  The vector $(1,2)$ (written as a column vector) is an eigenvector for eigenvalue $-1/20$; the vector $(1,-2)$, again written as a column vector, is an eigenvector for eigenvalue $-3/20$. 
Theory tells us that the general solution $(x,y)$ is given by
$$(x,y)=Ce^{-t/20}(1,2)+De^{-3t/20}(1,-2)$$
where $C$ and $D$ are constants.
Less compactly, 
$$x=Ce^{-t/20}+De^{-3t/20} \qquad \text{and}\qquad y=2Ce^{-t/20}-2De^{-3t/20}.$$
From the fact that $x(0)=60$ we get $C+D=60$. From $y(0)=0$ we get $2C-2D=0$.  So $C=D=30$, and therefore
$$x=30e^{-t/20}+30e^{-3t/20} \qquad \text{and}\qquad y=60e^{-t/20}-60e^{-3t/20}.$$  
Now we know everything, so we can answer any question.  We were asked when there are equal percentages of salt in both tanks.  The tanks are of equal size, so equal percentage happens when the amounts of salt are equal, that is, at the time $t$ when $x=y$. Thus
$$30e^{-t/20}+30e^{-3t/20}=60e^{-t/20}-60e^{-3t/20}.$$
This simplifies to $3e^{-3t/20}=e^{-t/20}$, then to $e^{t/10}=3$. Take the natural logarithm of both sides: $t=10\ln 3\approx 10.98$.
