Q:

Linear Systems of Differential Equations - Container Mixing Problem

So the questions goes something like this:
I have two containers, 1 and 2. Pure water enters container 1 at a rate of 3 liters per minute, while the well-stirred mixture is leaving container 2 at the same rate. There are 100 liters in each container, and the well-stirred mixture flows from container 1 to container 2 at a rate of 4 liters per minute and flows back from container 2 to container 1 at a rate of 1 liter per minute. Initially there are 16 kilograms of salt in container 1 and 4 kilograms in container 2. Let $x(t)$ be the amount of salt in container 1 after $t$ minutes and $y(t)$ be the amount of salt in kilograms in container 2 after $t$ minutes.
Now there were several questions about this problem which included setting up the system of linear differential equations governing the process, indicating the IC, and solving the IVP, which I was able to solve without a hitch. But I am stuck on this particular problem regarding this system. It asks:
1.) How long will it take the containers to hold an equal amount of salt and what will that amount of salt be?
I found the unique solution to the problem with the given IC's which is:
$x_1(t) = 9e^{-0.02t}+7e^{-0.06t}\\$
and
$x_2(t) = 18e^{-0.02t}-14e^{-0.06t}$
I do not want the answer I am only looking for some guidance as to which direction I should take this. Would I set these equation equal to each other?? Something else???
Your help is greatly appreciated!!
THANK YOU!

A:

Let's see where your mistake was, because I think there was one. First note that the volumes are held constant, so that simplifies matters.
Input into container 1: $1 \frac{L}{\text{min}} \cdot (y \text{ kg }) \cdot \frac{1}{100 L} = \frac{y}{100} \frac{\text{kg}}{\text{min}}$.
Output from container 1: $4 \frac{L}{\text{min}} \cdot (x \text{ kg }) \cdot \frac{1}{100 L} = \frac{4x}{100} \frac{\text{kg}}{\text{min}}$.
Input into container 2 = output from container 1.
Output from container 2: the flow out of the system is $3 \frac{L}{\text{min}} \cdot \frac{y \text{ kg }}{100 L} = \frac{3y}{100} \frac{\text{kg}}{\text{min}}$. There is additionally the input into container 1, which is $\frac{y}{100}$. So overall you have
$$\frac{dx}{dt} = -\frac{4x}{100} + \frac{y}{100} \\
\frac{dy}{dt} = \frac{4x}{100} - \frac{4y}{100}$$
The usual eigenvalue-eigenvector method then gives a solution of the form
$$\begin{bmatrix} x \\ y \end{bmatrix} = c_1 e^{-0.02t} \begin{bmatrix} 1 \\ 2 \end{bmatrix} + c_2 e^{-0.06t} \begin{bmatrix} -1 \\ 2 \end{bmatrix}$$
As a reality check, this still conserves volume, because as written container 1 is losing 3 liters each minute, but this is precisely what is being replaced by the input water.
Solving for $c_1,c_2$ amounts to solving 
$$\begin{bmatrix} 1 & -1 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 16 \\ 4 \end{bmatrix}$$
which gives $c_1=9,c_2=-7$. So you have
$$\begin{bmatrix} x \\ y \end{bmatrix} = 9 e^{-0.02t} \begin{bmatrix} 1 \\ 2 \end{bmatrix} - 7 e^{-0.06t} \begin{bmatrix} -1 \\ 2 \end{bmatrix} \\
= \begin{bmatrix} 9 e^{-0.02t} + 7 e^{-0.06t} \\ 18 e^{-0.02t} - 14 e^{-0.06t} \end{bmatrix}$$
Checking we have $x(0)=9-(-7)=9+7=16$ and $y(0)=18-14=4$ as desired. Now you can set these two equal to each other and you should actually get a solution.
Ultimately you lost track of a minus sign, which got you $9e^{-0.02t} - 7 e^{-0.06t}$ instead of $+$. As for solving your problem, by putting the $e^{-0.02t}$ terms on one side and the $e^{-0.06t}$ terms on the other you get
$$9e^{-0.02t}=21e^{-0.06t}.$$
Take logarithms:
$$\ln(9)-0.02t=\ln(21)-0.06t.$$
Now it is just a problem of finding the intersection of two lines. You should get $25 \ln \left ( \frac{7}{3} \right )$.

