How to find concentration of a solution, differential equation problem A tank initially contains $100 \text{gal}$ of brine, a solution of salt and water, whose salt concentration is $0.5 \frac{\text{lb}}{\text{gal}}$. Brine whose salt concentration is $2 \frac{\text{lb}}{\text{gal}}$ flows into the tank at the rate of $3\frac{ \text{gal}}{ \text{min}}$. The mixture flows out at the rate of $2\frac{ \text{gal}}{ \text{min}}$. Assume the salt is uniformly distributed throughout the mixture. Find the salt concentration of the brine at the end of $30$ minutes. 
Letting $x $ be the amount of salt at any time $t$
I got
$$
\frac{\text{dx}}{\text{dt}}=6-\frac{2x}{100+t}
$$ 
$$
x = \frac{6[10,000t+100t^2+\frac{t^3}{3}]}{(100+t)^2}
$$
What is the meaning of concentration of the brine? Is it $\frac{\text{dx}}{\text{dt}}$?
 A: (NOTE: My first answer was wrong. This is a corrected answer.)
Concentration is the amount of solute (salt in this case) divided by the volume of the solution (brine in this case). So assuming that the unit of $x$, the amount of salt, is pounds, the concentration of the brine in the tank at the start is
$$\frac x{100}\frac{\text{lb}}{\text{gal}}$$
However, the volume in the tank does not remain at $100$ gallons. The inflow of liquid has the rate $3\frac{\text{lb}}{\text{gal}}$ while the outflow has the rate $2 \frac{\text{lb}}{\text{gal}}$, so the net inflow is $1 \frac{\text{lb}}{\text{gal}}$. So the volume of the tank at time $t$ is not $100$ but rather $100+t$.
Therefore the concentration of salt in the tank at time $t$ is
$$\frac x{100+t}\frac{\text{lb}}{\text{gal}}$$
Because of that, your differential equation is correct. The rate of flow of salt into the tank is $2 \frac{\text{lb}}{\text{gal}}\cdot 3\frac{ \text{gal}}{ \text{min}}=6\frac{\text{lb}}{\text{min}}$. The rate of flow of salt out of the tank is the concentration of the brine times the rate of flow out, namely $\frac{x}{100+t}\cdot 2\frac{ \text{lb}}{ \text{min}}$. Therefore, your equation is
$$\frac{dx}{dt}=6-\frac{2x}{100+t}$$ 
with the initial value
$$x(0)=100\cdot 0.5$$
How did you get the correct equation without understanding the concentration? Anyway, the meaning of $\frac{dx}{dt}$ is the rate of change of the mass of salt in the tank, not the concentration.
A: Oh, yumyumyum, a brine tank problem! The first question was about concentration: this is the amount of solute per volume of solution, so here it is that $\frac x{100+t}$ that you already knew to put in the outflow term. @Rory Daulton is making the mistake of neglecting the changing volume of brine in the tank, @Jimmy Kudo is forgetting about the outflow from the tank, and @shoestringfries has the right differential equation, but is not integrating it properly. Let's copy that equation:
$$\frac{dx}{dt}=6-\frac{2x}{100+t}$$
This is first-order linear, usually written as
$$\frac{dx}{dt}+\frac2{100+t}\cdot x=6$$
The integrating factor is
$$\mu=e^{\int\frac2{100+t}dt}=e^{2\ln(100+t)}=(100+t)^2$$
So
$$\frac d{dt}\left((100+t)^2x\right)=(100+t)^2\frac{dx}{dt}+2(100+t)x=6(100+t)^2$$
Integrating,
$$(100+t)^2x=2(100+t)^3+C$$
At $t=0$, $100^2(100)(0.5)=2(100)^3+C$, so $C=-\frac32(100)^3$.
Then at $t=30$,
$$\frac x{t+100}=2-\frac{\frac32(100)^3}{(130)^3}=\frac{2894}{2197}\,\,[lb/gal]$$
A: Let $t$ be time in minutes
$$We\ know\ that,\qquad Concentration= \frac{amount\ of \ solute}{volume\ of\ solution}$$

$At \ t=0\min\\
Amount\ of\ brine=(0.5\frac{lb}{gal})(100 gal) =50lb
\\Amount\ of\ solution\ in\ tank=100gal$

$At\ time=t\ min
\\Amount\ of\ solute\ incoming=(2\frac{lb}{gal})(3\frac{gal}{min})t=6t\ lb
\\Volume\ of\ solution\ incoming\ in\ tank=3t\ gal
\\Amount\ of\ solute\ outgoing=\frac{(50+6t)(2t)}{100+3t}\ lb
\\Volume\ of\ solution\ outgoing=2t\ gal
\\Amount\ of\ solute\ remaining=\frac{(50+6t)(100+t)}{100+3t}\ lb
\\Volume\ of\ solution\ remaining=(100+t)\ gal$

$$Concentration\ of\ brine\ solution\ at\ time\ t\ minutes =\frac{50+6t}{100+3t}\frac{lb}{gal}
\\ At\ t=30\ minutes
\\Concentration\ of\ brine\ solution =230/190\frac{lb}{gal}$$
I hope that I didn't make any mistake, if you find any mistake, please comment or edit.Question doesn't belong to mathematics, its from Mole Concept of Physical Chemistry
EDIT:- Explanation of amount of outgoing solute considering Concentration to be constant in whole process of out-coming of the solution $$Concentration\ of\ solution = \frac{50+6t}{100+3t} 
\\ Amount\ of\ solute\ outgoing=\frac{(50+6t)(2t)}{100+3t}
\\ Amount\ of\ solute\ remaining=1-\frac{(50+6t)(2t)}{100+3t}$$   
