Differential Equation In Context of A Concentration Puzzle I've got a word problem that involves solving a differential equation. I'll organize this in 3 parts: The problem itself, what I've done so far, and then where I'm stuck. I'm writing the entire thing for the sake of completeness, but I believe you can skip to part 3 pretty safely; I'm essentially just having an issue with a differential equation, not the setup of the word problem.
1. The Problem
A tank contains 150 000 L of water with 20kg of acid dissolved in it. Water containing 3kg/1000L enters the tank at a rate of 2000L/hour. The solution is kept thoroughly mixed and there is also an exit drain, where 2000L/hour of the solution is drained from (leaving the overall volume constant). Find a differential equation for the amount of acid at time $t, t>0$, with the initial condition $x(0) = 20$.
2. What I've Done So Far
$s(t) = $ Acid in tank (kg) at time $t$ (hours).
$s'(t)$ = Rate of acid in - Rate of acid out
Rate of acid in = $\frac{3kg}{1000L} * \frac{2000L}{hour} = 6\frac{kg}{hour}$
Rate of acid out = $\frac{s(t)kg}{150 000L} * \frac{2000L}{hour} = \frac{1}{75}\frac{kg}{hour}$
∴ $s'(t) = 6 - \frac{s(t)}{75}$ 
Rewrite as $\frac{ds}{dt} = 6 - \frac{1}{75}s$
Now solve differential equation.
3. Where I'm Stuck
I'm a little unclear on how to solve $\frac{ds}{dt} = 6 - \frac{1}{75}s$. I can't find a way to write it in standard form, and it's unclear how I should be trying to separate the variables. I'm also unsure if this is a format where I can use the integrating factor $e^{\int P(x)dx},$ because that would involve being able to convert to standard form. Would anyone be able to point out where I should be going from here in order to arrive at a finished differential equation? Any help appreciated.
 A: If you're solving a problem like this, you are not just doing an exercise in deriving a solution to a differential equation. (If you are, then you can find the method here, for example.) 
Seeing that this is a simple linear differential equation, rewrite it as 
$$\frac{ds}{dt} + \frac{1}{75}s = 6$$ 
If $s'(t)=\frac{ds}{dt}$, then $s(t)=\int s'(t)dt + k$, where k is a constant of integration.
From that, substitute into the first equation, getting $$s'(t)+\frac{1}{75}(\int s'(t)dt + k)=6$$
You can equate like terms and solve $$s'(t)+\frac{1}{75}\int s'(t)dt=0 \text{ and } \frac{k}{75}=6$$ 
Now you should recognize that $s(t)$ is of the form $s(t)=ae^{bx}+k$. (If you don't, see the referenced link for an explanation.) So $s'(t) = abe^{bx}$. Plug this into the formula above:
$$s'(t)+\frac{1}{75}\int s'(t)dt=0 \implies abe^{bx} + \frac{1}{75}ae^{bx}=0$$
You can solve for the constants $b$ and $k$. There is a solution for any value of $a$.
You didn't say what level of understanding you are at, but at some point, you will know how to solve this using Laplace transforms, as suggested by JeanMarie in the comments. 
