Differential equation for the amount of chemical in the pond at any time > A pond intitally contains $1\,000\,000\,\rm gal$ of water and an unknown amount of an undesirable chemical. Water containing $0.01\,\rm g$ of this chemical per gallon flows into the pond at a rate of $300\,\rm gal/h$. The mixture flows out at the same rate, so the amount of water in the pond remains constant. Assume that the chemical is uniformly distributed throughout the pond. Write a differential equation for the amount of chemical in the pond at any time. The answer key says the answer is $$\frac{\mathrm{d}q}{\mathrm{d}t} = 300(10^{-2}-10^{-6}q)$$ where $q$ is the amount of chemical in gallons. I don't get how they got this, and also how, if there is an unknown amount of chemical in the pond given the rate at which the amount of chemical changes per hour, can we tell the amount of chemical?

Some basics about differential equations: when we solve a linear differential equation with a first derivative, we always get one arbitrary constant to be determined. For example: $y' = 1$ means that $y = x + C$. That constant will be determined by some conditions that are given to you, usually initial conditions (in this case, like the amount of chemical present before the mixing). So you can set up the differential equation and solve it, but you won't get the full answer. (You still get a lot of information, though.)

The problem should be tackled like this:

$$\frac{\mathrm{d}q}{\mathrm{d}t} = \text{rate of inflow} - \text{rate of outflow}$$

Calculate flows by multiplying the rate of flow of water with how much chemical there is per gallon. Inflow is easy. Outflow is too, if you realize that the amount of chemical, q, is spread out over the full volume of water.