A tank contains 500L of salty water. Pure water is pumped in at a rate of 10 L/sec, and the the mixture is pumped out at a rate of 15L/ sec. If the concentration of salt is 5g/L initially, form an equation of amount of salt, s, at t seconds.

This is a word problem with the task of forming a first order differential equation (or DE), and involves two major aspects that need to be worked out: the volume of water in the tank at any given point, and the amount of salt in the mixture at that same point. The water volume in the tank is constantly changing, as water is both being pumped in and out of the tank. Hence the expression for the volume of water can be written as 500 (the initial) + 10t(as 10 L are pumped in every second) - 15t (as 15L are pumped out), = 500-5t. The second aspect, the amount of salt at any given time, can be considered in terms of the rate of change of salt amount, or change in salt concentration. The salt concentration is very simply amount of salt/volume. As no salt is being added, only removed when being pumped out, the change in salt concentration is -15s/V, or -15s/500-5t, which can be simplified.
Hence, ds/dt = -3s/100-t, and hence ds/dt + 3s/100-t = 0. This is a first order homogeneous DE, and, using an integrating factor, e^{integral of (3/100-t) dt} , the equation becomes (100-t)^-3 ds/dt + 3y(100-t)^-4 (as the integral of 3/100-t is -3ln|100-t|). Applying the product rule and implicit differentiation, the differential of y(100-t)^-3 is the expression on the LHS, and hence integrating both sides of the DE gives this expression, with the final answer being y(100-t)^-3= C (as the integral of 0 is 0, and the constant of integration can hence be placed on the RHS for simplicity). Using the initial boundary condition (as an initial concentration of 5g/L of salt means there is 5X500 grams of salt initially), the value of the constant of integration can then be found as 1/400. The final equation formed is hence y(100-t)^-3= 1/400, or y = 1/400(100-t)³.