 So let's take a look at an important application of differential equations, known as Newton's Law of Cooling. So a classic example of a differential equation involves something known as Newton's Law of Cooling. And we make the following assumptions. If an object's temperature, capital T, is a function of time t, and the rate of change of temperature is proportional to the difference in temperature between the object and its environment, then we can, at least hope to, find the object's temperature, t, at any time, t. So let's consider our setup. First, the rate of change of temperature is the derivative dT over dt. If we assume it's proportional to the difference in temperature between the object and its environment, we have dt over dt is k times t minus t sub a, where k is our constant of proportionality, and t sub a is the ambient temperature. In other words, the temperature of its environment. So let's consider this our first real problem in differential equations. Solve dt over dt is k times t minus t sub a. So notice that the lower case variable, t, doesn't actually appear here, so this is a separable differential equation, almost by default. So let's separate our variables. We want to get all the capital T's over onto the left and all the lower case T's over onto the right. Now let's anti-differentiate both sides. Over on the right-hand side, remember, k is just a constant. So the anti-derivative of k dt is just going to be kt plus whatever our constant is going to be. On the left, we'll use a u substitution with u equals t minus t sub a. And remember, t sub a is a constant, so du is going to be dt, and we can rewrite our left-hand side as then find the anti-derivative. And since we've already included the constant of anti-differentiation over on the right, we don't need to introduce a new one over on the left. And very important, put things back where you found them. u is t minus t sub a, so we'll replace and get our anti-derivatives. Now we do actually want to solve for t, so when we solve for t, we get. And again, the important thing to remember is that any algebra you do to a constant reduces a constant, which means that this e to power c is going to be another constant, which we can just call c. And so we get our final equation. And so now all we need to know how to use this is to find these constants c, k, and t sub a. For example, Quig-Quig's coffee brews its coffee at 100 degrees Celsius and allows it to cool to a drinkable 54 degrees Celsius. When the ambient temperature is 20 degrees Celsius, it takes five minutes for the temperature to drop to 80 degrees Celsius. How long will it take before the coffee is drinkable? And the correct answer is never, because at 100 degrees Celsius, you're boiling the water and you shouldn't do that to poor defenseless coffee grounds. But we're probably interested in a mathematical answer to this question, so Newton's law of cooling applies. Let t be the temperature of the coffee in degrees Celsius and little t be the time in minutes. The Newton's law of cooling says the rate of change of the temperature is proportional to the difference in temperature between the coffee and the ambient environment, which in this case is 20 degrees Celsius. So that's our differential equation. We'll solve this differential equation, producing our function of temperature where there are two unknown constants, c and k. So how can we find those unknown constants? Well, let's start off with some of the things we know. Since the initial temperature of the coffee is 100 degrees Celsius, then we have our temperature equals 100 when t is equal to 0. So we can substitute those values into our function, then solve, and we find that c is equal to 80. Since we also know that after five minutes, the coffee is at 80 degrees Celsius, then t equals 80 when little t is equal to 5. So we can substitute those values into our formula, then solve for k. And if we reduce this to a decimal, we find that k is about negative 0.05754. And so that gives us our function. And at this point, we essentially have an algebra problem. We want to know little t when temperature, big t, is 54. And so we solve our equation, equals means replaceable. So capital T equals 54 will replace and solve for little t. And after all the dust settles, we find it's approximately 14.87. And since we're letting t be the time in minutes, we know that the temperature will be 54 degrees Celsius after about 14.87 minutes, and it still won't be drinkable.