Topic 7.6 key terms
- separation of variables
- Solving dy/dx = g(x)h(y) by putting every y (with dy) on one side and every x (with dx) on the other, then integrating both sides. It works because of the chain rule.
- separable differential equation
- A differential equation whose right side factors into a function of x times a function of y, such as dy/dx = xy + x = x(y + 1).
- general solution
- The family of all solutions of a differential equation, written with a constant of integration C. For dy/dx = -x/y it is x^2 + y^2 = 2C with 2C > 0: each circle gives two solutions, its top half and its bottom half.
- particular solution
- The one solution that passes through a given point (the initial condition), found by using that point to solve for C.
- initial condition
- A known point on the solution, such as y(0) = 3 or H(0) = 196. Use it right after integrating to find the constant.
- constant of integration
- The + C that appears when you integrate. One is enough. Put it in before using the initial condition; leaving it out on the AP exam costs the last points of a separation part.
- domain restriction
- A particular solution lives on one open interval containing the initial input, where the solution and the differential equation are defined (y = sqrt(9 - x^2) on -3 < x < 3).
- accumulation form of a particular solution
- For dy/dx = f(x) with y(a) = y0, the particular solution is y = y0 + the integral from a to x of f(t) dt.
- exponential growth and decay model
- If the rate of change of a quantity is proportional to its size, dy/dt = ky and y = y0 e^(kt): growth when k > 0, decay when k < 0.
- Newton's law of cooling
- The temperature T of an object in surroundings at temperature A changes at a rate proportional to the gap: dT/dt = -k(T - A) with k > 0, so T = A + (T0 - A)e^(-kt).
- half-life
- The time for a decaying quantity to fall to half its amount. From e^(-kT) = 1/2, k = ln 2 / T (plutonium-238: T = 87.7 years).
- equilibrium solution
- A constant solution, such as y = 0 for dy/dt = ky or H = 4 (the air temperature) for Newton's law. Check for it whenever you divide by an expression that could be zero.
Transcript
763 words · 3 parts
He measured fire with a clock
In 1701, Isaac Newton wanted to know how hot a fire really gets, and he had a problem, because no thermometer he owned could survive that kind of heat. His answer was to pull a glowing block of iron out of the fire and set it down outside in a steady wind.
He couldn’t measure it yet, so he timed little beads of metal on top, each freezing at its own temperature, as the glow faded, and only once the iron was cool enough could his thermometer read it. Those cool readings gave Newton somewhere to start, and his rule about how fast heat escapes let him run the clock backward, working out how hot the iron had been at every moment, all the way back to the fire.
By the end of this lesson, you’ll be able to do what Newton did. I’ll hand you one reading from a cooling iron, and you’ll turn one rule about how it cools into a formula. Then you’ll run time backward to the exact minute it came out of the fire. To get there, let’s start with Newton’s rule itself.
Newton’s rule is about the rate
Here’s the iron right out of the fire. In our version of his experiment, on Newton’s heat scale, where your body is about twelve, it starts at a hundred ninety-six, and the night air is at four. What matters is the gap between them, how much hotter the iron is than the air, and right now that gap is a hundred ninety-two.
Newton’s rule says the iron loses heat at a rate proportional to that gap, so with a huge gap it’s cooling fast, about twenty-seven of his degrees every minute. The arrow is always the same fraction of the gap, so they shrink together, and five minutes later the gap is half as big, ninety-six, and the iron cools half as fast.
Let’s keep going until the gap is down to forty-eight, a quarter of where it started. How fast do you think it’s cooling now? Drag the arrow to the length you predict.
In symbols, Newton’s rule says dH dt, the rate of change of the heat, equals negative k times H minus four, the gap above the air. Here k is a positive constant that depends on the iron and the wind, and the minus sign says the heat is going down whenever the iron is hotter than the air.
An equation about a rate of change like this one is a differential equation. We’ve used slope fields to sketch solutions like these, but now let’s find an exact formula, and the trick is called separation of variables.
Separate the variables
Newton’s equation has a few moving parts, so let’s learn the trick on a friendlier one first, dy dx equals negative x over y. The trouble is that y shows up on both sides, so we can’t just integrate the right side, because it still has the unknown y in it. I want the y’s beside dy and the x’s beside dx, so let’s multiply both sides by y and by dx, and we get y dy equals negative x dx.
Why can we move dy and dx around? The chain rule says the derivative of y squared over two is y times dy dx, and multiplying our equation by y says that’s negative x, so integrating both sides is exactly what the separated line says. Now each side has only one variable in it, which means we can integrate both sides.
On the left that’s y squared over two, and on the right negative x squared over two, plus C. One constant is enough, since a constant on each side would just combine into one. Multiply everything by two and move the x squared across, and we get x squared plus y squared equals two C. Look at that, it’s a circle centered at the origin! Every positive value of C gives a different circle, and each one fits the slopes in this field.
That whole family is the general solution, because a differential equation like this one doesn’t have just one solution. It has infinitely many, two for every positive value of C, each circle’s top half and its bottom half.
The method has four steps, and these are the moves we’ll keep coming back to. Separate the variables, integrate both sides with a single plus C, use the starting point to find the constant, and then solve for y. Before we use it, let’s make sure you can spot when it works. Which of these three differential equations can you separate?