Topic 7.7 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
473 words · 1 parts
One point picks one curve
Now, which circle is ours? Let’s find the solution that passes through the point zero, three. Plug the point straight into the equation we just got, before solving for anything. Zero squared plus three squared is nine, so two C is nine.
Out of that whole family, exactly one circle passes through zero, three, the circle with radius three. We call the solution through our starting point a particular solution. Solving for y gives plus or minus the square root of nine minus x squared, the top and bottom halves of the circle, and each half is its own solution, because a solution has to be a function.
Our point has y equal to positive three, so we keep the top half. Now try following our curve all the way out to x equals three. There y reaches zero, and the differential equation would divide by zero, and the same thing happens at negative three, so the solution only lives between negative three and three.
That domain restriction is part of the answer, so the particular solution is y equals the square root of nine minus x squared, on the open interval from negative three to three. Let’s check it, the way you should every time. Its derivative is negative x over the square root, which is negative x over y, and at zero it gives three.
Now you pick the start. Drag a starting point on the y-axis whose solution is the bottom half of the circle with radius two.
Let’s see whether you can use all four steps on a new equation. Solve dy dx equals three x squared over two y, with y of zero equal to three, and then tell me y of three.
There’s one kind of differential equation where you don’t need to separate anything at all, when dy dx depends only on x and you know y at one starting x, which we’ll call a. Then the particular solution is the starting value plus everything it accumulates from there, so y equals y nought plus the integral from a to x of f of t, dt.
At x equals a the integral is zero, so you get y nought, and by the Fundamental Theorem of Calculus its derivative is exactly f of x. This form is especially useful when none of our usual functions can write the antiderivative down. If dy dx is e to the negative x squared and y of zero is three, the answer is three plus the integral from zero to x of e to the negative t squared, d t. That’s an exact answer, though no simpler formula exists.
Let’s see if you can build one. Which of these is the solution of dy dx equals the square root of one plus x cubed, with y of two equal to five?