An equation is autonomous when its right-hand side mentions only the unknown, y′ = f(y), and never the time. That one restriction makes the slope field identical on every vertical line, lets a solution be slid sideways into another solution, and turns the question “where does the equation have a constant solution?” into “where is f(y) zero?”. Those roots are the equilibria, and the lab finds them exactly.
Core ideas
Three ideas, in the order they depend on each other. The first is a definition and the other two follow from it by arithmetic.
y′ = y² − 1 and y′ = 3 − y are autonomous: how fast y changes depends on where y is and on nothing else. y′ = t·y and y′ = y + t are not, because the clock appears on the right. The word is not decoration. Everything this course does — the phase line, the classification of equilibria, the sketches — works because the clock is absent.
If f(c) = 0, the constant function y = c has y′ = 0 and the right side f(c) = 0, so substituting it leaves a residual of exactly zero at every t. That is the same check as in “Checking a Proposed Solution”, and it needs no solving. The equilibria of y′ = f(y) are therefore the roots of f, found by algebra alone.
The slope at (t, y) is f(y), whatever t is. Look up one vertical line of the field and you have seen them all, and every horizontal line carries one slope from end to end. A solution slid to the right or the left by any amount therefore meets the same slopes and is again a solution.
The material
The definition, the two things it buys, and how the lab finds the equilibria exactly.
A first-order equation is autonomous when it can be written y′ = f(y): the right side is a function of y alone.
An equilibrium is a number y* with f(y*) = 0. The constant function y = y* is then a solution on every t.
Whether an equation is autonomous is read off its right-hand side, and the test is only whether t appears. Equations from earlier courses sort cleanly. Exponential growth, y′ = k·y, is autonomous. So is logistic growth in Separable Equations, Growth and Decay, and so is Newton's law of cooling, y′ = −k·(y − A), whose only equilibrium is the surrounding temperature A. The equation y′ = 2t of Differential Equations and Euler's Method is not.
In “Slope Fields” a field was drawn by evaluating f(t, y) at grid points. When f is a function of y alone, the value at (1, y) is the value at (5, y): the column of slopes above t = 1 is the column above t = 5, and all that varies is the height. The lab therefore draws one line, the y axis, and puts on it everything the field knows.
If y(t) solves y′ = f(y) and c is any constant, then y(t − c) solves it too.
Put w(t) = y(t − c). By the chain rule, w′(t) = y′(t − c), since the inner function t − c has slope 1.
And y′(t − c) = f(y(t − c)) = f(w(t)), because y satisfies the equation at every argument, t − c included. So w′ = f(w). The step that fails for y′ = t·y is the last one: the right side would become (t − c)·y(t − c), which is not t·w(t).
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Write the right side as a polynomial in y, factor it if it factors, and read off the roots. The lab does this exactly, with the rational roots first and then the roots of what is left when that is a quadratic.
Not every equilibrium is a fraction. For y′ = y² − y − 1 the lab prints (1 ± √5)/2, which is exact, and in its status line the same two numbers rounded, ≈ 1.61803 and ≈ −0.618034. The square root is carried as a surd for the same reason the rest of this Subject carries fractions: a decimal equilibrium would make the check “is f zero there?” impossible to state exactly.
A solution that is at an equilibrium at one instant is there at every instant, before and after. Through each point there is exactly one solution, which “Blow-Up and the Interval of Existence” states in words and this course uses without proof. The constant function is already a solution through the point (t, y*), so it is the solution through that point, and no other curve can pass through it.
Cooling is y′ = −(1/2)·(y − 20), with the room at 20. The right side is zero exactly when y − 20 is zero, so y = 20 is the only equilibrium.
Substitute: the left side of the constant y = 20 is 0, and the right side is −(1/2)·(20 − 20) = 0. A cup that is already at room temperature stays at room temperature at every later time, and at every earlier one.
Interactive
Equilibria, the phase line, and where solutions go
Method
Four moves. The first decides whether the method applies at all.
If t appears on the right side, the equation is not autonomous and nothing in this course applies to it directly. If only y appears, name the right side f(y) and carry on.
Expand brackets and move everything to one side so that f(y) stands alone. Pull out common factors first, and keep them: a factor of y is a root.
Rational roots as fractions, and a leftover quadratic by the quadratic formula, kept as a surd such as (1 ± √5)/2. Round only to report a size, and say that you rounded.
For y = c, the left side y′ is 0 and the right side is f(c). They agree exactly when c is a root, which is the check and also the reason the roots are the answer.
The right side has no t, so the equation is autonomous, with f(y) = y² − 1. Factor, set to zero, then test the constants.
So y = 1 and y = −1 are solutions for all time. Every other solution has a slope that is positive above 1, negative between −1 and 1, and positive below −1 — and by uniqueness none of them ever touches either constant. Which way each one moves is the subject of the next lesson.
For a rehearsal, take y′ = y³ − y and find its three equilibria by factoring y·(y − 1)·(y + 1), then say what the slope is at y = 2, for t = 0 and for t = 100.
You should be able to test an equation for a t on the right side, factor f(y), give every root as a fraction or a surd, verify each by substitution as a constant, and explain why one vertical line of the slope field is all the field there is.
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