The method of “Separable Equations” integrates y·y′ as though it were the derivative of something, and it is: of y²/2, by the chain rule. Read backwards, the chain rule is the whole justification. Differentiating the implicit relation gives the equation back, which is also how a solution is checked. The lab's check view performs that differentiation exactly, line by line, and prints whether the result equals the right-hand side.
Core ideas
Three ideas. The first is the chain rule from an earlier course; the second turns it round; the third says what the shorthand with the differentials is and is not.
If G′ = g and y is a function of t, then G(y(t)) has the derivative g(y)·y′. For G(y) = y²/2 that is y·y′. The factor y′ is not decoration: it is the rate at which the inside function moves, and it is why the y′ stays with the y.
The left side g(y)·y′, viewed as a function of t, is the derivative of G(y(t)). The right side h(t) is the derivative of H(t). Two functions with equal derivatives on an interval differ by a constant, which is the C.
∫g dy = ∫h dt is a convenient way to write the result. In this course dy/dt means y′, one symbol, and nothing is cancelled to get from the equation to the relation. The chain rule is what is being used, and it holds whether or not you ever write a differential.
The material
The chain rule once more, the statement that justifies the method with its proof, and three relations checked by differentiating them.
“The Chain Rule” in Rates of Change and the Derivative was checked exactly on polynomials: if u = y(t) and F is a function of u, the rate of F(y(t)) is F′(y) times y′. Here F is an antiderivative G, so G′ = g, and the rate is g(y)·y′. Nothing about y is assumed beyond having a derivative.
Suppose G′ = g and H′ = h, and y is differentiable on an interval. Then g(y)·y′ = h(t) on that interval exactly when G(y(t)) − H(t) is constant on it.
By the chain rule, the rate of G(y(t)) − H(t) is g(y)·y′ − h(t).
That is zero for every t in the interval exactly when the equation holds. A function whose rate is zero on an interval does not change on it — the fact behind the constant of integration in Accumulation and the Integral — so G(y(t)) − H(t) = C, one number.
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So the relation G(y) = H(t) + C is not a heuristic. Along a solution it holds with some constant, and the proof also runs the other way: a differentiable y that satisfies the relation satisfies the equation, wherever g(y) makes sense. That second direction is what a check uses.
Take the relation y² = t² + 4. Differentiate both sides with respect to t, remembering that y depends on t: the left side gives 2y·y′ by the chain rule, the right side gives 2t. So 2y·y′ = 2t, which is y·y′ = t, the equation we started with.
The explicit form is checked the same way, with one more use of the chain rule for the square root. The check view of the lab does the differentiation with exact fractions: it shows G′ = g, H′ = h, and, where the relation can be solved for y, the composite G(y(t)) differentiated. The tile reads equal when the result is the equation and a dash when there is nothing to compare.
For (1/y²)·y′ = t the relation is −1/y = t²/2 − 1. The derivative of −1/y with respect to t is y′/y², by the chain rule, and the derivative of t²/2 − 1 is t. So y′/y² = t, which is (1/y²)·y′ = t.
For 3y²·y′ = 2t + 1 the relation is y³ = t² + t + 1. The derivative of y³ is 3y²·y′ and the derivative of the right side is 2t + 1, so the equation comes straight back. The lab does not solve this relation for y, which would take a cube root, and the check does not need it to.
Whenever a lesson in this course reports a solution, this is the test: differentiate it and compare. It catches a lost sign, a missing factor of y′ and a wrong constant, and it needs no knowledge of how the answer was found.
Interactive
Both sides integrated exactly
Method
Four moves, and none of them needs to know how the answer was found.
Every y is a function of t, so each term picks up a factor y′ from the chain rule: y² becomes 2y·y′ and 1/y becomes −y′/y².
It is a function of t alone, so this is an ordinary derivative, and the constant disappears.
Equal after rearranging means the relation is consistent with the equation. Unequal means a sign or a factor was lost, and the difference tells you which.
Substitute (t₀, y₀) into the relation. The derivative check says nothing about the constant, and this is the line that catches a wrong C.
The solution of y·y′ = t through y(0) = 2 from “Separable Equations”, checked both as a relation and as a formula. The first line is the one the lab's table prints for the first preset.
Two checks, two different jobs. The derivative check confirms that the relation's rate is the equation's rate; the point check confirms the constant. Either one alone leaves room for a wrong answer, and together they leave none.
The same relation also holds for y = −√(t² + 4), which has the same square and the opposite sign. Both pass the derivative check, and only one passes the point check for a given initial value. That is the subject of the next lesson.
You should be able to state the chain rule for G(y(t)), say why it justifies integrating g(y)·y′, differentiate a relation in y and t with the factor y′ in the right places, and explain why the derivative check cannot catch a wrong constant.
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