Some first-order equations can be written so that the left side is a function of y times y′ and the right side is a function of t alone. For those, each side has an antiderivative of its own, the two are set equal with a single constant, and an initial value fixes that constant exactly. The lab does the integration in exact fractions and prints the implicit relation between y and t. The one thing to watch is what “separate” means: y′ stays with the y, not with the t.
Core ideas
Three ideas, in the order you use them. The first is a test, the second is the integration, and the third is the only place an initial value enters.
An equation is separable when it can be written with a function of y multiplying y′ on one side and a function of t alone on the other. y′ = t/y becomes y·y′ = t by multiplying through by y. Nothing is moved across the equals sign and left behind: the factor y′ travels with the y it multiplies, and that is what the next lesson shows to be the whole point.
Call an antiderivative of g by the name G and one of h by H. The relation G(y) = H(t) + C holds along a solution. Each side could carry its own constant, but their difference is a single number, so one C is enough, and writing it on the right is only a habit.
If the solution passes through (t₀, y₀) then C = G(y₀) − H(t₀). With the antiderivatives in exact fractions, C is an exact fraction too, and the lab prints it. Before the constant is fixed the answer is a whole family of curves; after, it is one.
The material
The definition, the four moves, two more equations through the same moves, and one equation the moves cannot reach.
A first-order equation that can be written g(y)·y′ = h(t), where g is a function of y only and h is a function of t only.
An antiderivative of g is a function G with G′ = g, as in Accumulation and the Integral. The lab accepts a g that is a polynomial in y, or exactly 1/y or 1/y^2, and an h that is a polynomial in t.
Take y′ = t/y. Multiply both sides by y and the equation reads y·y′ = t: the left side is g(y) = y times y′, and the right side is h(t) = t. The antiderivatives are G(y) = y²/2 and H(t) = t²/2, so along any solution y²/2 = t²/2 + C. That is a relation between y and t, not yet a formula for y, and “Implicit and Explicit Solutions” is about what to do with it.
The lab does moves two to four. Its first tile shows G(y), its second H(t), its third the constant, and the fourth the relation with the constant in it, scaled so that the leading coefficient on the y side is 1. Its first preset is the equation above, and the tiles print y²/2, t²/2, 2 and y² = t² + 4.
y′ = t·y² separates as (1/y²)·y′ = t, so g(y) = 1/y², G(y) = −1/y (the rate of −1/y is 1/y², from “The Derivative of 1/t”) and H(t) = t²/2. With y(0) = 1 the constant is C = −1/1 − 0 = −1, and the relation is −1/y = t²/2 − 1; the lab also prints the explicit form y = 2/(2 − t²).
The division by y² is only allowed where y ≠ 0. The constant function y = 0 also satisfies y′ = t·y², and the separated form cannot produce it. An initial value of 1 never goes near it, so nothing is lost here, but the method has no way to hand back a solution it divided away.
3y²·y′ = 2t + 1 has G(y) = y³ and H(t) = t² + t. With y(0) = 1 the constant is C = 1³ − 0 = 1, so y³ = t² + t + 1.
Here the relation can be solved for y as a cube root, which is a formula. The lab marks it as implicit anyway, and “Implicit and Explicit Solutions” says when an explicit form exists.
y′ = t + y has no separated form. The right side is a sum, and a sum of a function of t and a function of y cannot be written as h(t) over g(y) for any such pair. Nothing in the lab will rescue it, and a later course solves it by another method. y′ = t² + y² fails for the same reason.
Interactive
Both sides integrated exactly
Method
Four moves. The first is the one that decides whether the rest apply.
Get the equation to g(y)·y′ = h(t). Write down any value of y you divided by zero at, because a constant solution there is not recovered by the later steps.
G from g with respect to y, H from h with respect to t. Differentiate each answer once in your head; a lost sign here survives to the end.
G(y) = H(t) + C. Put the constant on the t side, and do not add a second one to the y side.
C = G(y₀) − H(t₀), in that order. Substitute the point back into the relation to confirm it holds.
The equation is already separated, with g(y) = y and h(t) = t. Every figure below is exact, and the lab's first preset prints the same four results.
The relation is what the method delivers. It is not yet a function: for each t it allows two values of y, one positive and one negative, and the initial value y(0) = 2 is what will pick between them. That choice, and where it can be made, is the work of “Implicit and Explicit Solutions”.
Before that, there is a question the method has so far dodged: why is it allowed to integrate y·y′ as though it were an ordinary function of one variable? The next lesson answers it, and the answer is the chain rule.
You should be able to say whether an equation separates, integrate each side exactly, write one relation with one constant, take the constant from the initial value in the right order, and name the value of y you divided by.
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