When the rule comes apart
You already know what a first-order equation looks like: dy/dx tells you the slope at every point, and you are hunting for the function whose slope obeys that rule everywhere. Most of the time you cannot just stare at it and integrate. But sometimes the rule is unusually friendly — the right-hand side splits into a piece that depends only on x times a piece that depends only on y. That is a separable equation, and it is the first kind we can crack with nothing but integration.
Concretely, a separable equation can be written as dy/dx = g(x) h(y). Think of g(x) as the part the clock controls and h(y) as the part the current state controls. A radioactive sample decays at a rate proportional to how much is left: dy/dx = -k y. Here g(x) = -k (a constant — the clock part is trivial) and h(y) = y. A cooling cup follows dy/dx = -k (y - T): again the x-part is just a constant, the y-part carries all the action. The test is purely mechanical: can you factor the right side into an x-only times a y-only? If yes, you are in business.
Shuffle, then integrate both sides
The method has an almost suspiciously simple shape. Starting from dy/dx = g(x) h(y), gather everything with y on the left and everything with x on the right: dy / h(y) = g(x) dx. Now each side is a clean one-variable integral. Integrate both — the left over y, the right over x — and you have tied y and x together in one equation. This is the move that gives the method its name: separation of variables.
- Check it separates: write the right side as g(x) h(y). If you cannot factor it that way, this method does not apply — stop here.
- Separate: move dy and all y's to one side, dx and all x's to the other, giving dy / h(y) = g(x) dx.
- Integrate both sides — left in y, right in x. Add a single arbitrary constant C on one side; you never need two.
- Solve for y if you can. Often you can, but sometimes you must leave the answer tangled — that is an implicit general solution, and it is still a perfectly good answer.
Why is one constant enough? Each integral honestly produces its own constant, but they live on opposite sides of an equals sign, so you can subtract one from the other and roll them into a single C. Take dy/dx = -k y: separating gives dy/y = -k dx, integrating gives ln|y| = -k x + C, and exponentiating gives y = A e^(-k x) with A = e^C a fresh constant. That is exponential growth and decay, derived from scratch — not memorized, but built.
What dividing quietly throws away
Here is the quiet trap. To separate, you divided by h(y) — and dividing is only legal when h(y) is not zero. Every value of y where h(y) = 0 is a place you were forbidden to divide, and those are precisely the constant solutions the method silently drops. For dy/dx = -k y we divided by y, which is illegal at y = 0; and indeed y = 0 (the empty sample, staying empty forever) is a genuine solution that the formula y = A e^(-k x) can only reach in the limit A -> 0. The full guide on this is next, but learn to flinch at the division now.
The fix is a one-line habit: before dividing by h(y), find every root of h(y) = 0 and write those constant functions down as solutions of their own. They are called the lost constant solutions. The logistic equation dy/dx = r y (1 - y) has h(y) = r y (1 - y), which vanishes at both y = 0 and y = 1; separating loses both, yet they are real (extinction and carrying capacity). Skipping this step does not make your algebra wrong — it makes your answer incomplete.
Pinning down the one curve you want
Integrating gives a whole family of curves, one for each value of C — a stack of decay curves, each starting from a different amount. To single out the one solution your problem actually describes, you supply one extra fact: a starting point, like y = 5 when x = 0. Plug it in, solve for C, and the family collapses to a single curve. An equation plus such a starting fact is an initial value problem; for a separable equation it is a separable IVP.
dy/dx = -k y, y(0) = 5 separate: dy/y = -k dx integrate: ln|y| = -k x + C general: y = A e^(-k x) apply IC: 5 = A e^0 -> A = 5 solution: y = 5 e^(-k x)
One honest caveat before you trust the picture: the solution you build is only guaranteed on an interval around the starting point, not for all x. The formula might blow up, or h(y) might hit a forbidden zero, and the genuine solution stops there. That window is the interval of validity, and reading it off the formula — spotting where a denominator vanishes or a logarithm's argument turns negative — is part of solving the problem honestly, not an afterthought.
When the answer refuses to untangle
Sometimes, after integrating both sides, you simply cannot solve for y in terms of x — the two are knotted together in one relation, like y^2/2 + y = x^3/3 + C. Do not panic and do not force it. A relation that ties x and y together and is satisfied by the solution curve is an implicit general solution. It still draws the right curves, still answers initial-value questions (plug in the starting point, get C), and is often the truest form the answer can take. Demanding an explicit y = (formula in x) is sometimes asking for something that does not exist in elementary functions.
This connects to a much larger and humbling truth of the whole subject: the overwhelming majority of differential equations have no closed-form solution at all. Separable equations are precious precisely because they are among the rare cases where pure integration wins. When integration fails — and it usually does — we lean instead on numerical methods to compute solutions step by step, and on qualitative methods (slope fields, phase lines) to understand their shape without ever writing a formula. Separation is the first rung; it teaches the moves, but it is not the whole ladder.