1. Recognising a Separable First-Order ODE
Definition
A first-order ODE is separable if it can be written in the form
$$\frac{dy}{dx} = g(x)\,h(y)$$That is, the right-hand side is a product of a function of $x$ alone and a function of $y$ alone.
The Recognition Craft — A Gallery of Disguises
Separability is not always obvious. You may need to factor first — this is where your algebra skills cash in.
| Given Form | Algebra Step | Separated Form $g(x)\cdot h(y)$ |
|---|---|---|
| $\dfrac{dy}{dx}=x^2+x^2 y^2$ | Factor out $x^2$ | $x^2(1+y^2)$ |
| $\dfrac{dy}{dx}=\dfrac{xy+2x}{y-1}$ | Factor numerator: $x(y+2)$ | $x\cdot\dfrac{y+2}{y-1}$ |
| $\dfrac{dy}{dx}=e^{x+y}$ | Exponent law: $e^{x+y}=e^x e^y$ | $e^x\cdot e^y$ |
| $\dfrac{dy}{dx}=\dfrac{6x^2}{2y+\cos y}$ | Already separated | $6x^2\cdot\dfrac{1}{2y+\cos y}$ |
| $\dfrac{dy}{dx}=x+y$ | Cannot factor as product | NOT separable |
⚠ Callback to Algebra: The factoring techniques you drilled — common factors, exponent laws, grouping — are exactly the tools that reveal separability. Without them the disguise holds.
2. The Method — Separate, Integrate, Solve
The Mechanism, Stated Honestly
Starting from $\dfrac{dy}{dx}=g(x)\,h(y)$, we herd each variable to its own side together with its own differential:
$$\frac{1}{h(y)}\,dy = g(x)\,dx$$Then integrate both sides independently:
$$\int \frac{1}{h(y)}\,dy = \int g(x)\,dx$$The differential notation $dy$ and $dx$ earns its keep here: it lets us treat the derivative symbol as a ratio of differentials for the purpose of separation — the bookkeeping promise of the Leibniz notation, redeemed in full.
Why Only One Constant?
Officially, each integral produces its own constant: