Get the y's with dy, the x's with dx: A first-order ODE is separable when you can write it as dy/dx = f(x)·g(y) — an x-part times a y-part.
Separate, then integrate
Treat dy/dx like a fraction
Divide both sides by g(y) and multiply by dx.
Group the variables
All the y's sit with dy and all the x's sit with dx (see below).
Integrate each side separately
Work out each integral on its own.
Add one constant
You only need one constant of integration — put a single +C on the right.
After the split: $$\frac{1}{g(y)}\,dy = f(x)\,dx$$
IB-style question — separate and integrate
A quantity changes so that its rate of change is the product of x and y.
Find the general solution of dy/dx = xy.
Step by step
- It is separable: f(x) = x, g(y) = y. Divide by y and multiply by dx so the y's go left, the x's go right.
- Integrate each side. The left gives ln|y|; the right gives x²/2. Combine the two constants into one +C.
- Make y the subject. Exponentiate both sides; eC is just another positive constant, call it A (and A can carry the ± sign).
Final answer
General solution y = A ex²/2, where A is an arbitrary constant.
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A known point turns the general solution into THE solution: An initial (or boundary) condition is one known point the curve must pass through, written like y = 2 when x = 0.
General solution
- Has an unknown constant (+C or A).
- It's a whole family of curves.
Particular solution
- Substitute the known point to solve for the constant.
- This picks out the one particular curve through that point.
Tip: Substitute the point as soon as you have the general solution — the algebra is usually cleaner before you rearrange.
IB-style question — particular solution
Given dy/dx = 2x(y + 1), and y = 3 when x = 0.
Find y in terms of x.
Step by step
- Separate: the y-part is (y + 1). Divide by (y + 1), multiply by dx.
- Integrate both sides. Left: ln|y + 1|. Right: x². One +C.
- Use the condition y = 3, x = 0 to find C (substitute now, before rearranging).
- Rearrange: exponentiate. ln(y+1) = x² + ln4, so y + 1 = ex²·eln4 = 4ex².
- Make y the subject.
Final answer
y = 4ex² − 1. (Check: at x = 0, y = 4 − 1 = 3 ✓.)
IB-style question — Newton's cooling style
A cup of tea cools so that dT/dt = −0.1(T − 20), where T is temperature and t is time. Initially T = 80.
Find T in terms of t.
Step by step
- Separate. The y-part (here T-part) is (T − 20).
- Integrate both sides.
- Use T = 80 at t = 0: ln(60) = C.
- Rearrange: T − 20 = e−0.1t·eln60 = 60e−0.1t.
Final answer
T = 20 + 60e−0.1t. (As t → ∞, T → 20 — room temperature.)