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.
The trick: treat dy/dx like a fraction, divide both sides by g(y) and multiply by dx, so all the y's sit with dy and all the x's sit with dx:
$$\frac{1}{g(y)}\,dy = f(x)\,dx$$
Then integrate each side separately. You only need one constant of integration — put a single +C on the right.
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: The general solution has an unknown constant (+C or A) — it's a whole family of curves. An initial (or boundary) condition is one known point the curve must pass through, written like y = 2 when x = 0.
Substitute that 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.)