Separable Differential Equations (AI HL)
Some differential equations can be pulled apart so that every \(y\) term sits on one side and every \(x\) term sits on the other, letting you integrate each side on its own. This page covers that separation-of-variables method, from rearranging \(\dfrac{dy}{dx}\) through to using an initial condition to pin down the constant, with worked examples and the mistakes that lose the most marks. It's part of the broader Differential Equations topic.
11 questions on this sub-topic.
The method
Covered under IB syllabus reference AHL5.14: setting up a differential equation from a context, and solving it by separation of variables. There is no formula-booklet entry for this - it's a technique you apply, not a formula you look up.
Separate
\(\dfrac{dy}{dx} = f(x)g(y) \implies \displaystyle\int \dfrac{1}{g(y)}\,dy = \int f(x)\,dx\)
Every \(y\) (and \(dy\)) moves to one side, every \(x\) (and \(dx\)) to the other, before either side is integrated.
Solve for the constant
General solution \(\to\) substitute a given point \((x_0, y_0)\) \(\to\) particular solution
A general solution has an unknown constant \(C\). An initial condition such as \(y(0)=3\) turns it into one specific, particular solution.
Need the full syllabus wording and the rest of the differential-equations toolkit, including Euler's method and slope fields? See Differential Equations.
Worked examples
Solve \(\dfrac{dy}{dx} = \dfrac{x}{y}\) given \(y(0) = 3.\)
Worked solution
\(\int y\,dy = \int x\,dx\) M1 A1
\(\tfrac{y^2}{2} = \tfrac{x^2}{2} + c \Rightarrow y^2 = x^2 + C.\) A1
\(y(0) = 3 \Rightarrow C\) M1
\(= 9.\) A1
\(y = \sqrt{x^2 + 9}.\) A1
Solve \(\dfrac{dy}{dx} = y\cos x\) given \(y(0) = 2.\)
Worked solution
\(\int \tfrac1y\,dy = \int \cos x\,dx\) M1 A1
\(\ln|y| = \sin x + c\) A1
\(\Rightarrow y = Ae^{\sin x}.\) A1
\(y(0) = 2 \Rightarrow A = 2.\) M1
\(y = 2e^{\sin x}.\) A1
Common mistakes
- Dropping the constant of integration. Every separable equation needs a \(+c\) written in as soon as you integrate - without it there's nothing for the initial condition to solve for, and you can't reach a particular solution.
- Forgetting the modulus in \(\ln|y|\). When \(\tfrac{1}{y}\) is integrated the result is \(\ln|y|\), not \(\ln y\). Dropping the absolute value can quietly produce a solution that's undefined for the values you actually need.
- Applying the initial condition too early. Substitute the given point only once the equation has been fully rearranged into \(y = \ldots\) (or as close to it as the algebra allows) - substituting mid-rearrangement risks losing the constant or mixing it into the wrong side.
Ready to practise properly?
10 separable-differential-equation questions, marked instantly like the real exam.
Quick answers
How do you solve a separable differential equation?
Rearrange the equation so every \(y\) term (including \(dy\)) is on one side and every \(x\) term (including \(dx\)) is on the other, then integrate both sides separately and add a constant of integration.
How do you find the value of the constant C?
Substitute the given initial condition, such as \(y(0)=3\), into the general solution once it is fully rearranged, then solve the resulting equation for \(C\).