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Pure · Integration

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Differential equations

Solving first-order separable differential equations by separating the variables and integrating both sides, finding a particular solution from a boundary condition, and modelling rates of change.

Pure · Integration

Differential equations

Solving first-order separable differential equations by separating the variables and integrating both sides, finding a particular solution from a boundary condition, and modelling rates of change.

Why it works

A differential equation is an equation involving a derivative — it tells you the rate at which something changes rather than its value directly. Solving it means recovering the original relationship yy from dydx\dfrac{dy}{dx}, which is exactly what integration does.

The type you can solve at A-level is separable: the right-hand side factorises into a part in xx and a part in yy, dydx=f(x) g(y).\frac{dy}{dx} = f(x)\,g(y). The trick is to get all the yy's (with dydy) on one side and all the xx's (with dxdx) on the other — separating the variables — then integrate both sides: ∫1g(y) dy=∫f(x) dx.\int \frac{1}{g(y)}\,dy = \int f(x)\,dx. Treating dydx\dfrac{dy}{dx} as a fraction to move dxdx and dydy around is legitimate here (it's substitution in disguise). One arbitrary constant +c+c on one side is enough.

This gives the general solution — a whole family of curves, one for each value of cc. A boundary condition (a known pair of values, e.g. "y=3y = 3 when x=0x = 0") pins down cc and selects the single particular solution that fits.

Modelling. Many real processes have a rate proportional to the current amount: dydt=ky.\frac{dy}{dt} = ky. Separating, ∫1y dy=∫k dt\int \tfrac1y\,dy = \int k\,dt gives ln⁡∣y∣=kt+c\ln|y| = kt + c, so y=Aekty = Ae^{kt} — exponential growth (k>0k > 0) or decay (k<0k < 0). This single equation models populations, radioactive decay, cooling and compound interest; the constants AA and kk come from the conditions given in the problem.

Harder separations. Once separated, each side is just an integral — and it can need any technique you've met. A factor like 4xx2+4\dfrac{4x}{x^2 + 4} wants the reverse chain rule, an xexx e^x wants integration by parts, and a logistic model dPdt=kP(M−P)\frac{dP}{dt} = kP(M - P) — growth that levels off at a capacity MM — needs partial fractions to integrate 1P(M−P)\dfrac{1}{P(M-P)}. These are common in modelling questions, where the long-term behaviour (t→∞t \to \infty) is read straight off the solution.