First-order differential equations and separation of variables (QCE Specialist Mathematics Unit 4)

What this dot point is asking

QCAA wants you to set up a first-order differential equation from a worded rates-of-change situation, solve it by separation of variables, apply an initial condition to find the particular solution, and interpret the result. This dot point underpins much of IA1 modelling and is examined in IA3 and the external assessment, where growth, decay, cooling and logistic models recur.

The answer

What a differential equation says

A differential equation relates a quantity to its rate of change. The statement "the rate of growth is proportional to the current amount" becomes

Solving means finding the function $y(t)$ whose derivative matches the equation. The general solution contains an arbitrary constant; an initial condition pins it down to a particular solution.

Separation of variables

If a first-order equation can be written as $\dfrac{dy}{dx} = f(x)g(y)$, separate the variables so each side involves only one variable, then integrate both sides:

Treating $\dfrac{dy}{dx}$ as a ratio of differentials is a valid shortcut for this technique. Include a single constant of integration, then solve algebraically for $y$ where possible.

Exponential growth and decay

The equation $\dfrac{dy}{dt} = ky$ separates to $\int \frac{1}{y}\,dy = \int k\,dt$, giving $\ln|y| = kt + C$ and hence

Positive $k$ gives growth; negative $k$ gives decay (for example radioactive decay, where the half-life is $t_{1/2} = \dfrac{\ln 2}{|k|}$).

Newton's law of cooling

The rate of cooling is proportional to the temperature difference from the surroundings $T_s$:

Substituting $u = T - T_s$ reduces this to exponential decay, giving $T = T_s + (T_0 - T_s)e^{-kt}$. The temperature approaches the ambient $T_s$ as $t \to \infty$.

The logistic model

When growth is limited by a carrying capacity $M$, the model is

Growth is near-exponential when $P$ is small and slows to zero as $P$ approaches $M$. Solving by separation (using partial fractions on the left) produces an S-shaped curve that levels off at $P = M$. The logistic model is a favourite IA1 modelling context because it captures realistic saturation.

General versus particular solutions

Integrating a first-order equation introduces exactly one arbitrary constant, and the family of curves this produces is the general solution. An initial or boundary condition (a known value of $y$ at a particular $x$ or $t$) fixes the constant and selects the single particular solution that fits the situation. The order matters: keep the constant through the algebra, apply the condition once you have isolated it, and only then simplify. Applying the condition too early, or absorbing it into an exponential before evaluating, is the source of most lost marks.

Velocity, acceleration and the inverse-tangent integral

Specialist Mathematics often phrases differential equations in a kinematics context, where acceleration is $\dfrac{dv}{dt}$ or $v\dfrac{dv}{dx}$. Separating such an equation can produce standard integrals beyond logarithms, in particular $\displaystyle\int \dfrac{1}{a^2 + v^2}\,dv = \dfrac{1}{a}\arctan\dfrac{v}{a} + C$ and $\displaystyle\int \dfrac{1}{\sqrt{a^2 - v^2}}\,dv = \arcsin\dfrac{v}{a} + C$. Recognising which standard form the separated integrand matches is the planning step, and choosing $a$ to match the constant under the square or the constant added to $v^2$ completes it.

Checking a solution

Always verify a solution by substituting it back into the original equation. Differentiate your $y(t)$ and confirm it reproduces the stated rate, and check that the initial condition is satisfied. A second sanity check is the long-term behaviour: a decay model should tend to zero, a cooling model should tend to the ambient temperature, and a logistic model should tend to the carrying capacity. If the long-run behaviour contradicts the physical setup, a sign error in $k$ or a dropped constant is the usual cause.