Forming first-order differential equations from a context, solving them by separation of variables, finding particular solutions from initial conditions, and interpreting the solution in modelling.

What this dot point is asking

OCR wants you to form a first-order differential equation from a described rate of change (often "rate proportional to ..."), solve it by separating the variables and integrating, use an initial or boundary condition to find the particular solution, and interpret the result in a modelling context such as population growth, radioactive decay or Newton's law of cooling.

The answer

Forming a differential equation

A differential equation links a quantity to its rate of change. The key phrases translate directly: "the rate of change of $y$" is $\dfrac{dy}{dt}$, "proportional to $y$" multiplies by a constant $k$, and a decreasing quantity gives a negative constant.

Separation of variables

A separable equation $\dfrac{dy}{dx} = f(x)g(y)$ is solved by gathering all the $y$ terms on one side and all the $x$ terms on the other, then integrating both sides.

Particular solutions

The general solution contains an arbitrary constant. An initial condition (a known value at a known time) pins it down to a particular solution.

Examples in context

A growth model

The proportional-rate model $\dfrac{dP}{dt} = kP$ always integrates to exponential growth $P = P_0 e^{kt}$. Two data points determine both $P_0$ and $k$. Recognising this shape lets you go straight to the form and just fit the constants.

A cooling model

Newton's law of cooling says a body cools at a rate proportional to the difference between its temperature and that of its surroundings. The solution always approaches the surrounding temperature $\theta_0$ as $t \to \infty$, because the exponential term decays to zero. Reading off this long-run value is a common final part of a cooling question.

Interpreting and checking a solution

A modelling question usually ends by asking you to interpret the solution: the long-run value, the time to reach a target, or whether the model is realistic. You can also check a candidate solution by differentiating it and confirming it satisfies the original equation, which is a quick way to catch an algebra slip. Note too that a proportional-rate model predicts unbounded growth, so it is only sensible over a limited time; real populations level off, which is why such models are stated to hold "for small $t$" or "while resources are plentiful".

Try this

Q1. Solve $\dfrac{dy}{dx} = \dfrac{y}{x}$ for $x > 0$. [3 marks]

• Cue. $\frac{1}{y}\,dy = \frac{1}{x}\,dx$ gives $\ln|y| = \ln|x| + c$, so $y = Ax$.

Q2. The number $N$ decays as $\dfrac{dN}{dt} = -0.2N$ with $N = 100$ at $t = 0$. Find $N$ at $t = 5$. [3 marks]

• Cue. $N = 100e^{-0.2t}$, so at $t = 5$, $N = 100e^{-1} \approx 36.8$.