Solve first-order differential equations by direct integration and by separating variables, find particular solutions from conditions, and interpret solutions in context

What this dot point is asking

SEAB wants you to solve first-order differential equations by direct integration and by separating variables, find a particular solution using a given condition, and interpret the solution in a modelling context such as growth, decay or cooling.

The answer

Direct integration

If $\dfrac{dy}{dx} = \mathrm{f}(x)$ (the right side depends only on $x$), integrate directly:

Separation of variables

If $\dfrac{dy}{dx} = \mathrm{g}(x)\mathrm{h}(y)$, separate the variables to opposite sides and integrate:

This is the central technique. Treat $\dfrac{dy}{dx}$ as a ratio of differentials for the rearrangement, gather all $y$ with $dy$ and all $x$ with $dx$, then integrate both sides (one constant suffices).

General and particular solutions

Integration introduces an arbitrary constant, giving the general solution (a family of curves). A boundary or initial condition fixes the constant, giving the particular solution for the specific situation.

Modelling growth and decay

The equation $\dfrac{dy}{dt} = ky$ (rate proportional to amount) separates to $y = Ae^{kt}$: exponential growth for $k > 0$, decay for $k < 0$. Newton's law of cooling $\dfrac{d\theta}{dt} = -k(\theta - \theta_s)$ separates similarly, giving an exponential approach to the surrounding temperature.

Using partial fractions to separate

When separation produces a rational function of $y$ on the left, partial fractions often make it integrable. The logistic model $\dfrac{dP}{dt} = kP(1 - P)$ separates to $\displaystyle\int \dfrac{1}{P(1 - P)}\,dP = \int k\,dt$, and the left integrand splits as $\dfrac{1}{P} + \dfrac{1}{1 - P}$, each a logarithm. This links the differential-equations work directly to the partial-fractions and integration techniques: a separable equation is only as solvable as the integrals it produces, so recognising when partial fractions are needed is part of the method.

Sketching the family of solution curves

The general solution of a first-order equation is a family of curves, one for each value of the constant, and a boundary condition selects the single curve through a given point. Sketching a few members of the family, then highlighting the particular solution, is a common H2 task that checks understanding rather than algebra. For $\dfrac{dy}{dx} = ky$, the family is a set of exponential curves $y = Ae^{kx}$ of different heights, and the condition fixes which one passes through the stated point. Reading the constant as "which curve in the family" makes the general-versus-particular distinction concrete.

Examples in context

Example 1. Radioactive decay. A sample with $\dfrac{dN}{dt} = -\lambda N$ decays as $N = N_0 e^{-\lambda t}$; the half-life follows from setting $N = \tfrac{1}{2}N_0$, the standard nuclear-physics result obtained by separation of variables.

Example 2. Cooling coffee. A cup of coffee obeying Newton's law of cooling approaches room temperature exponentially, with the time constant from $k$; solving the differential equation predicts the temperature at any later time.

Try this

Q1. Solve $\dfrac{dy}{dx} = 3x^2$ with $y = 1$ at $x = 0$. [2 marks]

• Cue. $y = x^3 + C$, condition gives $C = 1$, so $y = x^3 + 1$.

Q2. Solve $\dfrac{dy}{dx} = ky$ in general form. [2 marks]

• Cue. Separate to $\frac{1}{y}dy = k\,dx$; $y = Ae^{kx}$.

Q3. State the difference between a general and a particular solution. [2 marks]

• Cue. The general solution contains an arbitrary constant (a family of curves); the particular solution fixes it using a given condition.