Forming differential equations from a rate of change, solving first-order equations by separating the variables, and applying them to growth, decay and mechanics.

What this dot point is asking

WJEC wants you to form a first-order differential equation from a described rate of change, solve it by separating the variables and integrating, find a particular solution using a given condition, and apply the method to growth, decay and mechanics. This is where the integration techniques of Unit 3 meet real modelling.

The answer

Forming a differential equation

The skill is translating a verbal description of a rate into symbols.

Solving by separating the variables

A first-order separable equation can be written with all the $y$ terms on one side and all the $x$ (or $t$) terms on the other, then integrated.

A single constant of integration suffices (combine the two). A general solution contains the constant; a particular solution fixes it from a given condition such as an initial value. When the right-hand side after integrating the $y$ side gives a logarithm, exponentiate both sides to make $y$ the subject, and the constant becomes a multiplicative factor: $\ln|y| = kx + c$ becomes $y = A\mathrm{e}^{kx}$ where $A = \mathrm{e}^c$ absorbs the constant.

Applications

Examples in context

Example 1. Newton's law of cooling. A body cools so that $\dfrac{d\theta}{dt} = -k(\theta - \theta_0)$, where $\theta_0$ is room temperature. Separating and integrating gives $\theta - \theta_0 = A\mathrm{e}^{-kt}$, so the temperature decays exponentially towards room temperature. The model explains why a hot drink cools fast at first and then slowly.

Example 2. A draining tank. Water drains so that the rate of fall of depth is proportional to the square root of the depth: $\dfrac{dh}{dt} = -k\sqrt{h}$. Separating gives $\displaystyle\int h^{-1/2}\,dh = -\int k\,dt$, so $2\sqrt{h} = -kt + c$. The separable method handles a non-exponential rate just as cleanly.

Try this

Q1. Form a differential equation for a population growing at a rate proportional to its size $P$. [1 mark]

• Cue. $\dfrac{dP}{dt} = kP$ for a positive constant $k$.

Q2. Solve $\dfrac{dy}{dx} = 3x^2 y$, leaving the general solution in terms of a constant. [3 marks]

• Cue. Separate: $\dfrac{1}{y}dy = 3x^2 dx$, integrate to $\ln|y| = x^3 + c$, so $y = A\mathrm{e}^{x^3}$.

Q3. A quantity decays so that $\dfrac{dN}{dt} = -2N$, with $N = 100$ at $t = 0$. Find $N$. [2 marks]

• Cue. $N = 100\mathrm{e}^{-2t}$.