Formulation and solution of first-order differential equations including those solvable by direct integration and by separation of variables, the use of initial conditions to find particular solutions, and the interpretation of solutions in modelling contexts

What this dot point is asking

VCAA wants you to recognise and solve first-order differential equations, principally those solvable by direct integration (where $\frac{\mathrm{d}y}{\mathrm{d}x}$ is a function of $x$ alone) and by separation of variables (where the equation can be written with all the $y$ terms on one side and all the $x$ terms on the other). You also need to apply an initial condition to pin down the arbitrary constant and produce a particular solution, then interpret that solution in a modelling context.

What a differential equation is

A differential equation is an equation involving a function and its derivatives. A first-order equation involves only the first derivative. A solution is a function that satisfies the equation; the general solution contains an arbitrary constant and so describes a whole family of curves, while a particular solution fixes the constant using extra information (an initial or boundary condition).

The order of business is: identify the type, integrate to obtain the general solution, then apply the condition.

Solution by direct integration

When the derivative is a function of the independent variable only,

the solution is found by antidifferentiating once:

Second derivatives are handled by integrating twice, introducing two constants. This is the simplest case and uses the integration techniques from elsewhere in Unit 4.

Solution by separation of variables

Many equations have the derivative depending on both variables in a product or quotient form:

Treat $\frac{\mathrm{d}y}{\mathrm{d}x}$ as a ratio of differentials, gather all $y$ terms with $\mathrm{d}y$ and all $x$ terms with $\mathrm{d}x$:

then integrate both sides. A single constant of integration suffices (combine the two). Where convenient you can leave the solution in implicit form, or solve for $y$ explicitly if the question asks.

Applying initial conditions

The general solution carries an arbitrary constant $c$. An initial condition of the form $y = y_0$ when $x = x_0$ is substituted into the general solution to solve for $c$, giving the particular solution. In modelling problems the condition is usually a known starting value, such as a population or temperature at time $t = 0$.

Interpreting solutions in modelling

Many first-order equations model change. For example, exponential growth or decay arises from $\frac{\mathrm{d}N}{\mathrm{d}t} = kN$, whose separable solution is $N = N_0 e^{kt}$; the constant $k > 0$ gives growth and $k < 0$ gives decay. Newton's law of cooling, $\frac{\mathrm{d}T}{\mathrm{d}t} = -k(T - T_s)$, models a body cooling towards the surrounding temperature $T_s$. When interpreting, state the units, the long-term behaviour (limiting value as $t \to \infty$ where relevant), and whether the solution's domain is restricted by the physical situation.

Examples in context

Example 1. Direct integration. $\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2$ with $y(0) = 1$ gives $y = 2x^3 + c$, and $c = 1$, so $y = 2x^3 + 1$.

Example 2. Exponential decay. $\frac{\mathrm{d}N}{\mathrm{d}t} = -0.2N$ with $N(0) = 50$ separates to $\ln N = -0.2t + c$, giving $N = 50 e^{-0.2t}$; as $t \to \infty$, $N \to 0$.

Try this

Q1. Solve $\frac{\mathrm{d}y}{\mathrm{d}x} = \cos x$ with $y(0) = 2$. [2 marks]

• Cue. $y = \sin x + c$; $c = 2$, so $y = \sin x + 2$.

Q2. Solve $\frac{\mathrm{d}y}{\mathrm{d}x} = 2xy$ with $y(0) = 1$, $y > 0$. [3 marks]

• Cue. Separate: $\frac{1}{y}\mathrm{d}y = 2x\,\mathrm{d}x$, so $\ln y = x^2 + c$; $c = 0$, giving $y = e^{x^2}$.

Q3. A tank's water height satisfies $\frac{\mathrm{d}h}{\mathrm{d}t} = -3$ with $h(0) = 12$. When is the tank empty? [2 marks]

• Cue. $h = 12 - 3t$; empty when $h = 0$, so $t = 4$.