Separable differential equations (SACE Stage 2 Specialist Mathematics)

Syllabus dot point

“Solve first-order separable differential equations and find particular solutions using initial conditions.”

8 min readUpdated 2 Jun 2026

Quick answer

A separable differential equation can be written $\dfrac{dy}{dx} = f(x)\,g(y)$ . Separate the variables so all the $y$ terms (with $dy$ ) are on one side and all the $x$ terms (with $dx$ ) on the other, then integrate both sides. This gives the general solution with one arbitrary constant; an initial condition pins down that constant to give the particular solution.

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What this dot point is asking
The separation method
A worked general-then-particular solution
Recognising a separable equation
Exponential growth and decay
General versus particular solutions
Exam-style questions

What this dot point is asking

You need to solve first-order separable differential equations and use initial conditions to find particular solutions.

The separation method

If $\dfrac{dy}{dx} = f(x)\,g(y)$ , divide by $g(y)$ and multiply by $dx$ to gather like variables:

\frac{1}{g(y)}\,dy = f(x)\,dx.

Integrate both sides:

\int \frac{1}{g(y)}\,dy = \int f(x)\,dx.

A single constant of integration

C

on one side captures the whole family of solutions.

One constant, applied once

You only need one arbitrary constant. Combine the constants from both integrals into a single $C$ on the right-hand side. Then apply the initial condition to solve for $C$ before, or after, rearranging for $y$ - whichever is cleaner.

A worked general-then-particular solution

Separable equation with an initial condition

Solve $\dfrac{dy}{dx} = \dfrac{2x}{y}$ given $y = 3$ when $x = 0$

Step 1: Separate variables

Rearrange so all $y$ terms are on the left with $dy$ and all $x$ terms are on the right with $dx$ . Multiplying both sides by $y$ achieves this.

y\,dy = 2x\,dx.

Step 2: Integrate both sides

Each side now involves only one variable, so we integrate independently and combine the constants of integration into a single $C$ on the right.

\int y\,dy = \int 2x\,dx \implies \frac{y^2}{2} = x^2 + C.

Step 3: Apply the initial condition to find $C$

Substituting the known values $y = 3$ and $x = 0$ into the general solution pins down the arbitrary constant.

\frac{3^2}{2} = 0 + C \implies C = \frac{9}{2}.

Step 4: Rearrange for $y$

Substitute the value of $C$ back in, multiply through, and take the positive square root, choosing the branch consistent with the initial condition $y = 3 > 0$ .

So $\dfrac{y^2}{2} = x^2 + \dfrac{9}{2}$ , that is $y^2 = 2x^2 + 9$ . Taking the root and choosing the branch through $(0, 3)$ (positive):

y = \sqrt{2x^2 + 9}.

Recognising a separable equation

Not every first-order equation is separable. An equation is separable only if the right side can be written as a product of a function of $x$ alone and a function of $y$ alone, $\dfrac{dy}{dx} = f(x)g(y)$ . For instance $\dfrac{dy}{dx} = xy + x = x(y + 1)$ is separable (it factors), whereas $\dfrac{dy}{dx} = x + y$ is not, because the sum cannot be split into a single $x$ -factor times a single $y$ -factor. Spotting whether a factorisation exists is the first decision, and SACE problems are designed so that the equation either is already separated or factors cleanly once you look for the common factor.

Exponential growth and decay

The equation $\dfrac{dy}{dt} = ky$ is separable and central to modelling:

\int \frac{1}{y}\,dy = \int k\,dt \implies \ln|y| = kt + C \implies y = A e^{kt},

where

A = e^{C}

is set by the initial value. Positive

k

gives growth; negative

k

gives decay. This is the backbone of population, radioactive-decay and cooling models.

An exponential decay solution

Solve $\dfrac{dN}{dt} = -0.05N$ with $N = 200$ at $t = 0$

Step 1: Separate variables

To separate, divide both sides by $N$ so the $N$ term moves to the left. This is the standard technique for any equation where the rate is proportional to the quantity itself.

\frac{1}{N}\,dN = -0.05\,dt.

Step 2: Integrate both sides and exponentiate

Integrating each side gives a logarithm on the left and a linear function on the right. Exponentiating both sides removes the logarithm and introduces a positive constant $A = e^C$ .

\int \frac{1}{N}\,dN = \int -0.05\,dt \implies \ln|N| = -0.05t + C.

Exponentiate:

N = A e^{-0.05t}

, where

A = e^{C}

Step 3: Apply the initial condition $N = 200$ at $t = 0$

Substituting $t = 0$ makes the exponential equal to 1, so $A$ is read off directly. Apply $N = 200$ at $t = 0$ : $200 = A e^{0} = A$ , so $A = 200$ . Hence

N = 200 e^{-0.05t}.

Watch the sign of the square root

When the solution involves $y^2 = \ldots$ , taking the square root introduces a sign choice. Use the initial condition to decide whether $y$ is the positive or negative root, and state which branch you keep.

General versus particular solutions

Integrating a separable equation produces a general solution containing one arbitrary constant - a whole family of curves, one for each value of the constant. An initial or boundary condition selects the single member of that family passing through the given point, the particular solution. SACE questions almost always supply a condition such as " $y = 3$ when $x = 0$ ", and full marks require you to apply it and report the specific solution, not just the general form. A good habit is to find the general solution first, simplify it, and only then substitute the condition to evaluate the constant.

Common errors

Integrating before separating: You cannot integrate $\dfrac{dy}{dx}$ with respect to $x$ while $y$ terms are mixed in. Separate first so each side has one variable.
Forgetting the constant of integration: Without $+C$ there is no family of solutions and no way to apply the initial condition.
Adding the constant on both sides: One combined constant suffices; writing $C_1$ and $C_2$ separately just clutters the work.
Dropping the modulus when integrating $1/y$: $\int \dfrac{1}{y}\,dy = \ln|y| + C$ . The modulus matters before you exponentiate.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2023-style3 marks

Calculator-free. The rate of change of the area of a shadow is

\dfrac{dA}{dt} = -2t + 150\cos t\sin^2 t

, where

A

is in square metres and

t

in hours. At sunrise

t = 0

and

A = 375

. Solve for

A

in terms of

t

Show worked answer →

The right side depends only on $t$ , so integrate directly.

$\int -2t\,dt = -t^2$ . For $\int 150\cos t\sin^2 t\,dt$ , let $u = \sin t$ , $du = \cos t\,dt$ , giving $150\int u^2\,du = 50\sin^3 t$ . [1 mark]

So $A = -t^2 + 50\sin^3 t + C$ . [1 mark]

Apply $A = 375$ at $t = 0$ : $375 = 0 + 0 + C$ , so $C = 375$ . Hence $A = -t^2 + 50\sin^3 t + 375$ . [1 mark]

SACE 2022-style3 marks

Calculator-free. Solve

\dfrac{dy}{dx} = \dfrac{2x}{y}

given

y = 3

when

x = 0

Show worked answer →

Separate: $y\,dy = 2x\,dx$ . Integrate both sides:

\frac{y^2}{2} = x^2 + C.

[1 mark]

Apply $y = 3$ at $x = 0$ : $\dfrac{9}{2} = C$ . [1 mark]

So $y^2 = 2x^2 + 9$ , and taking the positive branch through $(0, 3)$ gives $y = \sqrt{2x^2 + 9}$ . [1 mark]

SACE 2021-style3 marks

Calculator-free. Solve

\dfrac{dN}{dt} = -0.05N

with

N = 200

t = 0

Show worked answer →

Separate and integrate: $\int \dfrac{1}{N}\,dN = \int -0.05\,dt$ , so $\ln|N| = -0.05t + C$ . [1 mark]

Exponentiate: $N = A e^{-0.05t}$ . [1 mark]

Apply $N = 200$ at $t = 0$ : $A = 200$ , so $N = 200e^{-0.05t}$ . [1 mark]

Practise this

Sources & how we know this

Stage 2 Specialist Mathematics subject outline (2025 onwards) — SACE Board
Stage 2 Specialist Mathematics: external assessment — SACE Board

What this dot point is asking

The separation method

A worked general-then-particular solution

Solve dydx=2xy\dfrac{dy}{dx} = \dfrac{2x}{y}dxdy​=y2x​ given y=3y = 3y=3 when x=0x = 0x=0

Step 1: Separate variables

Step 2: Integrate both sides

Step 3: Apply the initial condition to find CCC

Step 4: Rearrange for yyy

Recognising a separable equation

Exponential growth and decay

Solve dNdt=−0.05N\dfrac{dN}{dt} = -0.05NdtdN​=−0.05N with N=200N = 200N=200 at t=0t = 0t=0

Step 1: Separate variables

Step 2: Integrate both sides and exponentiate

Step 3: Apply the initial condition N=200N = 200N=200 at t=0t = 0t=0

General versus particular solutions

Exam-style questions

Practise this

Sources & how we know this

Solve $\dfrac{dy}{dx} = \dfrac{2x}{y}$ given $y = 3$ when $x = 0$

Step 3: Apply the initial condition to find $C$

Step 4: Rearrange for $y$

Solve $\dfrac{dN}{dt} = -0.05N$ with $N = 200$ at $t = 0$

Step 3: Apply the initial condition $N = 200$ at $t = 0$