SACE Board SACE 2022-style practice: Calculator-free. Solve dydx = 2xy given y = 3 when x = 0.

Separate: y\,dy = 2x\,dx. Integrate both sides: $(y^2)/(2) = x^2 + C.$ [1 mark] Apply y = 3 at x = 0: 92 = 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]

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SASpecialist MathematicsSyllabus dot point

How do we solve a differential equation by separating the two variables onto opposite sides?

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

Solving first-order separable differential equations by separating variables and integrating both sides, applying initial conditions to find the particular solution, and handling the constant of integration.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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.

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}.

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.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20233 marksCalculator-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 20223 marksCalculator-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 20213 marksCalculator-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]