2023 HSC Q12 HSC: Solve the differential equation $\frac{dy}{dx} = \frac{x}{y}$ with $y(0) = 2$.

NSWMaths Extension 1Syllabus dot point

How do we solve a first-order differential equation when the variables separate?

Solve separable first-order differential equations of the form $\frac{dy}{dx} = f(x) g(y)$ by separating variables and integrating both sides

A focused answer to the HSC Maths Extension 1 dot point on separable differential equations. The separation-of-variables method, integration on both sides, applying initial conditions, and standard models including Newton's law of cooling, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy8 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to solve a first-order differential equation of the form $\frac{dy}{dx} = f(x) g(y)$ by separating the variables, integrating each side, and applying any given initial condition to find the constant of integration.

The answer

The separation method

For an equation $\frac{dy}{dx} = f(x) g(y)$ :

Rewrite as $\frac{dy}{g(y)} = f(x) \, dx$ .
Integrate both sides: $\int \frac{dy}{g(y)} = \int f(x) \, dx + C$ .
Solve for $y$ in terms of $x$ (or leave in implicit form if cleaner).
Apply any initial condition $y(x_0) = y_0$ to determine $C$ .

If $g(y_0) = 0$ , then $y = y_0$ is a constant solution (a fixed point of the equation). This must be checked separately because the separation step would have divided by zero.

When separation is possible

A first-order ODE separates if and only if you can write $\frac{dy}{dx} = f(x) g(y)$ for some functions $f$ and $g$ . Examples that separate:

IMATH_18 .
IMATH_19 .
IMATH_20 .

Examples that do not separate:

IMATH_21 (sum, not product).
IMATH_22 (cannot factor cleanly).

Extension 1 problems are always separable; non-separable equations belong to Extension 2.

The constant of integration

Always include a constant after integration. Usually it is cleaner to combine the constants from both sides into one $C$ on the right.

If the question gives an initial condition, substitute to find $C$ . If not, leave the answer in terms of $C$ .

Newton's law of cooling

A body cools toward an ambient temperature $T_a$ at a rate proportional to the temperature difference:

\frac{dT}{dt} = -k (T - T_a),

where $k > 0$ .

Separate: $\frac{dT}{T - T_a} = -k \, dt$ .

Integrate: $\ln |T - T_a| = -k t + C$ .

So $T - T_a = A e^{-k t}$ for some constant $A$ , giving

T(t) = T_a + (T_0 - T_a) e^{-k t}.

The temperature approaches $T_a$ as $t \to \infty$ .

Implicit versus explicit solutions

Sometimes the integrated equation cannot be solved cleanly for $y$ . In that case, leave it implicit. For example,

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

This is the family of circles $x^2 + y^2 = C$ , which is the implicit solution.

Worked example

Basic separable

Solve $\frac{dy}{dx} = 3 x^2 y$ .

$\frac{dy}{y} = 3 x^2 \, dx$ .

$\ln |y| = x^3 + C$ , so $y = A e^{x^3}$ (with $A = \pm e^C$ ).

With initial condition

Solve $\frac{dy}{dx} = -2 x y^2$ with $y(0) = 1$ .

$\frac{dy}{y^2} = -2 x \, dx$ .

$-\frac{1}{y} = -x^2 + C$ , so $\frac{1}{y} = x^2 - C = x^2 + D$ (with $D = -C$ ).

Apply $y(0) = 1$ : $\frac{1}{1} = 0 + D$ , so $D = 1$ .

$\frac{1}{y} = x^2 + 1$ , so $y = \frac{1}{x^2 + 1}$ .

Trigonometric right side

Solve $\frac{dy}{dx} = \frac{\sin x}{y}$ .

$y \, dy = \sin x \, dx$ .

$\frac{y^2}{2} = -\cos x + C$ , so $y^2 = -2 \cos x + C'$ (with $C' = 2 C$ ).

Newton's cooling

A cup of coffee at $90^\circ$ C is placed in a room at $20^\circ$ C. After $10$ minutes the temperature is $60^\circ$ C. Find the temperature after $30$ minutes.

$T(t) = 20 + 70 e^{-k t}$ . At $t = 10$ : $60 = 20 + 70 e^{-10 k}$ , so $\frac{40}{70} = e^{-10 k}$ , giving $k = -\frac{1}{10} \ln \frac{4}{7} \approx 0.0560$ .

At $t = 30$ : $T = 20 + 70 e^{-0.0560 \cdot 30} = 20 + 70 e^{-1.68} \approx 20 + 70 \cdot 0.186 \approx 33^\circ$ C.

Logistic-like

Solve $\frac{dy}{dx} = y(1 - y)$ .

$\frac{dy}{y(1 - y)} = dx$ .

Partial fractions: $\frac{1}{y(1 - y)} = \frac{1}{y} + \frac{1}{1 - y}$ .

$\int \frac{dy}{y(1 - y)} = \ln |y| - \ln |1 - y| = \ln \left| \frac{y}{1 - y} \right|$ .

So $\ln \left| \frac{y}{1 - y} \right| = x + C$ , giving $\frac{y}{1 - y} = A e^x$ .

Solve: $y = \frac{A e^x}{1 + A e^x}$ , the logistic curve.

Implicit form is enough

Solve $\frac{dy}{dx} = -\frac{y}{x}$ with $y(1) = 2$ .

$\frac{dy}{y} = -\frac{dx}{x}$ .

$\ln |y| = -\ln |x| + C$ , so $\ln |x y| = C$ , so $x y = A$ .

At $(1, 2)$ : $A = 2$ . So $y = \frac{2}{x}$ .

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2023 HSC Q124 marksSolve the differential equation

\frac{dy}{dx} = \frac{x}{y}

with

y(0) = 2

Show worked answer →

Separate variables: $y \, dy = x \, dx$ .

Integrate both sides: $\frac{y^2}{2} = \frac{x^2}{2} + C$ , so $y^2 = x^2 + C'$ (writing $C'$ for $2 C$ ).

Apply $y(0) = 2$ : $4 = 0 + C'$ , so $C' = 4$ .

$y^2 = x^2 + 4$ , so $y = \sqrt{x^2 + 4}$ (positive root because $y(0) = 2 > 0$ and continuity).

Markers reward the separation, the integration on both sides, applying the initial condition to find the constant, and choosing the correct sign for the square root.