Year 12 Maths Extension 1 (2027) Further applications of calculus

Solve dy/dx = f(x)g(y)

Q: What is separation of variables?

A method for dy by dx equals f of x times g of y: rewrite as one over g of y dy equals f of x dx and integrate both sides.

Q: When can you separate variables?

When the right side factors into a function of x times a function of y.

Q: How many constants do you need?

Just one; combine both integration constants into a single C.

Q: When do you apply the initial condition?

After integrating, to determine the constant C.

20 practice questions 2 video lessons Theory + worked examples

Master separation of variables for NSW Year 12 Mathematics Extension 1. When a derivative factors into a function of x times a function of y, the variables can be separated and each side integrated.

You will learn to separate the variables, integrate both sides, and apply an initial condition to find the particular solution through a given point — the key method for solving separable differential equations in Extension 1.

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Practice questions

Every question with a fully worked solution.

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Separable differential equations Watch
Separable First Order Differential Equations - Basic Introduction Watch

Theory

When $\frac{d y}{d x} = f (x) g (y)$ , the variables separate: $\int \frac{1}{g (y)} d y = \int f (x) d x$ . This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-05) integrates both sides and applies an initial condition.

When $\frac{d y}{d x} = f (x) g (y)$ , the variables separate: gather the $y$ 's with $d y$ and the $x$ 's with $d x$ , then integrate both sides.

This gives $\int \frac{1}{g (y)} d y = \int f (x) d x$ . One constant $C$ suffices; put it on the $x$ -side, then solve for $y$ and apply any initial condition.

Sometimes you must rearrange first so each side has a single variable — for example, $\sec x \frac{d y}{d x} = y$ becomes $\frac{d y}{d x} = y \cos x$ .

NESA link. Part of the Year 12 Further applications of calculus focus area, outcome ME1-12-05 ("applies calculus to solve problems involving polynomials, further rates of change, areas and volumes and differential equations") with MAO-WM-01. A syllabus example: given $\frac{d y}{d x} = - \frac{x}{y e^{x^{2}}}$ , find the solution through $(0, 1)$ .

Separate, integrate both sides, add

C

, then solve for

y

\frac{d y}{d x} = x y, y (0) = 2

gives the particular solution

y = 2 e^{x^{2} / 2}

\frac{d y}{d x} = f (x) g (y) ⟹ \int \frac{1}{g (y)} d y = \int f (x) d x .

integral of 1/g(y) dy = integral of f(x) dx

One constant is enough. Combine the two constants of integration into a single $C$ on the $x$ -side, solve for $y$ explicitly where possible, and apply the initial condition last.

How to separate variables

Rearrange so $\frac{d y}{d x}$ is a product $f (x) g (y)$ .
Divide by $g (y)$ and multiply by $d x$ : $\frac{1}{g (y)} d y = f (x) d x$ .
Integrate both sides, with one constant $C$ .
Solve for $y$ and apply any initial condition.

Example 1 — A quotient

Find the general solution of

\frac{d y}{d x} = \frac{x}{y}

Solution

$\int y d y$	$=$	$\int x d x$
$\frac{y^{2}}{2}$	$=$	$\frac{x^{2}}{2} + C_{1} \Rightarrow y^{2} = x^{2} + C$

y^2 = x^2 + C

$y^{2} = x^{2} + C$ .

Example 2 — Giving arctan

Find the general solution of

\frac{d y}{d x} = 2 x (1 + y^{2})

Solution

$\int \frac{d y}{1 + y^{2}}$	$=$	$\int 2 x d x$
$\tan^{- 1} y$	$=$	$x^{2} + C \Rightarrow y = \tan (x^{2} + C)$

y = tan(x^2 + C)

$y = \tan (x^{2} + C)$ .

Example 3 — With a condition

Solve

\frac{d y}{d x} = x y

given

y (0) = 2

Solution

$\ln \| y \|$	$=$	$\frac{x^{2}}{2} + C$
$y (0) = 2$	$\Rightarrow$	$\ln 2 = C$

y = 2 e^(x^2/2)

$y = 2 e^{x^{2} / 2}$ .

Example 4 — A trig factor

Solve

\frac{d y}{d x} = y^{2} \cos x

given

y (0) = 1

Solution

$- \frac{1}{y}$	$=$	$\sin x + C$
$y (0) = 1$	$\Rightarrow$	$C = - 1$

y = 1/(1 - sin x)

$y = \frac{1}{1 - \sin x}$ .

Common pitfalls

Not fully separating. Each side must contain a single variable before integrating — rearrange if needed.

Two constants. One

C

suffices; put it on the

x

-side.

Leaving it implicit. Solve for

y

explicitly where possible.

Applying the condition early. Integrate first, then use the initial condition to fix

C

Frequently asked questions

What is separation of variables?

A method for $\frac{d y}{d x} = f (x) g (y)$ : rewrite as $\frac{1}{g (y)} d y = f (x) d x$ and integrate both sides.

When can you separate variables?

When the right side factors into a function of $x$ times a function of $y$ .

How many constants do you need?

Just one — combine both integration constants into a single $C$ .

Do you always solve for y?

Solve explicitly where possible; otherwise leave the relation implicit.

When do you apply the initial condition?

After integrating, to determine the constant $C$ .

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Solve dy/dx = f(x)g(y)

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How to separate variables

Common pitfalls

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Theory