Separation of variables | Year 12

01 / LEARN

The idea, step by step

Separate variables with care

An equation is separable if it can be written $d y / d x = f (x) g (y)$ . Where $g (y) \neq 0$ , rearrange to $1 / g (y) d y = f (x) d x$ and integrate each side with its own variable. A single combined arbitrary constant is enough after integration.

Dividing by $g (y)$ can lose solutions. Before dividing, solve $g (y) = 0$ and check whether the resulting constant functions satisfy the original equation. State these equilibria alongside the nonconstant family when finding all solutions.

Choose a branch and verify

After integration, solve for $y$ if possible. The identity $\int 1 / y d y = ln ∣ y ∣ + C$ is valid away from zero. Exponentiation gives a signed multiplicative constant; do not accidentally restrict all solutions to positive values.

Use initial data to fix the constant and choose the interval containing the initial point on which the expression is defined. Verify by differentiation. A solution with a finite denominator-zero point cannot be continued through that point using the same formula.

A common trap

Divide by y and conclude zero is not a solution.

Instead: Check the original equation at y=0 before division; a valid equilibrium may be lost by dividing.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Solve $y^{'} = 2 x y$ with $y (0) = 3$ .

For $y \neq 0$ , separate $d y / y = 2 x d x$ .
Integrate: $ln ∣ y ∣ = x^{2} + C$ , so $y = A e^{x^{2}}$ .
$y (0) = 3$ gives $A = 3$ . The separate equilibrium $y = 0$ does not meet this initial condition.

y = 3 e^{x^{2}}

Example 2

Worked example

Solve $y^{'} = y^{2}$ with $y (0) = 2$ .

Separate: $y^{- 2} d y = d x$ . Integrate $- 1 / y = x + C$ .
At $x = 0$ , $C = - 1 / 2$ , so $y = 2 / (1 - 2 x)$ .
The interval containing zero stops at $x = 1 / 2$ .

y = 2 / (1 - 2 x)

x < 1 / 2

Check your understanding

Solve $y^{'} = - 3 y$ , $y (0) = 5$ .

Show answer

$y = 5 e^{- 3 x}$

Separation gives $ln ∣ y ∣ = - 3 x + C$ and the initial value fixes the multiplier.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Solve $y^{'} = 4 x y$ with $y (0) = 2$ .
2 marks
Solve $y^{'} = x / y$ with $y (0) = 3$ .
2 marks
Solve $y^{'} = (1 + y^{2})$ with $y (0) = 0$ . State the interval around zero.
2 marks
Find all constant solutions of $y^{'} = x (y - 2) (y + 1)$ .
2 marks
Solve $y^{'} = - y^{2}$ with $y (0) = 1$ .
2 marks
Solve $y^{'} = y / x$ with $y (1) = - 2$ , on $x > 0$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Solve $y^{'} = 4 x y$ with $y (0) = 2$ .
2 marks
$d y / y = 4 x d x$ gives $ln ∣ y ∣ = 2 x^{2} + C$ .
Initial data gives multiplier 2.
$y = 2 e^{2 x^{2}}$
Solve $y^{'} = x / y$ with $y (0) = 3$ .
2 marks
$y d y = x d x$ gives $y^{2} = x^{2} + C$ .
$C = 9$ ; the positive branch matches $y (0) = 3$ .
$y = \sqrt{x^{2} + 9}$
Solve $y^{'} = (1 + y^{2})$ with $y (0) = 0$ . State the interval around zero.
2 marks
$d y / (1 + y^{2}) = d x$ gives $arctan y = x + C$ .
$C = 0$ and tangent is finite only between its adjacent poles.
$y = tan x$ , $- π / 2 < x < π / 2$ .
Find all constant solutions of $y^{'} = x (y - 2) (y + 1)$ .
2 marks
A constant solution has zero derivative.
Both factors give values making the right side zero for every $x$ .
$y = 2$ and $y = - 1$ .
Solve $y^{'} = - y^{2}$ with $y (0) = 1$ .
2 marks
$y^{- 2} d y = - d x$ gives $- 1 / y = - x + C$ .
$C = - 1$ , giving the result and domain.
$y = 1 / (x + 1)$ on $x > - 1$ .
Solve $y^{'} = y / x$ with $y (1) = - 2$ , on $x > 0$ .
2 marks
$d y / y = d x / x$ gives $ln ∣ y ∣ = ln x + C$ .
$y = A x$ and $A = - 2$ ; retaining a signed multiplier is essential.
$y = - 2 x$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics topic 6.1–6.2, with the approved school-level rates, differential-equation and motion applications.

Official curriculum source ↗

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