Separation of variables | Year 12

01 / LEARN

Understand the idea.

Separate variables with care

A separable equation has the form $y^{'} = f (x) g (y)$ . Where $g (y) \neq 0$ , rearrange it to $\frac{1}{g (y)} d y = f (x) d x$ and integrate each side in its own variable. Combine the two arbitrary constants into one.

Before dividing, inspect $g (y) = 0$ . Its roots may give constant solutions that division would remove. Check those functions in the original equation and retain them when finding all solutions, even if the given initial condition later rules them out.

Choose a branch and verify

After integration, solve for $y$ if possible and apply the initial condition. From $ln ∣ y ∣ = H (x) + C$ , exponentiation permits a signed multiplier in $y = A e^{H (x)}$ ; replacing it automatically with a positive constant would lose negative solutions.

Choose the maximal interval containing the initial point where the original equation and the solution formula make sense. A denominator zero may stop continuation at a finite input. Differentiate the explicit solution and verify both the equation and initial value.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

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

Worked solution

1
Check the divided-out solution.
$y = 0 \Rightarrow y^{'} = 0 = 2 x \cdot 0$
The zero function is an equilibrium, but it fails the specified initial value 3.
2
Separate and integrate the nonzero branch.
$\frac{d y}{y} = 2 x d x, ln ∣ y ∣ = x^{2} + C$
On a nonzero solution branch, dividing by y separates the variables.
3
Exponentiate and use the initial value.
$y = A e^{x^{2}}, 3 = A e^{0} \Rightarrow A = 3$
A signed multiplier represents the branches before the initial condition selects the positive one.
4
Verify the solution.
$\begin{array}{c} y = 3 e^{x^{2}}, y^{'} = 6 x e^{x^{2}} = 2 x y \\ y (0) = 3 \end{array}$
The formula is defined for every real x and satisfies both requirements.

Answer

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

EXAMPLE 02

Worked example

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

Worked solution

1
Separate the nonzero solution.
$y^{- 2} d y = d x, - \frac{1}{y} = x + C$
The initial value is nonzero; the equilibrium y equals zero does not meet it.
2
Apply the initial condition.
$- \frac{1}{2} = 0 + C, - \frac{1}{y} = x - \frac{1}{2}$
Substituting y(0) equals 2 determines the integration constant.
3
Solve and select the initial interval.
$y = \frac{2}{1 - 2 x}, x < \frac{1}{2}$
The denominator vanishes at one half; the connected interval containing zero lies to its left.
4
Check the derivative.
$y^{'} = \frac{4}{{(1 - 2 x)}^{2}} = y^{2}, y (0) = 2$
The rational function satisfies the differential equation and initial value on the stated interval.

Answer

$y = 2 / (1 - 2 x)$ on $x < 1 / 2$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

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

Separating gives $\frac{d y}{y} = - 3 d x$ , hence $ln ∣ y ∣ = - 3 x + C$ and $y = A e^{- 3 x}$ . The condition $y (0) = 5$ fixes $A = 5$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Solve $y^{'} = 4 x y$ with $y (0) = 2$ .
2 marks
Solve $y^{'} = \frac{x}{y}$ 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 $y^{'} = \frac{y}{x}$ 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
$\frac{d y}{y} = 4 x d x$ gives $ln ∣ y ∣ = 2 x^{2} + C$ .
Initial data gives multiplier $2$ .
$y = 2 e^{2 x^{2}}$
Solve $y^{'} = \frac{x}{y}$ 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
$\frac{d y}{1 + y^{2}} = d x$ gives $arctan y = x + C$ .
$C = 0$ and tangent is finite only between its adjacent poles.
$y = tan x$ , $- \frac{π}{2} < x < \frac{π}{2}$ .
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 $- \frac{1}{y} = - x + C$ .
$C = - 1$ , giving the result and domain.
$y = \frac{1}{x + 1}$ on $x > - 1$ .
Solve $y^{'} = \frac{y}{x}$ with $y (1) = - 2$ , on $x > 0$ .
2 marks
$\frac{d y}{y} = \frac{d x}{x}$ 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 ↗

In this lesson

Understand the idea.

Separate variables with care

Choose a branch and verify

See the method in action.

Worked example

Check the divided-out solution.

Separate and integrate the nonzero branch.

Exponentiate and use the initial value.

Verify the solution.

Worked example

Separate the nonzero solution.

Apply the initial condition.

Solve and select the initial interval.

Check the derivative.

Try it yourself.

Check your understanding

Keep practising