Learn7357Paper 1 · PureIntegrationSeparating the variables

Integration · Subtopic 11 of 13

Separating the Variables

Gather every $y$ with $dy$ on one side and every $x$ with $dx$ on the other, integrate, and add one constant. The method that solves almost every Paper 3 differential equation.

Cambridge 9709 Paper 3Edexcel YMA01 WMA14Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 35 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Recognise when $\dfrac{dy}{dx}=f(x)\,g(y)$ can be separated.
Separate, integrate both sides and include one arbitrary constant.
Rearrange to $y=\ldots$ , handling logarithms and exponentials correctly.
Use a condition to choose one curve from the family of solutions.

Pause and think

$\dfrac{dy}{dx}=2xy$ . If you divide both sides by $y$ , what is the left-hand side the derivative of?

Section 2 of 14

Core idea

Start hereFrom a rule about change to a formula

"Grows in proportion to its size": $\frac{dy}{dx}=y$ . Which functions do that? We could guess $e^x$ , but guessing doesn't scale. If the $x$ 's and $y$ 's can be pulled to opposite sides, we can integrate each side and find every solution.

The big idea

If $\frac{dy}{dx}=f(x)g(y)$ : separate to $\int\frac1{g(y)}dy=\int f(x)\,dx$ , integrate both sides, add one constant. A given point fixes the constant (particular solution).

1. The method

Why separating is legitimate

$\frac1{g(y)}\frac{dy}{dx}=f(x)$
Divide by $g(y)$ .
$\int\frac1{g(y)}\frac{dy}{dx}dx=\int f(x)\,dx$
Integrate both sides with respect to $x$ .
$\int\frac1{g(y)}dy=\int f(x)\,dx$
Substitution rule: $\frac{dy}{dx}dx$ becomes $dy$ .

Worked example $\frac{dy}{dx}=y$

Find the general solution.

Goal: $y$ in terms of $x$ .
Strategy: Separate.

$\int\frac1y\,dy=\int1\,dx$
Separate.
$\ln|y|=x+c$
Integrate.
$|y|=e^{x+c}=e^ce^x\Rightarrow y=Ae^x$
Un-log; $A=\pm e^c$ .

Answer

y=Ae^x

✓ Check

\frac{d}{dx}Ae^x=Ae^x=y

✓.

dy/dx = y: the family y = A eˣ

A = -3.0

Every curve in the family obeys

\dfrac{dy}{dx}=y

. The value of

A

is fixed by one known point.

Worked example $\frac{dy}{dx}=xy$ , $y(0)=2$

Find the particular solution.

Goal: $y$ .
Strategy: Separate; find $c$ ; rearrange.

$\int\frac{dy}y=\int x\,dx\Rightarrow\ln y=\tfrac12x^2+c$
Integrate.
$\ln2=c$
Condition.
$y=e^{x^2/2+\ln2}=2e^{x^2/2}$
Un-log.

Answer

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

✓ Check

y'=2xe^{x^2/2}=xy

✓;

y(0)=2

✓.

Worked example $\frac{dy}{dx}=\frac{y^2}{2x^2}$ through $(1,1)$

Find $y$ .

Goal: $y$ explicitly.
Strategy: Separate; negative powers.

$\int y^{-2}dy=\int\tfrac12x^{-2}dx$
Separate.
$-\frac1y=-\frac1{2x}+c$
Integrate.
$-1=-\tfrac12+c\Rightarrow c=-\tfrac12$
Point.
$\frac1y=\frac{x+1}{2x}\Rightarrow y=\frac{2x}{x+1}$
Rearrange.

Answer

y=\frac{2x}{x+1}

✓ Check

y(1)=1

✓.

Common misconceptionTwo constants One constant is enough:

c_2-c_1

is just another constant.

Pause and think

Solve $\frac{dy}{dx}=\frac{\cos x}y$ , $y(0)=2$ .

Summary

Separate, integrate, one constant.
Use the condition, then rearrange.

General solution: Contains an arbitrary constant.
Particular solution: One curve fixed by a condition.

Solutions of dy/dx = y: one curve for each starting value

Separating:

\int\tfrac1y\,dy=\int1\,dx

, so

\ln y=x+c

and

y=Ae^x

. A boundary condition chooses one curve from the family.

HistoryLeibniz and the fraction that is not a fraction. Gottfried Wilhelm Leibniz invented the notation

\dfrac{dy}{dx}

and, around 1691, the method of separating the variables. His notation is so well designed that moving

dx

across "like a fraction" gives the right answer, even though

\dfrac{dy}{dx}

is really a limit, not a fraction.

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