Learn9MA0Paper 1 · Pure Mathematics 1IntegrationParticular solutions with harder integrals

Integration · Subtopic 13 of 14

Particular Solutions with Harder Integrals

Exam differential equations are integration questions in disguise. Once the variables are separated, one side often needs partial fractions, integration by parts or a trigonometric identity. Bring all of Paper 3 integration to bear, then pin down the constant.

Cambridge 9709 Paper 3Edexcel YMA01 WMA14Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 45 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Separate variables where one side needs partial fractions, by parts, a substitution or a trigonometric identity.
Find the constant from a condition and give $y$ explicitly in terms of $x$ .
Evaluate a particular solution at a point, and describe how $y$ behaves for large $x$ .
Recognise where a solution stops existing.

Pause and think

To solve $\dfrac{dy}{dx}=\dfrac{y}{(x+1)(x+2)}$ you need $\displaystyle\int\frac{dx}{(x+1)(x+2)}$ . Which technique from earlier in Paper 3 does that need?

Section 2 of 14

Core idea

Start hereSeparation needs all your integration skills

Separating is the easy part. The integrals that appear can need partial fractions, substitution, by parts or trig identities. This lesson is where all the integration techniques work together.

The big idea

After separating, use whatever integration method the integral needs. Write the constant as $\ln A$ to combine logs neatly. Find the constant before rearranging when that's easier.

1. Partial fractions

Worked exampleLogistic growth: $\frac{dx}{dt}=x(2-x)$ , $x(0)=1$

Solve for $x$ .

Goal: $x$ in terms of $t$ .
Strategy: Separate; partial fractions; combine logs.

$\int\frac{dx}{x(2-x)}=\int dt$
Separate.
$\frac1{x(2-x)}=\frac12\left(\frac1x+\frac1{2-x}\right)$
Partial fractions.
$\tfrac12[\ln x-\ln(2-x)]=t+c$
$\int\frac{dx}{2-x}=-\ln(2-x)$ .
$t=0,\ x=1:\ c=0$
$\ln1=0$ .
$\ln\frac x{2-x}=2t\Rightarrow\frac x{2-x}=e^{2t}\Rightarrow x=\frac{2e^{2t}}{1+e^{2t}}$
Rearrange.

Answer

x=\frac{2e^{2t}}{1+e^{2t}}=\frac2{1+e^{-2t}}

;

x\to2

t\to\infty

✓ Check

t=0

\frac22=1

✓.

2. The ln A trick

The particular solution y = 2x/(x + 1)

The curve passes through the given point and levels off towards

y=2

for large

x

Worked example $\frac{dy}{dx}=\frac{y+1}{x}$

Find the general solution.

Goal: $y$ .
Strategy: Write $c=\ln A$ .

$\int\frac{dy}{y+1}=\int\frac{dx}x$
Separate.
$\ln(y+1)=\ln x+\ln A=\ln(Ax)$
Constant as a log.
$y=Ax-1$
Un-log.

Answer

y=Ax-1

✓ Check

y'=A=\frac{Ax}x=\frac{y+1}x

✓.

3. By parts and trig

Worked example $\frac{dy}{dx}=xe^{x-y}$

Find the general solution.

Goal: $y$ .
Strategy: $e^{x-y}=e^xe^{-y}$ ; separate; by parts.

$\int e^ydy=\int xe^xdx$
Separate.
$e^y=xe^x-e^x+c$
By parts.
$y=\ln(xe^x-e^x+c)$
Take ln.

Answer

y=\ln(xe^x-e^x+c)

✓ CheckDifferentiate

e^y=xe^x-e^x+c

e^yy'=xe^x

⇒

y'=xe^{x-y}

✓.

Common misconception $e^{\ln x+c}\neq x+e^c$

e^{\ln x+c}=e^{\ln x}e^c=Ax

Pause and think

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

Summary

All integration methods apply.
$\ln A$ trick.

Logistic model: Growth that levels off at a limit.

HistoryThe Bernoulli family business. Jacob and Johann Bernoulli, brothers from Basel, solved many of the first differential equations in the 1690s, often by separating variables after clever substitutions. They were fierce rivals: when Johann set a famous problem in 1696 (the curve of fastest descent), Newton is said to have solved it overnight.

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