Mathematics 9709 · Pure Mathematics 3

Solve a given DE and rearrange to the requested form

This is the standard exam question, worth about 6 to 11 marks on nearly every P3 paper.
You are given a differential equation and a boundary condition. You must solve it and rearrange to the exact form asked for.
Same steps as the last topic, just a different thing to integrate.

How to spot it: you get a DE plus a condition plus a form to aim for (“ $y$ in terms of $x$ ”, “ $y^2$ in terms of $x$ ”, “ $t$ in terms of $x$ ”) $\;\Rightarrow\;$ separate, integrate, find $C$ , rearrange.

The order that matches the mark scheme

Always find the constant first and rearrange last. You plug into the simple $\ln$ form, not a messy rearranged one.
It is the order the mark scheme rewards.

Separate so each side has one variable. (the first mark, a gate)
Integrate each side with the right standard integral. (the bulk of the marks)
Substitute the condition to evaluate $C$ now.
Rearrange to the exact requested subject (un-log, take the right root, isolate).

The integration itself (turning $\sec^2 3y$ into $\frac13\tan 3y$ , $\frac1y$ into $\ln|y|$ , $e^{4x}$ into $\frac14 e^{4x}$ ) is the integration chapter's job, so we do not re-teach it here.
Go back to that chapter if a side looks unfamiliar.

A trig side and an exponential side

Worked example

9709/33 Oct–Nov 2023 Q8: $e^{4x}\dfrac{dy}{dx}=\cos^2 3y$ , $y=0$ when $x=2$ [7]

To get each variable on its own side, divide by $\cos^2 3y$ (which makes a $\sec^2$ ) and by $e^{4x}$ (which makes an $e^{-4x}$ ).

Separate: $\sec^2 3y\,dy=e^{-4x}\,dx$ .
Integrate each side. Left: $\int\sec^2 3y\,dy=\tfrac13\tan 3y$ . Right: $\int e^{-4x}\,dx=-\tfrac14 e^{-4x}$ . So $\tfrac13\tan 3y=-\tfrac14 e^{-4x}+C$ .
Find $C$ : put $y=0,\;x=2$ . Since $\tan 0=0$ : $0=-\tfrac14 e^{-8}+C$ , so $C=\tfrac14 e^{-8}$ .
Rearrange to $y$ : $\tan 3y=\tfrac34\big(e^{-8}-e^{-4x}\big)$ , then $3y=\tan^{-1}(\cdots)$ .

Answer

y=\tfrac13\tan^{-1}\!\Big[\tfrac34\big(e^{-8}-e^{-4x}\big)\Big]

Checked by substitution. The mark-scheme order here is exactly: separate (B1), the two integrated terms (M1/A1 each), substitute the condition (M1), final form (A1).

Common mistake

\int\sec^2 a y\,dy=\tfrac{1}{a}\tan a y

and

\int e^{ax}\,dx=\tfrac{1}{a}e^{ax}

: it is easy to forget the

\tfrac1a

. For

\cos^2 3y

the factor is

\tfrac13

; for

e^{-4x}

it is

-\tfrac14

The other everyday side: a ½ln from a fraction in x

Not every $x$ -side is a clean $e^{ax}$ .
A very common one is a fraction whose top is (a multiple of) the derivative of the bottom. That integrates straight to a log, with no partial fractions.

How to spot it: a side like $\dfrac{x}{1+x^2}$ , whose top is half the derivative of the bottom $\;\Rightarrow\;$ $\displaystyle\int\frac{x}{1+x^2}\,dx=\tfrac12\ln(1+x^2)$ . No splitting needed; the half comes from the missing $2$ .

Worked example

9709/31 May–Jun 2022 Q4: $\dfrac{dy}{dx}=\dfrac{xy}{1+x^2}$ , $y=2$ when $x=0$ ; simplified $y$ in terms of $x$ [7]

Both sides are easy: a $\tfrac1y$ on the left, the $\tfrac{x}{1+x^2}$ log-pattern on the right.

Separate: $\dfrac{1}{y}\,dy=\dfrac{x}{1+x^2}\,dx$ .
Integrate each side. Left: $\ln y$ (the question implies $y>0$ ). Right: top is half the derivative of $1+x^2$ , so $\tfrac12\ln(1+x^2)$ . Thus $\ln y=\tfrac12\ln(1+x^2)+C$ .
Find $C$ : at $x=0,\;y=2$ , $\ln 2=\tfrac12\ln 1+C=C$ , so $C=\ln 2$ .
Un-log and simplify. $\ln y=\ln 2+\tfrac12\ln(1+x^2)=\ln\!\big(2(1+x^2)^{1/2}\big)$ , so the logs cancel.

Answer

y=2\sqrt{1+x^2}

Checked against the mark scheme. The word “simplified” tells you to actually do the $e^{\ln(\ldots)}$ step; stopping at $\ln y=\ldots$ loses the final A1. (9709/31 M/J 2022 Q4)

When the tail asks for an exact value at a point

Worked example

9709/32 Feb–Mar 2022 Q9: $(x+1)(3x+1)\dfrac{dy}{dx}=y$ , $y=1$ when $x=1$ ; exact $y$ at $x=3$ [9]

The $y$ -side gives a clean $\ln$ , and the $x$ -side has a factorised bottom, so split it into partial fractions with the Ch7 method.

Separate: $\dfrac{1}{y}\,dy=\dfrac{1}{(x+1)(3x+1)}\,dx$ .
Decompose the right side (Ch7): $\dfrac{1}{(x+1)(3x+1)}=-\dfrac{1/2}{x+1}+\dfrac{3/2}{3x+1}$ ( $A=-\tfrac12,\;B=\tfrac32$ ).
Integrate: $\ln|y|=-\tfrac12\ln|x+1|+\tfrac12\ln|3x+1|+C$ ( $\int\frac{3/2}{3x+1}=\frac12\ln|3x+1|$ ).
Find $C$ : at $x=1,\;y=1$ , $\ln 1=0$ , so $0=-\tfrac12\ln 2+\tfrac12\ln 4+C$ , giving $C=-\tfrac12\ln 2$ .
Substitute $x=3$ : $\ln y=-\tfrac12\ln 4+\tfrac12\ln 10-\tfrac12\ln 2=\tfrac12\ln\tfrac{10}{8}=\tfrac12\ln\tfrac54$ .

Answer

y=\sqrt{\tfrac54}=\dfrac{\sqrt5}{2}

Checked. Notice the answer is an exact surd; a decimal here would lose the final accuracy mark.

The mark-scheme pitfalls, banked

Common mistake

Four ways this question type loses marks:

Constant outside the exponential. $y=A\,e^{(\ldots)}$ , never $y=e^{(\ldots)}+C$ (find $C$ at the $\ln$ stage).
Decimal instead of exact. If the answer is a surd or an $\ln$ , leave it exact. A rounded decimal loses the accuracy mark.
Wrong subject. If it asks for $y^2$ in terms of $x$ , stop at $y^2$ ; if $t$ in terms of $x$ , get $t$ on its own.
Sign lost on a square root. When you square-root, the condition tells you which sign to keep, so do not guess.

Your turn— tap to reveal the worked answer (9709/33 Oct–Nov 2020 Q8)

$x\dfrac{dy}{dx}=(1-2x^2)y$ for $x>0$ , $y=1$ when $x=1$ ; find $y$ in terms of $x$ . Separate to $\frac1y\,dy=\big(\frac1x-2x\big)dx$ ; integrate to $\ln y=\ln x-x^2+C$ ; at $x=1$ , $0=0-1+C$ so $C=1$ ; un-log. (9709/33 O/N 2020 Q8)

Answer

y=x\,e^{\,1-x^2}

Your turn— tap to reveal the worked answer (9709/31 May–Jun 2025 Q10)

$\dfrac{dy}{dx}=\dfrac{x^3+5x^2-2x-15}{6y(x^2-3)}$ , $y=2$ when $x=2$ ; find $y^2$ in terms of $x$ . Separating gives $6y\,dy=\frac{x^3+5x^2-2x-15}{x^2-3}\,dx$ . The right side is top-heavy, so divide first: the quotient is $x+5$ with remainder $x$ , so it becomes $x+5+\dfrac{x}{x^2-3}$ (do not drop the remainder; that $\frac{x}{x^2-3}$ integrates to a log). Integrate: $3y^2=\tfrac12 x^2+5x+\tfrac12\ln(x^2-3)+C$ . At $x=2$ : $12=2+10+\tfrac12\ln 1+C$ , so $C=0$ . Stop at $y^2$ , the form asked for; do not square-root. (9709/31 M/J 2025 Q10)

Answer

y^2=\tfrac16 x^2+\tfrac{5}{3}x+\tfrac16\ln(x^2-3)

PreviousWhat a differential equation is, and how to solve one NextWhen the integral needs partial fractions

Solve a given DE and rearrange to the requested form

This is the standard exam question, worth about 6 to 11 marks on nearly every P3 paper.
You are given a differential equation and a boundary condition. You must solve it and rearrange to the exact form asked for.
Same steps as the last topic, just a different thing to integrate.

The order that matches the mark scheme

Always find the constant first and rearrange last. You plug into the simple $\ln$ form, not a messy rearranged one.
It is the order the mark scheme rewards.

Separate so each side has one variable. (the first mark, a gate)
Integrate each side with the right standard integral. (the bulk of the marks)
Substitute the condition to evaluate $C$ now.
Rearrange to the exact requested subject (un-log, take the right root, isolate).

The integration itself (turning $\sec^2 3y$ into $\frac13\tan 3y$ , $\frac1y$ into $\ln|y|$ , $e^{4x}$ into $\frac14 e^{4x}$ ) is the integration chapter's job, so we do not re-teach it here.
Go back to that chapter if a side looks unfamiliar.

A trig side and an exponential side

Worked example

9709/33 Oct–Nov 2023 Q8: $e^{4x}\dfrac{dy}{dx}=\cos^2 3y$ , $y=0$ when $x=2$ [7]

To get each variable on its own side, divide by $\cos^2 3y$ (which makes a $\sec^2$ ) and by $e^{4x}$ (which makes an $e^{-4x}$ ).

Separate: $\sec^2 3y\,dy=e^{-4x}\,dx$ .
Integrate each side. Left: $\int\sec^2 3y\,dy=\tfrac13\tan 3y$ . Right: $\int e^{-4x}\,dx=-\tfrac14 e^{-4x}$ . So $\tfrac13\tan 3y=-\tfrac14 e^{-4x}+C$ .
Find $C$ : put $y=0,\;x=2$ . Since $\tan 0=0$ : $0=-\tfrac14 e^{-8}+C$ , so $C=\tfrac14 e^{-8}$ .
Rearrange to $y$ : $\tan 3y=\tfrac34\big(e^{-8}-e^{-4x}\big)$ , then $3y=\tan^{-1}(\cdots)$ .

Answer

y=\tfrac13\tan^{-1}\!\Big[\tfrac34\big(e^{-8}-e^{-4x}\big)\Big]

Checked by substitution. The mark-scheme order here is exactly: separate (B1), the two integrated terms (M1/A1 each), substitute the condition (M1), final form (A1).

Common mistake

\int\sec^2 a y\,dy=\tfrac{1}{a}\tan a y

and

\int e^{ax}\,dx=\tfrac{1}{a}e^{ax}

: it is easy to forget the

\tfrac1a

. For

\cos^2 3y

the factor is

\tfrac13

; for

e^{-4x}

it is

-\tfrac14

The other everyday side: a ½ln from a fraction in x

Not every $x$ -side is a clean $e^{ax}$ .
A very common one is a fraction whose top is (a multiple of) the derivative of the bottom. That integrates straight to a log, with no partial fractions.

Worked example

9709/31 May–Jun 2022 Q4: $\dfrac{dy}{dx}=\dfrac{xy}{1+x^2}$ , $y=2$ when $x=0$ ; simplified $y$ in terms of $x$ [7]

Both sides are easy: a $\tfrac1y$ on the left, the $\tfrac{x}{1+x^2}$ log-pattern on the right.

Separate: $\dfrac{1}{y}\,dy=\dfrac{x}{1+x^2}\,dx$ .
Integrate each side. Left: $\ln y$ (the question implies $y>0$ ). Right: top is half the derivative of $1+x^2$ , so $\tfrac12\ln(1+x^2)$ . Thus $\ln y=\tfrac12\ln(1+x^2)+C$ .
Find $C$ : at $x=0,\;y=2$ , $\ln 2=\tfrac12\ln 1+C=C$ , so $C=\ln 2$ .
Un-log and simplify. $\ln y=\ln 2+\tfrac12\ln(1+x^2)=\ln\!\big(2(1+x^2)^{1/2}\big)$ , so the logs cancel.

Answer

y=2\sqrt{1+x^2}

Checked against the mark scheme. The word “simplified” tells you to actually do the $e^{\ln(\ldots)}$ step; stopping at $\ln y=\ldots$ loses the final A1. (9709/31 M/J 2022 Q4)

When the tail asks for an exact value at a point

Worked example

9709/32 Feb–Mar 2022 Q9: $(x+1)(3x+1)\dfrac{dy}{dx}=y$ , $y=1$ when $x=1$ ; exact $y$ at $x=3$ [9]

The $y$ -side gives a clean $\ln$ , and the $x$ -side has a factorised bottom, so split it into partial fractions with the Ch7 method.

Separate: $\dfrac{1}{y}\,dy=\dfrac{1}{(x+1)(3x+1)}\,dx$ .
Decompose the right side (Ch7): $\dfrac{1}{(x+1)(3x+1)}=-\dfrac{1/2}{x+1}+\dfrac{3/2}{3x+1}$ ( $A=-\tfrac12,\;B=\tfrac32$ ).
Integrate: $\ln|y|=-\tfrac12\ln|x+1|+\tfrac12\ln|3x+1|+C$ ( $\int\frac{3/2}{3x+1}=\frac12\ln|3x+1|$ ).
Find $C$ : at $x=1,\;y=1$ , $\ln 1=0$ , so $0=-\tfrac12\ln 2+\tfrac12\ln 4+C$ , giving $C=-\tfrac12\ln 2$ .
Substitute $x=3$ : $\ln y=-\tfrac12\ln 4+\tfrac12\ln 10-\tfrac12\ln 2=\tfrac12\ln\tfrac{10}{8}=\tfrac12\ln\tfrac54$ .

Answer

y=\sqrt{\tfrac54}=\dfrac{\sqrt5}{2}

Checked. Notice the answer is an exact surd; a decimal here would lose the final accuracy mark.

The mark-scheme pitfalls, banked

Common mistake

Four ways this question type loses marks:

Constant outside the exponential. $y=A\,e^{(\ldots)}$ , never $y=e^{(\ldots)}+C$ (find $C$ at the $\ln$ stage).
Decimal instead of exact. If the answer is a surd or an $\ln$ , leave it exact. A rounded decimal loses the accuracy mark.
Wrong subject. If it asks for $y^2$ in terms of $x$ , stop at $y^2$ ; if $t$ in terms of $x$ , get $t$ on its own.
Sign lost on a square root. When you square-root, the condition tells you which sign to keep, so do not guess.

Your turn— tap to reveal the worked answer (9709/33 Oct–Nov 2020 Q8)

Answer

y=x\,e^{\,1-x^2}

Your turn— tap to reveal the worked answer (9709/31 May–Jun 2025 Q10)

Answer

y^2=\tfrac16 x^2+\tfrac{5}{3}x+\tfrac16\ln(x^2-3)