Edexcel Pure Paper 2 2024 Question 12 Solution

Edexcel Pure Paper 2 2024 Question 12 – Differential Equations Model

❓ The Question

🧠 Before you start

This one doesn’t stay in one place.

It starts off looking like a partial fractions question, which is fine. Then suddenly you’re separating variables. Then logs appear. And right at the end you’re thinking about what the model actually means.

So it’s not hard in one go — it’s just a bit… layered.

If anything, the risk is forgetting why you’re doing each step.

✏️ Working

Part (a)

You’re asked to split:

$\frac{1}{V(25 – V)}$

There’s not much choice here — it’s partial fractions.

So write it as:

$\frac{A}{V} + \frac{B}{25 – V}$

Multiply everything through:

$1 = A(25 - V) + BV$

Expand it:

$1 = 25A - AV + BV$

Which is:

$1 = 25A + (B - A)V$

Now just match terms.

From the constant:

$25A = 1 \Rightarrow A = \frac{1}{25}$

From the $V$ term:

$B - A = 0$

So:

$B = \frac{1}{25}$

So that part becomes:

$\frac{1}{25V} + \frac{1}{25(25 – V)}$

Nothing surprising there.

Part (b)

Now the equation:

$\frac{dV}{dt} = \frac{1}{10}V(25 – V)$

Rearrange first. Don’t jump ahead.

$\frac{1}{V(25 – V)} dV = \frac{1}{10} dt$

Now use what you just found:

$\left(\frac{1}{25V} + \frac{1}{25(25 – V)}\right)dV = \frac{1}{10}dt$

At this point, it’s just integration — but worth slowing down slightly.

Pull out the $\frac{1}{25}$ :

$\frac{1}{25}\left(\int \frac{1}{V} dV + \int \frac{1}{25 – V} dV\right)$

The first one is straightforward:

$\ln V$

The second one is where people slip — there’s a negative:

$-\ln(25 – V)$

So together:

$\frac{1}{25}[\ln V – \ln(25 – V)] = \frac{t}{10} + C$

Multiply through:

$\ln\left(\frac{V}{25 – V}\right) = \frac{5t}{2} + C$

Now use the starting value.

At $t = 0$ , $V = 20$ :

$\ln\left(\frac{20}{5}\right) = \ln 4$

So:

$C = \ln 4$

Now go back and use it.

We want when $V = 24$ :

$\ln\left(\frac{24}{1}\right) = \frac{5t}{2} + \ln 4$

Rearrange:

$\ln 24 – \ln 4 = \frac{5t}{2}$

$\ln 6 = \frac{5t}{2}$

So:

$t = \frac{2}{5}\ln 6$

Convert it:

$\approx 0.716 \text{ hours}$

$\approx 43.0 \text{ minutes}$

Part (c)

Go back to:

$\ln\left(\frac{V}{25 – V}\right) = \frac{5t}{2} + C$

Rewrite it:

$\frac{V}{25 – V} = Ae^{\frac{5t}{2}}$

If you rearrange this properly, it becomes:

$V = \frac{A}{e^{-kt} + B}$

That’s the form they want.

Part (d)

Now just think about what happens over time.

From:

$\frac{dV}{dt} = \frac{1}{10}V(25 – V)$

As $V$ gets closer to 25, that $(25 - V)$ term gets smaller.

So the growth slows… and eventually stops.

So the model levels off at:

$V = 25$

🎯 Final answers

(a)
$\frac{1}{25V} + \frac{1}{25(25 – V)}$

(b)
$43.0 \text{ minutes}$

(d)
$25$

$\frac{16}{27} – \frac{136}{27}e^{-3}$

⚠️ What tends to happen here

It’s usually not one big mistake.

More like:

missing the negative in the log
not using the initial condition properly
or getting to the log stage and not finishing it

💡 One small thing

If you see something like $\frac{1}{V(25 – V)}$ , it’s not random.

It’s almost always there to make separation work.

🚀 If this felt a bit long

That’s normal.

This question jumps between ideas, so it doesn’t feel as smooth as others.

Working through more of these helps support for A level maths, especially when different topics start linking together.

And if this type of question still feels inconsistent, it can help to work with a maths tutor and go through a few similar ones slowly.

🔗 Next Steps

← Question 11 Solution
→ Question 13 Solution

👨‍🏫Author Bio

S Mahandru focuses on helping students build confidence with longer, multi-step questions by keeping methods clear and manageable.

❓ Frequently Asked Questions

📌Why partial fractions first?

So the variables can be separated.

📌Why logs appear?

Because of integrating $\frac{1}{V}$ .

📌Why minutes?

That’s what the question asks for.

📌What does the limit mean?

The maximum value the model approaches.