A2 June 2025 Q7

EdexcelCurrent spec11 marksDifferential Equations

7. The concentration, \(P\text{ mg m}^{-3}\), of a pollutant in a reservoir, \(t\) days after the pollutant entered the reservoir, is modelled by the differential equation

\[\frac{1}{P}\frac{\mathrm{d}P}{\mathrm{d}t} = 4tP^2 - 1 \qquad \text{(I)}\]

(a) Show that the transformation \(x = \dfrac{1}{P^2}\) transforms equation (I) into the equation\[\frac{\mathrm{d}x}{\mathrm{d}t} - 2x = -8t \qquad \text{(II)}\] (3)

Given that \(P = 0.5\) when \(t = 0\)

(b) solve differential equation (II) to show that, according to the model,\[P^2 = \frac{1}{4t + 2 + k\mathrm{e}^{2t}}\]where \(k\) is a constant to be determined. (6)

Given that the concentration of the pollutant in the reservoir, 3 days after the pollutant entered the reservoir, was \(0.034\text{ mg m}^{-3}\)

Mark scheme (a)

Scheme	Marks	AO
Examples: \(x = \dfrac{1}{P^2} \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} = -\dfrac{2}{P^3}\dfrac{\mathrm{d}P}{\mathrm{d}t}\) or \(\dfrac{\mathrm{d}P}{\mathrm{d}t} = \dfrac{\mathrm{d}P}{\mathrm{d}x} \times \dfrac{\mathrm{d}x}{\mathrm{d}t} = -\dfrac{P^3}{2}\dfrac{\mathrm{d}x}{\mathrm{d}t}\) or \(P = x^{-\frac{1}{2}} \Rightarrow \dfrac{\mathrm{d}P}{\mathrm{d}t} = -\dfrac{1}{2}x^{-\frac{3}{2}}\dfrac{\mathrm{d}x}{\mathrm{d}t}\) or \(\dfrac{\mathrm{d}P}{\mathrm{d}x} = -\dfrac{1}{2}x^{-\frac{3}{2}} \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{\mathrm{d}x}{\mathrm{d}P} \times \dfrac{\mathrm{d}P}{\mathrm{d}t} = -2x^{\frac{3}{2}} \times \left(4tP^3 - P\right)\) or \(x = \dfrac{1}{P^2} \Rightarrow xP^2 = 1 \Rightarrow P^2\dfrac{\mathrm{d}x}{\mathrm{d}t} + 2Px\dfrac{\mathrm{d}P}{\mathrm{d}t} = 0\) or \(x = \dfrac{1}{P^2} \Rightarrow P^2 = \dfrac{1}{x} \Rightarrow 2P\dfrac{\mathrm{d}P}{\mathrm{d}t} = -\dfrac{1}{x^2}\dfrac{\mathrm{d}x}{\mathrm{d}t}\)	M1	3.1a
\(\dfrac{1}{P} \times -\dfrac{P^3}{2}\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{4t}{x} - 1 \Rightarrow -\dfrac{1}{2x}\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{4t}{x} - 1\) or \(x^{\frac{1}{2}} \times -\dfrac{1}{2}x^{-\frac{3}{2}}\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{4t}{x} - 1 \Rightarrow -\dfrac{1}{2x}\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{4t}{x} - 1\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -2x^{\frac{3}{2}} \times \left(4tP^3 - P\right) = -2x^{\frac{3}{2}} \times \left(\dfrac{4t}{x^{\frac{3}{2}}} - \dfrac{1}{x^{\frac{1}{2}}}\right)\) or \(\dfrac{1}{P} \times -\dfrac{P}{2x}\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{4t}{x} - 1 \Rightarrow -\dfrac{1}{2x}\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{4t}{x} - 1\) or \(\dfrac{1}{P} \times -\dfrac{1}{2Px^2}\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{4t}{x} - 1 \Rightarrow -\dfrac{1}{2x}\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{4t}{x} - 1\)	M1	1.1b
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} - 2x = -8t\ *\) cso	A1*	2.1
	(3)

Notes

M1: Identifies and applies a correct strategy for the differentiation. This may be seen when \(\dfrac{\mathrm{d}P}{\mathrm{d}t}\) is substituted into the equation.

M1: Substitutes into the given differential equation and proceeds to an equation in \(x\) and \(t\) only.

A1*: Correct proof with sufficient working shown and no errors, cso

Mark scheme (b)

Scheme	Marks	AO
\(I = \mathrm{e}^{\int -2\,\mathrm{d}t} = \mathrm{e}^{-2t}\)	B1	2.2a
\(x\mathrm{e}^{-2t} = \displaystyle\int -8t\mathrm{e}^{-2t}\,\{\mathrm{d}t\}\)	M1	1.1b
\(= 4t\mathrm{e}^{-2t} - \displaystyle\int 4\mathrm{e}^{-2t}\,\{\mathrm{d}t\}\)	M1	3.1a
\(= 4t\mathrm{e}^{-2t} + 2\mathrm{e}^{-2t} + k\)	A1	1.1b
\(x\mathrm{e}^{-2t} = 4t\mathrm{e}^{-2t} + 2\mathrm{e}^{-2t} + k \Rightarrow \dfrac{1}{0.5^2} = 2 + k \Rightarrow k = \ldots\{2\}\) or \(x\mathrm{e}^{-2t} = 4t\mathrm{e}^{-2t} + 2\mathrm{e}^{-2t} + k \Rightarrow x = 4t + 2 + k\mathrm{e}^{2t} \Rightarrow \dfrac{1}{P^2} = 4t + 2 + k\mathrm{e}^{2t}\) \(P = 0.5, t = 0 \Rightarrow \dfrac{1}{0.5^2} = 2 + k \Rightarrow k = \ldots\{2\}\)	M1	3.4
\(\dfrac{1}{P^2} = 4t + 2 + 2\mathrm{e}^{2t} \Rightarrow P^2 = \dfrac{1}{4t + 2 + 2\mathrm{e}^{2t}}\) cso	A1	2.1
	(6)

Notes

B1: Deduces the correct integrating factor

M1: Applies their integrating factor \(I\) to obtain \(Ix = \displaystyle\int \pm 8It\,\{\mathrm{d}t\}\) condone missing \(\mathrm{d}t\)

M1: Recognises that integration by parts is required and applies this correctly to reach the form \(= At\mathrm{e}^{-2t} - \displaystyle\int B\mathrm{e}^{-2t}\,\{\mathrm{d}t\}\)

A1: Correct integration including an arbitrary constant

M1: Uses the conditions given in the model to find the constant of integration. This may be seen before rearranging to get \(P^2 = \ldots\), condone a slip

A1: Correct equation with no errors seen, cso, missing \(\mathrm{d}t\) during working would lose this mark

Note that the first 4 marks in (b) can also be obtained as follows:

B1: AE: \(m - 2 = 0 \Rightarrow m = 2 \Rightarrow x = A\mathrm{e}^{2t}\) (Correct CF)

M1: PI: \(x = at + b \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} = a\) (Selects the correct PI form and differentiates)

M1: \(a - 2at - 2b = -8t \Rightarrow a = \ldots(4), b = \ldots(2)\) (Substitutes and compares coefficients to find \(a\) and \(b\))

A1: \(x = 4t + 2 + A\mathrm{e}^{2t}\) (Correct expression)

Mark scheme (c)

Scheme	Marks	AO
\(t = 3 \Rightarrow P^2 = \dfrac{1}{14 + 2\mathrm{e}^6} \Rightarrow P = 0.03490326\ldots\)	M1	3.4
This value is close to 0.034 so the model is reliable. Between 0.0323 and 0.0357 good model Between 0.0306 and 0.0323 or 0.0357 and 0.0374 either a good or bad model with reasoning Less than 0.0306 or more than 0.0374 bad model	A1ft	3.5a
	(2)
	(11 marks)

Notes

M1: Uses the model with \(t = 3\) to find \(P\), where \(P^2 \gt 0\) or compares their value of \(P^2\) with \(0.001156 = 0.034^2\)

A1ft: Compares their value of \(P\) with 0.034 and makes a suitable conclusion, making any comparison with \(P^2\) is A0