A2 June 2025 Paper 2 Q9
9 When light hits a certain photo-sensitive cell, at time \(t = 0\), there is an electrical response in the cell which is denoted by \(y(t)\) where both \(y\) and \(t\) are measured in suitable units. A student wishes to model this response, \(y\).
In an attempt to model \(y\), the student sets up the differential equation
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}t^2} + 6\dfrac{\mathrm{d}y}{\mathrm{d}t} + 9y = 10\mathrm{e}^{-3t} \quad (*)\)
which is subject to the following conditions.
- \(y = 0\) when \(t = 5\)
- \(y \geqslant 0\) for all \(t \geqslant 0\)
The cell can be considered to be operating properly if \(y \lt 4\) when \(t = 1\).
| Scheme | Marks | AO |
|---|---|---|
| \(y = A\mathrm{e}^{-3t} + Bt\mathrm{e}^{-3t}\) \(\Rightarrow y' = -3A\mathrm{e}^{-3t} + B\mathrm{e}^{-3t} - 3Bt\mathrm{e}^{-3t}\) \(\Rightarrow y'' = 9A\mathrm{e}^{-3t} - 3B\mathrm{e}^{-3t} - 3B\mathrm{e}^{-3t} + 9Bt\mathrm{e}^{-3t}\) | M1 | 3.1a |
| \(\therefore y'' + 6y' + 9y =\) \(9A\mathrm{e}^{-3t} - 3B\mathrm{e}^{-3t} - 3B\mathrm{e}^{-3t} + 9Bt\mathrm{e}^{-3t}\) \(- 18A\mathrm{e}^{-3t} + 6B\mathrm{e}^{-3t} - 18Bt\mathrm{e}^{-3t} + 9A\mathrm{e}^{-3t}\) \(+ 9Bt\mathrm{e}^{-3t} = 0\) (as required for a CF.) | A1 | 2.1 |
| [2] |
Notes
M1: Differentiating given CF twice using product rule. Allow sign errors in either derivative (but coefficients must be correct).
Condone use of e.g. \(x\) instead of \(t\) for this mark but not a mixture.
A1: Cancelling or combining terms need not be seen, but all must be present before \(= 0\).
Condone arguments leading to e.g. \(0 = 0\) provided all terms correct.
| Scheme | Marks | AO |
|---|---|---|
| PI: Try \(y = \alpha t^2\mathrm{e}^{-3t}\) | B1 | 1.2 |
| \(y' = 2\alpha t\mathrm{e}^{-3t} - 3\alpha t^2\mathrm{e}^{-3t}\) \(y'' = 2\alpha\mathrm{e}^{-3t} - 6\alpha t\mathrm{e}^{-3t} - 6\alpha t\mathrm{e}^{-3t} + 9\alpha t^2\mathrm{e}^{-3t}\) \((= 2\alpha\mathrm{e}^{-3t} - 12\alpha t\mathrm{e}^{-3t} + 9\alpha t^2\mathrm{e}^{-3t})\) | M1* M1dep* | 1.1 1.1 |
| \(y'' + 6y' + 9y = 2\alpha\mathrm{e}^{-3t} - 12\alpha t\mathrm{e}^{-3t} + 9\alpha t^2\mathrm{e}^{-3t} + 12\alpha t\mathrm{e}^{-3t} - 18\alpha t^2\mathrm{e}^{-3t} + 9\alpha t^2\mathrm{e}^{-3t} = 10\mathrm{e}^{-3t}\) | M1 | 1.1 |
| \(\Rightarrow 2\alpha\mathrm{e}^{-3t} = 10\mathrm{e}^{-3t}\) \(\Rightarrow \alpha = 5 \Rightarrow\) GS is \((A + Bt)\mathrm{e}^{-3t} + 5t^2\mathrm{e}^{-3t}\) | A1 | 3.3 |
| \((A + Bt + 5t^2)\mathrm{e}^{-3t} = 0\) when \(t = 5\) \(\Rightarrow A + 5B + 125 = 0\) | M1 | 3.4 |
| \(A + Bt + 5t^2 \geq 0 \Rightarrow B^2 - 4 \times 5A = 0\) | M1 | 3.4 |
| \(\Rightarrow A = 125\) and \(B = -50\) \(\therefore\) PS is \(y = (5t^2 - 50t + 125)\mathrm{e}^{-3t}\) | A1 | 3.4 |
| [8] |
Notes
B1: Correct form for PI. Allow more general forms e.g. \(\alpha t^2\mathrm{e}^{-3t} + \cdots\) if subsequent coefficients \(\beta, \gamma\) etc. later found to be 0.
Condone use of e.g. \(x\) instead of \(t\) for this mark and the first 3 M marks (but no further).
M1*: Attempt to differentiate their \(y\) twice (finding \(y'\) and \(y''\)), using the product rule at least once.
Allow this mark if their PI is of the form \(\alpha p(t)\mathrm{e}^{-3t}\) where \(p(t)\) is any polynomial in \(t\) of degree \(\geq 1\).
M1dep*: Correctly differentiating their \(y\) to find \(y'\) and \(y''\). Condone sign errors.
Condone missing 3, extra 3 or 1/3 coming down but not \(t\) coming down and/or \(3t - 1\) as exponent.
M1: Substituting their \(y'\) and \(y''\) into the DE (with RHS seen or \(\mathrm{e}^{-3t}\) cancelled from both sides). Allow this mark following M0M0.
This mark may be implied by the next A1 (i.e. a correct GS).
A1: Full form of GS can be implied by later work.
M1: Substituting \(y = 0\) and \(t = 5\) into their GS and cancelling any exponential part properly.
May be implied by later working including \(\mathrm{e}^{-3t}\) if fully correct.
M1: Correctly interpreting and using the 2nd condition which, coupled with the 1st, must mean that the quadratic is a perfect square. May see \(3A + 14B = -325\) oe.
Or \(B + 10t = 0\) when \(t = 5\) (at what must be a minimum point). Or \(5\left(t^2 + \frac{Bt}{5}\right) + A = 5\left(t + \frac{B}{10}\right)^2 + A - \frac{B^2}{20} \Rightarrow A - \frac{B^2}{20} = 0\) since \(y = 0\).
A1: Must be \(y =\) or \(y(t) = \ldots\) May see \(5(t^2 - 10t + 25)\) or \(5(t - 5)^2\)
Allow SCB1 for \(A = 125\) and \(B = -50\) following M0, provided all 6 of the first marks are gained (7/8).
| Scheme | Marks | AO |
|---|---|---|
| (When \(t = 0\), \(y = 125\mathrm{e}^{-0} =\)) 125 | B1FT | 3.4 |
| [1] |
Notes
B1FT: FT their PS of the form \(p(t)\mathrm{e}^{kt}\) where \(p(t)\) is any polynomial of non-zero degree.
Must be a value (i.e. not an expression with \(A, B\), etc.) and must be \(\geq 0\).
| Scheme | Marks | AO |
|---|---|---|
| (According to the model) when \(t = 1\), \(y = 80\mathrm{e}^{-3} = 3.982965\ldots\) | M1 | 3.4 |
| This is \(\lt 4\) hence the cell can be inferred to be operating properly (but, as the value is very close to 4, there may be a margin of error in the model). | A1FT | 2.2b |
| [2] |
Notes
M1: Finding the value of their \(y\) when \(t = 1\). FT their PS of the form \(p(t)\mathrm{e}^{kt}\) where \(p(t)\) is any polynomial.
Must be a value (i.e. not an expression with \(A, B\), etc.)
A1FT: Sensible conclusion in context with correct comparison to 4 for their value (A0 for a value \(\lt 0\)). (i.e. \(\lt 4 \Rightarrow\) operating properly, \(\gt 4 \Rightarrow\) not operating properly). Condone ‘it is operating properly’.
Accept e.g.: ‘So it can be inferred that the cell is probably working well although we can’t be sure because it is close to 4 and we don’t know how well the model works’