A2 June 2024 Paper 1 Q14
14
(a) Find the general solution of the differential equation \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + \dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y = 12\mathrm{e}^{-x}\). [7]
You are given that \(y\) tends to zero as \(x\) tends to infinity, and that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) when \(x = 0\).
(b) Find the exact value of \(x\) for which \(y = 0\). [5]
| Scheme | Marks | AO |
|---|---|---|
| AE is \(\lambda^2 + \lambda - 2 = 0\) | M1 | 1.1 |
| \(\Rightarrow \lambda = -2, 1\) | A1 | 1.1 |
| CF is \(A\mathrm{e}^{-2x} + B\mathrm{e}^{x}\) | A1 | 1.1 |
| PI is \(y = C\mathrm{e}^{-x}\) | B1 | 1.1 |
| \(y' = -C\mathrm{e}^{-x},\ y'' = C\mathrm{e}^{-x} \Rightarrow (C - C - 2C)\mathrm{e}^{-x} = 12\mathrm{e}^{-x}\) | M1 | 2.1 |
| \(\Rightarrow C = -6\) | A1 | 1.1 |
| GS is \(y = A\mathrm{e}^{-2x} + B\mathrm{e}^{x} - 6\mathrm{e}^{-x}\) | A1 | 1.1 |
| [7] |
Notes
M1: Forming AE
M1: Attempt to differentiate their PI twice and substituting. Do not condone \(y = C\mathrm{e}^{-2x}\) or \(y = C\mathrm{e}^{x}\).
A1: Must see \(y =\)
| Scheme | Marks | AO |
|---|---|---|
| \(B = 0\) | B1FT | 3.1a |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -2A\mathrm{e}^{-2x} + B\mathrm{e}^{x} + 6\mathrm{e}^{-x}\) | M1* | 1.1 |
| \(A = 3\) (so \(\Rightarrow y = 3\mathrm{e}^{-2x} - 6\mathrm{e}^{-x}\)) | A1 | 1.1 |
| \(y = 0 \Rightarrow 3\mathrm{e}^{-x}(\mathrm{e}^{-x} - 2) = 0\) or \(3 - 6\mathrm{e}^{x} = 0\) | M1dep | 1.1 |
| \(\Rightarrow x = -\ln 2\) | A1 | 1.1 |
| [5] |
Notes
B1FT: Equating coefficient(s) of their \(\mathrm{e}^{kx}\) term(s) to 0 where \(k \gt 0\)
M1*: FT their GS. Accept \(\frac{\mathrm{d}y}{\mathrm{d}x} = -2A\mathrm{e}^{-2x} + 6\mathrm{e}^{-x}\) if \(B = 0\) already found. Allow a slip. Not implied by \(-2A + B + 6 = 0\).
A1: Do not FT.
M1dep: A complete method to solve their \(y = 0\) which leads to a solution
A1: or exact equivalent