A2 June 2025 Paper 1 Q4
4.
Given that \(y = 5\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 12\) when \(x = 0\)
| Scheme | Marks | AO |
|---|---|---|
| \(m^2 - 4m + 4 = 0 \Rightarrow m = \ldots\{2\}\) | M1 | 1.1b |
| \((y) = A\mathrm{e}^{2x} + Bx\mathrm{e}^{2x}\) | A1 | 1.1b |
| PI \(y = \lambda\mathrm{e}^{3x},\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = 3\lambda\mathrm{e}^{3x}\) and \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 9\lambda\mathrm{e}^{3x}\) \(9\lambda\mathrm{e}^{3x} - 4\left(3\lambda\mathrm{e}^{3x}\right) + 4\left(\lambda\mathrm{e}^{3x}\right) = 2\mathrm{e}^{3x}\) leading to \(\lambda = \ldots\) | M1 | 3.1a |
| \((y) = CF + PI\) | dM1 | 1.1b |
| \(y = A\mathrm{e}^{2x} + Bx\mathrm{e}^{2x} + 2\mathrm{e}^{3x}\) | A1 | 1.1b |
| (5) |
Notes
M1: Forms the correct auxiliary equation, leading to finding a value for \(m\).
Allow the use of any variable instead of \(m\) for this mark.
A1: Correct complementary function, oe such as \(y = (A + Bx)\mathrm{e}^{2x}\)
Do not need \(y =\) here but must be in terms of \(x\), but may be recovered later.
M1: A complete method to find the particular integral. Uses the correct form \(y = \lambda\mathrm{e}^{3x}\), differentiates twice and substitutes correctly into the differential equation to find a value for \(\lambda\ (= 2)\). May use a different variable for \(x\) here.
dM1: Dependent on the previous method mark. Finds the general solution by adding the particular integral (of the correct form) to their complementary function (which may not be of the correct form). Do not need \(y =\) here. May use a different variable for \(x\) but their variable must be consistent.
A1: Correct general solution. Must have \(y =\) here and must be in terms of \(x\)
| Scheme | Marks | AO |
|---|---|---|
| \(x = 0,\ y = 5 \Rightarrow 5 = A + 2 \Rightarrow A = \ldots\) | M1 | 1.1b |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2A\mathrm{e}^{2x} + B\mathrm{e}^{2x} + 2Bx\mathrm{e}^{2x} + 6\mathrm{e}^{3x}\) and uses their value of \(A\) and \(x = 0,\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = 12\) to find the value of \(B\) \(12 = 6 + B + 6 \Rightarrow B = \ldots\) | dM1 | 3.1a |
| \(y = 3\mathrm{e}^{2x} + 2\mathrm{e}^{3x}\) | A1 | 1.1b |
| (3) | ||
| (8 marks) |
Notes
M1: For a general solution of the form \(A\mathrm{e}^{\alpha x} + Bx\mathrm{e}^{\alpha x} + \text{``their PI''}\), uses the initial conditions, \(x = 0,\ y = 5\) to find a value for a constant.
(The correct form will be \(A\mathrm{e}^{mx} + Bx\mathrm{e}^{mx} + \lambda\mathrm{e}^{3x}\))
dM1: Dependent on the previous method mark.
Differentiates the general solution to achieve an expression of the form \(mA\mathrm{e}^{mx} + B\mathrm{e}^{mx} + mBx\mathrm{e}^{mx} + 3\lambda\mathrm{e}^{3x}\) using their \(m\) from part (a) where \(m \neq 1\)
Then uses the initial conditions to find a value for the other constant.
A1: Correct particular solution. Must have \(y =\) here and must be in terms of \(x\)