A2 October 2021 Paper 1 Q13
13 Find the general solution of the differential equation \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3y = 2\mathrm{e}^x\). [7]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3y = 2\mathrm{e}^x\) AE: \(\lambda^2 + 2\lambda - 3 = 0 \Rightarrow \lambda = -3, 1\) | M1 | 2.1 |
| CF: \(y = A\mathrm{e}^{-3x} + B\mathrm{e}^x\) | A1 | 2.1 |
| PI: \(y = Cx\mathrm{e}^x\) | M1 | 2.1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = C\left(\mathrm{e}^x + x\mathrm{e}^x\right)\) | A1 | 1.1 |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = C\left(2\mathrm{e}^x + x\mathrm{e}^x\right)\) | A1 | 1.1 |
| \(\Rightarrow C\left(2\mathrm{e}^x + x\mathrm{e}^x\right) + 2C\left(\mathrm{e}^x + x\mathrm{e}^x\right) - 3Cx\mathrm{e}^x = 2\mathrm{e}^x\) \(\Rightarrow 4C = 2 \Rightarrow C = \frac{1}{2}\) | M1 | 2.1 |
| GS: \(y = A\mathrm{e}^{-3x} + B\mathrm{e}^x + \frac{1}{2}x\mathrm{e}^x\) | A1 | 2.2a |
| [7] |