# Question 5:

| Answer | Marks | Guidance |
|--------|-------|----------|
| AE: $\lambda^2 + 2\lambda + 5 = 0$ | M1 | |
| $\lambda = -1 \pm 2i$ | A1 | |
| CF: $e^{-x}(A\cos 2x + B\sin 2x)$ | A1ft | Or $Ae^{-x}\cos(2x+\alpha)$ Must be in real form |
| PI: $y = ae^{-x}$ | B1 | If PI $y=axe^{-x}$, then max of M1,A1,A1, B0,M1,A0,A0 (since cannot be consistent) M1, M1, A1. Must have a constant in "their PI" |
| $ae^{-x} - 2ae^{-x} + 5ae^{-x} = e^{-x}$, $4a=1$ | M1 | Differentiate & substitute |
| $a = \frac{1}{4}$ | A1 | |
| GS: $y = e^{-x}\left(\frac{1}{4} + A\cos 2x + B\sin 2x\right)$ | A1ft | Must have "$y=$" |
| $\frac{dy}{dx} = -e^{-x}\left(\frac{1}{4} + A\cos 2x + B\sin 2x\right) + e^{-x}(-2A\sin 2x + 2B\cos 2x)$ | M1* | Differentiate their GS of form $y=e^{-x}(P + A\cos nx + B\sin nx)$ where $P$ is constant or linear term, $n$ not 0 or 1. Allow one error |
| $x=0, \frac{dy}{dx}=0 \Rightarrow -\left(\frac{1}{4}+A\right)+2B=0$ | *M1 | Use conditions. But M0 if leads to solution of $y=0$ |
| $x=0, y=0 \Rightarrow \frac{1}{4}+A=0$ | | |
| $A = -\frac{1}{4},\ B=0$ | A1ft | From their GS |
| $y = \frac{1}{4}e^{-x}(1-\cos 2x)$ | A1 | Must have '$y=$' and be in real form |
| | **[11]** | |

---