## Question 10(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $m^2 + 2m + 10 = 0 \Rightarrow m = -1 \pm 3i$ | M1 | Finds complementary function |
| $x = e^{-t}(A\cos 3t + B\sin 3t)$ | A1 | |
| $x = p\cos 3t + q\sin 3t$ | M1 | Finds particular integral |
| $\Rightarrow \dot{x} = -3p\sin 3t + 3q\cos 3t \Rightarrow \ddot{x} = -9p\cos 3t - 9q\sin 3t$ | A1 | |
| $(p + 6q)\cos 3t + (q - 6p)\sin 3t = 37\sin 3t$ | M1 | |
| $\Rightarrow p = -6,\quad q = 1$ | A1 | |
| $x = e^{-t}(A\cos 3t + B\sin 3t) - 6\cos 3t + \sin 3t$ | | |
| $x = 3$ when $t = 0 \Rightarrow A - 6 = 3 \Rightarrow A = 9$ | B1 | Uses initial conditions |
| $\dot{x} = e^{-t}(-3A\sin 3t + 3B\cos 3t) - e^{-t}(A\cos 3t + B\sin 3t) + 18\sin 3t + 3\cos 3t$ | M1 A1 | |
| $\dot{x} = 0$ when $t = 0 \Rightarrow 3B - A + 3 = 0 \Rightarrow B = 2$ | A1 | Obtains solution, AEF |
| $x = e^{-t}(9\cos 3t + 2\sin 3t) - 6\cos 3t + \sin 3t$ | | |
| **Total** | **10** | |

## Question 10(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| As $t \to \infty$, $e^{-t} \to 0 \Rightarrow x \approx -6\cos 3t + \sin 3t$ | B1 | Obtains limit |
| $\sin 3t - 6\cos 3t = \sqrt{37}\left(\dfrac{1}{\sqrt{37}}\sin 3t - \dfrac{6}{\sqrt{37}}\cos 3t\right)$ | M1 | Converts to $R\sin(3t - \phi)$ |
| $= \sqrt{37}\sin(3t - \tan^{-1}6)$ | A1 | AG |
| **Total** | **3** | |

## Question 11E(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2r+1)^2 - (2r-1)^2 = 8r$ | B1 | |
| $\Rightarrow 8\displaystyle\sum_{r=1}^{n} r = -1^2 + (2n+1)^2$ | M1 | Sums both sides and uses method of differences |
| $\Rightarrow \displaystyle\sum_{r=1}^{n} r = \dfrac{1}{2}n(n+1)$ | A1 | AG |
| **Total** | **3** | |

## Question 11E(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2r+1)^4 - (2r-1)^4 = \left((2r+1)^2 + (2r-1)^2\right)\left((2r+1)^2 - (2r-1)^2\right)$ | M1 | Uses difference of squares or expands |
| $= (8r^2 + 2)(8r) = 16(4r^3 + r)$ | A1 | |
| $\Rightarrow 4\displaystyle\sum_{r=1}^{n} r^3 + \sum_{r=1}^{n} r = -\left(1 - \dfrac{1}{2}\right)^4 + \left(n + \dfrac{1}{2}\right)^4$ | M1 | Sums both sides and uses method of differences |
| $4\displaystyle\sum_{r=1}^{n} r^3 = \left(n + \dfrac{1}{2}\right)^4 - \left(\dfrac{1}{2}\right)^4 - \dfrac{1}{2}n(n+1)$ | M1 | Uses formula for $\sum r$ |
| $= \left(\left(n+\dfrac{1}{2}\right)^2 - \left(\dfrac{1}{2}\right)^2\right)\left(\left(n+\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2\right) - \dfrac{1}{2}n(n+1) = n^2(n+1)^2$ | A1 | AG |
| **Total** | **5** | |

## Question 11E(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $S = \dfrac{1}{4}(2N+1)^2(2N+2)^2 = (2N+1)^2(N+1)^2$ | B1 | Uses formula for $\sum r^3$. AG |
| **Total** | **1** | |

## Question 11E(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $T = S - \displaystyle\sum_{r=1}^{N}(2r)^3 = (2N+1)^2(N+1)^2 - 2N^2(N+1)^2$ | M1 | Eliminates even terms from $S$ |
| $(N+1)^2(2N^2 + 4N + 1)$ | A1 | Accept $(N+1)^2(2N(N+2)+1)$ |
| **Total** | **2** | |

## Question 11E(v):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{S}{T} = \dfrac{(2N+1)^2}{2N^2 + 4N + 1} = \dfrac{4N^2 + 4N + 1}{2N^2 + 4N + 1}$ | M1 | Writes fraction as quadratic in $N$ divided by quadratic in $N$ |
| Converges to $2$ as $N \to \infty$ | A1 | |
| **Total** | **2** | |

## Question 11O(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = -1 \Rightarrow y^3 + 2y - 3 = 0 \Rightarrow (y-1)(y^2 + y + 3) = 0$ | M1 | Considers cubic polynomial in $y$ |
| There is only one real root ($= 1$) | A1 | |
| **Total** | **2** | |

## Question 11O(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $2x + 2\left(x\dfrac{dy}{dx} + y\right)$ | B1 | Differentiates LHS correctly |
| $= 3y^2\dfrac{dy}{dx}$ | B1 | Differentiates RHS correctly |
| $\Rightarrow (3y^2 - 2x)\dfrac{dy}{dx} = 2x + 2y$ | | |
| $x = -1, y = 1 \Rightarrow \dfrac{dy}{dx} = 0$ | B1 | Substitutes $(-1, 1)$. AG |
| **Total** | **3** | |

## Question 11O(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3y^2 - 2x)\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx}\left(6y\dfrac{dy}{dx} - 2\right) = 2 + 2\dfrac{dy}{dx}$ | M1 M1 | Differentiates equation again |
| $x = -1, y = 1, \dfrac{dy}{dx} = 0 \Rightarrow \dfrac{d^2y}{dx^2} = \dfrac{2}{5}$ | A1 | Substitutes $(-1,1)$ and $\dfrac{dy}{dx} = 0$. AG |
| **Total** | **3** | |

## Question 11O(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\displaystyle\int_{-1}^{0} x^n \dfrac{d^n y}{dx^n}\,dx = \left[x^n \dfrac{d^{n-1}y}{dx^{n-1}}\right]_{-1}^{0} - n\int_{-1}^{0} x^{n-1}\dfrac{d^{n-1}y}{dx^{n-1}}\,dx$ | M1 A1 | Integrates by parts |
| $= (-1)^{n+1}A_{n-1} - nI_{n-1}$ | A1 | AG |
| **Total** | **3** | |

## Question 11O(v):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_3 = (-1)^4 A_2 - 3I_2 = A_2 - 3(-A_1 - 2I_1) = \dfrac{2}{5} + 6I_1$ | M1 A1 | Uses reduction formulae |
| **Total** | **2** | |