## Question 10:

### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y' = x + x't$ | B1 | |
| $y'' = x' + x' + tx''$ | B1 | |
| Substitute correctly | B1 | AG |

### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Auxiliary equation: $m^2 + 9 = 0 \Rightarrow m = \pm 3i$ | M1 | Correct auxiliary |
| CF $= A\cos 3t + B\sin 3t$ | A1 | |
| PI $y = at^2 + bt + c$, so $y' = 2at + b$ and $y'' = 2a$ | M1 | Differentiate twice and substitute |
| $a = \frac{1}{3},\ b = 0,\ c = \frac{1}{27}$ | A1 | |
| $y = A\cos 3t + B\sin 3t + \frac{1}{3}t^2 + \frac{1}{27}$ | A1FT | Their CF + their PI both in correct form |
| $x = \frac{\pi}{9}$ when $t = \frac{\pi}{3}$ gives $A = \frac{1}{27}$ | B1 | |
| $y' = -3A\sin 3t + 3B\cos 3t + \frac{2}{3}t$ | M1 | Differentiating their $y$ of equivalent difficulty |
| $x' = \frac{2}{3}$ when $t = \frac{\pi}{3}$ gives $B = -\frac{\pi}{27}$ | A1 | |
| $x = \frac{\cos 3t - \pi\sin 3t + 9t^2 + 1}{27t}$ | A1 | AEF |

## Question 11E(i):

$I = \int_{-\pi/2}^{\pi/2} e^x \cos x \, dx = \left[e^x \cos x\right]_{-\pi/2}^{\pi/2} - \int_{-\pi/2}^{\pi/2} e^x(-)\sin x \, dx$ | **M1 A1** | The parts can be either way round.

$= \left[e^x \sin x\right]_{-\pi/2}^{\pi/2} - I$ | **M1** | Integrates by parts again.

$\Rightarrow 2I = e^{\pi/2} + e^{-\pi/2}$ AG | **A1** | Alt method: Integrate $\text{Re}(e^{ix})e^x$ by parts M1, M1, A1, A1

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## Question 11E(ii):

$I_n = \left[\frac{e^{2x}}{2}\cos^n x\right]_{-\pi/2}^{\pi/2} + \frac{n}{2}\int_{-\pi/2}^{\pi/2} e^{2x}\cos^{n-1}x\sin x \, dx$ | **M1** | Integrates by parts once.

$= 0 - \frac{n}{4}\int_{-\pi/2}^{\pi/2} e^{2x}\left(\cos^n x - (n-1)\sin^2 x \cos^{n-2} x\right)dx$ | **M1** | Integrates by parts again.

$\Rightarrow 4I_n = n(n-1)\int_{-\pi/2}^{\pi/2} e^{2x}\sin^2 x\cos^{n-2}x \, dx - nI_n$ | **A1** | Simplifies to answer. (AG)

$(n+4)I_n = n(n-1)\int_{-\pi/2}^{\pi/2} e^{2x}(1-\cos^2 x)\cos^{n-2}x \, dx$ | **M1** | For splitting $\sin^2 x$

$\Rightarrow (n+4)I_n = n(n-1)I_{n-2} - n(n-1)I_n$ | **M1** |

$\Rightarrow (n^2+4)I_n = n(n-1)I_{n-2}$ AG | **A1** |

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## Question 11E(iii):

$\left(\int_{-\pi/2}^{\pi/2} y^2 \, dx =\right) I_2 = \frac{1}{4}I_0 = \frac{1}{4}\left[\frac{e^{2x}}{2}\right]_{-\pi/2}^{\pi/2} = \frac{1}{8}(e^\pi - e^{-\pi})$ | **M1 A1** | Uses reduction formula to find $I_2$.

$\bar{y} = \frac{I_2}{2I} = \frac{1}{8}\left(\frac{e^\pi - e^{-\pi}}{e^{\pi/2}+e^{-\pi/2}}\right) (= 0.575)$ | **M1 A1** | Uses correct formula for $\bar{y}$.

| | **14** | |

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## Question 11O(i):

$\begin{pmatrix}1 & -2 & 0 & 0 \\ 2 & -5 & -3 & -2 \\ 0 & 5 & 15 & 10 \\ 2 & 6 & 18 & 8\end{pmatrix} \to \cdots \to \begin{pmatrix}1 & -2 & 0 & 0 \\ 0 & 1 & 3 & 2 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0\end{pmatrix}$ | **M1 A1** | Reduces to echelon or reduced row echelon form. At least 2 manipulations. Or finds 4 columns dependent, 3 columns independent.

$\dim V = \text{rank} = 3$, $c_1\ c_2\ c_3\ c_4$ | **A1** |

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## Question 11O(ii):

$c_4 = c_3 - c_2 - 2c_1 \Rightarrow \mathbf{v}_4 = \mathbf{v}_3 - \mathbf{v}_2 - 2\mathbf{v}_1$ | **M1 A1** | OE

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## Question 11O(iii):

$\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ | **B1** | OE

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## Question 11O(iv):

$\mathbf{v}_1 + \mathbf{v}_2 = M\begin{pmatrix}1\\1\\0\\0\end{pmatrix}$ so particular solution is $\begin{pmatrix}1\\1\\0\\0\end{pmatrix}$ | **M1 A1** | Finds particular solution.

$x - 2y = 0$, $y + 3z + 2t = 0$, $z + t = 0$ | **M1** | Finds basis for null space.

Basis for null space is $\left\{\begin{pmatrix}2\\1\\-1\\1\end{pmatrix}\right\}$ | **A1** | AEF

$\mathbf{x} = \begin{pmatrix}1\\1\\0\\0\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\-1\\1\end{pmatrix}$ | **M1 A1** | Finds general solution. Alt method: Set up equations M1, A1; Solve equations M1, A1: $x=$ etc. M1, A1 or using augmented matrix.

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## Question 11O(v):

Not closed under addition: $\begin{pmatrix}1\\0\\0\\0\end{pmatrix} + \begin{pmatrix}0\\1\\0\\0\end{pmatrix} = \begin{pmatrix}1\\1\\0\\0\end{pmatrix}$ | **M1** | Accept any valid reason with evidence.

So not a vector space. | **A1** |

| | **14** | |