## (i) (a)
$\frac{d^2x}{dt^2} = 3\frac{dx}{dt} - 2\frac{dy}{dt} = 3\frac{dx}{dt} - 2(y + 5x)$ | M1 | Differentiating $\frac{dx}{dt} = 3x - 2y$ and substituting in for $\frac{dy}{dt}$ from
$\frac{d^2x}{dt^2} = 3\frac{dx}{dt} - (3x - \frac{dx}{dt}) - 10x = 4\frac{dx}{dt} - 13x$ | A1 | Substituting for a second time to leave an equation in $x$, expanding and collecting.
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## (i) (b)
$m^2 - 4m + 13 = 0$ | M1 | Finding the auxiliary equation and attempting to solve
$m = 2 \pm 3i$ | A1 | Can be implied by correct form
$x = e^{2t}(A\sin 3t + B\cos 3t)$ | A1ft | 
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## (ii)
$\frac{d^2y}{dt^2} = \frac{dy}{dt} + 5\frac{dx}{dt} = \frac{dy}{dt} + 5(3x - 2y)$ | M1 | Differentiating $\frac{dy}{dt} = y + 5x$ and substituting in for $\frac{dx}{dt}$ from
$\frac{d^2y}{dt^2} = \frac{dy}{dt} + 3(\frac{dy}{dt} - y) - 10y = 4\frac{dy}{dt} - 13y$ | A1 | Substituting for a second time to leave an equation in $y$, expanding and collecting. No need to see $a = 4, b = -13$ or explicitly
Same form as for $x$, so $y = e^{2t}(C\sin 3t + D\cos 3t)$ | B1ft | FT on their answer to (i)(b) with two arbitrary constants
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## (iii)
$t = 0 \Rightarrow B = 4$ & $D = 5$ | M1 | Attempt to differentiate expression for $x$ or $y$ using the product rule (could have $B$ or $D$ substituted out)
$\frac{dy}{dt} = 2e^{2t}(C\sin 3t + D\cos 3t) + e^{2t}(3C\cos 3t - 3D\sin 3t)$ | M1 | Using the correct equation to set up equations in $C$ and $D$ (could have $B$ or $D$ substituted out)
$= y + 5x = e^{2t}(C\sin 3t + D\cos 3t) + 5e^{2t}(A\sin 3t + B\cos 3t)$ | M1 |
Comparing coefficients gives $2C - 3D = C + 5A$ $2D + 3C = D + 5B$ So $A = -2, C = 5$ | A1 |
Particular solutions are $x = 2e^{2t}(-\sin 3t + 2\cos 3t)$ and $y = 5e^{2t}(\sin 3t + \cos 3t)$ | A1 |
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## (iv)
$x = 0 \Rightarrow \tan 3T = 2 \Rightarrow T = 0.369$ | B1 | Setting $x = 0$ to and rearrange to an equation in $\tan 3T$
[1]

## (iv)
E.g. The model has continuous quantities, but the number of rabbits can only take integer values. | E1 |
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