# Question 8 (Differential Equations):

## Part (a):

| Working | Mark | Guidance |
|---------|------|----------|
| $\dfrac{d^2w}{dt^2} = \dfrac{5}{2}\left(\dfrac{dw}{dt} - \dfrac{ds}{dt}\right)$ or $\dfrac{ds}{dt} = \dfrac{dw}{dt} - \dfrac{2}{5}\dfrac{d^2w}{dt^2}$ | B1 | Differentiates first equation with respect to $t$ correctly |
| $\dfrac{ds}{dt} = \dfrac{dw}{dt} - \dfrac{2}{5}\dfrac{d^2w}{dt^2} \Rightarrow \dfrac{dw}{dt} - \dfrac{2}{5}\dfrac{d^2w}{dt^2} = \dfrac{2}{5}w - 90e^{-t}$ | M1 | Substitutes $\dfrac{ds}{dt}$ into derivative |
| $2\dfrac{d^2w}{dt^2} - 5\dfrac{dw}{dt} + 2w = 450e^{-t}$ | A1* | Printed answer; achieved with no errors |

## Part (b):

| Working | Mark | Guidance |
|---------|------|----------|
| $2m^2 - 5m + 2 = 0 \Rightarrow m = \ldots$ | M1 | Forms and solves auxiliary equation |
| $m = 2, \tfrac{1}{2}$ | A1 | Correct roots |
| $(w) = Ae^{\alpha t} + Be^{\beta t}$ | M1 | Forms complementary function from their roots |
| $(w) = Ae^{0.5t} + Be^{2t}$ | A1 | Correct CF |
| Try $w = ke^{-t} \Rightarrow \dfrac{dw}{dt} = -ke^{-t} \Rightarrow \dfrac{d^2w}{dt^2} = ke^{-t}$; $2ke^{-t} + 5ke^{-t} + 2ke^{-t} = 450e^{-t} \Rightarrow k = \ldots$ | M1 | Correct PI form; finds both derivatives; substitutes |
| $w = \text{their CF} + 50e^{-t}$, i.e. $w = Ae^{0.5t} + Be^{2t} + 50e^{-t}$ | A1ft | Dependent on all three M marks |

## Part (c):

| Working | Mark | Guidance |
|---------|------|----------|
| $s = w - \dfrac{2}{5}\dfrac{dw}{dt} = Ae^{0.5t} + Be^{2t} + 50e^{-t} - \dfrac{2}{5}\left(\dfrac{A}{2}e^{0.5t} + 2Be^{2t} - 50e^{-t}\right)$ | M1 | Substitutes part (b) answer for $w$ and derivative into first equation |
| $s = \dfrac{4A}{5}e^{0.5t} + \dfrac{B}{5}e^{2t} + 70e^{-t}$ | A1 | Correct simplified equation |

## Part (d):

| Working | Mark | Guidance |
|---------|------|----------|
| $65 = A + B + 50$, $85 = \dfrac{4A}{5} + \dfrac{B}{5} + 70 \Rightarrow A = \ldots, B = \ldots$ (NB $A=20$, $B=-5$) | M1 | Uses $t=0$, $w=65$, $s=85$ to form simultaneous equations |
| $w = 0 \Rightarrow 20e^{0.5t} - 5e^{2t} + 50e^{-t} = 0$ | dM1 | Sets $w = 0$; dependent on previous M1 |
| $e^{3t} - 4e^{1.5t} - 10 = 0$ or a multiple | A1 | Processes indices correctly to 3-term quadratic in $e^{1.5t}$ |
| $e^{1.5t} = \dfrac{4 \pm \sqrt{4^2 - 4(1)(-10)}}{2}$ | M1 | Solves 3TQ to reach $e^{pt} = q$ |
| $1.5t = \ln\!\left(\dfrac{4+\sqrt{56}}{2}\right)$ | M1 | Correct use of logarithms; $pt = \ln q$ where $q > 0$; rejects other solution |
| $T = \dfrac{2}{3}\ln\!\left(\dfrac{4+\sqrt{56}}{2}\right) \approx 1.165$ | A1 | awrt 1.165 |

## Part (e):

| Working | Mark | Guidance |
|---------|------|----------|
| E.g. either population becomes negative which is not possible; or when white-clawed crayfish have died out the model will not be valid | B1 | Any valid contextual reason |

## Question (e):

| Answer | Mark | Guidance |
|--------|------|----------|
| Suggests a suitable limitation of the model, e.g. not valid when negative population | B1 | Any mention of other factors such as does not take into account e.g. other predators, fishing, disease, lack of food etc is B0 |

**Note:** The final 3 marks only can be implied by a correct answer following the correct 3-term quadratic equation in terms of $e^{1.5t}$