# Question 6:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = e^x \Rightarrow \frac{dt}{dC} = e^x\frac{dx}{dC}$ or $\frac{dC}{dx} = t\frac{dC}{dt}$ or $\frac{dC}{dt} = e^{-x}\frac{dC}{dx}$ or $\frac{dC}{dt} = \frac{1}{t}\frac{dC}{dx}$ | M1 | Uses $t = e^x$ to obtain correct equation in terms of $\frac{dC}{dx}$, $\frac{dC}{dt}$ and $t$ (or $e^x$) or reciprocals |
| E.g. $\frac{dC}{dx} = t\frac{dC}{dt} \Rightarrow \frac{d^2C}{dx^2} \times \frac{dx}{dt} = t\frac{d^2C}{dt^2} + \frac{dC}{dt}$ | dM1, A1 | Differentiates again correctly with product rule and chain rule; needs fully correct calculus allowing sign errors only |
| $\frac{d^2C}{dx^2} \times \frac{1}{t} = t\frac{d^2C}{dt^2} + \frac{1}{t}\frac{dC}{dx} \Rightarrow t^2\frac{d^2C}{dt^2} = \frac{d^2C}{dx^2} - \frac{dC}{dx}$ | dM1 | Shows substitution clearly into DE to form new equation; dependent on first M mark and having two terms for second derivative |
| $t^2\frac{d^2C}{dt^2} - 5t\frac{dC}{dt} + 8C = \frac{d^2C}{dx^2} - \frac{dC}{dx} - 5\frac{dC}{dx} + 8C$ | | |
| $\frac{d^2C}{dx^2} - 6\frac{dC}{dx} + 8C = e^{3x}$ * | A1* | Fully correct proof with no errors |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $m^2 - 6m + 8 = 0 \Rightarrow m = 2, 4$ | M1 | Forms and solves quadratic auxiliary equation |
| $(C =)\ Ae^{4x} + Be^{2x}$ | A1ft | Correct form for CF; solutions must be distinct and real |
| PI is $C = ke^{3x}$ | B1 | Deduces correct form for PI |
| $\frac{dC}{dx} = 3ke^{3x},\ \frac{d^2C}{dx^2} = 9ke^{3x} \Rightarrow 9k - 18k + 8k = 1 \Rightarrow k = -1$ | M1 | Differentiates PI of correct form, substitutes into DE to find $k$ |
| $C = Ae^{4x} + Be^{2x} - e^{3x}$ | A1 | Correct GS for $C$ in terms of $x$ (must be seen explicitly unless implied by subsequent work) |
| $t = e^x \Rightarrow C = \ldots$ | M1 | Links solution of DE (II) to solution of model to find concentration at time $t$ |
| $C = At^4 + Bt^2 - t^3$ | A1 | Deduces correct GS for concentration |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = 6, C = 0 \Rightarrow 1296A + 36B - 216 = 0$ | M1 | Uses conditions $t = 6$, $C = 0$ to form equation in $A$ and $B$; acceptable to use $C$ in terms of $x$ using $x = \ln 6$ when $C = 0$ |
| $\frac{dC}{dt} = 4At^3 + 2Bt - 3t^2 \Rightarrow -36 = 864A + 12B - 108$ | M1 | Uses condition $\left(t=6, \frac{dC}{dt}=-36\right)$ to form another equation in $A$ and $B$; note: not acceptable to use $\frac{dC}{dx} = -36$ with $x = \ln 6$ |
| $A = 0,\ B = 6 \Rightarrow C = 6t^2 - t^3$ | A1 | Correct equation connecting $C$ with $t$ |
| $\frac{dC}{dt} = 12t - 3t^2 = 0 \Rightarrow t = 4 \Rightarrow C = \ldots$ | ddM1 | Uses suitable method to find maximum concentration; solves $\frac{dC}{dt} = 0$ for $t$ and substitutes; dependent on both previous method marks |
| $C = 6(4)^2 - (4)^3 = 32\ \mu\text{g}\text{L}^{-1}$ | A1 | Obtains $32\ \mu\text{g}\text{L}^{-1}$; units required (accept micrograms per litre, $\mu$g/L, $\mu$g/$l$, $\mu$g$l^{-1}$) |

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