Separable differential equations and modelling – A-level Maths exam questions with mark scheme

1StandardCalculator5 marks

Given that $y=0$ when $x=0$, solve the differential equation $$\frac{\mathrm{d}y}{\mathrm{d}x}=(2x+1)\mathrm{e}^{-y}$$ giving your answer in the form $y=\mathrm{f}(x)$. [5]

Separates variables: $\displaystyle\int\mathrm{e}^{y}\,\mathrm{d}y=\int(2x+1)\,\mathrm{d}x$M1
$\mathrm{e}^{y}=x^{2}+x\ (+c)$A1
Uses $x=0,\ y=0$ to find $c$ (must have included a constant)M1
$c=1$A1
$y=\ln(x^{2}+x+1)$A1
Answer: $y=\ln(x^{2}+x+1)$
Examiner tip: Find $c$ before taking logs – writing $y=\ln(x^{2}+x)+c$ is a different (wrong) function.

2HardCalculator8 marks

A cup of tea is placed in a room kept at a constant temperature of $20^\circ$C. The temperature of the tea, $\theta\,^\circ$C, $t$ minutes after it is made, is modelled by $$\frac{\mathrm{d}\theta}{\mathrm{d}t}=-k(\theta-20)$$ where $k$ is a positive constant. Initially the temperature of the tea is $90^\circ$C.

(a) Show that $\theta=20+70\mathrm{e}^{-kt}$. [4]

After 5 minutes the temperature of the tea is $60^\circ$C.

(b) Find the exact value of $k$, giving your answer in the form $a\ln b$ where $a$ and $b$ are rational. [2]

(a) Separates: $\displaystyle\int\frac{1}{\theta-20}\,\mathrm{d}\theta=\int-k\,\mathrm{d}t$M1
$\ln(\theta-20)=-kt+c$A1
Uses $t=0,\ \theta=90$ to find $c=\ln70$ (or $A=70$ in $\theta-20=A\mathrm{e}^{-kt}$)M1
Removes logs correctly to reach $\theta=20+70\mathrm{e}^{-kt}$ csoA1*
(b) $60=20+70\mathrm{e}^{-5k}\Rightarrow\mathrm{e}^{-5k}=\tfrac47$ and takes logsM1
$k=\tfrac15\ln\tfrac74$ (accept $-\tfrac15\ln\tfrac47$)A1
(c) $30=20+70\mathrm{e}^{-kt}\Rightarrow t=\dfrac{\ln7}{k}$ with their $k$M1
$17$ minutes (from $t=17.39\ldots$)A1
Answer: (b) $k=\tfrac15\ln\tfrac74$ (c) 17 minutes
Examiner tip: Include the constant of integration as soon as you integrate – adding it after removing logs gives the wrong form.

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