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A-Level Pure 4 · Edexcel IAL WMA14 · Topic pack

Topic pack · Differential equations แพ็กฝึกเฉพาะหัวข้อ · สมการเชิงอนุพันธ์

10 questions 10 ข้อ 55 marks 55 คะแนน 70 min 70 นาที Separable first-order & modelling, easy → hard

Ten questions on one topic, ordered easy → hard. The last three are full exam difficulty. Open the mark schemes only after a real attempt. สิบข้อหัวข้อเดียว เรียงง่าย → ยาก สามข้อสุดท้ายคือระดับข้อสอบจริง เปิดมาร์คสกีมหลังลองทำจริงเท่านั้น

⏱ 70:00 Scoreคะแนนรวม: –
Q1. 3 marks

Find the general solution of $\dfrac{dy}{dx}=2xy$, where $y>0$, giving your answer in the form $y=\mathrm{f}(x)$.

Mark schemeshow ▾
M1 separate variables: $\displaystyle\int\frac{dy}{y}=\int 2x\,dx$  A1 $\ln y=x^2+c$  A1 $y=Ae^{x^2}$ (where $A=e^{c}>0$).
Q2. 4 marks

Given that $\dfrac{dy}{dx}=\dfrac{6x^2}{y}$, $y>0$, and that $y=3$ when $x=1$, find the particular solution in the form $y^2=\mathrm{f}(x)$.

Mark schemeshow ▾
M1 separate: $\displaystyle\int y\,dy=\int 6x^2\,dx$  A1 $\tfrac{1}{2}y^2=2x^3+c$  M1 substitute $x=1$, $y=3$: $\tfrac92=2+c$, $c=\tfrac52$  A1 $y^2=4x^3+5$ $\left(\text{so } y=\sqrt{4x^3+5}\right)$.
Q3. 4 marks

The number of bacteria $N$ in a culture increases at a rate proportional to the number present. (a) Write down a differential equation for $N$ in terms of the time $t$. (b) Solve your equation to find the general solution for $N$ in terms of $t$.

Mark schemeshow ▾
(a) B1 $\dfrac{dN}{dt}=kN$, $k>0$ (translation of the words into an ODE).
(b) M1 separate: $\displaystyle\int\frac{dN}{N}=\int k\,dt$  A1 $\ln N=kt+c$  A1 $N=Ae^{kt}$.
Q4. 5 marks

A population of bacteria satisfies $\dfrac{dN}{dt}=kN$, where $t$ is the time in hours. Initially $N=500$, and after 6 hours $N=2000$. (a) Show that $N=500e^{kt}$ and find the exact value of $k$, giving also its value to 3 significant figures. (b) Find the number of bacteria after 9 hours.

Mark schemeshow ▾
(a) B1 general solution $N=Ae^{kt}$ with $N(0)=500 \Rightarrow N=500e^{kt}$ ∎  M1 $2000=500e^{6k} \Rightarrow e^{6k}=4$  A1 $k=\tfrac16\ln 4$ $(=0.231$ to 3 s.f.$)$.
(b) M1 $N=500e^{9k}=500\times 4^{3/2}$  A1 $N=4000$.
Q5. 5 marks

The mass $m$ grams of a radioactive substance decreases at a rate proportional to the mass present. Initially the mass is 80 g; after 10 years the mass is 60 g. (a) Write down a differential equation for $m$, and solve it to find $m$ in terms of the time $t$ years. (b) Find, to 3 significant figures, the value of $t$ when the mass is 40 g.

Mark schemeshow ▾
(a) B1 $\dfrac{dm}{dt}=-km$, $k>0$  M1 separating gives $m=80e^{-kt}$; use $m(10)=60$: $e^{-10k}=\tfrac34$  A1 $k=\tfrac{1}{10}\ln\tfrac43$, so $m=80e^{-\frac{t}{10}\ln\frac43}$.
(b) M1 $40=80e^{-kt} \Rightarrow kt=\ln 2$  A1 $t=\dfrac{10\ln 2}{\ln\frac43}=24.1$ years (3 s.f.).
Q6. 6 marks

Given that $\dfrac{dy}{dx}=xy+x$, $y>-1$, and that $y=2$ when $x=0$, find $y$ in terms of $x$, giving your answer in the form $y=\mathrm{f}(x)$.

Mark schemeshow ▾
M1 factorise and separate: $\dfrac{dy}{dx}=x(y+1)$, so $\displaystyle\int\frac{dy}{y+1}=\int x\,dx$  A1 $\ln(y+1)=\tfrac12 x^2+c$  M1 substitute $(0,2)$  A1 $c=\ln 3$  M1 exponentiate: $y+1=3e^{x^2/2}$  A1 $y=3e^{x^2/2}-1$.
Q7. 6 marks

A cup of coffee cools in a room of constant temperature $20\,^\circ$C. The rate of decrease of the coffee's temperature $\theta\,^\circ$C is proportional to $(\theta-20)$. Initially $\theta=80$, and after 5 minutes $\theta=60$. (a) Write down a differential equation for $\theta$ in terms of the time $t$ minutes. (b) Solve it to show that $\theta=20+60e^{-kt}$ and find the exact value of $k$. (c) Find, to 3 significant figures, the time at which the temperature reaches $40\,^\circ$C.

Mark schemeshow ▾
(a) B1 $\dfrac{d\theta}{dt}=-k(\theta-20)$, $k>0$.
(b) M1 $\displaystyle\int\frac{d\theta}{\theta-20}=-\int k\,dt$, $\ln(\theta-20)=-kt+c$  M1 use $\theta(0)=80$ and $\theta(5)=60$: $\theta-20=60e^{-kt}$, $e^{-5k}=\tfrac{40}{60}$  A1 $\theta=20+60e^{-kt}$ ∎ with $k=\tfrac15\ln\tfrac32$.
(c) M1 $40=20+60e^{-kt} \Rightarrow e^{-kt}=\tfrac13$  A1 $t=\dfrac{\ln 3}{k}=\dfrac{5\ln 3}{\ln\frac32}=13.5$ minutes (3 s.f.).
Q8. 7 marks

Water leaks from the bottom of a tank. The volume $V$ litres of water in the tank at time $t$ minutes satisfies $$\frac{dV}{dt}=-k\sqrt{V},\qquad k>0.$$ Initially the tank holds 400 litres, and after 30 minutes it holds 100 litres. (a) Solve the differential equation to show that $\sqrt{V}=20-\dfrac{kt}{2}$. (b) Find the value of $k$. (c) Find the time at which the tank is empty, and state the range of values of $t$ for which the model is valid.

Mark schemeshow ▾
(a) M1 separate: $\displaystyle\int V^{-1/2}\,dV=-\int k\,dt$  A1 $2\sqrt{V}=-kt+c$  M1 use $V(0)=400$: $c=2\sqrt{400}=40$  A1 $\sqrt{V}=20-\dfrac{kt}{2}$ ∎.
(b) M1 $V(30)=100$: $10=20-15k$  A1 $k=\tfrac23$  $\left(\text{so } V=\left(20-\tfrac{t}{3}\right)^2\right)$.
(c) B1 empty when $\sqrt{V}=0$: $t=\dfrac{40}{k}=60$ minutes; model valid for $0\le t\le 60$.
Q9. 7 marks

A metal rod is placed in an oven kept at a constant temperature of $100\,^\circ$C. The rate of increase of the rod's temperature $\theta\,^\circ$C is proportional to $(100-\theta)$. When the rod is placed in the oven, $\theta=20$; after 5 minutes, $\theta=60$. (a) Write down a differential equation for $\theta$ in terms of the time $t$ minutes. (b) Solve it to find $\theta$ in terms of $t$, giving the exact value of any constants. (c) Find the exact time at which $\theta=90$. (d) State the limiting value of $\theta$ as $t\to\infty$ and interpret it in the context of the model.

Mark schemeshow ▾
(a) B1 $\dfrac{d\theta}{dt}=k(100-\theta)$, $k>0$.
(b) M1 $\displaystyle\int\frac{d\theta}{100-\theta}=\int k\,dt$, $-\ln(100-\theta)=kt+c$  M1 use $\theta(0)=20$ and $\theta(5)=60$: $100-\theta=80e^{-kt}$, $e^{-5k}=\tfrac{40}{80}$  A1 $\theta=100-80e^{-kt}$ with $k=\tfrac15\ln 2$ $(=0.139$ to 3 s.f.$)$.
(c) M1 $90=100-80e^{-kt} \Rightarrow e^{-kt}=\tfrac18$  A1 $t=\dfrac{\ln 8}{k}=\dfrac{5\ln 8}{\ln 2}=15$ minutes exactly.
(d) B1 $\theta\to 100$: the rod's temperature approaches the oven temperature but never exceeds it.
Q10. 8 marks

The population $P$ (in thousands) of fish in a lake, $t$ years after the lake is stocked, is modelled by $$\frac{dP}{dt}=\frac{P(4-P)}{12},\qquad 0<P<4,$$ with $P=1$ when $t=0$. (a) Express $\dfrac{1}{P(4-P)}$ in partial fractions. (b) Hence show that $P=\dfrac{4}{1+3e^{-t/3}}$. (c) Find, to 3 significant figures, the time at which the population reaches 2000 fish. (d) State the limiting value of $P$ as $t\to\infty$ and interpret it in the context of the model.

Mark schemeshow ▾
(a) M1 $\dfrac{1}{P(4-P)}=\dfrac{A}{P}+\dfrac{B}{4-P}$, cover-up or compare coefficients  A1 $\dfrac{1}{4}\left(\dfrac1P+\dfrac{1}{4-P}\right)$.
(b) M1 separate: $\displaystyle\int\frac{dP}{P(4-P)}=\int\frac{dt}{12}$, so $\tfrac14\big(\ln P-\ln(4-P)\big)=\tfrac{t}{12}+c$, i.e. $\ln\dfrac{P}{4-P}=\dfrac{t}{3}+C$  M1 use $t=0$, $P=1$: $C=\ln\tfrac13$, so $\dfrac{P}{4-P}=\tfrac13 e^{t/3}$  A1 rearrange: $3P=(4-P)e^{t/3}$, so $P=\dfrac{4e^{t/3}}{3+e^{t/3}}=\dfrac{4}{1+3e^{-t/3}}$ ∎.
(c) M1 $2=\dfrac{4}{1+3e^{-t/3}} \Rightarrow 3e^{-t/3}=1 \Rightarrow t=3\ln 3$  A1 $t=3.30$ years (3 s.f.).
(d) B1 $P\to 4$: the population approaches a limiting value of 4000 fish, the long-term capacity of the lake.