Unit 4 · Topic 3.2 · Rates of change and differential equations
A differential equation tells you the gradient of a function, and your job is to find the function. Part 1 covers the forms you can solve by integrating. Part 2 applies them to the models the syllabus names: the logistic equation, radioactive decay and Newton's law of cooling. Slope fields have their own guide, Topic 3.3. Work through it in order, and tick off each skill once you can do it without help.
Only tick a skill when you could do it in an exam.
Part 1
Any equation that contains a derivative, such as $\frac{dy}{dx}$, is a differential equation. Solving one means finding $y$ itself.
The general solution contains a constant of integration, $c$, so it describes a whole family of curves. An initial condition, such as $y=3$ when $x=1$, picks out one curve from the family. That curve is the particular solution.
When the right-hand side has only $x$ in it, integrate both sides with respect to $x$.
For example, solve $\frac{dy}{dx}=6x^2-4$ given $y=3$ when $x=1$.
$$\begin{aligned}y&=2x^3-4x+c\\3&=2-4+c\ \Rightarrow\ c=5\\y&=2x^3-4x+5\end{aligned}$$
Sometimes the gradient is given in terms of $y$, for example $\frac{dy}{dx}=2y+2$. You can't integrate $2y+2$ with respect to $x$, because $y$ is the function you don't know yet. Instead, take the reciprocal of both sides, so that $x$ becomes a function of $y$.
$$\begin{aligned}\frac{dx}{dy}&=\frac{1}{2y+2}\\x&=\int\frac{1}{2y+2}\,dy=\frac12\ln|2y+2|+c\end{aligned}$$
The $\frac12$ comes from the 2 in front of $y$. Differentiate $\frac12\ln|2y+2|$ to check: you get $\frac12\times\frac{2}{2y+2}=\frac{1}{2y+2}$.
Rearranged, the general solution is $y=Ae^{2x}-1$. Taking the reciprocal assumes $2y+2\ne0$, so check the line $y=-1$ separately. Its gradient is $0$ everywhere, so it is a solution too.
Every grey curve is a solution of $\frac{dy}{dx}=2y+2$. Move the slider to choose the initial condition. The orange tangent at the starting point has the gradient the differential equation gives, and the initial condition picks out one curve.
At $y=-1$ the gradient is $0$, so the solution is a horizontal line. Taking the reciprocal divides by $2y+2$, which is $0$ here, so that method can't find this solution.
When the gradient is a function of $x$ multiplied by a function of $y$, collect every $y$ with $dy$ and every $x$ with $dx$. For example, solve $\frac{dy}{dx}=\left(x^2+1\right)e^y$.
$$\begin{aligned}\int e^{-y}\,dy&=\int\left(x^2+1\right)dx\\-e^{-y}&=\frac{x^3}{3}+x+c\end{aligned}$$
You only need one constant. Put it on the $x$ side.
This method covers the other two forms as well. Direct integration is the case $g(y)=1$, and taking the reciprocal is the case $f(x)=1$.
Practise
Determine the solution of the differential equation $\dfrac{dy}{dx}=\dfrac{\sin(2x)}{\cos(2x)}$ given $y=0$ when $x=\dfrac{\pi}{5}$.
$$y=\int\tan(2x)\,dx=-\frac12\ln|\cos(2x)|+c$$
$$0=-\frac12\ln\left|\cos\left(\tfrac{2\pi}{5}\right)\right|+c\ \Rightarrow\ c\approx-0.59$$
(A) and (B) have $-2$ in place of $-\frac12$. (D) has the wrong sign on the constant.
A particular solution to the differential equation $\dfrac{dy}{dx}=\dfrac{x}{\left(x^2+1\right)\tan(y)}$, where $x\ge0$ and $-\dfrac{\pi}{2}\lt y\le0$, passes through the origin.
Determine this solution in the form $x=f(y)$. Leave your answer in simplified form.
$$\int\tan(y)\,dy=\int\frac{x}{x^2+1}\,dx$$
$$-\ln|\cos(y)|=\frac12\ln\left(x^2+1\right)+c$$
$$(0,0):\quad-\ln(1)=\frac12\ln(1)+c\ \Rightarrow\ c=0$$
$$\begin{aligned}\ln\left(x^2+1\right)&=-2\ln|\cos(y)|=\ln\left(\sec^2(y)\right)\\x^2+1&=\sec^2(y)\end{aligned}$$
$$x^2=\sec^2(y)-1=\tan^2(y)\ \Rightarrow\ x=\pm\tan(y)$$
For $-\frac{\pi}{2}\lt y\le0$, $\tan(y)\le0$, but $x\ge0$. So take the negative.
$$x=-\tan(y)$$
$\int\tan(y)\,dy=-\ln|\cos(y)|$ is in the formula book. The identity $1+\tan^2(y)=\sec^2(y)$ is what gets you from the logarithms to a simple answer. This question is worked step by step in the complex unfamiliar collection.
A certain function $y=f(x)$, where $-\dfrac{\pi}{2}\lt x\lt\dfrac{\pi}{2}$, has a gradient of $\left(y^2-11y+30\right)\sec^2(x)$ and a $y$-intercept of 4.
Determine the value/s of $a$ such that $f(a)=a$.
$$\frac{dy}{dx}=(y-5)(y-6)\sec^2(x)\ \Rightarrow\ \int\frac{1}{(y-5)(y-6)}\,dy=\int\sec^2(x)\,dx$$
$$\frac{1}{(y-5)(y-6)}=\frac{1}{y-6}-\frac{1}{y-5}$$
$$\ln|y-6|-\ln|y-5|=\tan(x)+c$$
$$(0,4):\quad\ln\left|\frac{-2}{-1}\right|=\tan(0)+c\ \Rightarrow\ c=\ln(2)$$
At $(0,4)$, $\frac{y-6}{y-5}=2$ is positive, so the absolute value signs can go.
$$\begin{aligned}\frac{y-6}{y-5}&=2e^{\tan(x)}\\y-6&=2e^{\tan(x)}(y-5)\\y&=\frac{10e^{\tan(x)}-6}{2e^{\tan(x)}-1}\end{aligned}$$
$$\frac{10e^{\tan(a)}-6}{2e^{\tan(a)}-1}=a\ \Rightarrow\ a\approx-0.484$$
This is from the QCAA sample IA3, not a past external exam. There is no published marking guide, so this solution and the marks shown are my own. The solution through $(0,4)$ has a vertical asymptote where $2e^{\tan(x)}=1$, at $x\approx-0.606$. To the right of it, $y$ rises from $-\infty$ towards $5$, so it crosses the line $y=x$ exactly once.
This differential equation can be used to determine the current $I$ (amperes) at time $t$ (seconds) with voltage $V$ (volts) in an electric circuit containing a resistance $R$ (ohms):
$$k\frac{dI}{dt}+RI=V$$
where $k$, $R$ and $V$ are positive constants and $t\ge0$.
Assuming that there is no current in the electric circuit initially, show that the size of the current can never be greater than $\dfrac{V}{R}$.
$$k\frac{dI}{dt}=V-RI\ \Rightarrow\ \int\frac{k}{V-RI}\,dI=\int1\,dt$$
$$-\frac{k}{R}\ln|V-RI|=t+c$$
$$I=0,\ t=0:\quad c=-\frac{k}{R}\ln(V)\quad(V\gt0)$$
$$\ln|V-RI|=-\frac{R}{k}t+\ln(V)$$
$$V-RI=Ve^{-\frac{R}{k}t}$$
For all $t$, $Ve^{-\frac{R}{k}t}\gt0$, so $V-RI\gt0$, which gives $I\lt\frac{V}{R}$. The current can never be greater than $\frac{V}{R}$.
The $-\frac{k}{R}$ comes from the chain rule in reverse: the derivative of $V-RI$ with respect to $I$ is $-R$. As $t\to\infty$, the current approaches $\frac{V}{R}$, so this is the limit of the current. This question is worked step by step in the complex unfamiliar collection.
Part 2
The syllabus names three models: the logistic equation, radioactive decay and Newton's law of cooling. The exam gives you the differential equation, so you don't need to memorise it. Knowing its general form and what it measures helps you recognise what to do.
The logistic equation models growth that levels off, such as a population with limited food or space. It is used in the sciences and in economics.
$$\frac{dy}{dt}=y(a-by)\qquad\Rightarrow\qquad y=\frac{a}{b+Ce^{-at}}$$
While $y$ is small, the rate is close to $ay$, which is exponential growth. As $y$ gets close to $\frac{a}{b}$, the factor $a-by$ gets close to $0$ and the growth stops. So $y\to\frac{a}{b}$ as $t\to\infty$.
Do you need a separate method for this? Not really. The separation of variables steps in Part 1 will solve it. Recognising the form still saves you time in the exam: when the right-hand side is the population multiplied by a linear factor in the population, you know partial fractions are coming.
A quantity $m(t)$ that decays at a rate proportional to its size satisfies
$$\frac{dm}{dt}=-km$$
where $k$ is a positive constant, called the decay constant. The negative sign makes the rate negative, because $m$ is decreasing.
The half-life $h$ is the time it takes for half of the quantity to decay. From $\frac12=e^{-kh}$, you get $k=\frac{\ln(2)}{h}$.
Newton's law of cooling says the rate of change of an object's temperature is proportional to the difference between its temperature and the temperature of its surroundings, $T_m$.
$$\frac{dT}{dt}=-k\left(T-T_m\right),\qquad k\gt0$$
I use this form, with the negative sign written in and $k$ positive, all through this guide. When the object is hotter than the room, $T-T_m$ is positive and the rate is negative, so the object cools. Some books write $\frac{dT}{dt}=k\left(T-T_m\right)$ with $k\lt0$. That is the same model, so read the sign of $k$ from the question.
As $t\to\infty$, $e^{-kt}\to0$, so $T\to T_m$. The object gets closer and closer to room temperature but never reaches it.
The idea
The idea
Practise
The differential equation for which the solution is a logistic equation of the form $y=\dfrac{a}{b+Ce^{-at}}$, where $a$, $b$ and $C$ are constants, is
The logistic differential equation has the form $\frac{dy}{dt}=y(a-by)$.
$$0.25y(1-0.01y)=y(0.25-0.0025y)$$
(A) and (C) only involve $t$, so integrating them gives a polynomial. (B) has no factor of $y$ out the front, so its solution is an exponential that approaches $100$ with no S-shape.
A certain population can be approximately modelled by the differential equation
$$\frac{dP}{dt}=0.5P(1-0.2P)$$
where $P$ is the population in millions and $t$ is the number of years since 1 January 2025.
Given that the population on 1 January 2025 was estimated at 0.3 million, use a calculus approach to estimate the population on 1 January 2030.
$$\int\frac{1}{P(1-0.2P)}\,dP=\int0.5\,dt$$
$$\frac{1}{P(1-0.2P)}=\frac{1}{P}+\frac{0.2}{1-0.2P}$$
$$\ln|P|-\ln|1-0.2P|=0.5t+c$$
$$P=0.3,\ t=0:\quad c=\ln(0.3)-\ln(0.94)\approx-1.142$$
$$t=5:\quad\ln|P|-\ln|1-0.2P|=2.5-1.142=1.358$$
$$P\approx2.19$$
The estimated population on 1 January 2030 is about 2.2 million.
1 January 2030 is five years after the start, so $t=5$. The guide also accepts writing $1-0.2P$ as $\frac{5-P}{5}$ and using partial fractions on $\frac{1}{P(5-P)}$. Either way, this model levels off at 5 million.
The full walkthrough of 2025 Paper 2, Question 16 goes through it mark by mark.
For a certain experiment, the number of yeast cells, $N$, after $t$ hours in a test tube can be modelled by the differential equation
$$\frac{dN}{dt}=\frac{1}{1000}N(1000-N)\quad\text{for }t\ge0$$
A scientist commenced this experiment at 9:00 am on a certain day and observed that 100 yeast cells were present at this time.
The scientist predicted that the number of yeast cells would eventually exceed 1200.
Part a)
$$\int\frac{1000}{N(1000-N)}\,dN=\int1\,dt\ \Rightarrow\ \int\left(\frac{1}{N}+\frac{1}{1000-N}\right)dN=\int1\,dt$$
$$\begin{aligned}\ln|N|-\ln|1000-N|&=t+c\\\ln\left|\frac{N}{1000-N}\right|&=t+c\end{aligned}$$
Part b)
$$t=0,\ N=100:\quad\ln\left(\frac{100}{900}\right)=c\ \Rightarrow\ c=\ln\left(\frac19\right)$$
$$\frac{N}{1000-N}=e^{t+\ln\left(\frac19\right)}=\frac{e^t}{9}$$
$$\begin{aligned}9N&=1000e^t-Ne^t\\N\left(9+e^t\right)&=1000e^t\\N&=\frac{1000}{1+9e^{-t}}\end{aligned}$$
Part c)
$$900=\frac{1000}{1+9e^{-t}}\ \Rightarrow\ t\approx4.394\text{ hours}$$
4.394 hours is about 4 hours 24 minutes after 9:00 am, so the time of day is 1:24 pm.
Part d)
As $t\to\infty$, $e^{-t}\to0$, so $N\to1000$. The number of yeast cells never exceeds 1000.
$N$ can never reach 1200, so the scientist's prediction is not reasonable.
In a), the guide needs the line $\ln|N|-\ln|1000-N|=t+c$ or an equivalent step for the second mark. In b), it wants $N\left(9+e^t\right)=1000e^t$ or a similar step before the final form. For c), any time from 1:23 pm to 1:24 pm is accepted. By hand, $9e^{-t}=\frac19$ gives $t=\ln(81)$. This is the yeast in the animation above.
Practise
Radioactive decay and Newton's law of cooling haven't appeared directly in a QCAA external exam yet. I've written these three in the exam style, and the solutions and marks are my own.
The mass, $M$ (grams), of a radioactive isotope remaining after $t$ years can be modelled by the differential equation
$$\frac{dM}{dt}=-kM$$
where $k$ is a positive constant and $t\ge0$.
A sample of the isotope has an initial mass of 80 grams.
The half-life of the isotope is five years.
Part a)
$$\int\frac{1}{M}\,dM=-\int k\,dt\ \Rightarrow\ \ln(M)=-kt+c\quad(M\gt0)$$
$$t=0,\ M=80:\quad c=\ln(80)\ \Rightarrow\ M=e^{-kt+\ln(80)}=80e^{-kt}$$
Part b)
$$40=80e^{-5k}\ \Rightarrow\ e^{-5k}=\frac12$$
$$-5k=\ln\left(\frac12\right)\ \Rightarrow\ k=\frac{\ln(2)}{5}$$
Part c)
$$10=80e^{-kt}\ \Rightarrow\ e^{-kt}=\frac18\ \Rightarrow\ t=\frac{\ln(8)}{k}$$
$$t=\frac{3\ln(2)}{\ln(2)}\times5=15\text{ years}$$
Going from 80 g to 10 g is three halvings, so the answer to c) is three half-lives. That is a quick check.
Newton's law of cooling states that the rate of change of the temperature of an object is proportional to the difference between the temperature of the object and the temperature of its surroundings.
A cup of coffee is poured at 95 °C into a room held at a constant temperature of 22 °C. The temperature, $T$ (°C), of the coffee after $t$ minutes can be modelled by the differential equation
$$\frac{dT}{dt}=-k(T-22)$$
where $k$ is a positive constant and $t\ge0$.
The temperature of the coffee is 70 °C after five minutes.
A barista claims that the coffee will have cooled to room temperature within one hour of being poured.
Part a)
$$\int\frac{1}{T-22}\,dT=-\int k\,dt$$
$$\ln(T-22)=-kt+c\quad(T\gt22)$$
$$t=0,\ T=95:\quad c=\ln(73)\ \Rightarrow\ T-22=73e^{-kt}\ \Rightarrow\ T=22+73e^{-kt}$$
Part b)
$$70=22+73e^{-5k}\ \Rightarrow\ e^{-5k}=\frac{48}{73}$$
$$k=\frac15\ln\left(\frac{73}{48}\right)\approx0.0839$$
Part c)
$$T(15)=22+73e^{-15k}\approx42.8^{\circ}\text{C}$$
Part d)
$$T(60)=22+73e^{-60k}\approx22.5^{\circ}\text{C}$$
After one hour the coffee is still about 0.5 °C above room temperature. In this model it never reaches 22 °C, because $73e^{-kt}\gt0$ for every $t$. So the claim is not reasonable, although by then the coffee is very close to room temperature.
In c), $e^{-15k}=\left(e^{-5k}\right)^3=\left(\frac{48}{73}\right)^3$, which avoids rounding $k$. In d), I would also give the mark for deciding the claim is reasonable in practice, as long as you quote $T(60)$ and say that the model never quite reaches 22 °C.
A metal component is removed from a furnace and left to cool in a room held at a constant but unknown temperature. The temperature, $T$ (°C), of the component after $t$ minutes can be modelled by the differential equation
$$\frac{dT}{dt}=-k\left(T-T_r\right)$$
where $k$ is a positive constant and $T_r$ (°C) is the temperature of the room.
The temperature of the component is 100 °C when it is removed from the furnace, 60 °C after 10 minutes and 40 °C after 20 minutes.
Determine the temperature of the room and the exact value of $k$.
$$\int\frac{1}{T-T_r}\,dT=-\int k\,dt\ \Rightarrow\ \ln\left(T-T_r\right)=-kt+c$$
$$T=T_r+Ae^{-kt}$$
$$t=0,\ T=100:\quad A=100-T_r$$
$$\begin{aligned}60-T_r&=\left(100-T_r\right)e^{-10k}\\40-T_r&=\left(100-T_r\right)e^{-20k}\end{aligned}$$
Since $e^{-20k}=\left(e^{-10k}\right)^2$, square the first equation and divide it by the second.
$$\begin{aligned}\left(60-T_r\right)^2&=\left(100-T_r\right)\left(40-T_r\right)\\3600-120T_r&=4000-140T_r\\T_r&=20\end{aligned}$$
$$40=80e^{-10k}\ \Rightarrow\ k=\frac{\ln(2)}{10}$$
The room is at 20 °C.
The differences from room temperature are 80, 40 and 20 °C at equal 10-minute steps. Each one is half the one before, which is what an exponential model does over equal steps. That gives you a check on both answers.
Game plan
For the whole topic on one page, see Game plan 7 · Rates of change and differential equations in the External Game plan.
Practice
Ten questions of my own, worth 55 marks in total. Each one is labelled with the skill it practises, so if one goes wrong you know which part of the guide to go back to. Check whether each one is Technology-free or Technology-active before you start.
For a certain experiment, the number of bacteria, $N$, in a culture after $t$ hours can be modelled by the differential equation
$\frac{dN}{dt}=\frac{1}{500}N(500-N)\quad\text{for } t\ge0$
A scientist commenced this experiment at 8:00 am on a certain day and observed that 20 bacteria were present at this time.
Part a)
$\displaystyle \int\frac{500}{N(500-N)}\,dN=\int1\,dt\Rightarrow\int\left(\frac1N+\frac{1}{500-N}\right)dN=\int1\,dt$
$\ln|N|-\ln|500-N|=t+c\Rightarrow\displaystyle \ln\left|\frac{N}{500-N}\right|=t+c$
Part b)
$t=0$, $N=20$: $\displaystyle c=\ln\left(\frac{20}{480}\right)=\ln\left(\frac{1}{24}\right)=-\ln(24)$
$0<N<500$, so $\displaystyle \frac{N}{500-N}=\frac{e^t}{24}\Rightarrow24N=500e^t-Ne^t\Rightarrow N\left(24+e^t\right)=500e^t$
$\displaystyle N=\frac{500e^t}{24+e^t}=\frac{500}{1+24e^{-t}}$
Part c)
$\displaystyle 400=\frac{500}{1+24e^{-t}}\Rightarrow e^{-t}=\frac{1}{96}\Rightarrow t=\ln(96)\approx4.564$ hours
4.564 hours $\approx$ 4 hours 34 minutes after 8:00 am, i.e. 12:34 pm
Read 4.56 hours carefully. It is 4 hours and about 34 minutes, so 12:34 pm. Writing 4 hours 56 minutes loses the second mark of part c).
Consider the differential equation
$\frac{dy}{dx}=\frac{\sin(3x)}{\cos(3x)}$
for $-\frac{\pi}{6}<x<\frac{\pi}{6}$.
Part a)
$\displaystyle y=\int\frac{\sin(3x)}{\cos(3x)}\,dx$
$\displaystyle y=-\frac13\ln|\cos(3x)|+c$
Part b)
$\displaystyle 0=-\frac13\ln\left(\cos\left(\frac{\pi}{3}\right)\right)+c=-\frac13\ln\left(\frac12\right)+c\Rightarrow c=-\frac13\ln(2)$
For $-\tfrac{\pi}{6}<x<\tfrac{\pi}{6}$, $\cos(3x)>0$: $\displaystyle y=-\frac13\ln(\cos(3x))-\frac13\ln(2)=-\frac13\ln(2\cos(3x))$
Part c)
$\displaystyle y=-\frac13\ln\left(2\cos\left(\frac{\pi}{4}\right)\right)=-\frac13\ln\left(\sqrt2\right)=-\frac16\ln(2)$
The $\frac13$ comes from the $3x$. Multiplying by 3 instead loses the second mark of part a), but the later parts can still get follow-through marks.
The mass, $M$ (milligrams), of a radioactive isotope remaining after $t$ days can be modelled by the differential equation
$\frac{dM}{dt}=-kM$
where $k$ is a positive constant and $t\ge0$. A sample of the isotope has an initial mass of 120 milligrams.
The half-life of the isotope is eight days.
Part a)
$\displaystyle \int\frac1M\,dM=\int-k\,dt\Rightarrow\ln(M)=-kt+c$
$t=0$, $M=120$: $c=\ln(120)$
$\ln\left(\frac{M}{120}\right)=-kt\Rightarrow M=120e^{-kt}$
Part b)
$t=8$, $M=60$: $60=120e^{-8k}$
$\displaystyle e^{-8k}=\frac12\Rightarrow-8k=-\ln(2)\Rightarrow k=\frac{\ln(2)}{8}$
Part c)
$\displaystyle 15=120e^{-kt}\Rightarrow e^{-kt}=\frac18$
$\displaystyle -\frac{\ln(2)}{8}t=-3\ln(2)\Rightarrow t=24$ days
From 120 mg to 15 mg is three halvings, $120\to60\to30\to15$, so the answer is three half-lives. That reasoning gets both marks of part c) on its own.
Consider the differential equation
$\frac{dy}{dx}=2y+6$
where $y>-3$, and $y=1$ when $x=0$.
Part a)
$\displaystyle \frac{dx}{dy}=\frac{1}{2y+6}\Rightarrow\int1\,dx=\int\frac{1}{2y+6}\,dy$
$\displaystyle x=\frac12\ln(2y+6)+c$
$x=0$, $y=1$: $\displaystyle 0=\frac12\ln(8)+c\Rightarrow c=-\frac12\ln(8)$
$\displaystyle 2x=\ln\left(\frac{2y+6}{8}\right)\Rightarrow2y+6=8e^{2x}\Rightarrow y=4e^{2x}-3$
Part b)
$\displaystyle 9=4e^{2x}-3\Rightarrow e^{2x}=3\Rightarrow x=\frac12\ln(3)$
Leaving out the $\frac12$ gives $y=4e^{x}-3$, which doesn't satisfy the differential equation. That loses the second mark of part a), but you can still get the other marks.
A cup of tea is poured at $90^\circ$C in a room held at a constant temperature of $20^\circ$C. The temperature, $T$ ($^\circ$C), of the tea after $t$ minutes can be modelled by Newton's law of cooling
$\frac{dT}{dt}=-k(T-20)$
where $k$ is a positive constant and $t\ge0$.
The temperature of the tea is $60^\circ$C after ten minutes.
Part a)
$\displaystyle \int\frac{1}{T-20}\,dT=\int-k\,dt\Rightarrow\ln|T-20|=-kt+c$
$t=0$, $T=90$: $c=\ln(70)$
$T-20=70e^{-kt}\Rightarrow T=20+70e^{-kt}$
Part b)
$\displaystyle 60=20+70e^{-10k}\Rightarrow e^{-10k}=\frac47\Rightarrow k=\frac{1}{10}\ln\left(\frac74\right)\ (\approx0.0560)$
Part c)
$\displaystyle \frac{dT}{dt}=-k(50-20)=-30k\approx-1.68^{\circ}$C per minute (decreasing at $1.68^{\circ}$C per minute)
Part d)
$\displaystyle 30=20+70e^{-kt}\Rightarrow e^{-kt}=\frac17$
$\displaystyle t=\frac{\ln(7)}{k}\approx34.77$ minutes
In part c), use the differential equation with $T=50$. Leaving out the room temperature, or substituting $t=50$ into the solution, doesn't get the mark. In part d), carry the exact value of $k$ so rounding doesn't shift your answer.
Consider the differential equation
$\frac{dy}{dx}=e^{x-y}$
where $y=\ln(2)$ when $x=0$.
Part a)
$\displaystyle \frac{dy}{dx}=e^{x-y}=\frac{e^x}{e^y}\Rightarrow e^y\frac{dy}{dx}=e^x$
Part b)
$\displaystyle \int e^y\,dy=\int e^x\,dx\Rightarrow e^y=e^x+c$
$x=0$, $y=\ln(2)$: $2=1+c\Rightarrow c=1$
$e^y=e^x+1\Rightarrow y=\ln\left(e^x+1\right)$
Part c)
$y=\ln\left(e^{\ln(3)}+1\right)=\ln(4)\ (=2\ln(2))$
Taking logarithms term by term, to get $y=x$ or $y=x+1$, loses the third mark of part b). Neither one passes through $(0,\ln(2))$.
The number of people, $y$ (thousands), who have downloaded a new app $t$ weeks after its release can be modelled by the logistic function
$y=\frac{10}{1+4e^{-2t}}$
for $t\ge0$.
Part a)
$\displaystyle y=10\left(1+4e^{-2t}\right)^{-1}\Rightarrow\frac{dy}{dt}=\frac{80e^{-2t}}{\left(1+4e^{-2t}\right)^2}$
$\displaystyle 0.2y(10-y)=\frac{2}{1+4e^{-2t}}\left(10-\frac{10}{1+4e^{-2t}}\right)=\frac{2}{1+4e^{-2t}}\times\frac{40e^{-2t}}{1+4e^{-2t}}=\frac{80e^{-2t}}{\left(1+4e^{-2t}\right)^2}=\frac{dy}{dt}$
Part b)
$\displaystyle \frac{dy}{dt}=2y-0.2y^2\Rightarrow\frac{d}{dy}\left(\frac{dy}{dt}\right)=2-0.4y=0$
$y=5$, a maximum of the concave-down quadratic, so 5000 people
Part c)
$\displaystyle 5=\frac{10}{1+4e^{-2t}}\Rightarrow4e^{-2t}=1\Rightarrow t=\frac12\ln(4)=\ln(2)$ weeks
Completing the square, $2y-0.2y^2=5-0.2(y-5)^2$, also works in part b). At $y=10$ the rate is zero, which is where growth stops.
A drug is delivered into a patient's bloodstream through a drip at a constant rate. At the same time, the drug is removed from the bloodstream by the body. The amount of the drug, $A$ (mg), in the bloodstream $t$ hours after the drip is started can be modelled by the differential equation
$\frac{dA}{dt}=3-0.2A$
for $0\le A<15$.
There is no drug in the bloodstream when the drip is started.
Part a)
$\displaystyle \int\frac{1}{3-0.2A}\,dA=\int1\,dt$
$-5\ln|3-0.2A|=t+c$
$t=0$, $A=0$: $c=-5\ln(3)$
$\displaystyle \ln\left(\frac{3-0.2A}{3}\right)=-0.2t\Rightarrow3-0.2A=3e^{-0.2t}\Rightarrow A=15\left(1-e^{-0.2t}\right)$
Part b)
$\displaystyle 12=15\left(1-e^{-0.2t}\right)\Rightarrow e^{-0.2t}=0.2\Rightarrow t=5\ln(5)\approx8.05$ hours
Part c)
$A=15-15e^{-0.2t}$ and $15e^{-0.2t}>0$ for all $t$, so $A<15$; $A\to15$ as $t\to\infty$.
In part c), you can also argue from the differential equation: $\frac{dA}{dt}\gt0$ while $A\lt15$, and $\frac{dA}{dt}=0$ when $A=15$. Leaving out the $-5$ in part a) gives a negative amount and loses the second mark.
A particular solution to the differential equation
$\frac{dy}{dx}=x\left(y^2-1\right)$
for $y>1$.
passes through the point $(0,3)$.
Determine this solution in the form $y=f(x)$.
$\displaystyle \int\frac{1}{y^2-1}\,dy=\int x\,dx$
$\displaystyle \frac{1}{(y-1)(y+1)}=\frac{1}{2}\left(\frac{1}{y-1}-\frac{1}{y+1}\right)$
$\displaystyle \frac12\ln\left|\frac{y-1}{y+1}\right|=\frac{x^2}{2}+c$
$(0,3)$: $\displaystyle \frac12\ln\left(\frac24\right)=c\Rightarrow c=-\frac12\ln(2)$
$y>1$, so $\displaystyle \frac{y-1}{y+1}>0$: $\displaystyle \ln\left(\frac{y-1}{y+1}\right)=x^2-\ln(2)\Rightarrow\frac{y-1}{y+1}=\frac12e^{x^2}$
$2y-2=ye^{x^2}+e^{x^2}\Rightarrow y\left(2-e^{x^2}\right)=2+e^{x^2}\Rightarrow\displaystyle y=\frac{2+e^{x^2}}{2-e^{x^2}}$
$\displaystyle y=\frac{2+e^{x^2}}{2-e^{x^2}}$
Leaving out the $\frac12$ in the partial fractions gives an answer that doesn't satisfy the differential equation. That loses the second mark, but the rest can still follow through. The solution only exists for $|x|\lt\sqrt{\ln(2)}$, although you don't need to state that.
A certain population of native frogs can be modelled by the differential equation
$\frac{dP}{dt}=0.4P(1-0.05P)$
where $P$ is the population in thousands and $t$ is the number of years since 1 January 2026. The population on 1 January 2026 was estimated at 2 thousand.
An ecologist claims that, according to this model, the population will exceed 18 thousand on 1 January 2036.
Use a calculus approach to evaluate the reasonableness of the ecologist's claim. Justify your decision using mathematical reasoning.
$\displaystyle \frac{dP}{dt}=0.02P(20-P)\Rightarrow\int\frac{20}{P(20-P)}\,dP=\int0.4\,dt$, where $\displaystyle \frac{20}{P(20-P)}=\frac1P+\frac{1}{20-P}$
$\displaystyle \ln\left|\frac{P}{20-P}\right|=0.4t+c$
$t=0$, $P=2$: $\displaystyle c=\ln\left(\frac{2}{18}\right)=-\ln(9)$, so $\displaystyle \frac{P}{20-P}=\frac{e^{0.4t}}{9}\Rightarrow P=\frac{20}{1+9e^{-0.4t}}$
1 January 2036: $t=10$, $\displaystyle P(10)=\frac{20}{1+9e^{-4}}\approx17.17$ thousand
$17.17<18$, so the population will not exceed 18 thousand on 1 January 2036 (it reaches 18 thousand only when $t\approx10.99$ years).
The ecologist's claim is not reasonable.
Arguing only that the population levels off at 20 thousand, which is more than 18, doesn't answer the question about 2036 and loses the last two marks. Taking $P(10)$ straight from a technology solver doesn't get the first three marks.
Keep practising
2023 Paper 1, Question 18 and 2021 Paper 1, Question 18 are both worked step by step in the complex unfamiliar collection, alongside the other complex unfamiliar questions from every past paper.
Past questions and marking-guide steps are from the 2020–2025 QCAA Specialist Mathematics external assessments and sample assessment, © State of Queensland (QCAA) 2020–2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
© 2026 Tangent Tuition. This resource is licensed to you for your own study. Please don’t share it, send it to anyone else or post it anywhere. See the Terms of Use.
Spot a mistake on this page? Report it