Differential equations

The variables $x$ and $y$ obey the differential equation $\frac{dy}{dx} = x e^{x+y}$, and it is known that $y = 0$ when $x = 0$.

Feb/March 2016

The diagram shows a water tank with vertical sides and a horizontal rectangular base. The area of the base is $2\text{ m}^2$. When $t = 0$ the tank contains no water, and water starts to enter at a rate of $1\text{ m}^3$ per hour. At the same time, water also begins leaving through the base at a rate of $0.2\sqrt{h}\text{ m}^3$ per hour, where $h$ represents the depth of water in the tank after $t$ hours.

Feb/March 2017

Variables $x$ and $\theta$ are related by the differential equation $x\cos^2\theta\,\frac{dx}{d\theta} = 2\tan\theta + 1$, for $0 \le \theta < \frac{1}{2}\pi$ and $x > 0$. It is given that $x = 1$ when $\theta = \frac{1}{4}\pi$.

Feb/March 2018

The variables $x$ and $y$ are related by the differential equation $\frac{dy}{dx} = ky^3 e^{-x}$, where $k$ is a constant. It is known that $y = 1$ when $x = 0$, and that $y = \sqrt{e}$ when $x = 1$.

Feb/March 2019

The variables $x$ and $y$ are linked by the differential equation $\frac{dy}{dx} = \frac{1 + 4y^2}{e^x}$. It is known that $y = 0$ when $x = 1$.

Feb/March 2020

The variables $x$ and $y$ are linked by the differential equation $(1 - \cos x)\frac{dy}{dx} = y\sin x$. It is stated that $y = 4$ when $x = \pi$.

Feb/March 2021

The variables $x$ and $y$ are linked by the differential equation $(x + 1)(3x + 1)\frac{dy}{dx} = y$, with the condition that $y = 1$ when $x = 1$.

Feb/March 2022

The differential equation connecting $x$ and $y$ is $\frac{dy}{dx} = e^{3y} \sin^2 2x$. Also, $y = 0$ when $x = 0$.

Feb/March 2023

The variables $y$ and $\theta$ are related by $(1 + y)(1 + \cos 2\theta) \dfrac{dy}{d\theta} = e^{3y}$. It is also given that $y = 0$ when $\theta = \dfrac{1}{4}\pi$.

Feb/March 2024

The variables $x$ and $\theta$ are related by the differential equation $\dfrac{dx}{d\theta} = \left(\dfrac{1}{5}x + 1\right)\sin^2 2\theta$, with $x = 5$ when $\theta = 0$.

Feb/March 2025

If $y = 0$ when $x = 1$.

May/June 2010

The variables $x$ and $t$ satisfy the differential equation $e^{2t}\frac{dx}{dt} = \cos^2 x$, with $t \geq 0$. When $t = 0$, $x = 0$.

May/June 2010

Suppose that $x = 1$ when $t = 0$,

May/June 2010

The population of a particular bird species in a wooded region is monitored over a number of years. At time $t$ years, the bird count is $N$, with $N$ taken to be a continuous variable. The change in the bird population is represented by $\dfrac{dN}{dt} = \dfrac{N(1800 - N)}{3600}$. It is given that $N = 300$ when $t = 0$.

May/June 2011

For one curve, the gradient at a point $(x, y)$ is proportional to $xy$. At the point $(1, 2)$ the gradient is 4.

May/June 2011

In this chemical reaction, compound $X$ is produced from compounds $Y$ and $Z$. After $t$ seconds from the start of the reaction, the masses in grams of $X$, $Y$ and $Z$ are $x$, $10 - x$ and $20 - x$ respectively. At every instant, the rate at which $X$ is formed is proportional to the product of the masses of $Y$ and $Z$ present at that instant. When $t = 0$, $x = 0$ and $\frac{dx}{dt} = 2$.

May/June 2011

The variables $x$ and $y$ are linked by the differential equation $\frac{dy}{dx} = \frac{6xe^{3x}}{y^2}$. You are told that $y = 2$ when $x = 0$.

May/June 2012

The variables $x$ and $y$ satisfy the differential equation $\frac{dy}{dx} = e^{2x+y}$, with $y = 0$ when $x = 0$.

May/June 2012

In one chemical reaction, substance $A$ reacts with substance $B$. At $t$ seconds after the process begins, the masses in grams of $A$ and $B$ are $x$ and $y$ respectively. It is stated that $\frac{dy}{dt} = -0.6xy$ and $x = 5e^{-3t}$. When $t = 0$, $y = 70$.

May/June 2012

Liquid is entering a small tank that has a leak. At the beginning the tank contains no liquid, and after $t$ minutes the liquid volume in the tank is $V$ cm$^3$. The inflow rate is constant at $80$ cm$^3$ per minute. As a result of the leak, liquid is leaving the tank at a rate that, at any instant, is $kV$ cm$^3$ per minute, where $k$ is a positive constant.

May/June 2013

Write $\dfrac{1}{x^2(2x + 1)}$ as $\dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{2x + 1}$.

May/June 2013

The variables $x$ and $t$ are connected by the differential equation $t \frac{dx}{dt} = \frac{k - x^3}{2x^2}$, for $t > 0$, where $k$ is a constant. When $t = 1$, $x = 1$ and when $t = 4$, $x = 2$.

May/June 2013

The variables $x$ and $y$ satisfy the differential equation $\frac{dy}{dx} = \frac{6y e^{3x}}{2 + e^{3x}}$.

May/June 2014

At time $t$ years, the population of a country is $N$ millions. It is assumed that, at any instant, the rate at which $N$ increases is proportional to the product of $N$ and $(1 - 0.01N)$. When $t = 0$, $N = 20$ and $\frac{dN}{dt} = 0.32$.

May/June 2014

The variables $x$ and $\theta$ obey the differential equation $2\cos^2\theta\,\frac{dx}{d\theta} = \sqrt{2x + 1}$, and $x$ takes the value 0 when $\theta = \frac{1}{4}\pi$.

May/June 2014

Knowing that $y = 1$ when $x = 0$, solve the differential equation

May/June 2015

Let $x$ denote the population size at time $t$. If $x$ is treated as a continuous variable, it satisfies the differential equation $\frac{dx}{dt} = \frac{xe^{-t}}{k + e^{-t}}$, where $k$ is a positive constant.

May/June 2015

Let the number of micro-organisms in a population at time $t$ be represented by $M$. At any instant, the change in $M$ is assumed to obey the differential equation $\frac{dM}{dt} = k(\sqrt{M})\cos(0.02t)$, where $k$ is a constant and $M$ is treated as a continuous variable. It is given that when $t = 0$, $M = 100$.

May/June 2015

The variables $x$ and $y$ obey the differential equation $x\frac{dy}{dx} = y(1 - 2x^2)$, and the condition $y = 2$ when $x = 1$ is given.

May/June 2016

The variables $x$ and $\theta$ are linked by the differential equation $(3 + \cos 2\theta)\,\frac{dx}{d\theta} = x \sin 2\theta$, and the condition is that $x = 3$ when $\theta = \frac{1}{4}\pi$.

May/June 2016

The variables $x$ and $y$ obey the differential equation $\frac{dy}{dx} = e^{-2y}\tan^2 x$, for $0 \le x < \frac{\pi}{2}$, and you are told that $y = 0$ when $x = 0$.

May/June 2016

Write $\frac{1}{x(2x + 3)}$ as a partial fraction decomposition.

May/June 2017

In a certain chemical process, substance $A$ reacts with substance $B$ and reduces it. If the masses of $A$ and $B$ at time $t$ after the process begins are $x$ and $y$ respectively, it is given that $\frac{dy}{dt} = -0.2xy$ and $x = \frac{10}{(1+t)^2}$. At the start of the process, $y = 100$.

May/June 2017

During a particular chemical reaction, compound $A$ is produced from compound $B$. At time $t$ after the reaction begins, the masses of $A$ and $B$ are $x$ and $y$ respectively, and their total mass stays at $50$ for the whole reaction. At any moment, the rate at which the mass of $A$ increases is proportional to the mass of $B$ at that moment.

May/June 2017

In a particular chemical reaction, the quantity $x$ grams of a substance is falling. The differential equation connecting $x$ and $t$, where $t$ is the time in seconds since the reaction began, is $\frac{dx}{dt} = -k x \sqrt{t}$, where $k$ is a positive constant. It is stated that $x = 100$ at the beginning of the reaction.

May/June 2018

The diagram shows the tangent to the curve at $P$, where $P$ has coordinates $(x, y)$, and this tangent cuts the $x$-axis at $T$. $N$ is the point where the perpendicular from $P$ meets the $x$-axis. For every $x$, the curve has a positive gradient, and $TN = 2$.

May/June 2018

Write $\frac{1}{4 - y^2}$ as a partial fraction expression.

May/June 2018

Find the derivative of $\frac{1}{\sin^2 \theta}$ with respect to $\theta$.

May/June 2019

The variables $x$ and $y$ are linked by the differential equation $\frac{dy}{dx} = xe^{x+y}$. It is given that $y = 0$ when $x = 0$.

May/June 2019

The differential equation satisfied by the variables $x$ and $y$ is $(x + 1)\frac{dy}{dx} = y^2 + 5$, and the condition $y = 2$ applies when $x = 0$.

May/June 2019

A particular curve has gradient at a point $(x, y)$ proportional to $\dfrac{y}{x\sqrt{x}}$. It passes through the points $(1, 1)$ and $(4, e)$.

May/June 2020

The variables $x$ and $y$ obey the equation $y^2 = A e^{kx}$, where $A$ and $k$ are constants. The graph of $\ln y$ plotted against $x$ is a straight line, and it goes through the points $(1.5,\,1.2)$ and $(5.24,\,2.7)$, as shown in the diagram.

May/June 2020

The variables $x$ and $y$ are linked by the differential equation $\dfrac{dy}{dx} = \dfrac{y - 1}{(x + 1)(x + 3)}$. It is stated that $y = 2$ when $x = 0$.

May/June 2020

A tank of water has the form of a hemisphere. Its axis is vertical, the lowest point is $A$ and the radius is $r$, as shown in the diagram. The water depth at time $t$ is $h$. When $t = 0$ the tank is completely full, so the water depth is $r$. At this moment a tap at $A$ is opened and the water starts to leave at a rate proportional to $\sqrt{h}$. The tank is empty at time $t = 14$. When the depth is $h$, the volume of water in the tank is $V$. It is given that $V = \frac{1}{3}\pi(3rh^2 - h^3)$.

May/June 2020

The variables $x$ and $t$ are linked by the differential equation $\frac{dx}{dt} = x^2(1 + 2x)$, with the condition that $x = 1$ when $t = 0$.

May/June 2021

The curve has the property that the gradient at a point with coordinates $(x, y)$ is proportional to $\frac{y}{\sqrt{x+1}}$. It passes through the points with coordinates $(0, 1)$ and $(3, e)$.

May/June 2021

For the curve drawn in the diagram, the normal to the curve at the point $P$ with coordinates $(x, y)$ intersects the $x$-axis at $N$. Point $M$ is the perpendicular drop from $P$ to the $x$-axis. The curve is defined so that, for every value of $x$ in the range $0 \leq x < \frac{1}{2}\pi$, the area of triangle $PMN$ is $\tan x$.

May/June 2021

The differential equation $\frac{dy}{dx} = \frac{xy}{1 + x^2}$ is satisfied by the variables $x$ and $y$, and when $x = 0$, $y = 2$.

May/June 2022

The variables $x$ and $y$ are related by the differential equation $\frac{dy}{dx} = xe^{y-x}$, and $y=0$ when $x=0$.

May/June 2022

Let $N$ represent the number of insects in the population at $t$ days after observations begin. The change in the insect count is described by the differential equation $\frac{dN}{dt} = k N^{\frac{3}{2}} \cos 0.02t$, where $k$ is a constant and $N$ is regarded as a continuous variable. It is also given that $N = 100$ when $t = 0$.

May/June 2022

The variables $x$ and $y$ obey the differential equation $\cos 2x\,\frac{dy}{dx} = \frac{4\tan 2x}{\sin^2 3y}$, with $0 \le x < \frac{1}{4}\pi$. Also, $y = 0$ when $x = \frac{1}{6}\pi$.

May/June 2023

The variables $x$ and $y$ obey the differential equation $\frac{dy}{dx} = \frac{4 + 9y^2}{e^{2x+1}}$. It is also given that $y = 0$ when $x = 1$.

May/June 2023

The variables $x$ and $y$ are linked by the differential equation $\frac{dy}{dx} = \frac{y^2 + 4}{x(y + 4)}$ for $x > 0$. It is also stated that $x = 4$ when $y = 2\sqrt{3}$.

May/June 2023

A field contains 300 plants of one species, and every one of them may catch a particular disease. Let x be the number infected at time t after the first plant becomes infected. The rate at which x changes is proportional to the product of the number already infected and the number not yet infected. The variables x and t are regarded as continuous, and it is given that $\frac{dx}{dt} = 0.2$ and $x = 1$ when $t = 0$.

May/June 2024

If $y = \sec^3\theta$ is rewritten as $\frac{1}{\cos^3\theta}$, demonstrate that $\frac{dy}{d\theta} = 3\sin\theta\sec^4\theta$.

May/June 2024

The variables $x$ and $y$ obey the equation $ky = e^{cx}$, with $k$ and $c$ as constants. The plot of $ y$ against $x$ is a straight line that goes through the points $(2.80, 0.372)$ and $(5.10, 2.21)$, as the diagram shows.

May/June 2024

A container shaped as a cuboid has a square base with side $x$ and height $(10-x)$. It is stated that $x$ changes with time, $t$, where $t>0$. The container's volume is falling at a rate inversely proportional to $t$. When $t=\tfrac{1}{10}$, $x=\tfrac{1}{2}$ and the rate at which $x$ decreases is $\tfrac{20}{37}$.

May/June 2024

Determine the quotient and remainder when $x^3 + 5x^2 - 2x - 15$ is divided by $x^2 - 3$.

May/June 2025

The variables $x$ and $\theta$ are linked by the differential equation $\sin 2\theta\,\frac{dx}{d\theta} = (4x + 3)\cos 2\theta$, and $x = 0$ at $\theta = \tfrac{1}{12}\pi$.

May/June 2025

The variables $x$ and $y$ are linked by the differential equation $\sin 4y \frac{dy}{dx} = x \sin 2y \sin 3x$. It is also stated that $y = \frac{1}{12}\pi$ when $x = \frac{1}{2}\pi$.

May/June 2025

The variables $x$ and $y$ are related by the differential equation $(x^2 + 3)\frac{dy}{dx} = e^{3y}(x - 2)$. It is known that $y = 0$ when $x = 0$.

May/June 2025

A chemical reaction produces a certain substance. Let the mass of substance produced be $x$ grams after $t$ seconds from the start of the reaction. At every moment, the formation rate of the substance is proportional to $(20 - x)$. When $t = 0$, $x = 0$ and $\frac{dx}{dt} = 1.$

Oct/Nov 2010

A certain substance is produced during a chemical reaction. Let the mass of substance produced after $t$ seconds from the start of the reaction be $x$ grams. At every instant, the rate at which the substance is formed is proportional to $(20 - x)$. When $t = 0$, $x = 0$ and $\frac{dx}{dt} = 1$.

Oct/Nov 2010

A biologist is studying how a weed spreads across a particular region. At time $t$ weeks after the investigation begins, the area covered by the weed is $A \, \text{m}^2$. The biologist says that the rate at which $A$ increases is proportional to $\sqrt{(2A - 5)}$.

Oct/Nov 2010

A differential equation connects $x$ and $\theta$, with $0 < \theta < \frac{1}{2}\pi$; moreover, when $\theta = \frac{1}{12}\pi$, the value of $x$ is $0$.

Oct/Nov 2011

A differential equation links $x$ and $\theta$ as $\sin 2\theta \, \frac{dx}{d\theta} = (x + 1)\cos 2\theta$, with $0 < \theta < \frac{1}{2}\pi$. The condition $\theta = \frac{1}{12}\pi$ gives $x = 0$.

Oct/Nov 2011

In the experiment, the number of organisms at time $t$ days is represented by $N$, with $N$ taken to be a continuous variable. It is given that $\frac{dN}{dt} = 1.2e^{-0.02t}N^{0.5}$. At $t = 0$, the number of organisms present is 100.

Oct/Nov 2011

The variables $x$ and $y$ satisfy the differential equation $x \frac{dy}{dx} = 1 - y^2$. Also, when $x = 2$, $y = 0$.

Oct/Nov 2012

The variables $x$ and $y$ satisfy the differential equation $x\frac{dy}{dx} = 1 - y^2$. When $x = 2$, $y = 0$.

Oct/Nov 2012

A water-filled tank is shaped as a cone with vertex $C$. Its axis is vertical, and the semi-vertical angle is $60^\circ$, as the diagram shows. When $t = 0$, the tank is completely full and the water depth is $H$. At that moment, a tap at $C$ is turned on and water starts draining out. The volume of water in the tank falls at a rate proportional to $\sqrt{h}$, where $h$ denotes the water depth at time $t$. The tank is empty when $t = 60$.

Oct/Nov 2013

The water is held in a conical tank with vertex $C$. Its axis is vertical, and the semi-vertical angle is $60^\circ$, as the diagram indicates. When $t = 0$, the tank is completely filled and the water depth is $H$. At this moment, a tap at $C$ is opened so that water starts to leave the tank. The volume of water in the tank falls at a rate proportional to $\sqrt{h}$, where $h$ is the water depth at time $t$. The tank is empty at $t = 60$.

Oct/Nov 2013

One particular solution of the differential equation $3y^2 \frac{dy}{dx} = 4(y^3 + 1)\cos^2 x$ satisfies $y = 2$ when $x = 0$. The diagram gives a sketch of the graph of this solution for $0 \leq x \leq 2\pi$; stationary points are marked at $A$ and $B$.

Oct/Nov 2013

In one country, the government levies tax on every litre of petrol sold to motorists. The yearly revenue, in $R$ million dollars, depends on the tax rate $x$ dollars per litre. The relationship between $R$ and $x$ is represented by the differential equation $\frac{dR}{dx} = R\left(\frac{1}{x} - 0.57\right)$, with $R$ and $x$ treated as continuous variables. When $x = 0.5$, $R = 16.8$.

Oct/Nov 2014

In one country, the government levies tax on every litre of petrol bought by motorists. The annual revenue is $R$ million dollars when the tax rate is $x$ dollars per litre. The relationship between $R$ and $x$ is represented by the differential equation $\frac{dR}{dx} = R\left(\frac{1}{x} - 0.57\right)$, with $R$ and $x$ treated as continuous variables. When $x = 0.5$, $R = 16.8$.

Oct/Nov 2014

The variables $x$ and $y$ are linked by the differential equation $\dfrac{dy}{dx} = \frac{1}{5} x y^{\frac{1}{2}} \sin\!\left(\frac{1}{3}x\right)$.

Oct/Nov 2014

The variables $x$ and $\theta$ are related by the differential equation $\frac{dx}{d\theta} = (x + 2) \sin^2 2\theta$, and the condition $x = 0$ applies when $\theta = 0$.

Oct/Nov 2015

The variables $x$ and $\theta$ are linked by the differential equation $\frac{dx}{d\theta} = (x + 2) \sin^2 2\theta$, and $x = 0$ when $\theta = 0$ is given.

Oct/Nov 2015

Naturalists are running a wildlife reserve to raise the population of a rare plant species. The number of plants after $t$ years is represented by $N$, with $N$ taken as a continuous variable.

Oct/Nov 2015

A diseased soil infection is spreading across a field with total area $4\text{ km}^2$. After $t$ years, the infected region has area $x\text{ km}^2$, and its growth rate is described by $\frac{dx}{dt} = kx(4 - x)$, where $k$ is a positive constant. It is given that $x = 0.4$ when $t = 0$, and that $x = 2$ when $t = 2$.

Oct/Nov 2016

A broad field with area $4\,\text{km}^2$ is being affected by a soil disease. At time $t$ years, the diseased area is $x\,\text{km}^2$, and its growth rate is described by the differential equation $\frac{dx}{dt} = kx(4 - x)$, where $k$ is a positive constant. It is given that when $t = 0$, $x = 0.4$ and that when $t = 2$, $x = 2$.

Oct/Nov 2016

The diagram shows a moving point $P$ at $(x, y)$ and the point $N$, which is the foot of the perpendicular from $P$ to the $x$-axis. $P$ lies on a curve for which, whenever $x > 0$, the gradient of the curve is equal to the area of triangle $OPN$, where $O$ is the origin. The point $(0, 2)$ is on the curve.

Oct/Nov 2016

The variables $x$ and $y$ are linked by the differential equation $\frac{dy}{dx} = 4\cos^2 x\tan x$, for $0 \leq x < \frac{1}{2}\pi$, and $x = 0$ when $y = \frac{1}{4}\pi$.

Oct/Nov 2017

The variables $x$ and $y$ obey the differential equation $(x + 1)\frac{dy}{dx} = y(x + 2)$, and the condition $y = 2$ holds when $x = 1$.

Oct/Nov 2017

The variables $x$ and $y$ are linked by the differential equation $\frac{dy}{dx} = 4\cos^2 x\tan x$, for $0 \leq x < \frac{1}{2}\pi$, and $x = 0$ when $y = \frac{1}{4}\pi$.

Oct/Nov 2017

The differential equation satisfied by a general point $(x, y)$ on the curve is $x\frac{dy}{dx} = (2 - x^2)y$. The curve also passes through $(1, 1)$.

Oct/Nov 2018

For this curve, the gradient at a general point with coordinates $(x, y)$ is proportional to $\frac{y^2}{x}$. It also passes through the points with coordinates $(1, 1)$ and $(e, 2)$.

Oct/Nov 2018

Let $N$ denote the number of insects $t$ weeks after observations begin. The population is falling at a rate proportional to $Ne^{-0.02t}$. Both $N$ and $t$ are continuous variables, and when $t = 0$, $N = 1000$ and $\frac{dN}{dt} = -10$.

Oct/Nov 2019

The variables $x$ and $\theta$ are linked by the differential equation $\sin \frac{1}{2}\theta\, \frac{dx}{d\theta} = (x + 2)\cos \frac{1}{2}\theta$ for $0 < \theta < \pi$. It is given that $x = 1$ when $\theta = \frac{1}{3}\pi$.

Oct/Nov 2019

The variables $x$ and $t$ are linked by the differential equation $5\frac{dx}{dt} = (20 - x)(40 - x)$. You are told that $x = 10$ when $t = 0$.

Oct/Nov 2019

The coordinates $(x, y)$ of a typical point on the curve satisfy the differential equation $x \frac{dy}{dx} = (1 - 2x^2) y$, for $x > 0$. It is given that $y = 1$ when $x = 1$.

Oct/Nov 2020

The variables $x$ and $t$ are linked by the differential equation $e^{3t}\,\frac{dx}{dt} = \cos^2 2x$, for $t \geq 0$. It is given that $x = 0$ when $t = 0$.

Oct/Nov 2020

For a general point on the curve, the coordinates $(x, y)$ satisfy the differential equation $x\frac{dy}{dx} = (1 - 2x^2)y$, for $x > 0$. It is also given that $y = 1$ when $x = 1$.

Oct/Nov 2020

For $y = \ln(\ln x)$, show that $\frac{dy}{dx} = \frac{1}{x\ln x}$.

Oct/Nov 2021

The variables $x$ and $y$ are related by the differential equation $e^{2x} \frac{dy}{dx} = 4xy^2$, and the condition $y = 1$ is true when $x = 0$.

Oct/Nov 2021

A plantation with total area $20\text{ km}^2$ is being affected by a plant disease. At time $t$ years, the diseased area is $x\text{ km}^2$, and the rate at which $x$ increases is proportional to the ratio of the infected area to the area that is still unaffected. When $t = 0$, $x = 1$ and $\frac{dx}{dt} = 1$.

Oct/Nov 2021

In one chemical reaction, the quantity, $x$ grams, of a substance is rising. The differential equation linking $x$ and $t$, where $t$ is the time in seconds from the start of the reaction, is $\frac{dx}{dt} = kx e^{-0.1t}$, with $k$ a positive constant. It is also stated that $x = 20$ at the beginning of the reaction.

Oct/Nov 2022

The variables $x$ and $\theta$ satisfy the differential equation $x \sin^2 \theta \, \frac{dx}{d\theta} = \tan^2 \theta - 2 \cot \theta$, for $0 < \theta < \frac{1}{2}\pi$ and $x > 0$. It is also given that $x = 2$ when $\theta = \frac{1}{4}\pi$.

Oct/Nov 2022

A gardener is topping up an ornamental pool with water by means of a hose that supplies $30$ litres each minute. The pool starts off empty. If $t$ minutes have passed since the filling began, the amount of water in the pool is $V$ litres. There is a small leak in the pool, so water escapes at a rate of $0.01V$ litres per minute. The differential equation relating $V$ and $t$ has the form $\frac{dV}{dt} = a - bV$.

Oct/Nov 2022

The variables $x$ and $\theta$ are linked by the differential equation $\frac{x}{\tan \theta} \frac{dx}{d\theta} = x^2 + 3$. You are also told that $x = 1$ when $\theta = 0$.

Oct/Nov 2023

The variables $x$ and $y$ are linked by the differential equation $x^2 \frac{dy}{dx} + y^2 + y = 0$. Also given is $x = 1$ when $y = 1$.

Oct/Nov 2023