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PSLE Foundation Mathematics

<p>A particle moves back and forth in a horizontal line between two end points. A fixed origin $O$ is the midpoint between these end points. Its displacement $x$ metres from $O$ at time $t$ seconds is $x=4\sin(2t+\frac\pi3)$, where&nbsp;$0\leq t\leq\pi$.</p>
<ol class="abc-list"><li data-list-item-id="p3-tzc-q02-item01" data-list-item-path="p3-tzc-q02-item01"><div class="list-item-paragraph">Determine</div><ol class="i-list"><li data-list-item-id="p3-tzc-q02-sub01-item01" data-list-item-path="p3-tzc-q02-item01/p3-tzc-q02-sub01-item01"><div class="list-item-paragraph">the amplitude;<span class="inline-marks">[1]</span></div></li><li data-list-item-id="p3-tzc-q02-sub01-item02" data-list-item-path="p3-tzc-q02-item01/p3-tzc-q02-sub01-item02"><div class="list-item-paragraph">the initial displacement;<span class="inline-marks">[2]</span></div></li><li data-list-item-id="p3-tzc-q02-sub01-item03" data-list-item-path="p3-tzc-q02-item01/p3-tzc-q02-sub01-item03"><div class="list-item-paragraph">the value of $t$ when the particle first passes through $O$.<span class="inline-marks">[2]</span></div></li></ol></li><li data-list-item-id="p3-tzc-q02-item02" data-list-item-path="p3-tzc-q02-item02"><div class="list-item-paragraph">The acceleration $a$ at displacement $s$ satisfies $a=-n^2s$, where&nbsp;$n&gt;0$.</div><div class="list-item-paragraph">The displacement is $s=A\sin(nt+b)$, where&nbsp;$t\geq0$, $A,n&gt;0$ and $-\pi\leq b\leq\pi$.</div><div class="list-item-paragraph">By finding expressions for&nbsp;$\frac{\mathrm{d}s}{\mathrm{d}t}$ and $\frac{\mathrm{d}^2s}{\mathrm{d}t^2}$, verify that&nbsp;$s=A\sin(nt+b)$ satisfies $a=-n^2s$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="p3-tzc-q02-item03" data-list-item-path="p3-tzc-q02-item03"><ol class="i-list"><li data-list-item-id="p3-tzc-q02-sub03-item01" data-list-item-path="p3-tzc-q02-item03/p3-tzc-q02-sub03-item01"><div class="list-item-paragraph">Use the chain rule to show that&nbsp;$a=v\frac{\mathrm{d}v}{\mathrm{d}s}$, where&nbsp;$v$ is velocity.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="p3-tzc-q02-sub03-item02" data-list-item-path="p3-tzc-q02-item03/p3-tzc-q02-sub03-item02"><div class="list-item-paragraph">By solving $v\frac{\mathrm{d}v}{\mathrm{d}s}=-n^2s$, show that&nbsp;$v^2=n^2(A^2-s^2)$.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="p3-tzc-q02-sub03-item03" data-list-item-path="p3-tzc-q02-item03/p3-tzc-q02-sub03-item03"><div class="list-item-paragraph">Hence, or otherwise, find the particle’s maximum speed.<span class="inline-marks">[2]</span></div></li></ol></li><li data-list-item-id="p3-tzc-q02-item04" data-list-item-path="p3-tzc-q02-item04"><div class="list-item-paragraph">The continuous random variable $S$ denotes the particle’s displacement $s$ from $O$ at time $t$.</div><div class="list-item-paragraph">\[f(s)=\begin{cases}\frac1{\pi\sqrt{A^2-s^2}},&-A&lt;s&lt;A,\\0,&\text{otherwise}.\end{cases}\]</div><div class="list-item-paragraph">Show that&nbsp;$P(-\frac A2\leq S\leq\frac A2)=\frac13$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="p3-tzc-q02-item05" data-list-item-path="p3-tzc-q02-item05"><div class="list-item-paragraph">For $-A&lt;s&lt;A$, $f(s)$ can be expressed as $\frac{k}{|v(s)|}$, where $k&gt;0$ and $v(s)$ is the particle’s velocity at displacement $s$ from $O$.</div><div class="list-item-paragraph">Find the value of $k$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p3-tzc-q02-item06" data-list-item-path="p3-tzc-q02-item06"><ol class="i-list"><li data-list-item-id="p3-tzc-q02-sub06-item01" data-list-item-path="p3-tzc-q02-item06/p3-tzc-q02-sub06-item01"><div class="list-item-paragraph">Determine $E(S)$, justifying your answer.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="p3-tzc-q02-sub06-item02" data-list-item-path="p3-tzc-q02-item06/p3-tzc-q02-sub06-item02"><div class="list-item-paragraph">Interpret the result in the context of the particle’s motion.<span class="inline-marks">[1]</span></div></li></ol></li></ol>

A particle moves back and forth in a horizontal line between two end points. A fixed origin $O$ is the midpoint between these end points. Its displacement $x$ metres from $O$ at time $t$ seconds is $x=4\sin(2t+\frac\pi3)$, where $0\leq t\leq\pi$.

Solving Differential Equations by Separation of Variables

Given the Displacement Function

Derivatives of Trigonometric Functions

Probability Density Functions

Expectation and Variance of Continuous Random Variables

Second and Higher Derivatives

Product, Quotient and Chain Rules

Probabilities by Integrating a Density Function

IB AA HL May 2026 P3 TZC Q2

Techniques of Differentiation

Continuous Random Variables

Differential Equations

Kinematics

<p>Rowan is investigating the spread of an infection through a population of rabbits.</p>
<p>He models $N$, the number (in thousands) of infected rabbits in the population, at time $t$ days.</p>
<p>Rowan models the spread of the infection by the differential equation $\frac{\mathrm{d}N}{\mathrm{d}t}=k(-3+4N-N^2)$, where&nbsp;$t\geq0$ and $k\in\mathbb{R}^+$.</p>
<p>Initially, $1500$ rabbits are infected.</p>
<ol class="abc-list"><li data-list-item-id="p1-tza-q12-item01" data-list-item-path="p1-tza-q12-item01"><div class="list-item-paragraph">Express $\frac1{-3+4x-x^2}$ in the form $\frac{A}{x-1}+\frac{B}{3-x}$, where&nbsp;$A,B\in\mathbb{R}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p1-tza-q12-item02" data-list-item-path="p1-tza-q12-item02"><div class="list-item-paragraph">Hence, show that&nbsp;$\ln\left(\frac{N-1}{3-N}\right)=2kt-\ln p$, where&nbsp;$p\in\mathbb{R}^+$ is a constant to be determined.<span class="inline-marks">[8]</span></div></li><li data-list-item-id="p1-tza-q12-item03" data-list-item-path="p1-tza-q12-item03"><div class="list-item-paragraph">Find an expression for&nbsp;$N$ in the form $N=\frac{a+be^{-2kt}}{1+ce^{-2kt}}$, where&nbsp;$a,b,c\in\mathbb{Z}$.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="p1-tza-q12-item04" data-list-item-path="p1-tza-q12-item04"><div class="list-item-paragraph">According to Rowan’s model, the number of infected rabbits approaches a limit, $L$, over the long term.</div><div class="list-item-paragraph">Find the value of $L$.<span class="inline-marks">[2]</span></div></li></ol>

Rowan is investigating the spread of an infection through a population of rabbits.

He models $N$, the number (in thousands) of infected rabbits in the population, at time $t$ days.

Rowan models the spread of the infection by the differential equation $\frac{\mathrm{d}N}{\mathrm{d}t}=k(-3+4N-N^2)$, where $t\geq0$ and $k\in\mathbb{R}^+$.

Initially, $1500$ rabbits are infected.

Express $\frac1{-3+4x-x^2}$ in the form $\frac{A}{x-1}+\frac{B}{3-x}$, where $A,B\in\mathbb{R}$.

Hence, show that $\ln\left(\frac{N-1}{3-N}\right)=2kt-\ln p$, where $p\in\mathbb{R}^+$ is a constant to be determined.

Find an expression for $N$ in the form $N=\frac{a+be^{-2kt}}{1+ce^{-2kt}}$, where $a,b,c\in\mathbb{Z}$.

Integration Using Partial Fractions

Partial Fractions

IB AA HL May 2026 P1 TZA Q12

Polynomials, Cubic Equations and Partial Fractions

Techniques of Integration

<p>A particle moves back and forth in a horizontal line between two end points. A fixed origin $O$ is the midpoint between these end points. Its displacement $x$ metres from $O$ at time $t$ seconds is $x=4\sin(2t+\frac\pi3)$, where&nbsp;$0\leq t\leq\pi$.</p>
<ol class="abc-list"><li data-list-item-id="h2-motion-3" data-list-item-path="h2-motion-3"><div class="list-item-paragraph">Determine</div><ol class="i-list"><li data-list-item-id="h2-motion-4" data-list-item-path="h2-motion-3/h2-motion-4"><div class="list-item-paragraph">the amplitude;<span class="inline-marks">[1]</span></div></li><li data-list-item-id="h2-motion-6" data-list-item-path="h2-motion-3/h2-motion-6"><div class="list-item-paragraph">the initial displacement;<span class="inline-marks">[2]</span></div></li><li data-list-item-id="h2-motion-8" data-list-item-path="h2-motion-3/h2-motion-8"><div class="list-item-paragraph">the value of $t$ when the particle first passes through $O$.<span class="inline-marks">[2]</span></div></li></ol></li><li data-list-item-id="h2-motion-11" data-list-item-path="h2-motion-11"><div class="list-item-paragraph">More generally, a particle moves along a straight line with displacement $s=A\sin(nt+b)$ metres at time $t$ seconds, where&nbsp;$t\geq0$, $A,n&gt;0$ and $-\pi\leq b\leq\pi$.</div><div class="list-item-paragraph">Its velocity and acceleration are $v=\frac{\mathrm{d}s}{\mathrm{d}t}$ and $a=\frac{\mathrm{d}v}{\mathrm{d}t}$, respectively.</div><div class="list-item-paragraph">Find expressions for&nbsp;$\frac{\mathrm{d}s}{\mathrm{d}t}$ and $\frac{\mathrm{d}^2s}{\mathrm{d}t^2}$, and hence show that&nbsp;$a=-n^2s$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="h2-motion-15" data-list-item-path="h2-motion-15"><ol class="i-list"><li data-list-item-id="h2-motion-16" data-list-item-path="h2-motion-15/h2-motion-16"><div class="list-item-paragraph">Use the chain rule to show that&nbsp;$a=v\frac{\mathrm{d}v}{\mathrm{d}s}$, where&nbsp;$v$ is velocity.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="h2-motion-18" data-list-item-path="h2-motion-15/h2-motion-18"><div class="list-item-paragraph">The particle is instantaneously at rest when&nbsp;$s=A$. By separating the variables in $v\frac{\mathrm{d}v}{\mathrm{d}s}=-n^2s$, show that&nbsp;$v^2=n^2(A^2-s^2)$.</div></li><li data-list-item-id="h2-motion-20" data-list-item-path="h2-motion-15/h2-motion-20"><div class="list-item-paragraph">Hence, or otherwise, find the particle’s maximum speed.<span class="inline-marks">[2]</span></div></li></ol></li></ol>

Verifying Solutions of Differential Equations

Chain Rule

IB AA HL May 2026 P3 TZC Q2 (H2 adaptation)

<p>The spread of an infection through a population of rabbits is modelled by a differential equation.</p>
<p>At time $t$ days, $N$ thousand rabbits are infected.</p>
<p>The model is $\frac{\mathrm{d}N}{\mathrm{d}t}=k(-3+4N-N^2)$, where&nbsp;$t\geq0$ and $k$ is a positive constant.</p>
<p>Initially, $1500$ rabbits are infected.</p>
<ol class="abc-list"><li data-list-item-id="h2-rabbits-6" data-list-item-path="h2-rabbits-6"><div class="list-item-paragraph">Express $\frac1{-3+4x-x^2}$ in the form $\frac{A}{x-1}+\frac{B}{3-x}$, where&nbsp;$A,B\in\mathbb{R}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="h2-rabbits-8" data-list-item-path="h2-rabbits-8"><div class="list-item-paragraph">By separating the variables and using part (a), show that&nbsp;$\ln\left(\frac{N-1}{3-N}\right)=2kt-\ln p$, where&nbsp;$p$ is a positive constant to be determined. You may assume that&nbsp;$1&lt;N&lt;3$.<span class="inline-marks">[8]</span></div></li><li data-list-item-id="h2-rabbits-10" data-list-item-path="h2-rabbits-10"><div class="list-item-paragraph">Hence express $N$ explicitly in terms of $t$ and $k$, in the form $N=\frac{a+be^{-2kt}}{1+ce^{-2kt}}$, where&nbsp;$a$, $b$ and $c$ are integers to be determined.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="h2-rabbits-12" data-list-item-path="h2-rabbits-12"><div class="list-item-paragraph">According to the model, the number of infected rabbits approaches a limiting value, $L$, as&nbsp;$t$ becomes large.</div><div class="list-item-paragraph">Find the value of $L$.<span class="inline-marks">[2]</span></div></li></ol>

The spread of an infection through a population of rabbits is modelled by a differential equation.

At time $t$ days, $N$ thousand rabbits are infected.

The model is $\frac{\mathrm{d}N}{\mathrm{d}t}=k(-3+4N-N^2)$, where $t\geq0$ and $k$ is a positive constant.

Equilibrium Solutions and Long-Term Behaviour

IB AA HL May 2026 P1 TZA Q12 (H2 adaptation)

<p>The variables $x$ and $y$ satisfy the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=y^3-\frac{y}{x}$, where&nbsp;$x,y&gt;0$. It is given that&nbsp;$y=2$ when&nbsp;$x=1$.</p>
<p>Euler's method is a numerical method for estimating a solution of a differential equation. Over a small horizontal step of length $h$, it uses a straight line with the gradient given by the differential equation at the current estimated point.</p>
<p>The estimated change in $y$ is therefore&nbsp;$h\times\text{current gradient}$. Let&nbsp;$y_n$ be the estimate of $y$ at $x=x_n$, with $x_0=1$ and $y_0=2$. For this differential equation, the rule is</p>
<p>\[x_{n+1}=x_n+h,\qquad y_{n+1}=y_n+h\left(y_n^3-\frac{y_n}{x_n}\right).\]</p>
<p>At each step, calculate the gradient using the current values of $x$ and $y$, then add $h$ times this gradient to the current estimate of $y$. Recalculate the gradient at the new estimated point before repeating. Keep unrounded values in subsequent calculations.</p>
<ol class="abc-list"><li data-list-item-id="p2-tzc-q12-item01" data-list-item-path="p2-tzc-q12-item01"><div class="list-item-paragraph">Take $h=0.01$. First calculate the gradient at $(1,2)$. Use this gradient in the Euler formula to show that the first estimate, at $x_1=1.01$, is $y_1=2.06$.</div><div class="list-item-paragraph">Then calculate $y_2$, showing your substitution and giving its value to 5 decimal places. Continue the process to estimate $y$ when&nbsp;$x=1.05$, giving this estimate to 3 significant figures.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p2-tzc-q12-item02" data-list-item-path="p2-tzc-q12-item02"><div class="list-item-paragraph">Using the substitution $u=xy$, show that the differential equation can be written as&nbsp;$\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{u^3}{x^2}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p2-tzc-q12-item03" data-list-item-path="p2-tzc-q12-item03"><div class="list-item-paragraph">Hence, by separating the variables and using the given initial condition, solve the differential equation, giving $y$ as a function of $x$.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="p2-tzc-q12-item04" data-list-item-path="p2-tzc-q12-item04"><div class="list-item-paragraph">State the largest possible domain of this solution containing $x=1$, and its corresponding range.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p2-tzc-q12-item05" data-list-item-path="p2-tzc-q12-item05"><div class="list-item-paragraph">By considering the shape of the solution curve for&nbsp;$1\leq x\leq1.05$ and the straight-line steps in Euler's method, state whether the approximation in part (a) is an overestimate or an underestimate. Give a reason.<span class="inline-marks">[1]</span></div></li></ol>

The variables $x$ and $y$ satisfy the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=y^3-\frac{y}{x}$, where $x,y>0$. It is given that $y=2$ when $x=1$.

Euler's method is a numerical method for estimating a solution of a differential equation. Over a small horizontal step of length $h$, it uses a straight line with the gradient given by the differential equation at the current estimated point.

The estimated change in $y$ is therefore $h\times\text{current gradient}$. Let $y_n$ be the estimate of $y$ at $x=x_n$, with $x_0=1$ and $y_0=2$. For this differential equation, the rule is

\[x_{n+1}=x_n+h,\qquad y_{n+1}=y_n+h\left(y_n^3-\frac{y_n}{x_n}\right).\]

At each step, calculate the gradient using the current values of $x$ and $y$, then add $h$ times this gradient to the current estimate of $y$. Recalculate the gradient at the new estimated point before repeating. Keep unrounded values in subsequent calculations.

Euler's Method

Concavity of Graphs

Function Notation, Domain and Range

Solving Differential Equations by Substitution

Range of a Function

Product Rule

Domain of a Function

Sketching Solution Curves of Differential Equations

IB AA HL May 2026 P2 TZC Q12 [modified]

Functions

Applications of Differentiation

<p>A freshly brewed cup of tea, initially at a temperature of ${85^\circ\text{C}}$, is left out to cool in a room with a room temperature of ${25^\circ\text{C}}$. After ${20}$ minutes, the temperature of the tea is ${55^\circ\text{C}}$.</p>
<p>It is known that the rate of change of the temperature of the tea is proportional to the difference between the temperature of the tea and the room temperature.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Given that&nbsp;$\theta$ is the temperature of the tea $t$ minutes after the tea is left out to cool, show that&nbsp;$\theta = 25 + 60\mathrm{e}^{-kt}$, where&nbsp;$k$ is a constant to be determined.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Sketch the graph of $\theta$ against $t$.<span class="inline-marks">[2]</span></div></li></ol>

A freshly brewed cup of tea, initially at a temperature of ${85^\circ\text{C}}$, is left out to cool in a room with a room temperature of ${25^\circ\text{C}}$. After ${20}$ minutes, the temperature of the tea is ${55^\circ\text{C}}$.

It is known that the rate of change of the temperature of the tea is proportional to the difference between the temperature of the tea and the room temperature.

Newton’s Law of Cooling and Warming

2025 ACJC P1 Q7

<p>The variables $x$ and $y$ are related by the differential equation</p>
<p style="text-align: center">\[x^2 \frac{\mathrm{d}y}{\mathrm{d}x} - 2xy = 10x^2y - y^2 \quad \text{for}\hspace{0.5em} x &gt; 0.\]</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-paragraph">Using the substitution $y = ux^2$, for&nbsp;$x &gt; 0$, show that the differential equation can be reduced to</div><div class="list-item-paragraph" style="text-align: center">\[\frac{\mathrm{d}u}{\mathrm{d}x} = u(10 - u).\]<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-ii"><div class="list-item-paragraph">Hence show that the general solution for the differential equation</div><div class="list-item-paragraph" style="text-align: center">\[x^2 \frac{\mathrm{d}y}{\mathrm{d}x} - 2xy = 10x^2y - y^2 \quad \text{for}\hspace{0.5em} x &gt; 0\]</div><div class="list-item-paragraph">can be expressed in the form $y = \frac{10x^2}{Qe^{-10x} + 1}$, where&nbsp;$Q$ is an arbitrary constant.<span class="inline-marks">[6]</span></div></li><li data-list-item-id="item-iii"><div class="list-item-paragraph">Given that the graph of $y = \mathrm{f}(x)$ passes through the point with coordinates $\left(\frac{1}{10}, \frac{1}{10(1 - \mathrm{e}^{-1})}\right)$, find the equation of the graph of $y = \mathrm{f}(x)$.<span class="inline-marks">[2]</span></div></li></ol>

The variables $x$ and $y$ are related by the differential equation

\[x^2 \frac{\mathrm{d}y}{\mathrm{d}x} - 2xy = 10x^2y - y^2 \quad \text{for } x > 0.\]

2024 MI PRU2 P2 Q5

<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-paragraph">It is given that</div><div class="list-item-paragraph" style="text-align: center">\[\frac{\mathrm{d}y}{\mathrm{d}x} = y\left(2x + \tan^2 x\right) \text{ where}\hspace{0.5em} y \neq 0.\]</div><div class="list-item-paragraph">Find $y$ in terms of $x$.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-ii"><div class="list-item-paragraph">Hence, given that&nbsp;$y = -2$ when&nbsp;$x = 0$, find the particular solution to the differential equation in part <strong>(i)</strong>.<span class="inline-marks">[2]</span></div></li></ol>

2024 MI PRU2 P1 Q5

<ol class="abc-list"><li><div class="list-item-paragraph">Given that&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-2xy}{x^2 + y^2}$, use the substitution $y = vx$, where&nbsp;$v$ is a function of $x$ to show that\[x \frac{\mathrm{d}v}{\mathrm{d}x} = -\frac{3v + v^3}{1 + v^2}.\]</div></li><li><div class="list-item-content"><span class="list-item-text">Hence, show that the solution for the given differential equation is $3x^2 y + y^3 = c$ where&nbsp;$c$ is an arbitrary constant.</span></div></li></ol>

Hence, show that the solution for the given differential equation is $3x^2 y + y^3 = c$ where $c$ is an arbitrary constant.

Homogeneous equations

<p>A tank contains $500$ litres of brine in which $75$ kg of salt is dissolved. Brine containing $3$ kg of salt per $10$ litres of water flows in at a rate of $20$ litres per minute, and the mixture flows out at the same rate.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Show that, for&nbsp;$t \ge 0, \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{150 - x}{25}$, where&nbsp;$x$ is the amount of dissolved salt in the tank at time $t \ge 0$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">When will there be $125$ kg of dissolved salt in the tank?</span></div></li><li><div class="list-item-content"><span class="list-item-text">How much dissolved salt is in the tank after a long time?</span></div></li><li><div class="list-item-content"><span class="list-item-text">State an assumption used in the model.</span></div></li></ol>

A tank contains $500$ litres of brine in which $75$ kg of salt is dissolved. Brine containing $3$ kg of salt per $10$ litres of water flows in at a rate of $20$ litres per minute, and the mixture flows out at the same rate.

Show that, for $t \ge 0, \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{150 - x}{25}$, where $x$ is the amount of dissolved salt in the tank at time $t \ge 0$.

When will there be $125$ kg of dissolved salt in the tank?

How much dissolved salt is in the tank after a long time?

Mixing and Concentration Models

Assumptions and Limitations of Differential Equation Models

<p>The gradient of a curve at the point $(x, y)$ is proportional to $y^2$. <br>The curve passes through the points $(0, -2)$ and $(1, \frac{2}{3})$. <br>Find the equation of the curve.</p>

The gradient of a curve at the point $(x, y)$ is proportional to $y^2$. 
The curve passes through the points $(0, -2)$ and $(1, \frac{2}{3})$. 
Find the equation of the curve.

Modelling with Differential Equations

Word problems: recovering function

<p>As a tree grows, the rate of increase of its height, $h$ m, with respect to time, $t$ years after planting, is modelled by the differential equation</p>
<p style="text-align: center">\[\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{1}{10}\sqrt{\left(16 - \frac{1}{2}h\right)}\]</p>
<p>The tree is planted as a seedling of negligible height, so that&nbsp;$h=0$ when&nbsp;$t=0$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">State the maximum height of the tree, according to this model.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find an expression for&nbsp;$t$ in terms of $h$, and hence find the time the tree takes to reach half its maximum height.</span><span class="inline-marks">[5]</span></div></li></ol>

As a tree grows, the rate of increase of its height, $h$ m, with respect to time, $t$ years after planting, is modelled by the differential equation

\[\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{1}{10}\sqrt{\left(16 - \frac{1}{2}h\right)}\]

The tree is planted as a seedling of negligible height, so that $h=0$ when $t=0$.

State the maximum height of the tree, according to this model.

Find an expression for $t$ in terms of $h$, and hence find the time the tree takes to reach half its maximum height.

N2015 P2 Q1

<p>It is given that&nbsp;$2 \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x} - \ln x = y$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Use the substitution $y = \ln \left( \frac{u}{x} \right)$ to show that the differential equation can be reduced to $\frac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{f}(u)$, where the function $\mathrm{f}(u)$ is to be found.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Given that&nbsp;$y$ has a minimum value at $x = 3$, solve the differential equation $2 \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x} - \ln x = y$, to find the particular solution for&nbsp;$y$ in terms of $x$.</span><span class="inline-marks">[5]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Sketch the graph of this particular solution.</span><span class="inline-marks">[2]</span></div></li></ol>

It is given that $2 \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x} - \ln x = y$.

Use the substitution $y = \ln \left( \frac{u}{x} \right)$ to show that the differential equation can be reduced to $\frac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{f}(u)$, where the function $\mathrm{f}(u)$ is to be found.

Given that $y$ has a minimum value at $x = 3$, solve the differential equation $2 \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x} - \ln x = y$, to find the particular solution for $y$ in terms of $x$.

Sketch the graph of this particular solution.

Derivatives of Natural Logarithmic Functions

2025 VJC P1 Q11

<p>Welding fumes contain a dangerous amount of particulates. To protect workers in a workshop, an extraction device remove particulates in the air continuously. Let&nbsp;$V$ mg represent the mass of particulates in the air in the workshop at time $t$ min after the extraction device is activated.</p>
<p>Particulates are produced at a constant rate of $0.32$ mg/min. The rate at which particulates are removed is proportional to the mass of particulates in the air. When the mass of particulates in the air in the workshop is $18.75$ mg, the mass of particulates in the air increases at a rate of $0.12$ mg/min.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">Show that&nbsp;$\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{4}{375}(30-V)$.</span><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">Solve the differential equation, given that the workshop is initially free of particulates.</span><span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">Sketch the graph of $V$ against $t$.</span><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-iv"><div class="list-item-content"><span class="list-item-text">It is recommended that the mass of particulates in this workshop should be kept below $32$ mg. Comment on whether the extraction device is effective enough to meet the recommendation.</span><span class="inline-marks">[1]</span></div></li></ol>
<p>The extraction device becomes less efficient and now removes particulates at half its original rate.</p>
<ol class="i-list" start="5"><li data-list-item-id="item-v"><div class="list-item-content"><span class="list-item-text">Write down a new differential equation to model this.</span><span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-vi"><div class="list-item-content"><span class="list-item-text">The mass of particulates in the air in the workshop is said to reach a steady-state when it remains constant. Find the value of $V$ at steady-state.</span><span class="inline-marks">[1]</span></div></li></ol>

Welding fumes contain a dangerous amount of particulates. To protect workers in a workshop, an extraction device remove particulates in the air continuously. Let $V$ mg represent the mass of particulates in the air in the workshop at time $t$ min after the extraction device is activated.

Particulates are produced at a constant rate of $0.32$ mg/min. The rate at which particulates are removed is proportional to the mass of particulates in the air. When the mass of particulates in the air in the workshop is $18.75$ mg, the mass of particulates in the air increases at a rate of $0.12$ mg/min.

Show that $\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{4}{375}(30-V)$.

Solve the differential equation, given that the workshop is initially free of particulates.

It is recommended that the mass of particulates in this workshop should be kept below $32$ mg. Comment on whether the extraction device is effective enough to meet the recommendation.

The extraction device becomes less efficient and now removes particulates at half its original rate.

Write down a new differential equation to model this.

The mass of particulates in the air in the workshop is said to reach a steady-state when it remains constant. Find the value of $V$ at steady-state.

Modelling with Multiple Rates of Change

2025 SAJC P1 Q12

<p>A cafe roasts its own coffee beans and packages them to be sold. The amount of roasted coffee beans remaining in the cafe at time $t$ days is denoted by $x$ kg. The cafe produces roasted coffee beans at a fixed rate of $10$ kg/day, and sells the roasted coffee beans at a rate proportional to $x^2$ kg/day.</p>
<ol class="abc-list"><li data-list-item-id="item-cafe-a" data-list-item-path="item-cafe-a"><div class="list-item-paragraph">Given that there is no change in the amount of roasted coffee beans remaining in the cafe when there is $5$ kg of beans remaining, show that&nbsp;$\frac{\mathrm{d}x}{\mathrm{d}t} = 10 - \frac{2}{5}x^2$. Find the general solution of this differential equation. (You do not have to make $x$ the subject.)<span class="inline-marks">[3]</span></div></li></ol>
<p>Due to improvement in roasting capabilities of the cafe, it is now able to produce roasted coffee beans at the rate of $3x$ kg/day, and sells the roasted coffee beans at a rate proportional to $x^2$ kg/day.</p>
<ol class="abc-list" start="2"><li data-list-item-id="item-cafe-b" data-list-item-path="item-cafe-b"><div class="list-item-paragraph">Given that amount of roasted coffee beans is increasing at the rate of $6$ kg/day when the amount of roasted coffee beans remaining in the cafe is $6$ kg, write down the modified differential equation relating $\frac{\mathrm{d}x}{\mathrm{d}t}$ and $x$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-cafe-c" data-list-item-path="item-cafe-c"><div class="list-item-paragraph"><p>Given that the initial amount of roasted coffee beans in the cafe is $13.5$ kg, solve the differential equation in part <strong>(b)</strong>, expressing $x$ in terms of $t$. Deduce the amount of roasted coffee beans remaining in the cafe in the long run, and sketch the curve of $x$ against $t$.</p><span class="inline-marks">[6]</span></div></li><li data-list-item-id="item-cafe-d" data-list-item-path="item-cafe-d"><div class="list-item-paragraph"><p>Based on the curve in part <strong>(c)</strong>, explain in context why this might be favourable for the cafe owner.</p><span class="inline-marks">[1]</span></div></li></ol>

A cafe roasts its own coffee beans and packages them to be sold. The amount of roasted coffee beans remaining in the cafe at time $t$ days is denoted by $x$ kg. The cafe produces roasted coffee beans at a fixed rate of $10$ kg/day, and sells the roasted coffee beans at a rate proportional to $x^2$ kg/day.

Given that there is no change in the amount of roasted coffee beans remaining in the cafe when there is $5$ kg of beans remaining, show that $\frac{\mathrm{d}x}{\mathrm{d}t} = 10 - \frac{2}{5}x^2$. Find the general solution of this differential equation. (You do not have to make $x$ the subject.)

Due to improvement in roasting capabilities of the cafe, it is now able to produce roasted coffee beans at the rate of $3x$ kg/day, and sells the roasted coffee beans at a rate proportional to $x^2$ kg/day.

Given that amount of roasted coffee beans is increasing at the rate of $6$ kg/day when the amount of roasted coffee beans remaining in the cafe is $6$ kg, write down the modified differential equation relating $\frac{\mathrm{d}x}{\mathrm{d}t}$ and $x$.

Logistic Growth Models

2025 RVHS P1 Q13

<p>Scientists are studying the growth of a newly discovered biomass that entered Earth's atmosphere together with an asteroid. A sample of $100$ grams of the biomass was collected initially. It was found that on the next day, the sample grew to $110$ grams. The scientists observed that the rate of growth of biomass was proportional to the amount of biomass present. The amount of biomass at time $t$ days is denoted by $B$ grams.</p>
<ol class="abc-list"><li data-list-item-id="item-a-4" data-list-item-path="item-a-4"><div class="list-item-paragraph">Write down a differential equation relating $B$ and $t$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-b-4" data-list-item-path="item-b-4"><div class="list-item-paragraph">Solve the differential equation obtained in part <strong>(a)</strong>, expressing $B$ in terms of $t$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-c-4" data-list-item-path="item-c-4"><div class="list-item-paragraph">Sketch the graph of $B$ against $t$, and explain what would happen if the biomass was left on its own.<span class="inline-marks">[2]</span></div></li></ol>
<p>To limit the growth of the biomass, the scientist collected a second sample of $100$ grams of the biomass and introduced a nutrient depletion serum to the sample the moment it was collected. The rate of growth of the biomass for this second sample is modelled by the differential equation</p>
<p style="text-align: center">\[\frac{\mathrm{d}B}{\mathrm{d}t} = \frac{mB}{3 + 2t^2}\]</p>
<p>where&nbsp;$m$ is a constant. After $1$ day, the second sample grew to $101$ grams.</p>
<ol class="abc-list" start="4"><li data-list-item-id="item-d-4" data-list-item-path="item-d-4"><div class="list-item-paragraph">For this second sample, find $B$ in terms of $t$.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-e-4" data-list-item-path="item-e-4"><div class="list-item-paragraph">Determine the amount of the second biomass sample after a long time, leaving your answer to $5$ decimal places.<span class="inline-marks">[1]</span></div></li></ol>

Scientists are studying the growth of a newly discovered biomass that entered Earth's atmosphere together with an asteroid. A sample of $100$ grams of the biomass was collected initially. It was found that on the next day, the sample grew to $110$ grams. The scientists observed that the rate of growth of biomass was proportional to the amount of biomass present. The amount of biomass at time $t$ days is denoted by $B$ grams.

Write down a differential equation relating $B$ and $t$.

Sketch the graph of $B$ against $t$, and explain what would happen if the biomass was left on its own.

To limit the growth of the biomass, the scientist collected a second sample of $100$ grams of the biomass and introduced a nutrient depletion serum to the sample the moment it was collected. The rate of growth of the biomass for this second sample is modelled by the differential equation

\[\frac{\mathrm{d}B}{\mathrm{d}t} = \frac{mB}{3 + 2t^2}\]

where $m$ is a constant. After $1$ day, the second sample grew to $101$ grams.

For this second sample, find $B$ in terms of $t$.

Determine the amount of the second biomass sample after a long time, leaving your answer to $5$ decimal places.

Exponential Growth and Decay Models

2025 NYJC P1 Q12

<p>To prepare for the General Election, a political party studies the effectiveness of spreading information on Instantglam, a social media platform, by uploading a particular video. The number of Instantglam users who have viewed the video increases at a rate that is directly proportional to the number of Instantglam users who have yet to view the video. The number of Instantglam users who have viewed the video $t$ hours after the video is uploaded is $x$. The total number of Instantglam users is $P$, assumed to be fixed.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Write down a differential equation and show that&nbsp;$x = P(1 - e^{-kt})$, where&nbsp;$k$ is a positive constant.<span class="inline-marks">[4]</span></div></li></ol>
<p>It is known that&nbsp;$12$ hours after the video is uploaded, half the total number of Instantglam users have viewed the video.</p>
<ol class="abc-list" start="2"><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Estimate to the nearest hour, the time needed for&nbsp;$80\%$ of the total number of Instantglam users to have viewed the video.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Sketch the graph of $x$ against $t$.<span class="inline-marks">[2]</span></div></li></ol>

To prepare for the General Election, a political party studies the effectiveness of spreading information on Instantglam, a social media platform, by uploading a particular video. The number of Instantglam users who have viewed the video increases at a rate that is directly proportional to the number of Instantglam users who have yet to view the video. The number of Instantglam users who have viewed the video $t$ hours after the video is uploaded is $x$. The total number of Instantglam users is $P$, assumed to be fixed.

Write down a differential equation and show that $x = P(1 - e^{-kt})$, where $k$ is a positive constant.

It is known that $12$ hours after the video is uploaded, half the total number of Instantglam users have viewed the video.

2025 JPJC P2 Q1

<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Express $\frac{1}{Q(500 - Q)}$ in partial fractions.<span class="inline-marks">[2]</span></div></li></ol>
<ol class="abc-list" start="2"><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph"><p>A group of scientists is monitoring the population of a particular species of birds on an island. At time $t$ years after the start of the monitoring, the number of birds on the island is $Q$. The scientists observe that the birth rate is proportional to the bird's population and the birds are dying at a rate proportional to the square of the bird's population. There were $1000$ birds on the island when the scientists first started monitoring the population. They discover that the population remains unchanged when there are $500$ birds.</p></div><ol class="i-list"><li data-list-item-id="item-b-i" data-list-item-path="item-b/item-b-i"><div class="list-item-paragraph">Show that the differential equation relating $Q$ and $t$ is given by\[\frac{\mathrm{d}Q}{\mathrm{d}t} = kQ(500 - Q)\]where&nbsp;$k$ is a positive constant.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b-ii" data-list-item-path="item-b/item-b-ii"><div class="list-item-paragraph">Hence, solve the differential equation, expressing $Q$ in terms of $k$ and $t$.<span class="inline-marks">[5]</span></div></li></ol></li></ol>

Express $\frac{1}{Q(500 - Q)}$ in partial fractions.

2025 HCI P1 Q10

<p>A cylindrical container of height $h$ metres is originally completely filled with water. The water is flowing out through an opening at the base of the container into a pipe. The rate of flow of water through the opening at time $t$ seconds is proportional to the square root of the height of the water in the container, $y$ metres. Since the container is cylindrical, the amount of water in the container is also proportional to the height of water, so the differential equation relating $y$ and $t$ can be written as</p>
<p>\[\frac{\mathrm{d}y}{\mathrm{d}t} = -k\sqrt{y}\]</p>
<p>where $k$ is a positive constant.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Solve this differential equation to find $\sqrt{y}$ in terms of $t$, $h$ and $k$.</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence find the time taken for the container to empty, in terms of $h$ and $k$.</span><span class="inline-marks">[2]</span></div></li></ol>
<p>It is found that the water was draining too fast. To regulate the flow of water, a mechanical device is placed at the opening. This device slows the flow of water into the pipe by covering all or part of the opening, where the area covered varies with time. It is proposed to model this situation using the revised differential equation</p>
<p>\[\frac{\mathrm{d}y}{\mathrm{d}t} = -k\left( \frac{1 + \cos t}{2} \right)\sqrt{y},\]</p>
<p>where $k$ is the same constant as before.</p>
<ol class="abc-list" start="3"><li><div class="list-item-content"><span class="list-item-text">With reference to the possible values of $\cos t$, explain how the revised differential equation corresponds to a model with slower flow of water.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Given again that the container is originally completely filled with water, solve the revised differential equation to find $\sqrt{y}$ in terms of $t$, $h$ and $k$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">In the case where $h = 10$ and $k = 0.1$, find the ratio of the time taken in the revised model to the time taken in the original model for the container to empty. Give your answer to $3$ significant figures.</span><span class="inline-marks">[2]</span></div></li></ol>

A cylindrical container of height $h$ metres is originally completely filled with water. The water is flowing out through an opening at the base of the container into a pipe. The rate of flow of water through the opening at time $t$ seconds is proportional to the square root of the height of the water in the container, $y$ metres. Since the container is cylindrical, the amount of water in the container is also proportional to the height of water, so the differential equation relating $y$ and $t$ can be written as

\[\frac{\mathrm{d}y}{\mathrm{d}t} = -k\sqrt{y}\]

Solve this differential equation to find $\sqrt{y}$ in terms of $t$, $h$ and $k$.

Hence find the time taken for the container to empty, in terms of $h$ and $k$.

It is found that the water was draining too fast. To regulate the flow of water, a mechanical device is placed at the opening. This device slows the flow of water into the pipe by covering all or part of the opening, where the area covered varies with time. It is proposed to model this situation using the revised differential equation

\[\frac{\mathrm{d}y}{\mathrm{d}t} = -k\left( \frac{1 + \cos t}{2} \right)\sqrt{y},\]

where $k$ is the same constant as before.

With reference to the possible values of $\cos t$, explain how the revised differential equation corresponds to a model with slower flow of water.

Given again that the container is originally completely filled with water, solve the revised differential equation to find $\sqrt{y}$ in terms of $t$, $h$ and $k$.

In the case where $h = 10$ and $k = 0.1$, find the ratio of the time taken in the revised model to the time taken in the original model for the container to empty. Give your answer to $3$ significant figures.

2025 EJC P1 Q11

<p>It is given that&nbsp;$2xy\frac{\mathrm{d}y}{\mathrm{d}x}=x^2-y^2$ where&nbsp;$x&gt;0$, $y&lt;0$. Using the substitution $u=\frac{y}{x}$, show that the differential equation can be transformed to $\frac{2u}{1-3u^2}\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{1}{x}$. Hence find the general solution of $y$ in terms of $x$.<span class="inline-marks">[6]</span></p>

It is given that $2xy\frac{\mathrm{d}y}{\mathrm{d}x}=x^2-y^2$ where $x>0$, $y<0$. Using the substitution $u=\frac{y}{x}$, show that the differential equation can be transformed to $\frac{2u}{1-3u^2}\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{1}{x}$. Hence find the general solution of $y$ in terms of $x$.

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