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<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>

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

Derivatives of Trigonometric Functions

Second and Higher Derivatives

Verifying Solutions of Differential Equations

Chain Rule

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

Differential Equations

Techniques of Differentiation

<p>Using the techniques of differentiation, or otherwise, answer the following questions:</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Given that&nbsp;$y = \frac{x^2}{\mathrm{e}^x}$ and $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 2\frac{\mathrm{d}y}{\mathrm{d}x} = m \mathrm{e}^{-x} - y$, find the value of $m$.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Given that&nbsp;$y = a \mathrm{e}^{2x} + b \mathrm{e}^{-x}$ and $\frac{\mathrm{d}y}{\mathrm{d}x} + 4y = \mathrm{e}^{2x} - \mathrm{e}^{-x}$, find the values of $a$ and $b$.</div></li></ol>

Using the techniques of differentiation, or otherwise, answer the following questions:

Given that $y = \frac{x^2}{\mathrm{e}^x}$ and $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 2\frac{\mathrm{d}y}{\mathrm{d}x} = m \mathrm{e}^{-x} - y$, find the value of $m$.

Given that $y = a \mathrm{e}^{2x} + b \mathrm{e}^{-x}$ and $\frac{\mathrm{d}y}{\mathrm{d}x} + 4y = \mathrm{e}^{2x} - \mathrm{e}^{-x}$, find the values of $a$ and $b$.

Derivatives of Exponential Functions

Differentiation

<p>It is given that $y={{e}^{-x}}$, $y={{e}^{2x}}$ and $y={{e}^{-3x}}$ are solutions to the differential equation
</p>
<p style="text-align: center">$\frac{{{\mathrm{d}}^{3}}y}{\mathrm{d}{{x}^{3}}}+a\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}+b\frac{\mathrm{d}y}{\mathrm{d}x}+cy=0$.</p>
<p>Find the values of the constants $a$, $b$ and $c$.
</p>

It is given that $y={{e}^{-x}}$, $y={{e}^{2x}}$ and $y={{e}^{-3x}}$ are solutions to the differential equation


$\frac{{{\mathrm{d}}^{3}}y}{\mathrm{d}{{x}^{3}}}+a\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}+b\frac{\mathrm{d}y}{\mathrm{d}x}+cy=0$.

Find the values of the constants $a$, $b$ and $c$.


2013 NJC SH1 Promo Q1

<p>A curve has equation</p>
<p style="text-align: center">${y={{x}^{{{x}^{-1}}}}}$, ${x\in \mathbb{R}}$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Show that ${y}$ is a solution of the following differential equation.</div><div class="list-item-paragraph" style="text-align: center">${{{x}^{3}}\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}-x\left( 1-\ln x \right)\frac{\mathrm{d}y}{\mathrm{d}x}+\left( 3-2\ln x \right)y=0}$.</div></li><li><div class="list-item-paragraph">Show further that</div><div class="list-item-paragraph" style="text-align: center">${{{\left. \frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}} \right|}_{x=\mathrm{e}}}=-{{\mathrm{e}}^{{{\mathrm{e}}^{-1}}-3}}}$.</div></li></ol>

${y={{x}^{{{x}^{-1}}}}}$, ${x\in \mathbb{R}}$.

Second-Order Linear Differential Equations

Logarithmic Differentiation

Higher Derivatives by Implicit Differentiation

Techniques and Applications Of Differentiation Q1

Applications of Differentiation

<p>The variables $x,y$ and $z$ are connected by the following differential equations.</p>
<p style="text-align: center">$\frac{\mathrm{d}z}{\mathrm{d}x}=3-2z$  $\mathrm{(A)}$</p>
<p style="text-align: center">$\frac{\mathrm{d}y}{\mathrm{d}x}=z$           $\mathrm{(B)}$</p>
<ol class="i-list"><li><div class="list-item-paragraph">Given that $z<\frac{3}{2}$, solve equation $\mathrm{(A)}$ to find $z$ in terms of $x$.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">Hence find $y$ in terms of $x$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Use the result in part <strong>(ii)</strong> to show that</div><div class="list-item-paragraph" style="text-align: center">$\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}=a\frac{\mathrm{d}y}{\mathrm{d}x}+b$,</div><div class="list-item-paragraph">for constants $a$ and $b$ to be determined.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph"><strong>* Not in the latest 9758 H2 Math Syllabus</strong></div><div class="list-item-paragraph">The result in part <strong>(ii)</strong> represents a family of curves. Some members of the family are straight lines. Write down the equations of two of these lines. On a single diagram, sketch one of your lines together with a non-linear member of the family of curves that has your line as an asymptote.</div><p style="text-align: right;">[4]</p></li></ol>

The variables $x,y$ and $z$ are connected by the following differential equations.

$\frac{\mathrm{d}z}{\mathrm{d}x}=3-2z$  $\mathrm{(A)}$

$\frac{\mathrm{d}y}{\mathrm{d}x}=z$           $\mathrm{(B)}$

Sketching Exponential Functions and Solving Equations Graphically

Solving Differential Equations by Direct Integration

Asymptotes and Discontinuities

Sketching Solution Curves of Differential Equations

Order, General and Particular Solutions of Differential Equations

N2013 P1 Q10

Graphing Techniques

<p>An electrical circuit comprises a power source of $V$ volts in series with a resistance of $R$ ohms, a capacitance of $C$ farads and an inductance of $L$ henries. The current in the circuit, $t$ seconds after turning on the power, is $I$ amps and the charge on the capacitor is $q$ coulombs. The circuit can be used by scientists to investigate resonance, to model heavily damped motion and to tune into radio stations on a stereo tuner. It is given that&nbsp;$R,C$ and $L$ are constants, and that&nbsp;$I=0$ when&nbsp;$t=0$.</p>
<p>A differential equation for the circuit is $L\frac{\mathrm{d}I}{\mathrm{d}t}+RI+\frac{q}{C}=V$, where&nbsp;$I=\frac{\mathrm{d}q}{\mathrm{d}t}$.</p>
<ol class="i-list"><li data-list-item-id="item-1760098299435-a3aws34m7" data-list-item-path="item-1760098299435-a3aws34m7"><div class="list-item-paragraph"><p>Show that, under certain conditions on $V$ which should be stated, $L\frac{{{\mathrm{d}}^{2}}I}{\mathrm{d}{{t}^{2}}}+R\frac{\mathrm{d}I}{\mathrm{d}t}+\frac{I}{C}=0$</p><span class="inline-marks">[2]</span></div></li></ol>
<p>It is now given that the differential equation in part <strong>(i)</strong> holds for the rest of the question.</p>
<ol class="i-list" start="2"><li data-list-item-id="item-1760098313918-h7wkwhabo" data-list-item-path="item-1760098313918-h7wkwhabo"><div class="list-item-paragraph">Given that&nbsp;$I=\mathrm{A}t{{e}^{-\frac{Rt}{2L}}}$ is a solution of the differential equation, where&nbsp;$\mathrm{A}$ is a positive constant, show that&nbsp;$C=\frac{4L}{{{R}^{2}}}$.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-1760098320060-1e6k6z20a" data-list-item-path="item-1760098320060-1e6k6z20a"><div class="list-item-paragraph">In a particular circuit, $R=4,\,L=3$ and $C=0.75$. Find the maximum value of $I$ in terms of $A$, showing that this value is a maximum.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1760098320908-kl00x1xi4" data-list-item-path="item-1760098320908-kl00x1xi4"><div class="list-item-paragraph">Sketch the graph of $I$ against $t$.<span class="inline-marks">[2]</span></div></li></ol>

An electrical circuit comprises a power source of $V$ volts in series with a resistance of $R$ ohms, a capacitance of $C$ farads and an inductance of $L$ henries. The current in the circuit, $t$ seconds after turning on the power, is $I$ amps and the charge on the capacitor is $q$ coulombs. The circuit can be used by scientists to investigate resonance, to model heavily damped motion and to tune into radio stations on a stereo tuner. It is given that $R,C$ and $L$ are constants, and that $I=0$ when $t=0$.

A differential equation for the circuit is $L\frac{\mathrm{d}I}{\mathrm{d}t}+RI+\frac{q}{C}=V$, where $I=\frac{\mathrm{d}q}{\mathrm{d}t}$.

It is now given that the differential equation in part **(i)** holds for the rest of the question.

Product Rule

Problems Involving Maximisation and Minimisation

N2018 P1 Q10

<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text"></span></div><ol class="i-list"><li><div class="list-item-paragraph">Given $x$ and $y$ are related by the differential equation ${{x}^{2}}\frac{\mathrm{d}y}{\mathrm{d}x}+xy=k$ for $k\in \mathbb{R}$, show that $y=\frac{k\left( \ln x+\alpha \right)}{x}$ is a solution of the differential equation where $\alpha $ is an arbitrary constant.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Hence, show that $\left( {{\mathrm{e}}^{1-\alpha }},k{{\mathrm{e}}^{\alpha -1}} \right)$ is a stationary point of the curve $y=\frac{k\left( \ln x+\alpha \right)}{x}$.</div><p style="text-align: right;">[2]</p></li></ol></li><li><div class="list-item-paragraph">It is given that $x$ and $y$ are related by the differential equation $y\frac{\mathrm{d}y}{\mathrm{d}x}+x=\sqrt{{{x}^{2}}+{{y}^{2}}}$ and that $y=0$ when $x=-2$.</div><ol class="i-list"><li><div class="list-item-paragraph">By substituting $v={{x}^{2}}+{{y}^{2}}$, show that the differential equation can be written as $\frac{\mathrm{d}v}{\mathrm{d}x}=2\sqrt{v}$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find $v$ in terms of $x$ and hence show that ${{y}^{2}}=\mathrm{f}\left( x \right)$ where $\mathrm{f}\left( x \right)$ is to be determined.</div><p style="text-align: right;">[4]</p></li></ol></li></ol>

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