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<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

Differential Equations

<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>

Method of Separable Variables

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>

Given that \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-2xy}{x^2 + y^2}\), use the substitution \(y = vx\), where \(v\) is a function of \(x\) to show that\n\[x \frac{\mathrm{d}v}{\mathrm{d}x} = -\frac{3v + v^3}{1 + v^2}.\]

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

Modelling in Real-Life Problems

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 Tank Problem

<p>A curve passing through the points \((1,2)\) and \((\frac{1}{4},-10)\) has a gradient which is inversely proportional to \(x^2\). Find the equation of the curve.</p>

A curve passing through the points \((1,2)\) and \((\frac{1}{4},-10)\) has a gradient which is inversely proportional to \(x^2\). Find the equation of the curve.

<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.

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.

2025 VJC P1 Q11

Techniques of Differentiation

<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.

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"><div class="list-item-content"><span class="list-item-text">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><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"><div class="list-item-content"><span class="list-item-text">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><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-cafe-c"><div class="list-item-content"><span class="list-item-text">Given that the initial amount of roasted coffee beans in the cafe is \(13.5\) kg, solve the differential equation in part (b), 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 \(t\).</span><span class="inline-marks">[6]</span></div></li><li data-list-item-id="item-cafe-d"><div class="list-item-content"><span class="list-item-text">Based on the curve in part (c), explain in context why this might be favourable for the cafe owner.</span><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 \(x\).

Given that the initial amount of roasted coffee beans in the cafe is \(13.5\) kg, solve the differential equation in part (b), 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 \(t\).

Based on the curve in part (c), explain in context why this might be favourable for the cafe owner.

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><div class="list-item-content"><span class="list-item-text">Write down a differential equation relating \(B\) and \(t\).</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Solve the differential equation obtained in part (a), expressing \(B\) in terms of \(t\).</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Sketch the graph of \(B\) against \(t\), and explain what would happen if the biomass was left on its own.</span><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><div class="list-item-content"><span class="list-item-text">For this second sample, find \(B\) in terms of \(t\).</span><span class="inline-marks">[5]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Determine the amount of the second biomass sample after a long time, leaving your answer to \(5\) decimal places.</span><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 \(t\).

Solve the differential equation obtained in part (a), expressing \(B\) in terms of \(t\).

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.

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"><div class="list-item-content"><span class="list-item-text">Write down a differential equation and show that&nbsp;\(x = P(1 - e^{-kt})\), where&nbsp;\(k\) is a positive constant.</span><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"><div class="list-item-content"><span class="list-item-text">Estimate to the nearest hour, the time needed for&nbsp;\(80\)% of the total number of Instantglam users to have viewed the video.</span><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-c"><div class="list-item-content"><span class="list-item-text">Sketch the graph of \(x\) against \(t\).</span><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.

Estimate to the nearest hour, the time needed for \(80\)% of the total number of Instantglam users to have viewed the video.

2025 JPJC P2 Q1

<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Express \(\frac{1}{Q(500 - Q)}\) in partial fractions.</span><span class="inline-marks">[2]</span></div></li></ol>
<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>
<ol class="abc-list" start="2"><li><div class="list-item-content"><span class="list-item-text"></span></div><ol class="i-list"><li><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 \(k\) is a positive constant.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence, solve the differential equation, expressing \(Q\) in terms of \(k\) and \(t\).</span><span class="inline-marks">[5]</span></div></li></ol></li></ol>

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

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.

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 \(x\).

2025 DHS P1 Q1 

<p>A Koi pond farmer introduces a circular patch of duckweed at the center of his circular pond to enhance water purification and provide supplemental nutrition for his Koi, as shown in the figure below.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768639935103-yxe5g5n68ml.png?token=eyJraWQiOiJzdG9yYWdlLXVybC1zaWduaW5nLWtleV9iZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njg2Mzk5MzUxMDMteXhlNWc1bjY4bWwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4NTcxMzQ3MiwiZXhwIjoxNzg1NzEzNzcyfQ.ABlT2qLpyud_IN1Zt4FhD_R00DzektArodGyZL9C5D0" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>The duckweed grows over time \(t\) days, forming a continuous circular patch on the pond's surface. It is assumed that the duckweed forms a thin, floating layer with negligible thickness. The rate of increase of the area covered by duckweed is directly proportional to the area of the pond that remains uncovered by the duckweed.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Given that the radius of the pond is \(10\) m, show that the radius, \(r\) m of the circular patch of duckweed satisfies the differential equation \(\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{k(100 - r^2)}{2r}\), \(k &gt; 0\).</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">The initial radius of the circular patch of duckweed is \(1\) m. Solve the differential equation, expressing \(r\) in the form \(r = \sqrt{P - Qe^{-kt}}\), where&nbsp;\(P\) and \(Q\) are constants to be determined.</span><span class="inline-marks">[6]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Sketch the graph of \(r\) against \(t\).</span><span class="inline-marks">[3]</span></div></li></ol>

A Koi pond farmer introduces a circular patch of duckweed at the center of his circular pond to enhance water purification and provide supplemental nutrition for his Koi, as shown in the figure below.

The duckweed grows over time \(t\) days, forming a continuous circular patch on the pond's surface. It is assumed that the duckweed forms a thin, floating layer with negligible thickness. The rate of increase of the area covered by duckweed is directly proportional to the area of the pond that remains uncovered by the duckweed.

Given that the radius of the pond is \(10\) m, show that the radius, \(r\) m of the circular patch of duckweed satisfies the differential equation \(\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{k(100 - r^2)}{2r}\), \(k > 0\).

The initial radius of the circular patch of duckweed is \(1\) m. Solve the differential equation, expressing \(r\) in the form \(r = \sqrt{P - Qe^{-kt}}\), where \(P\) and \(Q\) are constants to be determined.

2025 CJC P1 Q9

<p>The curve \(C\), where&nbsp;\(y \ge 0\), is such that it satisfies the following differential equation</p>
<p style="text-align: center">\[(1+x^2)y \frac{\mathrm{d}y}{\mathrm{d}x} + 2xy^2 = 0,\]</p>
<p>where&nbsp;\(y = 1\) when&nbsp;\(x = 0\).</p>
<p>By using the substitution \(z = y^2\), solve the differential equation and find the value of \(y\) when&nbsp;\(x = 1\).<span class="inline-marks">[5]</span></p>

The curve \(C\), where \(y \ge 0\), is such that it satisfies the following differential equation

\[(1+x^2)y \frac{\mathrm{d}y}{\mathrm{d}x} + 2xy^2 = 0,\]

By using the substitution \(z = y^2\), solve the differential equation and find the value of \(y\) when \(x = 1\).

Solving by Substitution

2020 NYJC P1 Q7

<ol class="abc-list"><li><div class="list-item-paragraph">Consider the differential equation</div><div class="list-item-paragraph" style="text-align: center">\(\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{f}\left( \frac{y}{x} \right), \quad x &gt; 0.\)</div><div class="list-item-paragraph">Use the substitution \(y = vx\) to show that the general solution of this differential equation is</div><div class="list-item-paragraph" style="text-align: center">\(\displaystyle\int \frac{\mathrm{d}v}{\mathrm{f}(v) - v} = \ln x + \text{Constant}.\)<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Hence, or otherwise, solve the differential equation</div><div class="list-item-paragraph" style="text-align: center">\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^2 + 3xy + y^2}{x^2}, \quad x &gt; 0,\)</div><div class="list-item-paragraph">given that \(y = 1\) when \(x = 1\). Give your answer in the form \(y = \mathrm{g}(x)\).<span class="inline-marks">[10]</span></div></li></ol>

Consider the differential equation\n\n\[\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{f}\left( \frac{y}{x} \right), \quad x > 0.\]\n\nUse the substitution \(y = vx\) to show that the general solution of this differential equation is\n\n\[\displaystyle \displaystyle\int \frac{\mathrm{d}v}{\mathrm{f}(v) - v} = \ln x + \text{Constant}.\]

Hence, or otherwise, solve the differential equation\n\n\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^2 + 3xy + y^2}{x^2}, \quad x > 0,\]\n\ngiven that \(y = 1\) when \(x = 1\). Give your answer in the form \(y = \mathrm{g}(x)\).

2017 IB HL P3 Q4

<p>The variables \(w\), \(x\) and \(y\) are connected by the following differential equations:</p>
<p style="text-align: center">\(\frac{\mathrm{d}w}{\mathrm{d}x} = -\frac{3}{2}w - 2 \quad\quad (A)\)</p>
<p style="text-align: center">\(\frac{\mathrm{d}y}{\mathrm{d}x} = w \qquad\qquad\quad\,\,\, (B)\)</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Solve equation (A) to find \(w\) in terms of \(x\).</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence find \(y\) in terms of \(x\).</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Different arbitrary constants found in part <strong>(ii)</strong> yields different solution curves. Some of the solution curves are straight lines. Write down the equation of one of these lines. On a single diagram, sketch your line together with a non-linear solution curve that has your line as an asymptote, indicating clearly any axes intercepts.<span class="inline-marks">[3]</span></div></li></ol>

The variables \(w\), \(x\) and \(y\) are connected by the following differential equations:

\(\frac{\mathrm{d}w}{\mathrm{d}x} = -\frac{3}{2}w - 2 \quad\quad (A)\)

\(\frac{\mathrm{d}y}{\mathrm{d}x} = w \qquad\qquad\quad\,\,\, (B)\)

Solve equation (A) to find \(w\) in terms of \(x\).

Different arbitrary constants found in part (ii) yields different solution curves. Some of the solution curves are straight lines. Write down the equation of one of these lines. On a single diagram, sketch your line together with a non-linear solution curve that has your line as an asymptote, indicating clearly any axes intercepts.

2016 SAJC P1 Q8 [Modified]

<p>A swimming pool contains \(375000\) litres of pure water. Water containing \(s\) milligrams of free chlorine per litre flows into the pool at a rate of \(10\) litres per minute. The pool is also draining at a rate of \(10\) litres per minute, such that the volume of the water in the pool remains constant.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that the rate at which the mass of the free chlorine, \(x\) grams, is changing in the pool over time, \(t\) minutes, can be modelled by the differential equation<br>\[\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{375s - x}{37500},\]State a necessary assumption for the above model to be valid.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Given that no free chlorine is present in the pool initially, find \(x\) in terms of \(t\) and \(s\).</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find, in terms of \(s\), the mass of free chlorine present in the pool after one hour.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Sketch the graph of \(x\) against \(t\) which is relevant to the context, labelling the point(s) where the curve crosses the axes and the equation(s) of the asymptote(s). Hence determine the mass of free chlorine present in the pool after a long period in terms of \(s\).</span><span class="inline-marks">[3]</span></div></li></ol>
<p>To prevent health complications, the recommended safe level of free chlorine to be used is between \(375\) grams and \(1125\) grams in a pool which contains \(375000\) litres of water. Find the range of values of \(s\) that should be used.<span class="inline-marks">[2]</span></p>

A swimming pool contains \(375000\) litres of pure water. Water containing \(s\) milligrams of free chlorine per litre flows into the pool at a rate of \(10\) litres per minute. The pool is also draining at a rate of \(10\) litres per minute, such that the volume of the water in the pool remains constant.

Show that the rate at which the mass of the free chlorine, \(x\) grams, is changing in the pool over time, \(t\) minutes, can be modelled by the differential equation\n\[\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{375s - x}{37500},\]\nState a necessary assumption for the above model to be valid.

Given that no free chlorine is present in the pool initially, find \(x\) in terms of \(t\) and \(s\).

Find, in terms of \(s\), the mass of free chlorine present in the pool after one hour.

Sketch the graph of \(x\) against \(t\) which is relevant to the context, labelling the point(s) where the curve crosses the axes and the equation(s) of the asymptote(s). Hence determine the mass of free chlorine present in the pool after a long period in terms of \(s\).

To prevent health complications, the recommended safe level of free chlorine to be used is between \(375\) grams and \(1125\) grams in a pool which contains \(375000\) litres of water. Find the range of values of \(s\) that should be used.

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