Differential Equations

Differential Equations

In this topic, we are looking into transforming some of the real-life physical quantities into mathematical models. The goal is to understand “how fast” a certain quantity change with respect to time.

The typical procedure is to create a mathematical model relating the phenomenon in question. By separating the variables and then perform the integration, we will arrive at the general solution to the problem.

Given a set of initial conditions, we can then create a specific equation relating to the problem, namely the particular solution.

I have put together some of the questions I received in the comment section below. You can try these questions also to further your understanding on this topic.

To check your answer, you can look through the solutions that I have posted either in Youtube videos or Instagram posts.

You can subscribe, like or follow my youtube channel and IG account. I will keep updating my IG daily post, preferably.

Furthermore, you can find some examples and more practices below! =).

Try some of the examples below. You can look at the solution I have written below to study and understand the topic. Cheers ! =) .
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EXAMPLE:

$${\small 1.\enspace}$$ 9709/03/SP/20 – Specimen Paper 2020 Pure Maths 3 No 10
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In a chemical reaction, a compound X is formed from two compounds Y and Z.
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The masses in grams of X, Y and Z present at time t seconds after the start of the reaction are x, 10 – x, 20 – x respectively. At any time the rate of formation of X is proportional to the product of the masses of Y and Z present at the time. When t = 0, x = 0 and $${\small \ {\large\frac{\textrm{d}x}{\textrm{d}t}} \ = \ 2}$$.
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$${\small\hspace{1.2em}\left(\textrm{a}\right).\hspace{0.8em}}$$ Show that x and t satisfy the differential equation
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$$\hspace{3em} {\large \frac{\mathrm{d}x}{\mathrm{d}t }} \ = \ 0.01(10 \ – \ x)(20 \ – \ x)$$ .
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$${\small\hspace{1.2em}\left(\textrm{b}\right).\hspace{0.8em}}$$ Solve this differential equation and obtain an expression for x in terms of t.

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$${\small 2.\enspace}$$ 9709/32/M/J/16 – Paper 32 May June 2016 Pure Maths 3 No 6
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The variables $${\small x}$$ and $${\small \theta}$$ satisfy the differential equation
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$$\hspace{3em} (3 \ + \ \cos2\theta) \ {\large \frac{\mathrm{d}x}{\mathrm{d} \theta }} \ = \ x \sin 2\theta$$,
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and its given that $${\small x \ = \ 3}$$ when $${\small \theta \ = \ {\large \frac{1}{4}} \pi}$$.
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$${\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}$$ Solve the differential equation and obtain an expression for $${\small x}$$ in terms of $${\small \theta}$$.
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$${\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}$$ State the least value taken by $${\small x }$$.

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$${\small 3.\enspace}$$ 9709/03/SP/17 – Specimen Paper 03 2017 No 8
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The variables $${\small x}$$ and $${\small \theta}$$ satisfy the differential equation
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$$\hspace{3em} {\large \frac{\mathrm{d}x}{\mathrm{d} \theta }} \ = \ (x \ + \ 2) \ {\sin}^{2} 2\theta$$,
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and its given that $${\small x \ = \ 0}$$ when $${\small \theta \ = \ 0}$$. Solve the differential equation and calculate the value of $${\small x}$$ when $${\small \ \theta \ = \ {\large \frac{1}{4}} \pi }$$, giving your answer correct to 3 significant figures.

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$${\small 4.\enspace}$$ Compressed air is escaping from a container. The pressure of the air in the container at time t is P, and the constant atmospheric pressure of the air outside the container is A. The rate of decrease of P is proportional to the square root of the pressure difference (PA). Thus the differential equation connecting P and t is
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$$\hspace{3em} {\large\frac{\textrm{d}P}{\textrm{d}t}} = -k \ \sqrt{P \ – \ A}$$,
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$$\\[7pt]$$ where k is a positive constant.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the general solution of this differential equation.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Given that $${\small P \ = \ 5A}$$ when $${\small t \ = \ 0}$$ and that $${\small P \ = \ 2A}$$ when $${\small t \ = \ 2}$$, show that $${\small k \ = \ \sqrt{A}}$$.
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$${\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}$$ Find the value of t when $${\small P \ = \ A}$$.
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$${\small\hspace{1.2em}\left(d\right).\hspace{0.8em}}$$ Obtain an expression for P in terms of A and t.

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$${\small 5.\enspace}$$ The temperature of a quantity of liquid at time t is $${\small \theta}$$. The liquid is cooling an atmosphere whose temperature is constant and equal to A. The rate of decrease of $${\small \theta}$$ is proportional to the temperature difference $${\small (\theta \ – \ A)}$$. Thus $${\small \theta}$$ and t satisfy the differential equation
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$$\hspace{3em} {\large\frac{\textrm{d}\theta}{\textrm{d}t}} = -k \ (\theta \ – \ A)$$,
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$$\\[7pt]$$ where k is a positive constant.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the solution of this differential equation, given that $${\small \ \theta \ = \ 4A \ }$$ when $${\small t \ = \ 0}$$.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Given also that $${\small \theta \ = \ 3A}$$ when $${\small t \ = \ 1}$$, show that $${\small k \ = \ \ln{\large\frac{3}{2}}}$$.
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$${\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}$$ Find $${\small \theta}$$ in terms of A when $${\small t \ = \ 2}$$, expressing your answer in its simplest form.

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$${\small 6.\enspace}$$ The variables x and y are related by the differential equation
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$$\hspace{3em} {\large\frac{\textrm{d}y}{\textrm{d}x}} = {\large\frac{6x \ {\textrm{e}}^{3x}}{{y}^{2}}} \ .$$
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It is given that $${\small y \ = \ 2}$$ when $${\small x \ = \ 0}$$. Solve the differential equation and hence find the value of y when $${\small x \ = \ 0.5}$$, giving your answer correct to 2 decimal places.

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$${\small 7.\enspace}$$ In the diagram the tangent to a curve at a general point P with coordinates (x, y) meets the x-axis at T. The point N on the x-axis such that PN is perpendicular to the x-axis. The curve is such that, for all values of x in the interval $${\small 0 \lt x \lt {\large \frac{1}{2}} \pi}$$, the area of triangle PTN is equal to $${\small \tan x}$$, where x is in radians.
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$${\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}$$ Using the fact that the gradient of the curve at P is $${\small {\large \frac{PN}{TN}}}$$, show that
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$$\hspace{3em} {\large\frac{\textrm{d}y}{\textrm{d}x}} = {\large \frac{1}{2}} {y}^{2} \cot x \ .$$
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$${\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}$$ Given that $${\small y = 2}$$ when $${\small x = {\large \frac{1}{6} } \pi}$$, solve this differential equation to find the equation of the curve, expressing y in terms of x.

$${\small 8.\enspace}$$ The variables x and $${\small \theta}$$ satisfy the differential equation
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$$\hspace{3em} x \tan \theta {\large\frac{\textrm{d}x}{\textrm{d}\theta}} \ + \ \textrm{cosec}^{2} \theta = 0,$$
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for $${\small 0 \lt \theta \lt {\large \frac{1}{2}} \pi} \$$ and $$\ {\small x \gt 0}$$. It is given that $${\small x \ = \ 4}$$ when $${\small \theta \ = \ {\large \frac{1}{6}} \pi}$$. Solve the differential equation, obtaining an expression for $${\small x }$$ in terms of $${\small \theta }$$.

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$${\small 9.\enspace}$$ The variables x and y satisfy differential equation
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$$\hspace{3em} {\large\frac{\textrm{d}y}{\textrm{d}x}} = x \ \textrm{e}^{x+y}$$.
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It is given that $${\small y \ = \ 0}$$ when $${\small x \ = \ 0}$$.
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$${\small\hspace{1.2em}\textrm{i}.\hspace{0.8em}}$$ Solve the differential equation, obtaining $${\small y }$$ in terms of $${\small x }$$.
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$${\small\hspace{1.2em}\textrm{ii}.\hspace{0.8em}}$$ Explain why $${\small x }$$ can only take values that are less than 1.

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$${\small 10.\enspace}$$ 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 4(a), (b)
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The equation of a curve is $$y = x \ {\tan}^{-1} \big(\frac{1}{2}x\big)$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find $${\large \frac{\mathrm{d}y}{\mathrm{d}x}}$$.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ The tangent to the curve at the point where $$x = 2$$ meets the $$y$$-axis at the point with coordinates (0, $$p$$).
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Find $$p$$.

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$${\small 11.\enspace}$$ 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 7
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The variables $$x$$ and $$y$$ satisfy the differential equation
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$${\large \frac{\mathrm{d}y}{\mathrm{d}x}} = { \large\frac{y \ – \ 1}{(x \ + \ 1)(x \ + \ 3)} }$$.
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It is given that $$y = 2$$ when $$x = 0$$.
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Solve the differential equation, obtaining an expression for $$y$$ in terms of $$x$$.

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$${\small 12.\enspace}$$ 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 8(a), (b)
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A certain curve is such that its gradient at a point $$(x, y)$$ is proportional to $${\large \frac{y}{x\sqrt{x}} }$$. The curve passes through the points with coordinates $$(1, 1)$$ and $$(4, \mathrm{e})$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ By setting up and solving a differential equation, find the equation of the curve, expressing $$y$$ in terms of $$x$$.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Describe what happens to $$y$$ as $$x$$ tends to infinity.

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$${\small 13.\enspace}$$ 9709/32/F/M/20 – Paper 32 March 2021 Pure Maths 3 No 4(a), (b)
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The variables $$x$$ and $$y$$ satisfy the differential equation
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$$\hspace{1.2em} (1 \ – \ \cos x){\large\frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ y \sin x$$.
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It is given that $$y = 4$$ when $$x = \pi$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Solve the differential equation, obtaining an expression for $$y$$ in terms of $$x$$.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Sketch the graph of $$y$$ against $$x$$ for $$0 \lt x \lt 2 \pi$$.

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$${\small 14.\enspace}$$ 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 7(a), (b)
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For the curve shown in the diagram, the normal to the curve at the point $$P$$ with coordinates $$(x, y)$$ meets the $$x$$-axis at $$N$$. The point $$M$$ is the foot of the perpendicular from $$P$$ to the $$x$$-axis.
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The curve is such that for all values of $$x$$ in the interval $$0 \leq x \lt {\large \frac{1}{2} }\pi$$, the area of triangle $$PMN$$ is equal to $$\tan x$$.
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$${\small\hspace{1.2em}\left(a\right)\hspace{0.4em} (i). \hspace{0.6em} }$$ Show that $${\large \frac{MN}{y} } \ = \ {\large \frac{\mathrm{d}y}{\mathrm{d}x} }$$.
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$${\small\hspace{2.7em} (ii). \hspace{0.4em} }$$ Hence show that $$x$$ and $$y$$ satisfy the differential equation $${\large \frac{1}{2} } {y}^{2} {\large \frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ \tan x$$.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Given that $$y = 1 \$$ when $$x = 0$$, solve this differential equation to find the equation of the curve, expressing $$y$$ in terms of $$x$$.

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$${\small 15.\enspace}$$ 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 7
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A curve is such that the gradient at a general point with coordinates $$(x, y)$$ is proportional to $${\large \frac{y}{\sqrt{x \ + \ 1}} }$$.
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The curve passes through the points with coordinates $$(0, 1)$$ and $$(3, \mathrm{e})$$.
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By setting up and solving a differential equation, find the equation of the curve, expressing $$y$$ in terms of $$x$$.

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$${\small 16.\enspace}$$ 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 10
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The variables $$x$$ and $$t$$ satisfy the differential equation
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$$\hspace{1.2em} {\large \frac{\mathrm{d}x}{\mathrm{d}t} } \ = \ x^{2} (1 \ + \ 2x)$$,
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and $$x = 1 \$$ when $$t \ = \ 0$$.
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Using partial fractions, solve the differential equation, obtaining an expression for $$t$$ in terms of $$x$$.

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$${\small 17.\enspace}$$ 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 9
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The variables $$x$$ and $$y$$ satisfy the differential equation
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$$\hspace{1.2em} (x \ + \ 1)(3x \ + \ 1){\large \frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ y$$,
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and it is given that $$y = 1 \$$ when $$x \ = \ 1$$.
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Solve the differential equation and find the exact value of $$y$$ when $$x \ = \ 3$$, giving your answer in a simplified form.

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$${\small 18.\enspace}$$ 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 9
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The variables $$x$$ and $$y$$ satisfy the differential equation
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$$\hspace{1.2em} (x \ + \ 1)(3x \ + \ 1){\large \frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ y$$,
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and it is given that $$\ y \ = \ 1 \$$ when $$x \ = \ 1$$.
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Solve the differential equation and find the exact value of $$y$$ when $$\ x \ = \ 3$$, giving your answer in a simplified form.

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$${\small 19.\enspace}$$ 9709/12/O/N/19 – Paper 12 November 2019 Pure Maths 1 No 3
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A curve is such that
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$$\hspace{1.2em} {\large \frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ {\large \frac{k}{\sqrt{x} } }$$,
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where $$\ k \$$ is a constant.
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The points $$\ P \ (1, −1) \$$ and $$\ Q \ (4, 4) \$$ lie on the curve. Find the equation of the curve.

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PRACTICE MORE WITH THESE QUESTIONS BELOW!

$${\small 1.\enspace (\textrm{i}) \enspace}$$ The number of bacteria in a colony is increasing at a rate that is proportional to the square root of the number of bacteria present. Form a differential equation relating number of bacteria, x, to the time t.

$${\small \quad (\textrm{ii}) \enspace}$$ In another colony the number of bacteria, y, after time t minutes is modelled by the differential equation $${\large \frac{\mathrm{d}y}{\mathrm{d}t} \ = \ \frac{10000}{\sqrt{y}}}$$. Find y in terms of t, given that $${\small y = 900 }$$ when $${\small t = 0 }$$. Hence, find the number of bacteria after 10 minutes.

$${\small 2. \enspace}$$ A skydiver drops from a helicopter. Before she opens her parachute, her speed v m/s after time t seconds is modelled by the differential equation

$$\hspace{3em}{\large \frac{\mathrm{d}v}{\mathrm{d}t}} \ = \ 10 {\large {\textrm{e}}^{-\frac{1}{2}t}}$$

when $${\small t = 0 }$$, $${\small v = 0 }$$.

$${\small \quad (\textrm{i}) \hspace{0.7em}}$$ Find v in terms of t.

$${\small \quad (\textrm{ii}) \enspace}$$ According to this model, what is the speed of the skydiver in the long term?

She opens her parachute when her speed is 10 m/s. Her speed t seconds after this is w m/s and is modelled by the differential equation $${\large\frac{\mathrm{d}w}{\mathrm{d}t}} \ = \ -\frac{1}{2}(w \ – \ 4)(w \ + \ 5)$$.

$${\small \quad (\textrm{iii}) \hspace{0.3em}}$$ Express $${\large\frac{1}{(w \ – \ 4)(w \ + \ 5)}}$$ in partial fractions.

$${\small \quad (\textrm{iv}) \enspace}$$ Using this result show that

$$\hspace{3em}{\large \frac{w \ – \ 4}{w \ + \ 5}} \ = \ 0.4 {\large {\textrm{e}}^{-4.5t}}$$.

$${\small \quad (\textrm{v}) \hspace{0.7em}}$$ According to this model, what is the speed of the skydiver in the long term?

$${\small 3. \enspace}$$ The number of organisms in a population at time t is denoted by x. Treating x as a continuous variable, the differential equation satisfied by x and t is

$$\hspace{3em}{\large \frac{\mathrm{d}x}{\mathrm{d}t}} \ = \ {\large \frac{x {\textrm{e}}^{-t}}{k \ + \ {\textrm{e}}^{-t} } }$$,

where $${\small k}$$ is a positive constant.

$${\small \quad (\textrm{i}) \hspace{0.7em}}$$ Given that $${\small x = 10 }$$ when $${\small t = 0 }$$, solve the differential equation, obtaining a relation between x, k and t.

$${\small \quad (\textrm{ii}) \enspace}$$ Given also that $${\small x = 20 }$$ when $${\small t = 1 }$$, show that $$k = 1 \ – \ {\large \frac{2}{\textrm{e}}}$$.

$${\small \quad (\textrm{iii}) \hspace{0.3em}}$$ Show that the number of organisms never reaches 48, however large t becomes.

$${\small 4. \enspace}$$ In a model of the expansion of a sphere of radius r cm, it is assumed that, at time t seconds after the start, the rate of increase of the surface area of the sphere is proportional to its volume. When $${\small t = 0 }$$, $${\small r = 5 }$$ and $${\large \frac{\mathrm{d}r}{\mathrm{d}t}} = \ 2$$.

$${\small \quad (\textrm{i}) \hspace{0.7em}}$$ Show that r satisfies the differential equation

$$\hspace{3em}{\large \frac{\mathrm{d}r}{\mathrm{d}t}} \ = \ 0.08{r}^{2}$$.

$${\small \quad (\textrm{ii}) \enspace}$$ Solve this differential equation, obtaining an expression for r in terms of t.

$${\small \quad (\textrm{iii}) \hspace{0.3em}}$$ Deduce from your answer to part (ii) the set of values that t can take, according to this model.

$${\small 5. \enspace}$$ An underground storage tank is being filled with liquid as shown in the diagram. Initially the tank is empty. At time t hours after filling begins, the volume of liquid is V $${\small \mathrm{{m}^{3}}}$$ and the depth of liquid is h m. It is given that $${\small V = \ {\large \frac{4}{3}}{h}^{3} }$$.

The liquid is poured in at a rate of 20 $${\small \mathrm{{m}^{3}}}$$ per hour, but owing to leakage, liquid is lost at a rate proportional to $${\small {h}^{2} }$$. When $${\small h = 1 }$$, $${\large \frac{\mathrm{d}h}{\mathrm{d}t}} = \ 4.95$$.

$${\small \quad (\textrm{i}) \hspace{0.7em}}$$ Show that h satisfies the differential equation

$$\hspace{3em}{\large \frac{\mathrm{d}h}{\mathrm{d}t}} \ = \ {\large \frac{5}{{h}^{2}} } \ – \ {\large \frac{1}{20} }$$.

$${\small \quad (\textrm{ii}) \enspace}$$ Verify that

$$\hspace{3em} {\large \frac{20 {h}^{2}}{100 \ – \ {h}^{2}} } \ \equiv \ -20 \ + \ {\large \frac{2000}{(10 \ – \ h)(10 \ + \ h)} }$$.

$${\small \quad (\textrm{iii}) \hspace{0.3em}}$$ Hence, solve the differential equation in part (i), obtaining an expression for t in terms of h.

$${\small 6. \enspace}$$ The variables x and y are related by the differential equation

$$\hspace{3em}{\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ {\large \frac{6y {\textrm{e}}^{3x}}{2 \ + \ {\textrm{e}}^{3x} } }$$.

Given that $${\small y = 36 }$$ when $${\small x = 0 }$$, find an expression for y in terms of x.

$${\small 7. \enspace}$$ The variables x and y satisfy the differential equation

$$\hspace{3em} (x + 1) \ y \ {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ {y}^{2} \ + \ 5$$.

It is given that $${\small y = 2 }$$ when $${\small x = 0 }$$. Solve the differential equation obtaining an expression for $${\small {y}^{2} }$$ in terms of $${\small x }$$.

$${\small 8. \enspace}$$ The coordinates (x, y) of a general point on a curve satisfy the differential equation

$$\hspace{3em} x \ {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ {(2 \ – \ x)}^{2} \ y$$.

The curve passes through the point (1, 1). Find the equation of the curve, obtaining an expression for $${\small y }$$ in terms of $${\small x }$$.

$${\small 9. \enspace}$$ The variables x and y satisfy the differential equation

$$\hspace{3em} x \ {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ {(4 \ – \ y)}^{2}$$,

and $${\small y = 1 }$$ when $${\small x = 1 }$$. Solve the differential equation, obtaining an expression for $${\small {y}^{2} }$$ in terms of $${\small x }$$.

$${\small 10. \enspace}$$ The variables $${\small x}$$ and $${\small \theta}$$ satisfy the differential equation

$$\hspace{3em} x \ {\cos}^{2} \theta \ {\large \frac{\mathrm{d}x}{\mathrm{d} \theta }} \ = \ 2 \tan \theta \ + \ 1$$,

for $${\small 0 \leq \theta \lt {\large \frac{1}{2}} \pi} \$$ and $$\ {\small x \gt 0}$$. It is given that $${\small x \ = \ 1}$$ when $${\small \theta \ = \ {\large \frac{1}{4}} \pi}$$.

$${\small \quad (\textrm{i}) \hspace{0.7em}}$$ Show that

$$\hspace{3em} {\large \frac{\mathrm{d}}{\mathrm{d} \theta }}({\tan}^{2} \theta) \ = \ {\large \frac{2 \tan \theta}{{\cos}^{2} \theta} }$$.

$${\small \quad (\textrm{ii}) \enspace}$$ Solve the differential equation and calculate the value of $${\small x }$$ when $${\small \theta \ = \ {\large \frac{1}{3}} \pi}$$, giving your answer correct to 3 significant figures.

As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .