 ## Logarithmic and Exponential Functions

### Logarithmic and Exponential Functions

Logarithmic functions and exponential functions are closely related in such a way that they are inverses of each other.

To be able to solve the questions for this topic, some of the key points that will be needed are:

–   the definitions of log and exp functions to convert the form between the two

–   the use of laws of logarithms and laws of exponents

–   the relationship as the inverses to each other in graphing the two functions

–   transforming the exp functions as a straight-line equation in log functions

The definition of logarithmic function:

$$\hspace{3em} {\large y = a^{x} \ \Rightarrow \ \log_{a} y = x }$$

In words:

If $${\small y }$$ is base of $${\small a }$$ raised to the power of $${\small x }$$ then logarithm to the base of $${\small a }$$ of $${\small y }$$ is $${\small x}$$.

For example, if $${\small 8 = 2^3 \ \Rightarrow \ \log_{2} 8 = 3}$$

To solve a more complex problems in logarithmic and exponential functions, we can employ laws of logarithms and laws of exponents as our arsenals.

The laws of logarithms are:

$${\large 1.\hspace{2em} \log_{a} 1 = 0 }$$

$${\large 2.\hspace{2em} \log_{a} a = 1 }$$

$${\large 3.\hspace{2em} \log_{a} {bc} = \log_{a} {b} \ + \ \log_{a} {c} }$$

$${\large 4.\hspace{2em} \log_{a} ({\frac{b}{c}}) = \log_{a} {b} \ – \ \log_{a} {c} }$$

$${\large 5.\hspace{2em} \log_{a^n} {b}^{m} = (\frac{m}{n}) \times \log_{a} b }$$

$${\large 6.\hspace{2em} \log_{a} b = \frac{1}{\log_{b} a} }$$

$${\large 7.\hspace{2em} \log_{a} b = \frac{\log_{p} b}{\log_{p} a} }$$

$${\large 8.\hspace{2em} a^{\log_{a} b} = b}$$

And, the laws of exponents are:

$${\large 1.\hspace{2em} a^{0} = 1; \ a \neq 0 }$$

$${\large 2.\hspace{2em} a^{m} \times a^{n} = a^{m \ + \ n} }$$

$${\large 3.\hspace{2em} \frac{ a^{m} }{ a^{n} } = a^{m \ – \ n} }$$

$${\large 4.\hspace{2em} a^n \ \times \ b^n = {(ab)}^{n} }$$

$${\large 5.\hspace{2em} \frac{ a^{n} }{ b^{n} } = (\frac{a}{b})^{n}}$$

$${\large 6.\hspace{2em} {(a^{m})}^{n} = a^{mn} }$$

$${\large 7.\hspace{2em} \sqrt[n]{a^{m}} = a^{\frac{m}{n}} }$$

$${\large 8.\hspace{2em} \frac{1}{a^n} = a^{-n} }$$

The graph of logarithmic functions are shown as below: The logarithmic function $${\small y = \log_{a} x}$$ has a vertical asymptote at the y-axis and x-intercept at (1, 0).

For base: $${\small 0 < a < 1 }$$, the logarithmic function is a monotonic decreasing function.

For base: $${\small a > 1 }$$, the logarithmic function is a monotonic increasing function.

Meanwhile, the graph of exponential functions are shown below: The exponential function $${\small y = a^x}$$ has a horizontal asymptote at the x-axis and y-intercept at (0, 1).

For base: $${\small 0 < a < 1 }$$, the exponential function is a monotonic decreasing function.

For base: $${\small a > 1 }$$, the exponential function is a monotonic increasing function.

If we put both graphs of logarithmic and exponential functions together, it is easy to see that they are reflection of each other to the line y = x.  The exponential functions are widely use in many areas for example in describing the population growth or decay, time value of money with its depreciation or inflation nature and many others.

It is usually helpful to convert the exponential model to a linear model.

Let our exponential growth function be $$N = ka^t$$ where $$k$$ and $$a$$ are constants,

Taking logarithms to both side of the equation gives:

$$\hspace{2em} \log N = \log (ka^t)$$

$$\hspace{2em} \log N = \log a^t + \log k$$

$$\hspace{2em} \log N = \log a . t + \log k$$

Compare that to the straight-line equation: $$y = mx + c$$ with:

– the dependent variable $$y = \log N$$,

– the gradient $$m = \log a$$,

– the independent variable $$x = t$$ and

– the y-intercept $$= \log k$$

The y-intercept is the initial population or capital at $$t = 0$$.

The gradient represents how fast the population grown.

Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
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EXAMPLE:

$${\small 1.\enspace}$$ 9709/32/F/M/17 – Paper 32 March 2017 Pure Maths 3 No 1
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Solve the equation $${\ln (1 + 2^{x}) = 2}$$, giving your answer correct to 3 decimal places.
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$${\small 2.\enspace}$$ 9709/32/F/M/19 – Paper 32 March 2019 Pure Maths 3 No 1
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$${\small\hspace{1.2em}(\textrm{i}).\hspace{0.8em}}$$ Show that the equation $${ \log_{10} (x-4) = 2 \ – \ \log_{10} x}$$ can be written as a quadratic equation in $$x$$.
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$${\small\hspace{1.2em}(\textrm{ii}).\hspace{0.8em}}$$ Hence solve the equation $${ \log_{10} (x-4) = 2 \ – \ \log_{10} x }$$ giving your answer correct to 3 significant figures.
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$${\small 3.\enspace}$$ 9709/33/M/J/19 – Paper 33 June 2019 Pure Maths 3 No 1
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Use logarithms to solve the equation $$5^{3 \ – \ 2 x} = 4(7^{x})$$, giving your answer correct to 3 decimal places.
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$${\small 4.\enspace}$$ 9709/31/M/J/19 – Paper 31 June 2019 Pure Maths 3 No 2
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Showing all necessary working, solve the equation
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$$\hspace{2em} \ln (2x \ – \ 3) = 2 \ \ln x \ – \ \ln (x \ – \ 1)$$
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$${\small 5.\enspace}$$ 9709/32/F/M/20 – Paper 32 March 2020 Pure Maths 3 No 2
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Solve the equation $$\ln 3 + \ln (2x+5) = 2 \ \ln(x+2)$$.
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$${\small 6.\enspace}$$ 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 1
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Find the set of values of $$x$$ for which $$2(3^{1-2x}) \lt 5^{x}$$.
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$${\small 7.\enspace}$$ 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 2
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The variables $$x$$ and $$y$$ satisfy the equation $$y^2 = A {\mathrm{e}}^{kx}$$, where $$A$$ and $$k$$ are constants. The graph of $$\ln y$$ against $$x$$ is a straight line passing through the points (1.5, 1.2) and (5.24, 2.7) as shown in the diagram.
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Find the values of $$A$$ and $$k$$ correct to 2 decimal places.
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$${\small 8.\enspace}$$ 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 3
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$${\small (\textrm{a}).\hspace{0.8em}}$$ Show that the equation
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$$\hspace{2em} \ln (1 + \mathrm{e}^{−x}) + 2x = 0$$
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can be expressed as a quadratic equation in $$\mathrm{e}^{x}$$.
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$${\small (\textrm{b}).\hspace{0.8em}}$$ Hence solve the equation $$\ln (1 + \mathrm{e}^{−x}) + 2x = 0$$, giving your answer correct to 3 decimal places.
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$${\small 9.\enspace}$$ 9709/31/O/N/20 – Paper 31 Nov 2020 Pure Maths 3 No 4
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Solve the equation
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$$\hspace{2em} \log_{10} (2x + 1) = 2 \ \log_{10} (x+1) \ – \ 1$$.
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$${\small 10.\enspace}$$ 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 1
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Solve the equation $$\ln (x^3 − 3) = 3 \ln x − \ln 3$$. Give your answer correct to 3 significant figures.
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$${\small 11.\enspace}$$ 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 2
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Find the real root of the equation $$\frac{2\mathrm{e}^x \ + \ \mathrm{e}^{-x}}{2 \ + \ \mathrm{e}^x} = 3$$, giving your answer correct to 3 decimal places. Your working should show clearly that the equation has only one real root.
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$${\small 12.\enspace}$$ 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 3
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The variables $$x$$ and $$y$$ satisfy the equation $$x = A(3^{−y})$$, where $$A$$ is a constant.
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$${\small(\textrm{a}).\hspace{0.8em}}$$ Explain why the graph of $$y$$ against $$\ln x$$ is a straight line and state the exact value of the gradient
of the line.
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It is given that the line intersects the $$y$$-axis at the point where $$y$$ = 1.3.
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$${\small(\textrm{b}).\hspace{0.8em}}$$ Calculate the value of $$A$$, giving your answer correct to 2 decimal places.
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$${\small 13.\enspace}$$ 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 2
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Solve the equation $$4^x = 3 + 4^{−x}$$. Give your answer correct to 3 decimal places.
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$${\small 14.\enspace}$$ 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 3
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The variables $$x$$ and $$y$$ satisfy the equation $${\small {x}^{n}{y}^{2} \ = \ C }$$, where $$n$$ and $$C$$ are constants.
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The graph of $${\small \ln \ y }$$ against $${\small \ln \ x }$$ is a straight line passing through the points (0.31, 1.21) and (1.06, 0.91), as shown in the diagram.
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Find the value of $$n$$ and find the value of $$C$$ correct to 2 decimal places.
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PRACTICE MORE WITH THESE QUESTIONS BELOW!

$${\small 1.\enspace}$$ Solve the inequality $$|3^x – 5 | \lt 1$$, giving 3 significant figures in your answer.

$${\small 2. \enspace}$$ Given that $$x = 4(3^{-y})$$, express $$y$$ in terms of $$x$$.

$${\small 3. \enspace}$$ Using the substitution $$u = 3^{x}$$, or otherwise, solve, correct to 3 significant figures, the equation

$$\hspace{2em} 3^{x} = 2 + 3^{-x}$$.

$${\small 4. \enspace}$$ The variables $$x$$ and $$y$$ satisfy the relation $$3^y = 4^{x+2}$$.

$${\small\hspace{1.2em} (\textrm{a}).\hspace{0.8em}}$$ By taking logarithms, show that the graph of $$y$$ against $$x$$ is a straight line. Find the exact value of the gradient of this line.

$${\small\hspace{1.2em} (\textrm{b}).\hspace{0.8em}}$$ Calculate the $$x$$ co-ordinate of the point of intersection of this line with the line $$y = 2x$$, giving your answer correct to 2 decimal places.

$${\small 5. \enspace}$$ It is given that $$\ln(y + 5) – \ln y = 2 \ln x$$. Express $$y$$ in terms of $$x$$, in a form not involving logarithms.

$${\small 6. \enspace}$$ Given that $${(1.25)}^{x} = {(2.5)}^{y}$$, use logarithms to find the value of $$\frac{x}{y}$$ correct to 3 significant figures.

$${\small 7. \enspace}$$ Solve, correct to 3 significant figures, the equation

$$\hspace{2em} \mathrm{e}^x + \mathrm{e}^{2x} = \mathrm{e}^{3x}$$.

$${\small 8. \enspace}$$ The variables $$x$$ and $$y$$ satisfy the equation $$y = A(b^{–x})$$, where $$A$$ and $$b$$ are constants. The graph of $$\ln y$$ against $$x$$ is a straight line passing through the points (0, 1.3) and (1.6, 0.9), as shown in the diagram. Find the values of $$A$$ and $$b$$, correct to 2 decimal places.
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$${\small 9. \enspace}$$ Solve the equation $$\ln (2 + \mathrm{e}^{-x}) = 2$$, giving your answer correct to 2 decimal places.

$${\small 10.\enspace}$$ Two variable quantities $$x$$ and $$y$$ are related by the equation $$y = A{x}^{n}$$, where $$A$$ and $$n$$ are constants. The diagram shows the result of plotting $$\ln y$$ against $$\ln x$$ for four pairs of values of $$x$$ and $$y$$. Use the diagram to estimate the values of $$A$$ and $$n$$.
$$\\[1pt]$$ As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) . ## 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 =) .