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 ! =) .
\(\\[1pt]\)
EXAMPLE:
\(\\[1pt]\)
In a chemical reaction, a compound X is formed from two compounds Y and Z.
\(\\[1pt]\)
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 \({\small \ {\large\frac{\textrm{d}x}{\textrm{d}t}} \ = \ 2}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(\textrm{a}\right).\hspace{0.8em}}\) Show that x and t satisfy the differential equation
\(\\[1pt]\)
\(\hspace{3em} {\large \frac{\mathrm{d}x}{\mathrm{d}t }} \ = \ 0.01(10 \ – \ x)(20 \ – \ x) \) .
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(\textrm{b}\right).\hspace{0.8em}}\) Solve this differential equation and obtain an expression for x in terms of t.
\(\\[1pt]\)
\({\small 2.\enspace}\) 9709/32/M/J/16 – Paper 32 May June 2016 Pure Maths 3 No 6
\(\\[1pt]\)
The variables \({\small x}\) and \({\small \theta}\) satisfy the differential equation
\(\\[1pt]\)
\(\hspace{3em} (3 \ + \ \cos2\theta) \ {\large \frac{\mathrm{d}x}{\mathrm{d} \theta }} \ = \ x \sin 2\theta \),
\(\\[1pt]\)
and its given that \({\small x \ = \ 3}\) when \({\small \theta \ = \ {\large \frac{1}{4}} \pi}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}\) Solve the differential equation and obtain an expression for \({\small x}\) in terms of \({\small \theta}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\) State the least value taken by \({\small x }\).
\(\\[1pt]\)
\({\small 3.\enspace}\) 9709/03/SP/17 – Specimen Paper 03 2017 No 8
\(\\[1pt]\)
The variables \({\small x}\) and \({\small \theta}\) satisfy the differential equation
\(\\[1pt]\)
\(\hspace{3em} {\large \frac{\mathrm{d}x}{\mathrm{d} \theta }} \ = \ (x \ + \ 2) \ {\sin}^{2} 2\theta \),
\(\\[1pt]\)
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.
\(\\[1pt]\)
\({\small 4.\enspace}\) 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 (P – A). Thus the differential equation connecting P and t is
\(\\[1pt]\)
\(\hspace{3em} {\large\frac{\textrm{d}P}{\textrm{d}t}} = -k \ \sqrt{P \ – \ A}\),
\(\\[1pt]\)
\(\\[7pt]\) where k is a positive constant.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the general solution of this differential equation.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Given that \({\small P \ = \ 5A}\) when \({\small t \ = \ 0}\) and that \({\small P \ = \ 2A}\) when \({\small t \ = \ 2}\), show that \({\small k \ = \ \sqrt{A}}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Find the value of t when \({\small P \ = \ A}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(d\right).\hspace{0.8em}}\) Obtain an expression for P in terms of A and t.
\(\\[1pt]\)
\({\small 5.\enspace}\) 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
\(\\[1pt]\)
\(\hspace{3em} {\large\frac{\textrm{d}\theta}{\textrm{d}t}} = -k \ (\theta \ – \ A) \),
\(\\[1pt]\)
\(\\[7pt]\) where k is a positive constant.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the solution of this differential equation, given that \({\small \ \theta \ = \ 4A \ }\) when \({\small t \ = \ 0}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Given also that \({\small \theta \ = \ 3A}\) when \({\small t \ = \ 1}\), show that \({\small k \ = \ \ln{\large\frac{3}{2}}}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Find \({\small \theta}\) in terms of A when \({\small t \ = \ 2}\), expressing your answer in its simplest form.
\(\\[1pt]\)
\({\small 6.\enspace}\) The variables x and y are related by the differential equation
\(\\[1pt]\)
\(\hspace{3em} {\large\frac{\textrm{d}y}{\textrm{d}x}} = {\large\frac{6x \ {\textrm{e}}^{3x}}{{y}^{2}}} \ .\)
\(\\[1pt]\)
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.
\(\\[1pt]\)
\({\small 7.\enspace}\) 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.
\(\\[1pt]\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}\) Using the fact that the gradient of the curve at P is \({\small {\large \frac{PN}{TN}}}\), show that
\(\\[1pt]\)
\(\hspace{3em} {\large\frac{\textrm{d}y}{\textrm{d}x}} = {\large \frac{1}{2}} {y}^{2} \cot x \ .\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}\) 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
\(\\[1pt]\)
\(\hspace{3em} x \tan \theta {\large\frac{\textrm{d}x}{\textrm{d}\theta}} \ + \ \textrm{cosec}^{2} \theta = 0,\)
\(\\[1pt]\)
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 \({\small \theta }\).
\(\\[1pt]\)
\({\small 9.\enspace}\) The variables x and y satisfy differential equation
\(\\[1pt]\)
\(\hspace{3em} {\large\frac{\textrm{d}y}{\textrm{d}x}} = x \ \textrm{e}^{x+y}\).
\(\\[1pt]\)
It is given that \({\small y \ = \ 0}\) when \({\small x \ = \ 0}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{i}.\hspace{0.8em}}\) Solve the differential equation, obtaining \({\small y }\) in terms of \({\small x }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{ii}.\hspace{0.8em}}\) Explain why \({\small x }\) can only take values that are less than 1.
\(\\[1pt]\)
\({\small 10.\enspace}\) 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 4(a), (b)
\(\\[1pt]\)
The equation of a curve is \(y = x \ {\tan}^{-1} \big(\frac{1}{2}x\big) \).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find \( {\large \frac{\mathrm{d}y}{\mathrm{d}x}}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) The tangent to the curve at the point where \(x = 2\) meets the \(y\)-axis at the point with coordinates (0, \(p\)).
\(\\[1pt]\)
Find \(p\).
\(\\[1pt]\)
\({\small 11.\enspace}\) 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 7
\(\\[1pt]\)
The variables \(x\) and \(y\) satisfy the differential equation
\(\\[1pt]\)
\( {\large \frac{\mathrm{d}y}{\mathrm{d}x}} = { \large\frac{y \ – \ 1}{(x \ + \ 1)(x \ + \ 3)} } \).
\(\\[1pt]\)
It is given that \(y = 2\) when \(x = 0\).
\(\\[1pt]\)
Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).
\(\\[1pt]\)
\({\small 12.\enspace}\) 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 8(a), (b)
\(\\[1pt]\)
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 \( (4, \mathrm{e}) \).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) By setting up and solving a differential equation, find the equation of the curve, expressing \(y\) in terms of \(x\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Describe what happens to \(y\) as \(x\) tends to infinity.
\(\\[1pt]\)
\({\small 13.\enspace}\) 9709/32/F/M/20 – Paper 32 March 2021 Pure Maths 3 No 4(a), (b)
\(\\[1pt]\)
The variables \(x\) and \(y\) satisfy the differential equation
\(\\[1pt]\)
\(\hspace{1.2em} (1 \ – \ \cos x){\large\frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ y \sin x \).
\(\\[1pt]\)
It is given that \(y = 4\) when \(x = \pi \).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Sketch the graph of \(y\) against \(x\) for \( 0 \lt x \lt 2 \pi \).
\(\\[1pt]\)
\({\small 14.\enspace}\) 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 7(a), (b)
\(\\[1pt]\)
\(\\[1pt]\)
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.
\(\\[1pt]\)
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 \( \tan x \).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right)\hspace{0.4em} (i). \hspace{0.6em} }\) Show that \( {\large \frac{MN}{y} } \ = \ {\large \frac{\mathrm{d}y}{\mathrm{d}x} } \).
\(\\[1pt]\)
\({\small\hspace{2.7em} (ii). \hspace{0.4em} }\) Hence show that \(x\) and \(y\) satisfy the differential equation \( {\large \frac{1}{2} } {y}^{2} {\large \frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ \tan x \).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Given that \(y = 1 \ \) when \(x = 0\), solve this differential equation to find the equation of the curve, expressing \(y\) in terms of \(x\).
\(\\[1pt]\)
\({\small 15.\enspace}\) 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 7
\(\\[1pt]\)
A curve is such that the gradient at a general point with coordinates \( (x, y) \) is proportional to \( {\large \frac{y}{\sqrt{x \ + \ 1}} }\).
\(\\[1pt]\)
The curve passes through the points with coordinates \( (0, 1) \) and \( (3, \mathrm{e}) \).
\(\\[1pt]\)
By setting up and solving a differential equation, find the equation of the curve, expressing \(y\) in terms of \(x\).
\(\\[1pt]\)
\({\small 16.\enspace}\) 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 10
\(\\[1pt]\)
The variables \(x\) and \(t\) satisfy the differential equation
\(\\[1pt]\)
\(\hspace{1.2em} {\large \frac{\mathrm{d}x}{\mathrm{d}t} } \ = \ x^{2} (1 \ + \ 2x) \),
\(\\[1pt]\)
and \( x = 1 \ \) when \( t \ = \ 0 \).
\(\\[1pt]\)
Using partial fractions, solve the differential equation, obtaining an expression for \(t\) in terms of \(x\).
\(\\[1pt]\)
\({\small 17.\enspace}\) 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 9
\(\\[1pt]\)
The variables \(x\) and \(y\) satisfy the differential equation
\(\\[1pt]\)
\(\hspace{1.2em} (x \ + \ 1)(3x \ + \ 1){\large \frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ y \),
\(\\[1pt]\)
and it is given that \( y = 1 \ \) when \( x \ = \ 1 \).
\(\\[1pt]\)
Solve the differential equation and find the exact value of \(y\) when \(x \ = \ 3\), giving your answer in a simplified form.
\(\\[1pt]\)
\({\small 18.\enspace}\) 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 9
\(\\[1pt]\)
The variables \(x\) and \(y\) satisfy the differential equation
\(\\[1pt]\)
\(\hspace{1.2em} (x \ + \ 1)(3x \ + \ 1){\large \frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ y \),
\(\\[1pt]\)
and it is given that \( \ y \ = \ 1 \ \) when \( x \ = \ 1 \).
\(\\[1pt]\)
Solve the differential equation and find the exact value of \(y\) when \( \ x \ = \ 3 \), giving your answer in a simplified form.
\(\\[1pt]\)
\({\small 19.\enspace}\) 9709/12/O/N/19 – Paper 12 November 2019 Pure Maths 1 No 3
\(\\[1pt]\)
A curve is such that
\(\\[1pt]\)
\(\hspace{1.2em} {\large \frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ {\large \frac{k}{\sqrt{x} } } \),
\(\\[1pt]\)
where \( \ k \ \) is a constant.
\(\\[1pt]\)
The points \( \ P \ (1, −1) \ \) and \( \ Q \ (4, 4) \ \) lie on the curve. Find the equation of the curve.
\(\\[1pt]\)
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 =) .
I have a question about question 5 in the extra questions you put. To be specific, I do not understand how to do part 3 of that question. Please help.