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 \(\\[1pt]\)
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.
Check out my solution here:
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https://www.instagram.com/p/CMr1xAtlCUS/ \(\\[1pt]\)
Follow my instagram for daily updates on many more solutions on other topics too! \(\\[1pt]\)
\({\small 2.\enspace}\)
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 \({\small x }\).
Check out my solution here:
\(\\[1pt]\)
https://www.instagram.com/p/CML2XFVFRtO/ \(\\[1pt]\)
Follow my instagram for daily updates on many more solutions on other topics too! \(\\[1pt]\)
\({\small 3.\enspace}\)
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.
Check out my solution here:
\(\\[1pt]\)
https://www.instagram.com/p/CMVyt2yFy9x/ \(\\[1pt]\)
Follow my instagram for daily updates on many more solutions on other topics too! \(\\[1pt]\)
\({\small 4.\enspace}\) 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.
\({\small\hspace{1.2em}\left(a\right).\hspace{2em}} {\large\frac{\textrm{d}P}{\textrm{d}t}} = -k \ \sqrt{P \ – \ A}\)
\(\\[1pt]\)
Separation of variables,
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.9em}} {\large\frac{\textrm{d}P}{\sqrt{P \ – \ A}}} = -k \ \textrm{d}t \)
\(\\[20pt]{\small\hspace{1.2em} \displaystyle \int \frac{\textrm{d}P}{\sqrt{P \ – \ A}} = -k \ \int \textrm{d}t }\)
\(\\[1pt]\)
We can use direct substitution to solve the left hand side integral,
\(\\[1pt]\)
\(\\[7pt]\) Let \({\small \sqrt{P \ – \ A} = u}\)
\(\\[7pt] \hspace{2.1em}\) \({\small P \ – \ A = u^2}\)
\(\\[17pt] \hspace{3.2em}\) \({\small \mathrm{d}P = 2 \ u \ \mathrm{d}u}\)
\(\\[20pt]{\small\hspace{1.2em} \displaystyle \int \frac{2u}{\sqrt{u^2}}\mathrm{d}u = -kt \ + \ C_{1} }\)
\(\\[20pt]{\small\hspace{1.2em} \require{cancel} \displaystyle \int \frac{2 \ \cancel{u}}{\cancel{u}}\mathrm{d}u = -kt \ + \ C_{1} }\)
\(\\[20pt]{\small\hspace{1.2em} \displaystyle 2 \ \int \mathrm{d}u = -kt \ + \ C_{1} }\)
\(\\[20pt]{\small\hspace{1.2em} 2 \ u \ + \ C_{2} = -kt \ + \ C_{1} }\)
\(\\[20pt]{\small\hspace{1.2em} 2 \ u = -kt \ + \ C_{3} }\)
Revert u to P,
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em} 2 \ \sqrt{P \ – \ A} = -kt \ + \ C_{3} }\)
\(\\[20pt]{\small\hspace{1.2em} C_{1}, C_{2} \ \mathrm{and} \ C_{3}}\) are arbitrary constants.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{2em}}\) \({\small P \ = \ 5A}\) when \({\small t \ = \ 0}\)
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em} 2 \ \sqrt{5A \ – \ A} = -k \times 0 \ + \ C_{3} }\)
\(\\[20pt]{\small\hspace{1.2em} 2 \ \sqrt{4A} \ = \ C_{3} }\)
\(\\[20pt]{\small\hspace{1.2em} C_{3} \ = \ 4 \ \sqrt{A}}\).
\(\\[1pt]\)
\({\small\hspace{1.2em} P \ = \ 2A}\) when \({\small t \ = \ 2}\)
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em} 2 \ \sqrt{2A \ – \ A} = -k \times 2 \ + \ 4 \ \sqrt{A} }\)
\(\\[20pt]{\small\hspace{1.2em} 2 \ \sqrt{A} \ = -2k \ + \ 4 \ \sqrt{A} }\)
\(\\[20pt]{\small\hspace{1.2em} 2k = 2 \ \sqrt{A} }\)
\(\\[25pt]{\small\hspace{1.2em} k \ = \sqrt{A}}\).
\({\small\hspace{1.2em}\left(c\right).\hspace{2em}}\) \({\small P \ = \ A}\)
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em} 2 \ \sqrt{A \ – \ A} = -t \ \sqrt{A} \ + \ 4 \ \sqrt{A} }\)
\(\\[20pt]{\small\hspace{1.2em} 2 \times 0 = \ -t \ \sqrt{A} \ + \ 4 \ \sqrt{A} }\)
\(\\[20pt]{\small\hspace{1.2em} t \ \sqrt{A} \ = \ 4 \ \sqrt{A}}\)
\(\\[20pt]{\small\hspace{1.2em} t \ = \ 4 }\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(d\right).\hspace{2em}}\) expression for P
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em} 2 \ \sqrt{P \ – \ A} = -t \ \sqrt{A} \ + \ 4 \ \sqrt{A} }\)
\(\\[20pt]{\small\hspace{1.2em} 2 \ \sqrt{P \ – \ A} = (4 \ – \ t)\sqrt{A} }\)
\(\\[20pt] \hspace{1.2em}\) Square both sides of the equation,
\(\\[20pt]{\small\hspace{1.2em} {\big[2 \ \sqrt{P \ – \ A}\big]}^{2} = {\big[(4 \ – \ t)\sqrt{A}\big]}^{2} }\)
\(\\[20pt]{\small\hspace{1.2em} 4 \ (P \ – \ A) \ = \ {(4 \ – \ t)}^{2} \times A }\)
\(\\[20pt]{\small\hspace{1.2em} 4P \ – \ 4A \ = \ (16 \ – \ 8t \ + \ t^2) \times A}\)
\(\\[20pt]{\small\hspace{1.2em} 4P \ = \ 4A \ + \ 16A \ – \ 8At \ + \ A{t}^{2} }\)
\(\\[20pt] \hspace{1.2em}\) Divide both sides of the equation by 4,
\(\\[20pt]{\small\hspace{1.2em} P \ = \ 5A \ – \ 2At \ + \ {\large\frac{1}{4}}A{t}^{2} }\)
\(\\[1pt]\)
\({\small 5.\enspace}\) 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.
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}} {\large\frac{\textrm{d}\theta}{\textrm{d}t}} = -k \ (\theta \ – \ A) \)
\(\\[1pt]\)
Separation of variables,
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.9em}} {\large\frac{\textrm{d}\theta}{(\theta \ – \ A)}} = -k \ \textrm{d}t \)
\(\\[20pt]{\small\hspace{1.2em} \displaystyle \int \frac{\textrm{d}\theta}{(\theta \ – \ A)} = -k \ \int \textrm{d}t }\)
\(\\[20pt]{\small\hspace{1.2em} \ln \ (\theta \ – \ A) = -kt \ + \ C }\)
\(\\[10pt]\) Use the given initial condition to find C,
\(\\[10pt]{\small\hspace{1.2em} \theta \ = \ 4A}\) when \({\small t \ = \ 0}\)
\(\\[10pt]{\small\hspace{1.2em} \ln \ (4A \ – \ A) = -k \times 0 \ + \ C }\)
\(\\[10pt]{\small\hspace{4.8em} C = \ln 3A }\)
\(\\[10pt]\) Thus the solution of this differential equation is,
\(\\[20pt]\hspace{1.2em} \ln \ (\theta \ – \ A) = -kt \ + \ \ln 3A \)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em} \theta \ = \ 3A}\) when \({\small t \ = \ 1}\)
\(\\[1pt]\)
\(\\[15pt]{\small\hspace{1.2em} \ln \ (3A \ – \ A) = -k \times 1 \ + \ \ln 3A }\)
\(\\[15pt]{\small\hspace{1.2em} \ln 2A = -k \ + \ \ln 3A }\)
\(\\[15pt]{\small\hspace{2.7em} k = \ \ln 3A \ – \ \ln 2A}\)
\(\\[20pt]{\small\hspace{2.7em} \require{cancel} k = \ \ln \ {\large\frac{3\cancel{A}}{2\cancel{A}}} }\)
\(\\[20pt]{\small\hspace{2.7em} k = \ \ln \ {\large\frac{3}{2}} }\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) The value of \({\small \theta}\) in terms of A when \({\small t \ = \ 2}\)
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em} \ln \ (\theta \ – \ A) = \ -\ln\big({\large\frac{3}{2}}\big) \times 2 \ + \ \ln 3A }\)
\(\\[15pt]{\small\hspace{1.2em} \ln \ (\theta \ – \ A) = \ \ln 3A \ – \ \ln 3 }\)
\(\\[20pt]{\small\hspace{1.2em} \require{cancel} \ln \ (\theta \ – \ A) = \ \ln \Big({\large\frac{\cancel{3}A}{\cancel{3}}}\Big) }\)
\(\\[15pt]{\small\hspace{1.2em} \ln \ (\theta \ – \ A) = \ \ln A }\)
\(\\[15pt]{\small\hspace{2.9em} \theta \ – \ A = \ A }\)
\({\small\hspace{4.5em} \theta = \ 2A }\)
\(\\[1pt]\)
\({\small 6.\enspace}\) 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.
\(\hspace{1.2em}{\large\frac{\textrm{d}y}{\textrm{d}x}} = {\large\frac{6x \ {\textrm{e}}^{3x}}{{y}^{2}}}\)
\(\\[1pt]\)
Separation of variables,
\(\\[1pt]\)
\(\\[17pt]\hspace{1.5em} {\small {y}^{2} \ \mathrm{d}y \ = \ 6x \ {\textrm{e}}^{3x} \ \mathrm{d}x }\)
\(\\[20pt]{\small\hspace{1.2em} \displaystyle \int {y}^{2} \ \mathrm{d}y = 6 \int x{\textrm{e}}^{3x} \ \mathrm{d}x }\)
\(\\[17pt]{\small\hspace{1.2em} {\large\frac{1}{3}}{y}^{3} + \ C_{1} = 6 \displaystyle \int x{\textrm{e}}^{3x} \ \textrm{d}x }\)
To solve \( {\small \displaystyle \int x{\textrm{e}}^{3x} \ \textrm{d}x }\), we will use integration by parts.
\(\\[1pt]\)
\(\displaystyle \int {u\frac{{\mathrm{d}v}}{{\mathrm{d}x}}} \ \mathrm{d}x = uv \ – \int {\frac{{\mathrm{d}u}}{{\mathrm{d}x}}} v \ \mathrm{d}x \)
\(\\[1pt]\)
\(\\[10pt]\) Let \(u = x\) and \(v = \frac{1}{3}{\textrm{e}}^{3x}\)
\(\hspace{1.5em} {\small \textrm{d}u = \mathrm{d}x }\) and \({\small \mathrm{d}v = {\textrm{e}}^{3x} \ \mathrm{d}x }\)
\(\\[1pt]\)
Then,
\(\\[1pt]\)
\( \displaystyle \int x{\textrm{e}}^{3x} \ \mathrm{d}x = \frac{1}{3}x{\textrm{e}}^{3x} \ – \frac{1}{3} \int {\textrm{e}}^{3x} \ \mathrm{d}x \)
\(\\[1pt]\)
\( \displaystyle \int x{\textrm{e}}^{3x} \ \mathrm{d}x = \frac{1}{3}x{\textrm{e}}^{3x} \ – \frac{1}{9} {\textrm{e}}^{3x} \ + \ C_{2} \)
\(\\[1pt]\)
Let’s combine the result to get the general solution of the differential equation
\(\\[1pt]\)
\(\\[17pt]{\small\hspace{1.2em} {\large\frac{1}{3}}{y}^{3} + \ C_{1} = 6 \large[ {\small \frac{1}{3}x{\textrm{e}}^{3x} \ – \frac{1}{9} {\textrm{e}}^{3x} \ + \ C_{2}} \large] }\)
\(\\[20pt]{\small\hspace{1.2em} {y}^{3} = 6 x{\textrm{e}}^{3x} \ – 2 {\textrm{e}}^{3x} \ + \ C_{3} }\)
We can then solve the particular solution of the differential equation from the given initial condition, \({\small y \ = \ 2}\) when \({\small x \ = \ 0}\).
\(\\[1pt]\)
\(\\[17pt]{\small\hspace{1.2em} {2}^{3} = 6 . (0) . \ {\textrm{e}}^{(3 . (0))} \ – 2 {\textrm{e}}^{(3 . (0))} \ + \ C_{3} }\)
\(\\[17pt]{\small\hspace{1.2em} 8 = 0 \ – 2 \ + \ C_{3} }\)
\(\\[17pt]{\small\hspace{1.2em} C_{3} = 10 }\)
The particular solution of the differential equation is:
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em} {y}^{3} = 6 x{\textrm{e}}^{3x} \ – 2 {\textrm{e}}^{3x} \ + \ 10 }\)
\(\\[20pt]\)The y value when \({\small x \ = \ 0.5}\),
\(\\[17pt]{\small\hspace{1.2em} {y}^{3} = 6 . (0.5) . \ {\textrm{e}}^{(3 . (0.5))} \ – 2 {\textrm{e}}^{(3 . (0.5))} \ + \ 10 }\)
\({\small\hspace{1.2em} y \ \approx \ 2.44 }\)
\(\\[1pt]\)
\({\small 7.\enspace}\) 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.
\(\\[12pt] {\small\hspace{0.8em}\left(i\right).\hspace{0.8em}}\) Given that the coordinates of P is (x, y),
\(\\[15pt] {\small\hspace{3em} PN \ = \ y. }\)
\(\\[12pt]\) Also, area of \({\small \triangle PTN} \ = \ {\small {\large\frac{1}{2}} \times PN \times TN }\)
\(\\[17pt] {\small\hspace{6.1em} \tan x \ = \ {\large\frac{1}{2}} \times y \times TN }\)
\(\\[15pt]\) Re-arrange for TN,
\(\\[20pt] {\small\hspace{3em} TN \ = \ {\large\frac{2 \ \tan x}{y}} .}\)
\(\\[15pt]\) Since the gradient of the curve is \({\small {\large \frac{PN}{TN}}}\),
\(\\[20pt] {\small\hspace{3em} {\large \frac{\mathrm {d}y}{\mathrm {d}x}} \ = \ {\large\frac{PN}{TN}} .}\)
\(\\[20pt] {\small\hspace{3em} {\large \frac{\mathrm {d}y}{\mathrm {d}x}} \ = \ {\large\frac{y}{2 \tan x \ / \ y}} }\)
\(\\[20pt] {\small\hspace{3em} {\large \frac{\mathrm {d}y}{\mathrm {d}x}} \ = \ {\large\frac{{y}^{2}}{2 \tan x}} }\)
\(\\[15pt]{\small\hspace{3em} {\large \frac{\mathrm {d}y}{\mathrm {d}x}} \ = \ {\large \frac{1}{2}} {y}^{2} \cot x \ .}\)
\(\\[1pt]\)
\(\\[12pt] {\small\hspace{0.8em}\left(ii\right).\hspace{0.8em} {\large \frac{\mathrm {d}y}{\mathrm {d}x}} \ = \ {\large \frac{1}{2}} {y}^{2} \cot x }\)
\(\\[1pt]\)
Separation of variables,
\(\\[1pt]\)
\(\\[17pt]\hspace{1.5em} {\small {\large \frac{2}{{y}^{2}}} \ \mathrm{d}y \ = \ \cot x \ \mathrm{d}x }\)
\(\\[25pt]{\small\hspace{1.2em} 2 \displaystyle \int \frac{1}{{y}^{2}} \ \mathrm{d}y \ = \int \cot x \ \mathrm{d}x }\)
\(\\[25pt]{\small\hspace{1.2em} 2 \displaystyle \int {y}^{-2} \ \mathrm{d}y \ = \int \frac{\cos x}{\sin x} \ \mathrm{d}x }\)
\(\\[25pt]{\small\hspace{1.2em} \displaystyle \frac{-2}{y} \ + \ C_{1} = \int \frac{\cos x}{\sin x} \ \mathrm{d}x }\)
To solve \({\small \displaystyle \int \frac{\cos x}{\sin x} \ \mathrm{d}x }\), we will use the direct substitution method.
\(\\[1pt]\)
\(\\[12pt]\) Let \(u = \sin x\)
\(\\[20pt]\hspace{1.1em} \frac{\mathrm{d}u}{\mathrm{d}x} = \cos x \)
\(\hspace{1.1em} \mathrm{d}u = \cos x \ \mathrm{d}x \)
\(\\[1pt]\)
\(\\[15pt]\) Then,
\({\small \displaystyle \int \frac{\cos x}{\sin x} \ \mathrm{d}x \ = \ \int \frac{\mathrm{d}u}{u} }\)
\(\\[1pt]\)
\(\hspace{4.5em} = \ {\small \ln u \ + \ C_{2} }\)
\(\\[1pt]\)
Substitute u back to \({\small \sin x}\)
\(\\[1pt]\)
\(\hspace{4.5em} = \ {\small \ln \ \sin x \ + \ C_{2} } \)
\(\\[1pt]\)
and calculate the integration,
\(\\[1pt]\)
\(\\[25pt]{\small\hspace{2.1em} \displaystyle \frac{-2}{y} \ + \ C_{1} = \int \frac{\cos x}{\sin x} \ \mathrm{d}x }\)
\(\\[25pt]{\small\hspace{4em} {\large \frac{-2}{y}} \ = \ \ln \ \sin x + C_{3} }\)
We can then solve the particular solution of the differential equation using the condition, \({\small y \ = \ 2}\) when \({\small x \ = \ {\large \frac{\pi}{6} }}\).
\(\\[1pt]\)
\(\\[17pt]{\small\hspace{2.1em} \displaystyle \frac{-2}{2} \ = \ \ln \ \sin \frac{\pi}{6} + C_{3} }\)
\(\\[17pt]{\small\hspace{2.6em} -1 \ = \ \ln {\large \frac{1}{2} } \ + \ C_{3} }\)
\(\\[17pt]{\small\hspace{2.8em} C_{3} \ = \ \ln 2 \ – \ 1 }\)
The particular solution of the differential equation is:
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{2.1em} {\large \frac{-2}{y}} \ = \ \ln \ \sin x + \ \ln 2 \ – \ 1 }\)
\({\small\hspace{2.1em} {\large \frac{-2}{y}} \ = \ \ln \ (2\sin x) \ – \ 1 }\)
\(\\[1pt]\)
\({\small 8.\enspace}\) 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 \({\small \theta }\).
\(\hspace{1.9em} x \tan \theta{\large\frac{\textrm{d}x}{\textrm{d}\theta}} \ + \ \textrm{cosec}^{2} \theta = 0 \)
\(\\[1pt]\)
Separation of variables,
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{2.9em} x \ \tan \theta {\large\frac{\textrm{d}x}{\textrm{d}\theta}} \ = \ – \ \textrm{cosec}^{2} \theta }\)
\(\\[20pt]{\small\hspace{2.9em} x \ \textrm{d}x \ = \ -{\large\frac{ \textrm{cosec}^{2} \theta }{ \tan \theta }} \textrm{d}\theta }\)
\(\\[20pt]{\small\hspace{2.9em} x \ \textrm{d}x \ = \ -{\large\frac{ (1 \ / \ {\sin}^{2} \theta) }{ (\sin \theta \ / \cos \theta) }} \textrm{d}\theta }\)
\(\\[20pt]{\small\hspace{1.9em} \displaystyle \int x \ \textrm{d}x \ = \ – \ \int \frac{ \cos \theta \ \textrm{d}\theta }{ {\sin}^{3} \theta } }\)
\(\\[1pt]\)
We can use direct substitution to solve the right hand side integral,
\(\\[1pt]\)
\(\\[7pt]\) Let \({\small u = \sin \theta }\)
\(\\[17pt] \hspace{0.9em}\) \({\small \textrm{d}u = \cos \theta \ \textrm{d}\theta}\)
\(\\[20pt]{\small\hspace{1.2em} \displaystyle \int x \textrm{d}x \ = \ – \ \int \frac{\mathrm{d}u}{u^3}}\)
\(\\[20pt]{\small\hspace{1.9em} {\large\frac{1}{2}}x^2 \ = \ – \big({\large\frac{1}{-2}}\big) \ u^{-2} \ + \ C }\)
\(\\[20pt]{\small\hspace{2.7em} x^{2} \ = \ {\large\frac{1}{{u}^{2}}} \ + \ C_{1} }\)
Revert u to \({\small \sin \theta }\),
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{2.5em} x^{2} \ = \ {\large\frac{1}{ {\sin}^{2} \theta }} \ + \ C_{1} }\)
\({\small\hspace{1.2em} C \ \mathrm{and} \ C_{1}}\) are arbitrary constants.
\(\\[1pt]\)
\({\small\hspace{1.2em} x \ = \ 4}\) when \({\small \theta \ = \ {\large \frac{1}{6}} \pi}\),
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.3em} {4}^{2} \ = {\large \frac{1}{\sin^{2} {\large \frac{\pi}{6}}} } \ + \ C_{1} }\)
\(\\[20pt]{\small\hspace{1.2em} 16 \ = {\large \frac{1}{ {({\large\frac{1}{2}})}^{2} }} \ + \ C_{1} }\)
\(\\[15pt]{\small\hspace{1.2em} 16 \ = 4 \ + \ C_{1} }\)
\({\small\hspace{1.2em} C_{1} \ = \ 12}\).
\(\\[1pt]\)
Thus, the expression for \({\small x }\) in terms of \({\small \theta }\),
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em} x \ = \ \sqrt{ {\large\frac{1}{ {\sin}^{2} \theta }} \ + \ 12} }\)
Note that, the negative square root of \({\small x }\) is rejected since \( \ {\small x \gt 0}\).
\(\\[1pt]\)
\({\small 9.\enspace}\) 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.
\(\\[20pt]{\small\hspace{1.2em}\textrm{i}.\hspace{0.8em}}\hspace{1.2em}{\large\frac{\textrm{d}y}{\textrm{d}x}} \ = \ x \ \textrm{e}^{x+y}\)
\(\hspace{3.7em}{\large\frac{\textrm{d}y}{\textrm{d}x}} \ = \ x \ . \ \textrm{e}^{x} \ . \ \textrm{e}^{y}\)
\(\\[1pt]\)
Separation of variables,
\(\\[1pt]\)
\(\\[17pt]\hspace{2.7em} {\small \textrm{e}^{-y} \ \mathrm{d}y \ = \ x \ {\textrm{e}}^{x} \ \mathrm{d}x }\)
\(\\[20pt]{\small\hspace{1.9em} \displaystyle \int \textrm{e}^{-y} \ \mathrm{d}y \ = \ \int x \ {\textrm{e}}^{x} \ \mathrm{d}x }\)
\(\\[17pt]{\small\hspace{1.2em} – \ \textrm{e}^{-y} + \ C_{1} \ = \ \displaystyle \int x \ {\textrm{e}}^{x} \ \textrm{d}x }\)
To solve \( {\small \displaystyle \int x \ {\textrm{e}}^{x} \ \textrm{d}x }\), we will use integration by parts.
\(\\[1pt]\)
\(\displaystyle \int {u\frac{{\mathrm{d}v}}{{\mathrm{d}x}}} \ \mathrm{d}x = uv \ – \int {\frac{{\mathrm{d}u}}{{\mathrm{d}x}}} v \ \mathrm{d}x \)
\(\\[1pt]\)
\(\\[10pt]\) Let \(u = x\) and \(v = {\textrm{e}}^{x}\)
\(\hspace{1.5em} {\small \textrm{d}u = \mathrm{d}x }\) and \({\small \mathrm{d}v = {\textrm{e}}^{x} \ \mathrm{d}x }\)
\(\\[1pt]\)
Then,
\(\\[1pt]\)
\( \displaystyle \int x \ {\textrm{e}}^{x} \ \mathrm{d}x = x \ {\textrm{e}}^{x} \ – \ \int {\textrm{e}}^{x} \ \mathrm{d}x \)
\(\\[1pt]\)
\( \displaystyle \int x \ {\textrm{e}}^{x} \ \mathrm{d}x = x \ {\textrm{e}}^{x} \ – \ {\textrm{e}}^{x} \ + \ C_{2} \)
\(\\[1pt]\)
Let’s combine the result to get the general solution of the differential equation
\(\\[1pt]\)
\(\\[20pt]{\small\hspace{1.2em} – \ \textrm{e}^{-y} \ = \ x \ {\textrm{e}}^{x} \ – \ {\textrm{e}}^{x} \ + \ C }\)
We can then solve the particular solution of the differential equation from the given initial condition, \({\small y \ = \ 0}\) when \({\small x \ = \ 0}\).
\(\\[1pt]\)
\(\\[17pt]{\small\hspace{1.2em} -1 \ = \ 0 \ – \ 1 \ + \ C }\)
\(\\[17pt]{\small\hspace{1.6em} C \ = \ 0 }\)
Thus, the expression of \({\small y}\) in terms of \({\small x}\),
\(\\[1pt]\)
\(\\[15pt]{\small\hspace{1.2em} – \ \textrm{e}^{-y} \ = \ x \ {\textrm{e}}^{x} \ – \ {\textrm{e}}^{x} }\)
\(\\[15pt]{\small\hspace{1.9em} \textrm{e}^{-y} \ = \ {\textrm{e}}^{x} \ – \ x \ {\textrm{e}}^{x} }\)
\(\\[20pt]{\small\hspace{1.9em} \textrm{e}^{-y} \ = \ {\textrm{e}}^{x} ( 1 \ – \ x ) }\)
\(\\[12pt]\) Take the ln of both sides,
\(\\[15pt]{\small\hspace{1.2em} \ln{ \textrm{e}^{-y}} \ = \ \ln{ {\textrm{e}}^{x} ( 1 \ – \ x ) } }\)
\(\\[15pt]{\small\hspace{2.1em} – \ y \ = \ x \ + \ \ln{ ( 1 \ – \ x )} }\)
\( {\small\hspace{2.8em} y \ = \ – \ x \ – \ \ln{ ( 1 \ – \ x )} }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{ii}.\hspace{0.8em}}\) From the log definition, the logarithm of a number exist only if the number is greater than 0. Since \({\small y \ = \ – \ x \ – \ \ln{ ( 1 \ – \ x )} }\), for \({\small y}\) values to exist, the ln term in the expression of \({\small y}\) require that \({\small ( 1 \ – \ x ) \ \gt \ 0}\). Thus, \({\small x \ \lt \ 1}\).
\(\\[1pt]\)
\({\small 10.\enspace}\)
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 \(p\).
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\(\\[1pt]\)
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\({\small 11.\enspace}\)
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 \(x\).
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\(\\[1pt]\)
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\({\small 12.\enspace}\)
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.
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\(\\[1pt]\)
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\({\small 13.\enspace}\)
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 \( 0 \lt x \lt 2 \pi \).
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\(\\[1pt]\)
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\({\small 14.\enspace}\)
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 \(x\).
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\(\\[1pt]\)
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\({\small 15.\enspace}\)
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 \(x\).
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\(\\[1pt]\)
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\({\small 16.\enspace}\)
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 \(x\).
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\(\\[1pt]\)
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\({\small 17.\enspace}\)
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.
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\(\\[1pt]\)
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\({\small 18.\enspace}\)
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.
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\(\\[1pt]\)
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\({\small 19.\enspace}\)
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.
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\(\\[1pt]\)
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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 =) .
