In a city with a fixed population of $P$
people, the rate of change (with respect to time,
$t$,) of the
number $N$
of people with a certain contagious disease is proportional to the product of the
number who have the disease and the number who do not.
Which option is a differential equation which describes this situation? Exactly one option must
be correct)

*Choice (a) is correct!*

*Choice (b) is incorrect*

The rate of change
of $N$ is proportional
to the product of $N$
and $N-P$, not
their sum.

*Choice (c) is incorrect*

$P$ is a fixed
number. It is $N$
that is changing.

For which values of $C$
and $n$ (if any) is
$y=C{x}^{n}$ a solution of the differential
equation $x\frac{dy}{dx}-3y=0$ ? Exactly one
option must be correct)

*Choice (a) is incorrect*

$C=0$ does give the solution
$y=0$. However, there are
other possibilities.

*Choice (b) is correct!*

*Choice (c) is incorrect*

Try setting $n=3$ and
calculating $x\frac{dy}{dx}-3y$.

*Choice (d) is incorrect*

Using the
suggested value $y=C{x}^{n}$,
find $\frac{dy}{dx}$,
substitute it into the given differential equation and factorise.

Which function $P\left(t\right)$
is a solution to the differential equation
$\frac{dP}{dt}=P\left(1-P\right)$ ? Exactly one option
must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is correct!*

*Choice (d) is incorrect*

The size of a population $P$
is modelled by the differential equation

$$\frac{dP}{dt}=1.2P\left(1-\frac{P}{4200}\right).$$

For which values of $P$
is the population increasing? Exactly one option must be correct)
*Choice (a) is incorrect*

Remember that
$P$ is increasing if its
derivative is positive.

*Choice (b) is correct!*

*Choice (c) is incorrect*

These are
the values of $P$
for which $\frac{dP}{dt}$
is increasing.

*Choice (d) is incorrect*

Remember that $P$
is increasing if its derivative is positive.

Find the particular solution of the differential equation

$$\frac{dy}{dx}=xsin\phantom{\rule{0.3em}{0ex}}\left(3{x}^{2}\right),$$

given that $y\left(0\right)=0$. Exactly one
option must be correct)
*Choice (a) is incorrect*

Note that in this case, $y\left(0\right)=-\frac{1}{6}$.

*Choice (b) is incorrect*

Remember that
the derivative of $cosx$
is $-sinx$.

*Choice (c) is incorrect*

The question asks for a particular solution, not the general solution.

*Choice (d) is correct!*

Find the particular solution of the differential equation

$$\frac{dy}{dx}=\frac{1}{x}+x+1$$

passing through the point $\left(1,3\right)$.
For this particular solution, what is the value of
$y$ when
$x=2$ ? Exactly one option
must be correct)
*Choice (a) is correct!*

*Choice (b) is incorrect*

Check that your particular
solution is correct!

*Choice (c) is incorrect*

Check that your particular solution is correct!

*Choice (d) is incorrect*

Check that your particular solution is correct!

*Choice (e) is incorrect*

Which two functions below satisfy the differential equation

For example, choice (a) should be True.

For example, choice (b) should be False.

For example, choice (c) should be False.

For example, choice (d) should be False.

For example, choice (e) should be True.

$$\frac{dy}{dt}=2tant{sec}^{2}t+sin2t$$

. (Zero or more options can be correct)
*There is at least one mistake.*

For example, choice (a) should be True.

*There is at least one mistake.*

For example, choice (b) should be False.

*There is at least one mistake.*

For example, choice (c) should be False.

*There is at least one mistake.*

For example, choice (d) should be False.

*There is at least one mistake.*

For example, choice (e) should be True.

*Correct!*

*True**False**False**False**True*

Find the general solution of the differential equation

$$\frac{{d}^{2}y}{d{x}^{2}}=\frac{1+{cos}^{3}x}{{cos}^{2}x}$$

($C$ and
$D$
are arbitrary constants.) Exactly one option must be correct)
*Choice (a) is incorrect*

Antidifferentiation
gives $\frac{dy}{dx}=tanx+sinx+C$. Now do another
antidifferentiation!

*Choice (b) is incorrect*

Antidifferentiation gives $\frac{dy}{dx}=tanx+sinx+C$. Now
do another antidifferentiation!

*Choice (c) is incorrect*

Antidifferentiation gives $\frac{dy}{dx}=tanx+sinx+C$. Now
do another antidifferentiation!

*Choice (d) is incorrect*

Antidifferentiation gives $\frac{dy}{dx}=tanx+sinx+C$. Now
do another antidifferentiation!

*Choice (e) is correct!*

Find the particular solution of the second order differential equation
Antidifferentiation
gives $\frac{dy}{dt}=\frac{1}{2}{t}^{2}+{e}^{-t}+C$. Now find
the value of $C$ using
the condition ${y}^{\prime}\left(0\right)=0$.

$$\frac{{d}^{2}y}{d{t}^{2}}=t-{e}^{-t}$$

which satisfies the initial conditions $y\left(0\right)=0$
and ${y}^{\prime}\left(0\right)=0$. Exactly one option
must be correct)
*Choice (a) is incorrect*

Antidifferentiation gives $\frac{dy}{dt}=\frac{1}{2}{t}^{2}+{e}^{-t}+C$.
Now find the value of $C$
using the condition ${y}^{\prime}\left(0\right)=0$.

*Choice (b) is incorrect*

Antidifferentiation
gives $\frac{dy}{dt}=\frac{1}{2}{t}^{2}+{e}^{-t}+C$. Now find
the value of $C$ using
the condition ${y}^{\prime}\left(0\right)=0$.

*Choice (c) is correct!*

*Choice (d) is incorrect*

Antidifferentiation
gives $\frac{dy}{dt}=\frac{1}{2}{t}^{2}+{e}^{-t}+C$. Now find
the value of $C$ using
the condition ${y}^{\prime}\left(0\right)=0$.

*Choice (e) is incorrect*

Given that $x>0,$
find the particular solution of the differential equation

$$\frac{{d}^{2}y}{d{x}^{2}}=\frac{1-3{x}^{3}}{{x}^{2}}$$

satisfying the conditions ${y}^{\prime}\left(2\right)=1$
and $y\left(1\right)=3.$ Exactly one option
must be correct)
*Choice (a) is correct!*

*Choice (b) is incorrect*

Antidifferentiation
gives $\frac{dy}{dx}=-{x}^{-1}-\frac{3}{2}{x}^{2}+C$. Now find
the value of $C$ using
the condition ${y}^{\prime}\left(2\right)=1$.

*Choice (c) is incorrect*

Antidifferentiation
gives $\frac{dy}{dx}=-{x}^{-1}-\frac{3}{2}{x}^{2}+C$. Now find
the value of $C$ using
the condition ${y}^{\prime}\left(2\right)=1$.

*Choice (d) is incorrect*

Antidifferentiation
gives $\frac{dy}{dx}=-{x}^{-1}-\frac{3}{2}{x}^{2}+C$. Now find
the value of $C$ using
the condition ${y}^{\prime}\left(2\right)=1$.