MATH1003 Quizzes

Quiz 6: Introduction to Ordinary Differential Equations
Question 1 Questions
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)
 a) $\frac{dN}{dt}=KN\left(P-N\right)$ b) $\frac{dN}{dt}=K\left[N+\left(P-N\right)\right]$ c) $\frac{dP}{dt}=KN\left(P-N\right)$

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)
 a) $y=C{x}^{n}$ is a solution only if $C=0.$ b) Either $C=0$ (for any value of $n$), or $n=3$. c) There are no values of $C$ for which $y=C{x}^{n}$ is a solution. d) Not enough information has been provided to be able to answer the question.

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)
 a) $P\left(t\right)=\frac{1}{1+{e}^{t}}$ b) $P\left(t\right)=\frac{{e}^{t}}{1+{e}^{-t}}$ c) $P\left(t\right)=\frac{1}{1+{e}^{-t}}$ d) None of the above satisfies the equation.

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)
 a) $P>0$ b) $0 c) $P<2100$ d) Not enough information has been provided to be able to answer the question.

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)
 a) $y=\frac{-cos\phantom{\rule{0.3em}{0ex}}\left(3{x}^{2}\right)}{6}$ b) $y=\frac{cos\phantom{\rule{0.3em}{0ex}}\left(3{x}^{2}\right)-1}{6}$ c) $y=\frac{-cos\phantom{\rule{0.3em}{0ex}}\left(3{x}^{2}\right)}{6}+C$ d) $y=\frac{1-cos\phantom{\rule{0.3em}{0ex}}\left(3{x}^{2}\right)}{6}$

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)
 a) $\frac{11}{2}+ln2$ b) $\frac{7}{2}+ln2$ c) $4+ln2$ d) $\frac{5}{2}+ln2$ e) None of the above

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
$\frac{dy}{dt}=2tant{sec}^{2}t+sin2t$
. (Zero or more options can be correct)
 a) ${tan}^{2}t+{sin}^{2}t$ b) ${sec}^{2}t+{tan}^{2}t$ c) $2tant+sint$ d) $tant-sintcost$ e) ${sec}^{2}t-{cos}^{2}t$

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!
1. True
2. False
3. False
4. False
5. 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)
 a) $y=tanx+cosx+C$ b) $y=ln|cosx|+sinx+D$ c) $y=tanxsecx+cosx+Cx+D$ d) $ln\left(sinx\right)-cosx+Csinx+Dtanx$ e) $y=-ln|cosx|-cosx+Cx+D$

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
$\frac{{d}^{2}y}{d{t}^{2}}=t-{e}^{-t}$
which satisfies the initial conditions $y\left(0\right)=0$ and ${y}^{\prime }\left(0\right)=0$. Exactly one option must be correct)
 a) $y=\frac{1}{6}{t}^{3}-{e}^{-t}$ b) $y=\frac{1}{6}{t}^{3}-{e}^{-t}+1$ c) $y=\frac{1}{6}{t}^{3}-{e}^{-t}-t+1$ d) $y=\frac{1}{6}{t}^{3}+{e}^{-t}+t-1$ e) $y=\frac{1}{6}{t}^{3}+{t}^{2}+t$

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
Antidifferentiation gives $\frac{dy}{dt}=\frac{1}{2}{t}^{2}+{e}^{-t}+C$. Now find the value of $C$ using the condition ${y}^{\prime }\left(0\right)=0$.
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)
 a) $y=-lnx-\frac{1}{2}{x}^{3}+\frac{15}{2}x-4$ b) $y=-lnx+\frac{1}{2}{x}^{3}-\frac{15}{2}x+10$ c) $y=-lnx-\frac{1}{2}{x}^{3}-\frac{15}{2}x-4$ d) $y=lnx+\frac{2}{3}{x}^{3}+\frac{1}{3}x+2$

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 ${y}^{\prime }\left(2\right)=1$.