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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)
dN dt = KN(P N)
b)
dN dt = K[N + (P N)]
c)
dP dt = KN(P N)

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 = Cxn a solution of the differential equation xdy dx 3y = 0 ? Exactly one option must be correct)
a)
y = Cxn 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 = Cxn 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 xdy dx 3y.
Choice (d) is incorrect
Using the suggested value y = Cxn, find dy dx, substitute it into the given differential equation and factorise.
Which function P(t) is a solution to the differential equation dP dt = P(1 P) ? Exactly one option must be correct)
a)
P(t) = 1 1 + et
b)
P(t) = et 1 + et
c)
P(t) = 1 1 + et
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
dP dt = 1.2P 1 P 4200.
For which values of P is the population increasing? Exactly one option must be correct)
a)
P > 0
b)
0 < P < 4200
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 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
dy dx = xsin(3x2),
given that y(0) = 0. Exactly one option must be correct)
a)
y = cos(3x2) 6
b)
y = cos(3x2) 1 6
c)
y = cos(3x2) 6 + C
d)
y = 1 cos(3x2) 6

Choice (a) is incorrect
Note that in this case, y(0) = 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
dy dx = 1 x + x + 1
passing through the point (1,3). For this particular solution, what is the value of y when x = 2 ? Exactly one option must be correct)
a)
11 2 + ln2
b)
7 2 + ln2
c)
4 + ln2
d)
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
dy dt = 2tantsec2t + sin2t
. (Zero or more options can be correct)
a)
tan2t + sin2t
b)
sec2t + tan2t
c)
2tant + sint
d)
tant sintcost
e)
sec2t cos2t

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
d2y dx2 = 1 + cos3x cos2x
(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(sinx) cosx + Csinx + Dtanx
e)
y = ln|cosx| cosx + Cx + D

Choice (a) is incorrect
Antidifferentiation gives dy dx = tanx + sinx + C. Now do another antidifferentiation!
Choice (b) is incorrect
Antidifferentiation gives dy dx = tanx + sinx + C. Now do another antidifferentiation!
Choice (c) is incorrect
Antidifferentiation gives dy dx = tanx + sinx + C. Now do another antidifferentiation!
Choice (d) is incorrect
Antidifferentiation gives dy dx = tanx + sinx + C. Now do another antidifferentiation!
Choice (e) is correct!
Find the particular solution of the second order differential equation
d2y dt2 = t et
which satisfies the initial conditions y(0) = 0 and y(0) = 0. Exactly one option must be correct)
a)
y = 1 6t3 et
b)
y = 1 6t3 et + 1
c)
y = 1 6t3 et t + 1
d)
y = 1 6t3 + et + t 1
e)
y = 1 6t3 + t2 + t

Choice (a) is incorrect
Antidifferentiation gives dy dt = 1 2t2 + et + C. Now find the value of C using the condition y(0) = 0.
Choice (b) is incorrect
Antidifferentiation gives dy dt = 1 2t2 + et + C. Now find the value of C using the condition y(0) = 0.
Choice (c) is correct!
Choice (d) is incorrect
Antidifferentiation gives dy dt = 1 2t2 + et + C. Now find the value of C using the condition y(0) = 0.
Choice (e) is incorrect
Antidifferentiation gives dy dt = 1 2t2 + et + C. Now find the value of C using the condition y(0) = 0.
Given that x > 0, find the particular solution of the differential equation
d2y dx2 = 1 3x3 x2
satisfying the conditions y(2) = 1 and y(1) = 3. Exactly one option must be correct)
a)
y = lnx 1 2x3 + 15 2 x 4
b)
y = lnx + 1 2x3 15 2 x + 10
c)
y = lnx 1 2x3 15 2 x 4
d)
y = lnx + 2 3x3 + 1 3x + 2

Choice (a) is correct!
Choice (b) is incorrect
Antidifferentiation gives dy dx = x1 3 2x2 + C. Now find the value of C using the condition y(2) = 1.
Choice (c) is incorrect
Antidifferentiation gives dy dx = x1 3 2x2 + C. Now find the value of C using the condition y(2) = 1.
Choice (d) is incorrect
Antidifferentiation gives dy dx = x1 3 2x2 + C. Now find the value of C using the condition y(2) = 1.