menuicon

MATH1013 Quizzes

Quiz 10: First order linear differential equations
Question 1 Questions
Which of the following first order linear differential equations are in standard form ?
a)
xy(x) + y(x) x = e5x
 b)
dy dx + x2y = x3
c)
1 t dy dt + y = t5
 d)
y(t) + t4y = e4t

There is at least one mistake.
For example, choice (a) should be False.
There is at least one mistake.
For example, choice (b) should be True.
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 True.
Correct!
  1. False
  2. True
  3. False
  4. True
Put the first order linear differential equation
t2dx dt + x t = e2t
in standard form.
a)
dx dt + t3x = t2e2t
 b)
dx dt + t3x = e2t
c)
x2 dy dx + y x = e2x
 d)
t2dx dt + x t e2t = 0

Choice (a) is correct!
The standard form of a differential equation, for a function x(t), is x(t) + P(t)x(t) = R(t).
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Determine which of the following could be a multiplier (integrating factor) of the first order linear differential equation
dy dx + 2y x2 = x.
a)
e 2 3x3
 b)
e2 x
c)
e1 2x2
 d)
e1 x

Choice (a) is incorrect
Choice (b) is correct!
A multiplier is e 2 x2dx = e2 x.
Choice (c) is incorrect
Choice (d) is incorrect
Determine an integrating factor for the solution of the following first order linear differential equation
t3y(t) + 2y(t) = e3t.
a)
e2t
 b)
e1 3e3t
c)
et3e3tdt
 d)
et2

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Putting the equation into standard form, y(t) + 2t3y(t) = t3e3t, an integrating factor is e 2t3dx = et2 .
Find the general solution to the first order linear differential equation
dy dx 2y = e4x.
a)
y = 1 2e2x + C
 b)
y = 1 6e6x + C
c)
y = 1 2 + Ce2x
 d)
y = 1 2e4x + Ce2x

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
A multiplier is e2dx = e2x. Hence our equation becomes e2xdy dx 2e2xy = e2x. Thus d dx(e2xy) = e2x and e2xy = 1 2e2x + C which gives y = 1 2e4x + Ce2x.
Find the general solution to the first order linear differential equation
xy(x) + y(x) = x3.
a)
y(x) = 1 4x3 + C x
 b)
y(x) = 1 4x4 + C x
c)
y(x) = 1 4x3 + C
 d)
y(x)ex = x3exdx and solving this is outside the scope of this course.

Choice (a) is correct!
Putting the equation in standard form we get
y(x) + y(x) x = x2. A multiplier is thus e 1 xdx = eln x = x. Hence our equation becomes d dx(xy(x)) = x3 which upon integrating gives xy(x) = 1 4x4 + C so
y(x) = 1 4x3 + C x .
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find the general solution to the first order linear differential equation
dy dx 2y 1 1 + e2x = 0.
a)
y = 1 2e2x ln(1 + e2x) + Ce2x
 b)
y = 1 2e2x ln(1 + e2x) + Ce2x
c)
y = 1 2e2x ln 1 1 + e2x + Ce2x
 d)
y = 1 2e2x ln 1 1 + e2x + Ce2x

Choice (a) is incorrect
Choice (b) is correct!
A multiplier is e2dx = e2x hence
ye2x = e2x 1 + e2xdx, which implies ye2x = 1 2ln(1 + e2x) + C Hence so y = 1 2e2x ln(1 + e2x) + Ce2x.
Choice (c) is incorrect
Choice (d) is incorrect
Find the particular solution
etdy dt + etty = et2 (t + 1).
a)
y = e(t2+t) e1 2t2
 b)
y = e1 2t2 e(t2+t)
c)
y = e3 2t2+t e1 2 t2
 d)
y = 1 e(t2+t)

Choice (a) is incorrect
Choice (b) is correct!
Putting the equation in standard form we have dy dt + ty = e(t2+t) (t + 1) and a multiplier is etdt = e1 2 t2 . Hence we have ye1 2 t2 = e(1 2 t2+t) (t + 1)dt. Evaluating the integral by substitution we obtain
ye1 2 t2 = e1 2 t2+t + C so y = e(t2+t) + Ce1 2 t2 . At t = 0 we have y = 1 + C = 0 C = 1 and finally y = e1 2 t2 e(t2+t) .
Choice (c) is incorrect
Choice (d) is incorrect
Given that teatdt = 1 aeat(t 1 a) + C where a is a constant, find the particular solution of
y(t) + y(t) t = e5t
if y(1) = 0.
a)
y(t) = e5t 6 e6t 6
 b)
y(t) = e6t 6 (t 1 6) 5e6 36t
c)
y(t) = e5t 5t (t 1 5) 4e5 25t
 d)
y(t) = e5t 5 (t 1 5) + 3 50t

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
A multiplier is e1 tdt = t so d dt(ty(t)) = te5t and ty(t) = 1 5e5t(t 1 5) + C hence the general solution is
y(t) = 1 5te5t(t 1 5) + C t . Since y(1) = 1 5e5(4 5) + C = 0 we have C = 4 25e5 and then y(t) = e5t 5t (t 1 5) 4e5 25t.
Choice (d) is incorrect
Find the particular solution of the first order linear differential equation
t2dx dt + tx = t5
where x = 1 5 when t = 1.
a)
x = t4 5 .
 b)
x = t5 4 1 20t.
c)
x = et 5 (t 1 5) 4e 25.
 d)
x = et 5 (t 1 5) 1 t (1 5 4e 25).

Choice (a) is correct!
Putting the equation into standard form we have dx dt + 1 tx = t3 and a multiplier is t so we have tx = t4dt = 1 5t5 + C hence the general solution is
x = t4 5 + C t . Using x = 1 5 when t = 1 we have 1 5 + C 1 = 1 5 and C = 0 so that the particular solution is x = t4 5 .
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect