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MATH1013 Quizzes

Quiz 11: Second order linear differential equations with constant coefficients
Question 1 Questions
What is the auxiliary equation associated with the second order linear differential equation
d2x dt2 + 19dx dt 27x = 0?
a)
d2y dx2 + 19dy dx 27y = 0
 b)
d2y dt2 + 19dy dt 27y = 0
c)
k2 + 19k 27 = 0
 d)
k2 19k + 27 = 0

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Find the general solution of the second order linear differential equation
d2y dx2 dy dx 12y = 0.
a)
y = Ae4x + Be3x
 b)
y = Ae4x + Be3x
c)
y = Ae6x + Be2x
 d)
y = Ae6x + Be2x

Choice (a) is correct!
k2 k 12 = (k 4)(k + 3) = 0 hence k = 4 or k = 3 and the solution is y = Ae4x + Be3x.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find the general solution to the second order linear differential equation
2d2x dt2 8dx dt + 6x = 0.
a)
x = Ae(4+10)t + Be(410)t
 b)
x = Ae(4+10)t + Be(410)t
c)
x = Ae3t + Bet
 d)
x = Ae3t + Bet

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
2k2 8k + 6 = 2(k 3)(k 1) = 0 hence k = 3 or k = 1 and the solution is x = Ae3t + Bet.
Choice (d) is incorrect
Find the particular solution to the second order linear differential equation
d2y dt2 + 6dy dt + 8y = 0,
where y = 2 and dy dt = 10 when t = 0.
a)
y = Ae2t + Be4t
 b)
y = Ae2t + Be4t
c)
y = 9e2t + 7e4t
 d)
y = e2t 3e4t

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
k2 + 6k + 8 = (k + 2)(k + 4) = 0 hence k = 2 or k = 4 and the general solution is y = Ae2t + Be4t. Also dy dt = 2Ae2t 4Be4t. At t = 0 we have A + B = 2 and 2A 4B = 10 which gives A = 1 and B = 3.
Find the general solution to the second order linear differential equation
d2y dx2 14dy dx + 51y = 0.
a)
y = e7x(Asin2x + Bcos2x)
 b)
y = e2x(Asin7x + Bcos7x)
c)
y = Ae3x + Be17x
 d)
y = Ae3x + Be17x

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
k2 14k 51 = (k 17)(k + 3) = 0 hence k = 17 or k = 3 and the general solution is y = Ae3x + Be17x.
Find the particular solution to the second order linear differential equation
d2y dx2 6dy dx + 4y = 0
where y = 6 and dy dx = 18 when x = 0.
a)
y = 3e(3+5)x + 3e(35)x
 b)
y = 6e3x cos5x.
c)
y = e5x((6 25)sin3x + 6cos3x).
 d)
The initial conditions are inconsistent since the general solution is
y = Ae(3+5)x + Be(35)x.

Choice (a) is correct!
k2 6k 16 = 0 has solutions k = 3 ±5 hence the general solution is y = Ae(3+5)x + Be(35)x. Since y = 6 when x = 0 we have A + B = 6. Now
dy dx = (3 + 5)Ae(3+5)x + (3 5)Be(35)x
so we have 3(A + B) + 5(A B) = 18 which gives A = B = 3 and y = 3e(3+5)x + 3e(35)x.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find the general solution to the second order linear differential equation
d2y dx2 6dy dx + 13y = 0.
a)
y = e2x(Asin3x + Bcos3x)
 b)
y = e3x(Asin2x + Bcos2x)
c)
y = e2x(Asin3x + Bcos3x)
 d)
y = Aex + Be5x

Choice (a) is incorrect
Choice (b) is correct!
k2 6k + 13 = 0 has solutions k = 3 ± 21 hence y = e3x(Asin2x + Bcos2x).
Choice (c) is incorrect
Choice (d) is incorrect
Find the particular solution of the second order linear differential equation
d2y dt2 8dy dt + 25y = 0
where y = 3 and dy dt = 6 when t = 0.
a)
y = e4t(2sin3t 3cos3t)
 b)
y = e3t(2sin4t 3cos4t)
c)
y = e4t(2sin3t + 3cos3t)
 d)
y = e3t(2sin4t + 3cos4t)

Choice (a) is correct!
k2 8k + 25 = 0 has solutions k = 4 ± 31 hence y = e4t(Asin3t + Bcos3t). At t = 0 we have y = B = 3 and
dy dt = 4e4t(Asin3t + Bcos3t) + e4t(3Acos3t 3Bsin3t)
hence 12 + 3A = 6 and A = 2. The particular solution is thus
y = e4t(2sin3t 3cos3t).
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find the general solution to the second order linear differential equation
d2y dt2 25y = 0.
a)
y = Acos5t + Bsin5t
 b)
y = Ae5t + Be5t
c)
y = 5cost + 5sint
 d)
y = A + Be25t

Choice (a) is incorrect
Choice (b) is correct!
k2 25 = (k 5)(k + 5) = 0 hence k = 5 or k = 5 and the general solution is y = Ae5t + Be5t.
Choice (c) is incorrect
Choice (d) is incorrect
Find the particular solution to the second order linear differential equation
d2x dt2 6dx dt + 8x = 0,
where x = 3 and dx dt = 6 when t = 0.
a)
x = Ae2t + Be4t
 b)
x = Ae2t + Be4t
c)
x = 3 2e2t 3 2e4t
 d)
x = 3e2t

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
k2 6k + 8 = 0 has solutions k = 2, k = 4 hence the general solution is x = Ae2t + Be4t. Since x = 3 when t = 0 we have A + B = 3
dx dt = 2Ae2t + 4Be4t
so we have 2A + 4B = 6 which gives A = 3 and B = 0. Hence x = 3e2t.