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MATH1013 Quizzes

Quiz 12: First order linear differential equations and predator prey models
Question 1 Questions
Two populations, x = x(t) and y = y(t), satisfy the following differential equations x(t) = 2x(t) + 3y(t), y(t) = 5x(t) 4y(t). Eliminate y to find a second order linear differential equation in x such that x = x(t).
a)
x(t) + 2x(t) 23x(t) = 0
 b)
x(t) 2x(t) + 23x(t) = 0
c)
x(t) 2 3x(t) 7 3x(t) = 0
 d)
x(t) 6x(t) x(t) = 0

Choice (a) is correct!
x(t) = 2x(t) + 3y(t), therefore,x(t) = 2x(t) + 15x(t) 12y(t), = 2x(t) + 15x(t) 4x(t) + 8x(t) = 2x(t) + 23x(t), x(t) + 2x(t) 23x(t) = 0.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Two populations, x = x(t) and y = y(t), satisfy the following differential equations dx dt = 3x + 2y, dy dt = x + y. Eliminate y to find a second order linear differential equation in x such that x = x(t).
a)
d2x dt2 + 4dx dt x = 0
 b)
d2x dt2 4dx dt + 5x = 0
c)
d2x dt2 4dx dt + x = 0
 d)
d2x dt2 + 4dx dt 5x = 0.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
d2x dt2 = 3dx dt + 2dy dt , d2x dt2 = 3dx dt + 2x + 2y, d2x dt2 = 3dx dt + 2x + dx dt 3x = 0, d2x dt2 = 4dx dt x, d2x dt2 4dx dt + x = 0.
Choice (d) is incorrect
Let populations A and B be represented by x(t) and y(t) respectively at time t. Population A’s growth is 4 times itself and it is decreased by 2 times the population of B. Population B’s growth is 5 times population A and it is decreased by 3 times its own population. Which of the systems below model these populations ?
a)
x(t) = 4x(t) 2y(t),
y(t) = 5y(t) 3x(t).
 b)
x(t) = 4x(t) 2y(t),
y(t) = 5x(t) 3y(t).
c)
x(t) = 4x(t) + 2y(t),
y(t) = 5x(t) + 3y(t).
 d)
x(t) = 4x(t) + 2y(t),
y(t) = 5y(t) + 3x(t).

Choice (a) is incorrect
Choice (b) is correct!
This is a classic predator-prey relationship where population B is the predator and population A is the prey.
Choice (c) is incorrect
Choice (d) is incorrect
Let populations A and B be represented by x(t) and y(t) respectively at time t. Population As growth is 2 times itself and it is increased by 3 times the population of B. Population Bs growth is 3 times itself and it is increased by 4 times population A. Which of the systems below model these populations ?
a)
x(t) = 2x(t) 3y(t),
y(t) = 3y(t) 4x(t).
 b)
x(t) = 2x(t) + 3y(t),
y(t) = 3x(t) + 4y(t).
c)
x(t) = 2x(t) 3y(t),
y(t) = 4x(t) 3y(t).
 d)
x(t) = 2x(t) + 3y(t),
y(t) = 4x(t) + 3y(t).

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
This is a classic symbiotic relationship where each population depends on the other for growth.
Find the general solution to the system of linear differential equations x(t) = 4x(t) 2y(t), y(t) = 5x(t) 3y(t).
a)
x(t) = Ae2t + Bet,
y(t) = 3Ae2t + 3 2Bet.
 b)
x(t) = Ae2t + Bet,
y(t) = 5Ae2t + 5 2Bet.
c)
x(t) = Ae2t + Bet,
y(t) = 6Ae2t + 3 2Bet.
 d)
x(t) = Ae2t + Bet,
y(t) = Ae2t + 5 2Bet.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
First find a differential equation involving x(t) only : x(t) = 4x(t) 2y(t), x(t) = 4x(t) 10x(t) + 6y(t), = 4x(t) 10x(t) + 12x(t) 3x(t), = x(t) + 2x(t), x(t) x(t) 2x(t) = 0. The auxiliary equation is k2 k 2 = (k 2)(k + 1) = 0 and the general solution is x(t) = Ae2t + Bet, x(t) = 2Ae2t Bet, y(t) = 2x(t) 1 2x(t), = 2Ae2t + 2Bet Ae2t + 1 2Bet, = Ae2t + 5 2Bet.
Two populations x = x(t) and y = y(t) satisfy the following differential equations x(t) = 2x(t) + 3y(t), y(t) = 4x(t) + 3y(t). What is the solution to this system if x(0) = 3000 and y(0) = 500?
a)
x(t) = 3500e2t 500e3t,
y(t) = 500e3t.
 b)
x(t) = 9500 3 e2t 500 3 e3t,
y(t) = 500 3 e3t.
c)
x(t) = 1500e6t + 1500et,
y(t) = 2000e6t + 1500et.
 d)
x(t) = 1500e6t + 1500et,
y(t) = 2000e6t 1500et.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
We first find the differential equation for x(t) : x(t) = 2x(t) + 3y(t). therefore,x(t) = 2x(t) + 12x(t) + 9y(t), = 2x(t) + 12x(t) + 3x(t) 6x(t), = 5x(t) + 6x(t), therefore,x(t) 5x(t) 6x(t) = 0. It follows that the general solution is x(t) = Ae6t + Bet and y(t) = 4 3Ae6t Bet. Substituting the initial conditions gives A + B = 3000 and 4 3A B = 500 which gives A = B = 1500.
Two populations x = x(t) and y = y(t) satisfy the following differential equations dx dt = 3x + 2y, dy dt = 4x + y. Find the general solution to this system.
a)
x(t) = Ae(215)t + Be(2+15)t
y(t) = 5 15 2 Ae(215)t + 5 + 15 2 Ae(2+15)t
 b)
x(t) = Ae(215)t + Be(2+15)t,
y(t) = 1 15 2 Ae(215)t + 1 + 15 2 Ae(2+15)t.
c)
x(t) = Ae5t + Bet,
y(t) = Ae5t + 2Bet.
 d)
x(t) = Ae5t + Bet,
y(t) = 4Ae5t Bet.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
We first find the differential equation for x(t) : d2x dt2 = 3dx dt + 2dy dt , d2x dt2 3dx dt 8x 2y = 0, d2x dt2 3dx dt 8x dx dt + 3x = 0, d2x dt2 4dx dt 5x = 0. The auxiliary equation is k2 4k 5 = (k 5)(k + 1) = 0 hence the general solution is x(t) = Ae5t + Bet. therefore,dx dt = 5Ae5t Bet, y(t) = 1 2 dx dt 3 2x = 5 2Ae5t 1 2Bet 3 2Ae5t 3 2Bet = Ae5t 2Bet.
Choice (d) is incorrect
Find the particular solution to the system of linear differential equations x(t) = 7x(t) 12y(t), y(t) = 4x(t) 7y(t).
a)
x(t) = 3et 2et,
y(t) = 2et et.
 b)
x(t) = 2et et,
y(t) = 2 3et + 1 3et.
c)
x(t) = 11 7 sint + cost,
y(t) = 5 6sint + cost.
 d)
x(t) = 1 7e7t + 6 7e7t,
y(t) = e7t.

Choice (a) is correct!
We find the differential equation for x(t) : x(t) = 7x(t) 12y(t), therefore,x(t) = 7x(t) 48x(t) + 84y(t), = 7x(t) 48x(t) 7x(t) + 49x(t), = x(t), x(t) x(t) = 0. So x(t) = Aet + Bet and y(t) = 1 2Aet + 2 3Bet. Hence A + B = 1 and 1 2A + 2 3B = 1 which gives A = 2 and B = 3.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find the general solution to the system of linear differential equations x(t) = 2x(t) 3y(t), y(t) = 3x(t) + 2y(t).
a)
x(t) = Ae(2+17)t + Be(217)t,
y(t) = 3Ae2t + 3 2Bet.
 b)
x(t) = Ae(2+17)t + Be(217)t,
y(t) = 1 3Ae(2+17)t + 1 3Be(217)t.
c)
x(t) = e2t(Asin3t + Bcos3t),
y(t) = e2t((4 3A + B)sin3t (A 4 3B)cos3t).
 d)
x(t) = e2t(Asin3t + Bcos3t),
y(t) = e2t(Bsin3t Acos3t).

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
We find the differential equation for x(t) : x(t) = 2x(t) 3y(t), therefore,x(t) = 2x(t) 9x(t) 6y(t), = 2x(t) 9x(t) + 2x(t) 4x(t), = 4(t) 13x(t), therefore,x(t) 4x(t) + 13x(t) = 0. So the auxiliary equation k2 4k + 13 = 0 and its roots are k = 2 ± 31 and the general solution is x(t) = e2t(Asin3t + Bcos3t), x(t) = 2e2t(Asin3t + Bcos3t) + e2t(3Acos3t 3Bsin3t). y(t) = 2 3x(t) 1 3x(t) = e2t(Bsin3t Acos3t).
Find the particular solution to the system of linear differential equations x(t) = x(t) 2y(t), y(t) = 4x(t) 3y(t), where y(0) = 1 and y(π) = 2.
a)
x(t) = et(1 2sin2t 5 2cos2t),
y(t) = e2t(2sin2t + cos2t).
 b)
x(t) = et(1 2sin2t + 3 2cos2t),
y(t) = e2t(2sin2t + cos2t).
c)
x(t) = Ae(2+17)t + Be(217)t,
y(t) = 3Ae2t + 3 2Bet.
 d)
x(t) = Ae(2+17)t + Be(217)t,
y(t) = 1 3Ae(2+17)t + 1 3Be(217)t.

Choice (a) is incorrect
Choice (b) is correct!
x(t) = 2x(t) 3y(t). therefore,x(t) = 2x(t) 9x(t) 6y(t), = 2x(t) 9x(t) + 2x(t) 4x(t), = 4(t) 13x(t), therefore,x(t) = 4x(t) + 13x(t) = 0. The auxiliary equation is thus k2 4k 13 = 0 and its roots are k = 2 ± 31. The general solution is x(t) = e2t(Asin3t + Bcos3t). therefore,x(t) = 2e2t(Asin3t Bcos3t) + e2t(3Acos3t 3Bsin3t), y(t) = 2 3x(t) 1 3x(t) = 2e2t(Bsin3t Acos3t).
Choice (c) is incorrect
Choice (d) is incorrect