The University of Sydney
School of Mathematics and Statistics
Tutorial - Chapter 6
MATH1023: Multivariable Calculus and Modelling Semester 1, 2023
More General Differential Equations
1. (a) Find the general solution to y′′ − 8y′ + 16y = 0.
(b) Find a particular solution to y′′−8y′+16y = e4x. (Hint: try yp(x) = u(x)e4x
and then find u(x).)
(c) Find the general solution to y′′ − 8y′ + 16y = e4x.
2. Consider the following system of differential equations
x˙ = 2x,
y˙ = x− 3y.
(a) Find x(t) and y(t) by solving the equations in the system successively.
(b) Eliminate x to obtain a 2nd-order differential equation for y, and then find
y(t) and x(t).
3. Two species struggling to compete against each other in the same environment
have populations at time t of x(t) and y(t), satisfying the equations
x′(t) = 3x(t)− 4y(t) , y′(t) = −2x(t) + y(t) .
Find the second-order differential equation satisfied by x(t).
Hence find x(t) and y(t).
4. Two species are in a predator-prey relationship. One species, which numbers Y
individuals, eats the other, which numbers X individuals. Historically the numbers
of these species have been constant with X = 3000 and Y = 1500. After a severe
environmental disturbance the populations cease to be constant and start to change
with time.
Let x(t) and y(t) be the difference between the historically constant population
numbers and the new, changing population numbers X(t) and Y (t).
Then x(t) = X(t)− 3000 and y(t) = Y (t)− 1500 are the sizes of the perturbations
from the historically steady states.
The sizes of the perturbations are described by the following:
x′(t) = 3x(t)− 2y(t)
y′(t) = 4x(t)− y(t).
(a) Show that x′′(t)− 2x′(t) + 5x(t) = 0.
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(b) Find x(t) if x(0) = 100 and x′(0) = 100. (Take t = 0 to be the time at which
monitoring of the population sizes begins.)
(c) Hence find y(t).
(d) Sketch x(t) and y(t) and then X(t) and Y (t) as a function of t. What will
happen to the original populations?
REVIEW OF SINGLE VARIABLE CALCULUS - PART II
5. Compute the following limits.
a) lim
x→0
(esin(2x) − 2 +
√
x2 + 4) b) lim
x→3
x2 + 2x− 3
x2 − 4
c) lim
x→0
sin2 x
2x2
d) lim
x→0
x cos
(
1
x2
)
6. Is
f(x) =

x2 − x− 2
x− 2 if x ̸= 2
4 if x = 2
continuous at x = 2?
7. Find the global minimum and maximum values of f(x) = x3 +2x2− 4x+4 on the
closed interval [−3, 3].
Brief answers to selected exercises:
1. (a) y(x) = (Ax+B)e4x
(b) y(x) = 1
2
x2e4x
(c) y(x) =
(
x2
2
+ Ax+B
)
e4x
2. (a) x(t) = Ae2t, y(t) = A
5
e2t + Ce−3t.
(b) x(t) = 5Be2t, y(t) = Ae−3t +Be2t.
3. x = Ae5t +Be−t and y = −1
2
Ae5t +Be−t
4. (a) x(t) = 100et cos 2t
2
(b) y(t) = 100et(cos 2t+ sin 2t)
5. (a) 1 (b) 12
5
(c) 1
2
(d) 0
6. No.
7. On [−3, 3], f(x) has global minimum value 68
27
and global maximum value 37.