Engineering Mathematics 1: Solving Quadratic Roots and Equations

This document provides solutions to several engineering mathematics problems. It begins by determining the equation of a parabolic graph using its roots and verifying it by substitution. It then explores different techniques, such as the Newton-Raphson method, incremental method, and bisection method, to find the roots of a given equation. Finally, it addresses a second-order non-homogenous differential equation, detailing the process of finding both the complementary and particular solutions using the method of undetermined coefficients, culminating in a general solution for the equation. Desklib offers more solved assignments and past papers for students.

Engineering Mathematics 1
Engineering Mathematics
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Engineering Mathematics 2
Task 10
From the graph, the roots of the quadratic are -2 and 4 because the graph is cutting the x-axis
through these points.
So f(x) = (x + 2) (x – 4)
This can be expanded as:
f(x) = x2 – 2x – 8
To determine whether this is the correct equation of the parabolic graph, we substitute x = 0 to
see if the value of y = -8
f(0) = 0 – 0 – 8 = -8
Therefore the correct equation of the graph is
f(x) = x2 – 2x – 8
Alternatively, the equation of this graph can be determined using all the three points on the
general quadratic equation as follows:
y = ax² + bx + c
Substituting (-2, 0)
0 = a(-2)2 + b(-2) + c = 4a – 2b + c
Substituting (4, 0)
0 = a(4)2 + b(4) + c = 16a + 4b + c
Substituting (0, -8)

Engineering Mathematics 3
-8 = a(0)2 + b(0) + c = c
Therefore c = -8
Substituting c is the equation 4a – 2b + c = 0 gives 4a – 2b – 8 = 0
Substituting c in the equation 16a + 4b + c = 0 gives 16a + 4b – 8 = 0
These two can be solved simultaneously as follows:
(4a – 2b = 8) x 2
16a + 4b = 8
8a – 4b = 16
16a + 4b = 8
Adding the above equations gives:
24a = 24; a = 1
Substituting a = 1 into one of the equations to find b gives
16(1) + 4b = 8; 4b = -8; b = -2
So the quadratic equation of the graph becomes f(x) = x2 – 2x – 8
Domain are the x values while range are y values of the graph.
The domain of the graph are all real numbers −∞< x <∞
The range of the graph are the y values: range is y ≥−9
Task 11

Engineering Mathematics 4
Technique 1: Newton-Raphson method
This technique uses the formula xi+1 = f ( xi)
f ' ( xi), where root of the equation is any value of x that
gives zero as the solution to the equation
f(x) = x2 – 9
f'(x) = 2x
x0: f(0) = −9
0 = ∞
x1: f(1) = −8
2 = -4
x2: f(2) = −5
4 = -1.25
x3: f(3) = 0
6 = 0
x4: f(4) = 7
8 = 0.875
Since there is an iteration that has given the solution as zero, it is a root of the equation.
Hence the root is x = 3
Technique 2: Incremental method
Given the equation f(x) = x2 – 9 = 0, any value of x that satisfies this equation is a root of the
equation.
f(x) = x2 – 9 = 0

Engineering Mathematics 5
x = 0, f(x) = 0 – 9 = 0 = -9
x = 1, f(x) = 12 – 9 = 1 – 9 = -8
x = 2, f(x) = 22 – 9 = 4 – 9 = -5
x = 3, f(x) = 32 – 9 = 9 – 9 = 0
Since f(3) = 0, it means that 3 is a root of the equation x2 – 9 = 0
Technique 3: Bisection method
Let the root be between an interval [a, b]. For this to stand, f(a) and f(b) must be of opposite
signs. Let’s assume that a = 1.5 and b = 4.5. For the root c to be between these two numbers, f(a)
* f(b) < 0. This is proved as follows:
f(1.5) = 1.52 – 9 = -6.75
f(4.5) = 4.52 – 9 = 11.25
Therefore (-6.75) x (11.25) = -75.9375 < 0, meaning that the root lies between 1.5 and 4.5
First iteration, k = 0:
c 0= a 0+ b 0
2 = 1.5+4 .5
2 =3
f(3) = 32 – 9 = 9 – 9 = 0
This means that our first iteration has given us the root so we stop there.
Task 18
m d ² x
dt ² + c dx
dt +kx =F(t ) ………………………………………………………. (18.1)

Engineering Mathematics 6
This is a second order non-homogenous differential equation with constant coefficients (i.e. m, c
and x). The equation has a corresponding homogenous equation given as follows:
m d ² x
dt ² +c dx
dt +kx =0 …………………………………………………...………. (18.2)
The solution of this equation comprises of two parts: complementary solution and particular
solution. It can be expressed as follows:
x = xc + Xp …………...…………………………………………………………………….. (18.3)
Where xc = complementary solution and Xp = particular solution).
Therefore the main task here is to solve these two individual solutions.
Complementary solution is basically the homogenous solution of the non-homogenous equation
while particular solution is the non-homogenous solution of the non-homogenous equation
(Dawkins, 2018).
xc can be expressed as:
xc = B1x1 + B2x2 ………………………………………………………………………… (18.4)
Where B1 andBC2 are arbitrary constants and x1 and x2 are the homogenous solution of the non-
homogenous equation.
Since the non-homogenous equation in this case has constant coefficients, its complementary
equation (equation 18.4) is determined from the roots of its corresponding characteristic
polynomial. In this case, F(t) is the impressed oscillating force and it is equivalent to F0 sin ωt.
The corresponding homogenous equation is given as: mx'' + cx' + kx = 0

Engineering Mathematics 7
Dividing every term by m gives: x'' + c
m x' + k
m x=0, which gives:
n ²+ c
m B+ k
m=0; n=
−c
m ± √ ( c
m )
2
− 4 k
m
2
n 1=
−c
m + √ c ²
m² − 4 k
m
2
; n 2=
−c
m − √ c ²
m ² − 4 k
m
2
Xc = B1 cos + B2 sin
Xc=B 1cos
−c
m + √ c ²
m ² − 4 k
m
2 + B 2 sin
−c
m − √ c ²
m² − 4 k
m
2
The particular solution (Xp) can be determined using one of these two methods: method of
undetermined coefficients or the method of variation of parameters (Dawkins, 2017). Xp in this
task is determined using the method of undetermined coefficients. Since F(t) is equivalent to F0
sin ωt, it means that the term F(t) corresponds to the function K sin bx and the choice of Xp is
given as follows: Xp = A cos bt + B sin bt
Therefore Xp becomes:
Xp = At cos ωt + Bt sin ωt …………………………………………………………………. (18.5)
Xp' = A cos ωt – Aωt sin ωt + B sin ωt + Bωt cos ωt
Xp'' = - Aω sin ωt – Aω sin ωt - Aω²t cos ωt + Bω cos ωt + Bω cos ωt - Bω²t sin ωt
= -2Aω sin ωt - Aω²t cos ωt + 2Bω cos ωt - Bω²t sin ωt

Engineering Mathematics 8
Substituting the expressions of Xp, Xp' and Xp'' in the original equation x'' + c
m x' + k
m x=F 0 sin ωt
gives:
= -2Aω sin ωt - Aω²t cos ωt + 2Bω cos ωt - Bω²t sin ωt + (c/m)A cos ωt – (c/m)Aωt sin ωt +
(c/m)B sin ωt + (c/m)Bωt cos ωt + (k/m)At cos ωt + (k/m)Bt sin ωt = F 0
m sin ωt (multiplying
every term by m gives)
-2mAω sin ωt - Amω²t cos ωt + 2mBω cos ωt - Bmω²t sin ωt + cA cos ωt – cAωt sin ωt + cB sin
ωt + cBωt cos ωt + kAt cos ωt + kBt sin ωt = F 0 sin ωt
(-2mAω - Bmω²t – cAωt + cB + kBt) sin ωt + (- Amω²t + 2mBω + cA + cBωt + kAt) cos ωt =
F 0 sin ωt
-2mAω - Bmω²t – cAωt + cB + kBt = F0 …………………………………… (i)
- Amω²t + 2mBω + cA + cBωt + kAt = 0 ………………………………….. (ii)
The above two equations i and ii are simultaneous equations. Multiplying i by A and ii by B
gives
-2mA²ω - ABmω²t – cA²ωt + cAB + kABt = AF0 …………….. (iii)
+ 2mB²ω - ABmω²t + cB²ωt + cAB + kABt = 0 ……………….. (iv)
Subtracting iv from iii gives:
-2mA²ω – cA²ωt – 2mB²ω - cB²ωt = AF0; dividing every term by – gives:
2mA²ω + cA²ωt + 2mB²ω + cB²ωt = AF0
A²(2mω + cωt) + B²(2mω + cωt) = AF0

Engineering Mathematics 9
A²(2mω + cωt) = AFo; A= F 0
2 mω+cωt
B²(2mω + cωt) = 0; B = 0
Xp = F 0
2mω+ cωt t cos ωt
Hence the general equation X = Xc + Xp becomes:
X =B 1 cos
−c
m + √ c ²
m² − 4 k
m
2 + B 2sin
−c
m − √ c ²
m ² − 4 k
m
2 + F 0
2 mω+cωt t cos ωt
References
Dawkins, P., 2017. Mechanical vibrations. [Online]
Available at: http://tutorial.math.lamar.edu/Classes/DE/Vibrations.aspx
[Accessed 26 May 2018].
Dawkins, P., 2018. Nonhomogeneous Differential Equations. [Online]
Available at: http://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousDE.aspx
[Accessed 26 May 2018].

Engineering Mathematics 10