HCMC UNIVERSITY OF FINAL EXAMINATION ANSWERS
TECHNOLOGY AND EDUCATION SEMESTER 2 - ACADEMIC YEAR 2021-2022 Faculty of High Quality Training Course name:Higher Mathematics for Engineers 1 Program name: Group of Mathematics Course ID: MATH133101E
Question 1: (1 point) The augmented matrix of the system is
m 1 1 1
1 m 1 m
1 1 m m2
−→
1 1 m m2
1 m 1 m
m 1 1 1
−→
1 1 m m2
0 m-1 1-m m(1-m) 0 1-m 1-m2 1-m3
−→
1 1 m m2
0 m-1 1-m m(1-m)
0 0 (1-m)(m+2) (1-m)(m+1)2
(i) (0.5 point): If (1−m)(m + 2) 6= 0, i.e., m 6= 1 and m 6= −2, then the system has a unique solution given by
x=−m+ 1
m+ 2, y= 1
m+ 2, z = (m+ 1)2 m+ 2 .
(ii) (0.5 point): Ifm= 1, then the system has infinite many solution depending on two free variables
x= 1−α−β y =α
z =β
(iii) (0.5 point):If m=−2, then the system has no solution since the last equation 0z = 3.
Question 2: The equation is rewritten as the form dx
dt +kx=r, where the coefficients P(t) =k and Q(t) =r. We have
P(t)dt=
kdt=kt+C which implies an integrating factor of the DEs
I(t) = exp
P(t)dt
=ekt. (0.5 point) The general solution is given by
x(t) =e−kt
rektdt+C
=e−ktr
kektdt+C
= r
k +Ce−kt. (0.5 point) Since x(0) = 0 we have x(0) = rk+C = 0⇔C =−kr (0.5 point). Thus we obtain
x(t) = r
k 1−e−kt
⇒ lim
t→∞x(t) = r
k. (0.5 point)
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Question 3: The motion of the mass can be described by the DE md2x
dt2 =−kx−βdx dt, where k= 2, m= Wg = 18 and β = 1. And hence, we get
d2x
dt2 + 8dx
dt + 16x= 0. (0.25 point) By assumption we know that x(0) =−1 and x0(0) = 8. (0.25 point)
The general solution is given by
x(t) = c1e−4t+c2te−4t, c1, c2 ∈R. (0.5 point) Using the initial conditions we have
(c1 =−1
−4c1+c2 = 8 ⇔
(c1 =−1 c2 = 4 Thus x(t) = (4t−1)e−4t. (0.5 point)
(a) (0.25 point) The mass passes through the equilibrium position when x(t) = 0⇔t= 1
4.
(b) (0.25 point) Analyze the behavior of x(t), we imply that the mass attains its maximal displacement 1 ft at the initial time t= 0.
Question 4: We know that
L {y00(t)}=s2Y(s)−sy(0)−y0(0) =s2Y(s)−s.
Applying the Laplace transform both sides of the DE, we get
s2Y(s)−s+ 4Y(s) = L {sin(t)U(t−2π)}=e−2πsL {sin(t−2π)}= e−2πs
s2+ 1. (0.5 point) which implies
Y(s) = e−2πs 1
(s1+ 1)(s2+ 4) + s
s2+ 4. (0.5 point) Using the partial fraction method, one has
Y(s) = e−2πs 3
1
s1+ 1 − 1 s2+ 4
+ s
s2+ 4. (0.5 point) Taking the inverse Laplace transform, we arrive at
y(t) = L−1Y(s) = 1
3(sin(t−2π)−sin 2(t−2π))U(t−2π) + cos 2t
= 1
3(sint−sin 2t)U(t−2π) + cos 2t. (0.5 point) Question 5: We have
dv
dt = 32− 0.125
5 v2 =:f(t, v), v(0) = 0.
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(a) By the Euler’s method withh = 1 we have
v(xn)≈vn=vn−1+hf(tn−1, vn−1) = vn−1+ 32− 0.125 5 vn−12 , where tn=t0+nh,n = 1,2, ...(0.5 point)
tn 0 1 2 3 4 5
vn 0 32 38.4 33.536 37.4194 34.4109
(0.25 point)
(b) By the improved Euler’s method with h= 1 we have
v∗n=vn−1+hf(tn−1, vn−1) = vn−1+ 32− 0.125 5 vn−12 v(xn)≈vn=vn−1+h
2(f(tn−1, vn−1) +f(tn, v∗n)), where xn=x0+nh, n= 1,2, ... (0.5 point)
tn 0 1 2 3 4 5
vn 0 19.2 24.5588 27.5119 29.4569 30.8457
(0.25 point)
26/05/2022
Approved by program chair (signed and named)
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