HA NOI UNIVERSITY OF MINING AND GEOLOGY
Faculty of Mathematics
FINAL TEST
Subject: Math22B
Total time: 60 minutes

Batch 4/01 – Advance Program – HUMG
Full name: Nguyen KhacDuy
SOLUTION OF TEST CODE 01
Problem 1.
1.
y
1
1
x⇔y=C
1
1
x
dx
Let 1
x=u
du
dx
2
x
dx=2
u−1
du

1
1
x
dx
2
u−1
u
du
2
1
1
u
du=2
u
ln
u

⇔𝐲=𝟐
𝟏
𝐱
𝐥𝐧
𝟏
𝐱+𝐂
2. x
1
x
2
y
x
2−1
y+2x=0
y
1
x
2
y
x
1
x
2−2
1
x
2

⇒v
x
e
1
x
2
dx
x
1
x
2
e
1
x
2x
1
x
2
dx
e
ln
x
ln(1
x
2
x
1
x
2

⇒y
1
v
x
C
v
x
Q
x
dx
1
x
2
x
C
x
1
x
2−2
1
x
2
dx

⇒I
x
1
x
2−2
1
x
2
dx Let 1
x
2=u
du=2x
dx⇒I
1
u
2
du
1
u
1
1
x
2

⇒y
1
x
2
x
1
1
x
2+C)⇔𝐲
𝟏
𝐱(𝟏+𝐂
𝟏
𝐱
𝟐)
Problem 2.
1.
y+y
1
cos
2x
cos
2x
y+y
cos
2x−3
2

Step 1: Find
y
Homo

Consider the homogeneous equation:
y+y=0 ∀ General sol: y
e
kx


k
2+1=0
k
1=i
k
2=−i
𝐲
𝐇𝐨𝐦𝐨
𝐂
𝟏
𝐜𝐨𝐬
𝐱
𝐂
𝟐
𝐬𝐢𝐧
𝐱

Step 2: Find
C
1
x,
C
2(x) by Constant variable method
General sol: y
C
1(x
cos
x
C
2(x
sin
x


C
1
y
1
C
2
y
2=0
C
1
y
1
C
2
y
2′=f(x
C
1
cos
x
C
2
sin
x=0
C
1
sin
x
C
2
cos
x
cos
2x−3
2
C
1
C
2
tan
x


C
2
sin
x
cos
x
sin
x
cos
x
cos
2x−3
2
C
1
sin
x
cos
2x−3
2
C
2
cos
x
cos
2x−3
2


C
1
sin
x
dx
cos
2x
cos
2x
sin(2x−x
dx
cos
2x
cos
2x
sin
2x
cos
x
cos
2x
sin
x
dx
cos
2x
cos
2x


C
1−2
sin
2x
cos
x
2
cos
2x
cos
2x
sin
x
dx
cos
2x
cos
x
cos
2x


C
2
cos
x
dx
cos
2x
cos
2x
cos(2x−x
dx
cos
2x
cos
2x
cos
2x
cos
x
sin
2x
sin
x
dx
cos
2x
cos
2x


C
2
cos
2x
cos
x−2
sin
2x
2
cos
2x
sin
x
dx
cos
2x
sin
x
cos
2x
C
1
cos
x
cos
2x
k
1
C
2
sin
x
cos
2x
k
2

Conclusion: 𝐲
𝐜𝐨𝐬
𝐱
𝐜𝐨𝐬
𝟐𝐱
𝐤
𝟏
𝐜𝐨𝐬
𝐱
𝐬𝐢𝐧
𝐱
𝐜𝐨𝐬
𝟐𝐱