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 02
Problem 1.
1.
y
x
2
e
x⇔y=C
x
2
e
x
dx

Let
e
x
dx=dv
x
2=u
v
e
x
du=2x
dx
x
2
e
x
dx
x
2
e
x
e
x
2x
dx
x
2
e
x−2
e
x(x−1)
⇔𝐲
𝐞
𝐱
𝐱
𝟐−𝟐𝐱+𝟐+𝐂
2.
y+2xy=x
e−x
2⇒v
x
e
2x
dx
e
x
2

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

⇒𝐲
𝐞−𝐱
𝟐
𝐱
𝟐
𝟐+𝐂

Problem 2.
1.
y+y
tan
x

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
tan
x
C
1
C
2
tan
x


C
2
sin
x
cos
x
sin
x
cos
x
sin
x
cos
x
C
2
sin
x
C
1
sin
x
2
cos
x
cos
x
sec
x


C
1
sin
x
ln
sec
x
tan
x
k
1
C
2
cos
x
k
2

Conclusion: 𝐲
𝐬𝐢𝐧
𝐱
𝐥𝐧
𝐬𝐞𝐜
𝐱
𝐭𝐚𝐧
𝐱
𝐤
𝟏
𝐜𝐨𝐬
𝐱
𝐜𝐨𝐬
𝐱
𝐤
𝟐
𝐬𝐢𝐧
𝐱

2.
y+2
y+y=4
e−x

Step 1: Find
y
Homo

Consider the homogeneous equation:
y+2
y+y=0 ∀ General sol: y
e
kx


k
2+2k+1=0
k
1=
k
2=−1

𝐲
𝐇𝐨𝐦𝐨
𝐂
𝟏
𝐞−𝐱
𝐂
𝟐
𝐱
𝐞−𝐱

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


C
1
y
1
C
2
y
2=0
C
1
y
1
C
2
y
2′=f(x
C
1
e−x
C
2′x
e−x=0
C
1
e−x
C
2
x
e−x
C
2
e−x=4
e−x
C
2
e−x=4
e−x
C
1
C
2′x


C
1=−4x
C
2=4
C
1=−2
x
2
k
1
C
2=4x
k
2

Conclusion: 𝐲−𝟐
𝐱
𝟐
𝐤
𝟏
𝐞−𝐱+(𝟒𝐱
𝐤
𝟐)𝐱
𝐞−𝐱

Problem 3.

x
t+3
y
t+x
t=