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Người gửi: Hoàng Thị Hoa (trang riêng)
Ngày gửi: 19h:09' 07-07-2020
Dung lượng: 28.1 KB
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Nguồn:
Người gửi: Hoàng Thị Hoa (trang riêng)
Ngày gửi: 19h:09' 07-07-2020
Dung lượng: 28.1 KB
Số lượt tải: 0
Số lượt thích:
0 người
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 04
Problem 1.
1.
y
cos(x−y
Let x−y=z
dx
dy
dz
dy
dx
dz
dy
dx
cos
z
dx−dz
dx
cos
z⇔1
dz
dx
cos
z⇔1
cos
z
dz
dx
dx
dz
1
1
cos
z
x
1
2
sin
z
2
2⇔x=C
1
2
sin
z
2
2
dz⇔x=C
cot
z
2
⇒x=C
cot
x−y
2⇔𝐲=𝐱−𝟐
𝐭𝐚𝐧−𝟏
𝟏
𝐂−𝐱
2. 2ydx
y
2−6x
dy=0⇔2ydx
6x
y
2
dy
x
3x
y
y
2
x
3
y
x
y
2
⇒v
y
e−3
dy
y
e−3
ln
y
1
y
3⇒x
1
v
y
C
v
y
Q
y
dy
⇒x
y
3
C
1
y
3
y
2
dy
y
3
C−1
2
y
2
dy⇒𝐱
𝐲
𝟑
𝐂
𝟏
𝟐𝐲
Problem 2.
1.
y−y
e
x
e
x+1
Step 1: Find
y
Homo
Consider the homogeneous equation:
y−y=0 ∀ General sol: y
e
kx
k
2−1=0
k
1=−1
k
2=1
𝐲
𝐇𝐨𝐦𝐨
𝐂
𝟏
𝐞−𝐱
𝐂
𝟐
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
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
e
x=0
C
1
e−x
C
2
e
x
e
x
e
x+1−2
C
1
e−x
e
x
e
x+1
2
C
2
e
x
e
x
e
x+1
C
1
e
2x
2
e
x+1
C
2
1
2
e
x+1
Hint:Let
e
x=u
dx
du
u
C
1
1
2
e
x
ln
e
x+1
k
1
C
2
1
2
ln
e
x
e
x+1
k
2
Conclusion: 𝐲
𝟏
𝟐
𝐞
𝐱
𝐥𝐧
𝐞
𝐱+𝟏
𝐤
𝟏
𝐞−𝐱
𝟏
𝟐
𝐥𝐧
𝐞
𝐱
𝐞
𝐱+𝟏
𝐤
𝟐
𝐞
𝐱
2.
y−2
y+3y
e−x
cos
x
Step 1: Find
y
Homo
Consider the homogeneous equation:
y−2
y+3y=0 ∀ General sol: y
e
kx
k
2−2k+3=0
k
1=1+i
2,
k
2=1−i
2
𝐲
𝐇𝐨𝐦𝐨
𝐞
𝐱
𝐂
𝟏
𝐜𝐨𝐬
𝟐
𝐱
𝐂
𝟐
𝐬𝐢𝐧
𝟐
𝐱
Step 2: Find
y
Separate
We have α±iβ=−1±i aren’t sol of equation
k
2−2k+3=0
So, separate sol:
y
Separare
e
αx
H
x
cos
βx+L
x
sin
βx
⇒y
e−x
A
cos
x+B
sin
x
y
e−x−A+B
cos
x
A+B
sin
x
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 04
Problem 1.
1.
y
cos(x−y
Let x−y=z
dx
dy
dz
dy
dx
dz
dy
dx
cos
z
dx−dz
dx
cos
z⇔1
dz
dx
cos
z⇔1
cos
z
dz
dx
dx
dz
1
1
cos
z
x
1
2
sin
z
2
2⇔x=C
1
2
sin
z
2
2
dz⇔x=C
cot
z
2
⇒x=C
cot
x−y
2⇔𝐲=𝐱−𝟐
𝐭𝐚𝐧−𝟏
𝟏
𝐂−𝐱
2. 2ydx
y
2−6x
dy=0⇔2ydx
6x
y
2
dy
x
3x
y
y
2
x
3
y
x
y
2
⇒v
y
e−3
dy
y
e−3
ln
y
1
y
3⇒x
1
v
y
C
v
y
Q
y
dy
⇒x
y
3
C
1
y
3
y
2
dy
y
3
C−1
2
y
2
dy⇒𝐱
𝐲
𝟑
𝐂
𝟏
𝟐𝐲
Problem 2.
1.
y−y
e
x
e
x+1
Step 1: Find
y
Homo
Consider the homogeneous equation:
y−y=0 ∀ General sol: y
e
kx
k
2−1=0
k
1=−1
k
2=1
𝐲
𝐇𝐨𝐦𝐨
𝐂
𝟏
𝐞−𝐱
𝐂
𝟐
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
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
e
x=0
C
1
e−x
C
2
e
x
e
x
e
x+1−2
C
1
e−x
e
x
e
x+1
2
C
2
e
x
e
x
e
x+1
C
1
e
2x
2
e
x+1
C
2
1
2
e
x+1
Hint:Let
e
x=u
dx
du
u
C
1
1
2
e
x
ln
e
x+1
k
1
C
2
1
2
ln
e
x
e
x+1
k
2
Conclusion: 𝐲
𝟏
𝟐
𝐞
𝐱
𝐥𝐧
𝐞
𝐱+𝟏
𝐤
𝟏
𝐞−𝐱
𝟏
𝟐
𝐥𝐧
𝐞
𝐱
𝐞
𝐱+𝟏
𝐤
𝟐
𝐞
𝐱
2.
y−2
y+3y
e−x
cos
x
Step 1: Find
y
Homo
Consider the homogeneous equation:
y−2
y+3y=0 ∀ General sol: y
e
kx
k
2−2k+3=0
k
1=1+i
2,
k
2=1−i
2
𝐲
𝐇𝐨𝐦𝐨
𝐞
𝐱
𝐂
𝟏
𝐜𝐨𝐬
𝟐
𝐱
𝐂
𝟐
𝐬𝐢𝐧
𝟐
𝐱
Step 2: Find
y
Separate
We have α±iβ=−1±i aren’t sol of equation
k
2−2k+3=0
So, separate sol:
y
Separare
e
αx
H
x
cos
βx+L
x
sin
βx
⇒y
e−x
A
cos
x+B
sin
x
y
e−x−A+B
cos
x
A+B
sin
x
 




















