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Người gửi: Hoàng Thị Hoa (trang riêng)
Ngày gửi: 19h:09' 07-07-2020
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Nguồn:
Người gửi: Hoàng Thị Hoa (trang riêng)
Ngày gửi: 19h:09' 07-07-2020
Dung lượng: 27.7 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 03
Problem 1.
1.
y
cos
x
y
ln
y
ln
y
y
dy
sec
x
dx
ln
y
y
dy
sec
x
dx
𝐥𝐧
𝐲
𝟐
𝟐
𝐥𝐧
𝐬𝐞𝐜
𝐱
𝐭𝐚𝐧
𝐱+𝐂
2.
1
x
2
y−2xy(1
x
2
2
y
2xy
1
x
2=1
x
2
⇒v
x
e
2x
dx
1
x
2
e
ln(1
x
2
1
1
x
2
⇒y
1
v
x
C
v
x
Q
x
dx
1
x
2
C
1
1
x
2
1
x
2
dx
⇒𝐲=(𝟏
𝐱
𝟐)(𝐱+𝐂)
Problem 2.
1.
y+2
y+y=3
e−x
x+1
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
xe−x=0
C
1
e−x
C
2
xe−x
C
2
e−x=3
e−x
x+1
C
1
C
2′x=0
C
1
C
2
x
C
2′=3
x+1
C
1=−3x
x+1
C
2′=3
x+1
C
1
k
1−3x
x+1
dx
C
2
k
2
3
x+1
dx
Let
x+1
dx=dv
x=u
v
2
x+1
3
2
3
du=dx
I
1=−3
2x
x+1
3
2
3
2
x+1
3
2
3
dx
I
1=−3
2
3
x
x+1
3
2
4
15
x+1
5
2
2
5
x+1
3
2
4x−5+2x+x
C
1
2
5
x+1
3
2(2−3x
k
1
C
2=2(x+1
3
2
k
2
Conclusion: 𝐲−𝟐
𝟓
𝐱+𝟏
𝟑
𝟐(𝟑𝐱−𝟐
𝐤
𝟏
𝐞−𝐱+(𝟐(𝐱+𝟏
𝟑
𝟐
𝐤
𝟐)𝐱
𝐞−𝐱
2.
y−9
y+20y
x
2
e
4x
Step 1: Find
y
Homo
Consider the homogeneous equation:
y−9
y+20y=0 ∀ General sol: y
e
kx
k
2−9k+20=0
k
1=5,
k
2=4
𝐲
𝐇𝐨𝐦𝐨
𝐂
𝟏
𝐞
𝟓𝐱
𝐂
𝟐
𝐞
𝟒𝐱
Step 2: Find
y
Separate
Separate sol:
y
Separare=Q
x
x
e
4x
A
x
2+Bx+C
x
e
4x
A
x
3+B
x
2+Cx
e
4x
y
e
4x
4
A
x
3+B
x
2+Cx+3A
x
2+2Bx+C
y
e
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 03
Problem 1.
1.
y
cos
x
y
ln
y
ln
y
y
dy
sec
x
dx
ln
y
y
dy
sec
x
dx
𝐥𝐧
𝐲
𝟐
𝟐
𝐥𝐧
𝐬𝐞𝐜
𝐱
𝐭𝐚𝐧
𝐱+𝐂
2.
1
x
2
y−2xy(1
x
2
2
y
2xy
1
x
2=1
x
2
⇒v
x
e
2x
dx
1
x
2
e
ln(1
x
2
1
1
x
2
⇒y
1
v
x
C
v
x
Q
x
dx
1
x
2
C
1
1
x
2
1
x
2
dx
⇒𝐲=(𝟏
𝐱
𝟐)(𝐱+𝐂)
Problem 2.
1.
y+2
y+y=3
e−x
x+1
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
xe−x=0
C
1
e−x
C
2
xe−x
C
2
e−x=3
e−x
x+1
C
1
C
2′x=0
C
1
C
2
x
C
2′=3
x+1
C
1=−3x
x+1
C
2′=3
x+1
C
1
k
1−3x
x+1
dx
C
2
k
2
3
x+1
dx
Let
x+1
dx=dv
x=u
v
2
x+1
3
2
3
du=dx
I
1=−3
2x
x+1
3
2
3
2
x+1
3
2
3
dx
I
1=−3
2
3
x
x+1
3
2
4
15
x+1
5
2
2
5
x+1
3
2
4x−5+2x+x
C
1
2
5
x+1
3
2(2−3x
k
1
C
2=2(x+1
3
2
k
2
Conclusion: 𝐲−𝟐
𝟓
𝐱+𝟏
𝟑
𝟐(𝟑𝐱−𝟐
𝐤
𝟏
𝐞−𝐱+(𝟐(𝐱+𝟏
𝟑
𝟐
𝐤
𝟐)𝐱
𝐞−𝐱
2.
y−9
y+20y
x
2
e
4x
Step 1: Find
y
Homo
Consider the homogeneous equation:
y−9
y+20y=0 ∀ General sol: y
e
kx
k
2−9k+20=0
k
1=5,
k
2=4
𝐲
𝐇𝐨𝐦𝐨
𝐂
𝟏
𝐞
𝟓𝐱
𝐂
𝟐
𝐞
𝟒𝐱
Step 2: Find
y
Separate
Separate sol:
y
Separare=Q
x
x
e
4x
A
x
2+Bx+C
x
e
4x
A
x
3+B
x
2+Cx
e
4x
y
e
4x
4
A
x
3+B
x
2+Cx+3A
x
2+2Bx+C
y
e
 




















