Tài liệu phuong_trinh_vi_phan docx

F (x, y, y

,y

, ,y
(m)
)=0
y = y(x)
y = y(x)
I ⊂ R F
G R × R
m+1
y(x)=(y
1
(x), ,y
n
(x))
T
F
R
n
m m
F (x, y, y

)=0
F (x, y, z)
G ⊂ R
3
y

= f(x, y)
f D ⊂ R
2
e
y
+ y

2
cos x =1
y

2
− 2xy =lnx
∂
2
u
∂x
2
+
∂
2
u
∂y
2
=0
y : I → R
n
I =(a, b)
R m
I
F (x, y(x),y

(x),y

(x), ,y
(m)
)(x)=0 x ∈ I
y = y(x)
y = C
1
cos x + C
2
sin x
y

+ y =0
x =
x(t) y = y(t)
−
y

= y(α − βx),x

= x(γy − δ)
α, β, γ δ
y x
dy
dx
=
y(α −βx)
x(γy − δ)
(γy − δ)
y
dy =
(α − βx)
x
dx
γy − δ ln y = α ln x − βx+ C
C α = β =
γ =1,δ =2
1234
1
2
3
yy
z
z
X
−
y =
x
3
3
+ C
y

= x
2
y =
x
3
3
+1 y(0) = 1
y(x)
y(x
0
)=y
0
(x
0
,y
0
) ∈ D
y

= f (x, y),y(x
0
)=y
0
f D ⊂ R
2
y(x)
y(x)
y(x)=y
0
+

x
x
0
f(t, y(t))dt
C
1
I
− ¨o
I
− ¨o
y
0
(x)=y
0
y
k+1
(x)=y
0
+

x
x
0
f(t, y
k
(t))dt, k ∈ N
f
D = {(x, y)/|x − x
0
|≤a, |y −y
0
|≤b}
M := max
(x,y)∈D
|f(x, y)| h := min

a,
b
M

x ∈ I := [x
0
−
h, x
0
+ h]
|y
k
(x) −y
0
|≤b, k
y
k
D
x
0
− h ≤ x ≤ x
0
+ h
|y
k
− y
0
| =





x
x
0
f(t, y
k−1
(t))dt




≤

x
x
0
|f(t, y
k−1
(t))|dt ≤ M |x −x
0
|≤Mh ≤ b

y

= −y
2
y(0) = 1
y =
1
x +1
¨o y
0
=1 y
1
=1− x
y
2
=1−x + x
2
−
x
3
3
y
k
x
x
1234
Y (x)
2
Y (x)
0
Y (x)
4
Y (x)
1
Y (x)
3
− y

= −y
2
y(0) = 1
f(x, y) D ⊂ R
2
f
D y L
|f(x, y
1
) − f(x, y
2
)|≤L |y
1
− y
2
|, (x, y
1
), (x, y
2
) ∈ D
∂f
∂y
D
∂f
∂y




∂f
∂y




≤ M
f(x, y) y
f(x, y
1
) − f(x, y
2
)=(y
1
− y
2
)
∂f
∂y
[x, y
1
+ θ(y
2
− y
1
)]
f(x, y)
y
D = {(x, y)/ |x − x
0
|≤a, |y − y
0
|≤b}
I :=
[x
0
− h, x
0
+ h] h := min(a,
b
M
) M := max
(x,y)∈D
|f(x, y)|
I
|y
k+1
(x) −y
k
(x)|≤ML
k
|x −x
0
|
k+1
(k +1)!
, x ∈ I
k =0




x
x
0
f(t, y
k−1
(t))dt



≤ M |x − x
0
|
k − 1 x
0
≤ x ≤ x
0
+ h
|y
k+1
(x) − y
k
(x)| =





x
x
0
[f(t, y
k
(t)) − f(t, y
k−1
(t))] dt




≤

x
x
0
|f(t, y
k
(t)) − f(t, y
k−1
(t))|dt ≤ L

x
x
0
|y
k
(t) −y
k−1
(t)|dt
≤ L

x
x
0
|y
k
(t) − y
k−1
(t)|dt
≤ ML
k

x
x
0
|x −x
0
|
k
k!
dt = ML
k
|x −x
0
|
k+1
(k +1)!
x
0
− h ≤ x ≤ x
0
{y
k
(x)} I
|y
k+p
(x) − y
k
(x)|≤|y
k+p
(x) −y
k+p−1
(x)| + |y
k+p−1
(x) − y
k+p−2
(x)| + ···+ |y
k+1
(x) −y
k
(x)|
≤
M
L

(L |x − x
0
|)
k+p
(k + p)!
+ ···+
(L |x − x
0
|)
k+1
(k +1)!

≤
M
L

j≥k+1
(Lh)
j
j!

∞
j=0
(Lh)
j
j!
k {y
k
(x)}
I y(x) y(x)
y
k+1
(x)=y
0
+

x
x
0
f(t, y
k
(t))dt
{y
k
(x)} f D
{f(t, y
k
(t))} I f(t, y(t))
y(x)
z(x)
y(x) −z(x)=

x
x
0
[f(t, y(t)) − f(t, z(t))] dt
|y(x) −z(x)| =





x
x
0
[f(t, y(t)) − f(t, z(t))] dt




≤ 2M |x −x
0
|
k
|y(x) − z(x)|≤2ML
k
|x − x
0
|
k+1
(k +1)!
, x ∈ I
k −→ +∞ |y(x) − z(x)| =0 I
y(x) 
f(x, y) R
2
y

=2

|y|,y(0) = 0
y ≡ 0
y(x)=

(x −C)
2
x ≥ C
0 x ≤ C
y(x)=

0 x ≥ C
−(x − C)
2
x ≤ C
E d T : E → E
α ∈ (0, 1) x, y ∈ E
d(Tx,Ty) ≤ αd(x, y)
T
x
∗
∈ E
T (x
∗
)=x
∗
123-3 -2 -1
1
-1
y

=2

|y|,y(0) = 0
y = y(x) y = y(x)
(x
0
,y
0
)
(x
0
,y
0
) ∈ D
D ⊂ R
2
D y = y(x, C) C
(x
0
,y
0
) ∈ D C
C = ϕ(x
0
,y
0
)(∗)
ϕ
y = y(x, C) C (∗)
(x
0
,y
0
) D
ϕ(x, y)=C ϕ(x, y)
y

+ y =0 y(x)=Ce
−x
C
•
•
(x
0
,y
0
) ∈ D C
0
= ϕ(x
0
,y
0
) y = y(x, C
0
)
y(x) y

=3y + x y(0) = 1
y = −
x
3
−
1
9
+Ce
3x
C
1=y(0) = −
1
9
+ Ce
0
C =
10
9
y = −
x
3
−
1
9
+
10
9
e
3x
f(x, y) R
2
M(x, y)
k =
dy
dx
= f (x, y)
y = ϕ(x, C)(∗)
C
x
dy
dx
=
∂ϕ
∂x
(x, C)
(∗) (x, y) C = C(x, y)
C
y

=
∂ϕ
∂x
(x, C(x, y)) =: f(x, y)
–3
–2
–1
1
2
y(x)
–3 –2 –1 1 2 3
x
y

= −
y
x
y = Cx
2
x y

=2Cx C
y

=2
y
x
M(x)dx + N(y)dy =0

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