12.2 可分离变量的微分方程与一阶线性微分方程

,
dy =
dx
2xy
.
dy = 2xdx
y
dy
∫ = ∫ 2 xdx
y
,
ln y = x 2 + C1 ,
2
1 x
y = e C e
,
2
1 x
y = ±e C e
C
,
∴
y = Ce x2 .

.
dy x − y x +
+ cos = cos
dx 2 2
dy
dx
dy
dx
y
.
x − y x + y
+ cos − cos =
2 2
x y
+ 2 sin sin =
2 2
y − lncsc + cot
2
y
2
0,
0,
dy
y
2sin
2
∫ −∫
x
= sin dx,
2
x = 2cos
+ C .
2

.
k
>
0
20
1
tW
dW
= −kW
, W (0) = 100
dt
dW = −
dW
kdt,
,
W ∫ = ∫ − kdt
W
ln W = −kt
+
lnc
( QW > 0)
100

W
( t)
=
ce
−kt
,
W ( 0) = 100, c =
W ( 1) =
20
100
∴ W ( t)
= 100e
k
∴ 20 = 100e − ,
−kt
k
= ln5,
W ( t)
=
W
= 1
100e
ln100
t = ≈ 2.86
( )
ln5
−(ln5)
t
2
.86
1
.

.
.
1,
(1)
0,
(0)
1
]
[0,
0)
)
(
)(
(
=
=
+
≥
=
f
f
n
y
x
t
f
t
f
y
x
(t)
f
y =
(x)
f
1
0
)]
(
[
)
(
+
=
∫
n
x
x
f
k
dt
t
f
)
(
)]
(
1)[
(
)
( x
f
x
f
n
k
x
f
n ′
+
=
t
y
o
1,
)
(
)]
(
1)[
(
1
=
′
+
−
x
f
x
f
n
k
n
?
)
( =
x
f
1.
(1)
0,
0)
( =
= f
f

−1 dy
y = f (x),
n
k( n + 1) y = 1,
dx
n−1
∫ k(
n + 1) y dy =
n
∫
dx
y
k( n + 1) ⋅ = x + c,
y( 0) = 0, c = 0
n
n
y(1)
= 1, k =
n + 1
∴ y n = x f n ( x)
= x
Q
f
( x)
≥
0
∴ f ( x)
=
n x

1 e
. y′
+ y = .
x x
1
x
e
P ( x)
= , Q(
x)
= ,
x
x
1
⎛ x 1
−∫ dx
⎞
⎜ e ∫ dx
y = e x ⋅ + ⎟
⎜∫
e x dx C
x
⎟
⎝
⎠
⎛
x
−ln x e
⎞
= ⎜∫
ln x
e ⋅ e dx + C
⎟
⎛
x
1 e
=
⎝ x
⎠
⎜∫ ⋅ x dx + C
x ⎝ x
= 1 ( ∫ e dx + C )
x 1
= ( e
x + C
x
x
).
x
⎞
⎟
⎠

. f (x)
∫
x
f ( t)
dt
t + f ( t)
1 2
=
f
( x)
− 1,
f ( x)
f ( x).
x = 1 , f (1) =
d f ( x)
= f ′(
x)
2
dx x + f ( x)
y
y = f (x),
y
= y′
, y(1)
= 1
2
x + y
dx
= 1 ⋅ x + y x
dy y
1

dx
dy
− 1 ⋅
y
−
x
∫
=
y
1
1
( − ) dy ( ) dy
y
∫ −
y
+
x = e [ ∫ ye dy C]
ln y −ln
y
[ ye dy +
= e ∫
=
y ( y + C)
y( 1) = 1,
C = 0
∴
C]
1
= y [ ∫ y ⋅ dy + C]
y
x =
2
y
,
y =
x
f ( x)
= x.

)
[
∞
+
0
(t )
f
.
y
x
y
x
f
e
t
f
t
y
x
π t
d
d
2
1
)
(
2
2
2
2
4
2
2
4
∫∫
≤
+
⎟
⎠
⎞
⎜
⎝
⎛
+
+
=
(t ).
f
.

y
x
y
x
f
t
y
x
d
d
2
1
2
2
2
4
2
2
∫∫
≤
+
⎟
⎠
⎞
⎜
⎝
⎛
+
∫
∫
=
t
π
r
r
r
f
θ 2 0
2
0
d
)
2
1
(
d
∫
=
t
r
r
rf
2
0
)d
2
(
2π
∫
+
=
t
t
r
r
rf
e
t
f
2
0
4
)d
2
(
2
)
(
2
π
π
).
(
8
8
)
(
2
4
t
f
t
te
t
f
t
π
π
π
+
=
′

2
4
8
)
(
8
)
(
t
π
te
π
t
tf
π
t
f =
−
′
⎥⎦
⎤
⎢⎣
⎡
+
= ∫
∫
−
C
e
te
π
e
t
f
t
t
π
t
π
t
t
π
d
8
4
d
8
2
8
)
(
)
(4
2
4
2
C
t
π
e
t
π
+
=
1
1,
0)
( =
= C
f
.
1)
(4
)
(
2
4
2 t
e
t
t
f
π
π +
=

.
(
t = 0)
.
v(t),
: F = mg − kv
:
F = ma
dv
∴ m = mg − kv,
dt

g
v
m
k
t
v
=
+
d
d
1.
]
d
[
d
d
C
t
ge
e
v
t
m
k
t
m
k
+
=
∫
∫
∫
−
]
d
[ C
t
ge
e
t
m
k
t
m
k
+
= ∫
−
[ C]
e
k
mg
e
t
m
k
t
m
k
+
= − .
t
m
k
Ce
k
mg
−
+
=
k
mg
c
v| t −
=
=
= :
0
0
).
e
(1
t
m
k
k
mg
v
−
−
=
∴

2.
dv
m
dt
=
mg
−
kv
dv dt
∫ = ∫
mg − kv m
1
t
− ln( mg − kv)
= + c1 k
m
v| t =
= 0 c
= −
∴
v
0
=
mg
k
+ ce
−
v
k
m
t
=
( c
mg
k
mg
k
e
= −
(1 −
e
-kc1
k
−
k
m
t
)
).

.
y
3
y = f (x) y = x ( x ≥ 0)
PQ,
f (x).
∫
x
0
x
∫
0
f ( x)d
x
yd
x
=
x
=
3
x
−
3
−
y,
y ′ + y =
f ( x),
3x
2 ,
o
y
Q
P
x
y =
y =
f
x
3
x
(x)

y ′ + y =
2
3x
y
[ ] ∫
2
C x e∫
d x
3 d
∫ d x
+
= e
− x
=
−
Ce x ,
+
3x
2 − 6x
+
6,
y | 0 = 0 C = −6,
x=
y
x
− 2
= 3(
−2e
+ x − 2x
+
2).

. ϕ ( π ) = 1, ϕ ( x)
y
∫ [sin x −ϕ ( x)]
dx + ϕ ( x)
dy .
x
L
P
Q
∂P
∂Q
=
∂y
∂x
1
[sin x − ϕ ( x)]
= ϕ′
( x)
x
1 sinx
ϕ′ ( x)
+ ϕ(
x)
= , ϕ(
π ) =
x x
ϕ ( x)
=
1
?

ϕ(
x)
1
−∫ d x sin x ∫ d x
= e x [ ∫ e x d x +
x
1
C]
−ln
x sin x ln x
= e [ ∫ e d x + C]
x
1 sin x
1
= [ ∫ ⋅ xd
x + C]
= ( − cos x + C)
x x
x
ϕ( π ) = 1,
C = π − 1
∴
1
ϕ ( x)
= ( π − 1 − cos x)
x

. y ′ − 2 y = φ ( x),
⎧2,
x
φ ( x)
= ⎨
⎩0,
x
< 1
> 1
( −∞ , +∞ ) y = y(
x),
( −∞
,1) (1,
+∞
y(0)
= 0.
)
φ ( x)
, y = y(
x)
x = 1
y(
x)
.

x < 1, y′ − 2 y = 2,
y
∫ 2 d x − ∫
2 d
= e [ ∫ 2e
d x + C1]
x
2 x
− 2 x
= e [ ∫ 2e
d x + C1]
2 x
= C1 e − 1 ( x <
1)
y ( 0) = 0, C
1 = 1,
2 x
,
y = e − 1 ( x < 1)
x > 1, y′ − 2 y = 0,
y
2d
x
2 x
= C 2 e
∫
= C 2e
( x >
1)

2 x
2 x
lim C e = lim ( e − 1),
x → 1
+
C
2
= − e
2 1
y = (1 − e
y(
x)
x
x → 1
− 2
− 2
) e
−
2 x
C
2 2
2e
= e −
( x > 1)
= 1
y = = e
x
1
2
1 −
1,
( −∞ +
∞)
y
=
⎧
⎨
⎩(1
.
2
e x
−
1,
2
2 x
x
− e
− ) e x
≤
>
1
1

y′
=
cos
cos y
ysin 2 y −
xsin
.
y

d
d
∴
x
y
cos ysin 2 y − xsin
y
= = sin 2 y − x tan y,
cos y
d
d
x
y
+
( tan y) ⋅ x = sin 2 y,
x
=
e
ln cos
y
[ ]
−ln cos y
sin 2 y ⋅ e d y C
∫ +
⎡ 2sin ycos
y ⎤
= cos y⎢∫ d y + C⎥
= cos y[ C − 2cos y].
⎣ cos y ⎦

.
Σ
> 0
x
∫∫
Σ
=
−
− 0
d
d
d
)d
(
d
)d
(
2
y
x
z
e
x
z
x
xyf
z
y
x
xf
x
).
(
1,
)
(
lim
)
(0,
)
(
0
x
f
x
f
x
f
x
=
+∞
+
→
∫∫
Σ
−
−
= y
x
z
e
x
z
x
xyf
z
y
x
xf
x
d
d
d
)d
(
d
)d
(
0
2

( ) z
y
x
e
x
xf
x
f
x
xf
x
d
d
d
)
(
)
(
)
(
2
∫∫∫
Ω
−
−
+
′
±
=
Σ
Ω
"
"+
Σ
.
"
"
−
Σ
0)
(
0
)
(
)
(
)
(
2
>
=
−
−
+
′ x
e
x
xf
x
f
x
xf
x
0)
(
1
)
(
1)
1
(
)
( >
=
−
+
′ x
x
x
f
x
x
f
e 2 x

f
1
1
′
x
( x)
+ ( − 1) f ( x)
= e 2 ( x >
x
x
⎛ 1 ⎞ ⎡
⎛ 1 ⎞
∫ ⎜ 1 − ⎟ d x
∫ ⎜ − ⎟ x
=
⎝ x ⎠ ⎢
1
1 d
f x e ∫
2 x
e e
⎝ x
( )
⎠
d x + C
⎢ x
⎣
x
e ⎡ 1 ⎤
=
⎢∫
−
e 2 x x
xe d x + C
x ⎣ x
⎥
⎦
x
e x
= ( e + C )
x
0)
⎤
⎥
⎥
⎦

1,
)
(
lim
)
(
lim
2
0
0
=
+
=
+
+
→
→
x
Ce
e
x
f
x
x
x
x
0,
)
(
lim 2
0
=
+
+
→
x
x
x
Ce
e
1.
0
1 −
+
=
+ C
C
1).
(
)
( −
=
x
x
e
x
e
x
f

.
d y = ycos
x
d x
,
d y
y
∫ = ∫ cos x d x
ln
|
y | = sin x +
C
1`
y
=
=
± e
C
sin x + C 1
2
e
sin
x
=
± e
C
1
⋅
e
sin
x

sin C
C ± e 1
2
y=0, ,
y
sin
x
= Ce C.
,.
d y
y
cos
d
∫ = ∫ x x
ln | y | = sin x + ln | C |
y
=
Ce
sin
x

.
1,
, 1().
,
h()
t.
,
dV
Q = = 0.62⋅
S 2gh,
dt
o
h
h
h + dh

Q S = 1 cm 2 ,
∴ dV = 0.62 2ghdt,
dV = 0.62 2ghd
t (1)
[ t , t + d t ]
h h + d h (d h <
Q
r
dV
=
=
100
−π
r
2
−
2
d h,
(100
− h)
2
=
0),
r
o
200h
− h
2
h
h
,
h + dh
100
cm
∴
dV
=
−π
(200
h
−
h
2
)d
h
(2)
(1)(2):
− π(200h − h
2 ) dh = 0.62
2ghdt,

− π(200h − h
2 ) dh = 0.62
2ghdt,
π
3
dt
= − (200 h − h )dh,
0.62 2g
π 400 3 2 5
t = − ( h − h ) + C,
0.62 2g
3 5
Q h | t= 0 = 100,
π 14 5
∴ C = × × 10 ,
0.62 2g
15
π
5 3 3
t = (7×
10 −10
h + 3
4.65 2g
h
5
).

.
(2,3),
x
y
PQ − 1
y′
1 y
− =
y ′ 2x
y′
=
2x
− ,
y
y = − x + C,
2
∴ y +
∫
1 2 2
ydy = − 2xdx
∫
.
P(
x,
y)
Q
PQ
Q( −x,0)
o
y( 2) = 3, C =
2 2x
2 =
17.
y
P( x,
y)
17
2
x

.
sin
x
d
d
y
x
+
ycos
x
=
5sin
x
⋅
e
cos
x
d
d
y
x
+
ycot
x
=
5e
cos
x,
, P ( x)
= cot x,
Q ( x)
=
5e
cos x
,
y
−
=
∫ P(
x)d
x
e
∫
[
P(
x)d
x
]
Q(
x)
e d x C
∫ +

=
−∫ cot xd
x
e
∫
−ln|sin
x|
= e ∫
[ ]
cos x cotd x
5e
e d x C
∫ +
[
cosx
ln|sin x|
]
5 e e d x + C
[
x
]
5 e | sin x | d x
1 cos
= ∫ + C
| sin x |
[
x
]
5 e sin xd
x
1 cos
= ∫ + C
sin x
=
1
sin
x
[
cos
]
5⋅
e
x + C

. yd x + xd
y = sin yd
y
x, y,
d
d
x
y
+
x
y
,
sin y
=
y
1
P ( y)
= , Q(
y)
=
y
sin
y
y
.
x
−
=
∫ P(
y)d
y
e
∫
[
P(
y)d
y
]
Q(
y)
e d y C
∫ +

⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
⋅
= ∫
∫
−∫
C
y
e
y
y
e
y
y
y
y
d
sin
d
d
x
⎥
⎦
⎤
⎢
⎣
⎡
+
= ∫ C
y
y
y
y
y
d
|
|
sin
|
|
1
⎥
⎦
⎤
⎢
⎣
⎡
+
⋅
= ∫ C
y
y
y
y
y
d
sin
1
[ ].
cos
1
C
y
y
+
−
=

.
,
)
(
3
)
(
)
(
2
2
2
2
3
2
2
2
∫∫∫
≤
+
+
+
+
+
=
t
z
y
x
t
dV
z
y
x
f
t
f
t
f
).
(
,
0 t
t f
≥
3
2
0
0
2
0
sin
)
(
3
)
( t
dr
r
r
f
d
d
t
f
t
+
⋅
= ∫
∫
∫
ϕ
ϕ
θ π
π
3
0
2
0
)
(
)
cos
(
6 t
dr
r
f
r
t
+
⋅
−
= ∫
π
ϕ
π
3
0
2
)
(
12 t
dr
r
f
r
t
+
= ∫
π
?
)
( =
t
f
0.
(0)
,
3
)
(
12
)
(
2
2
=
+
=
′ f
t
t
f
t
t
f
π

f
t 2
12
2
∫ 12π t dt
t t
2 −∫ π t dt
∫0
( t)
= e
0
[ 3t
e
0
dt
+ 0]
=
e
4π
t
3 t
2 −4π
t
3
⋅
∫
0
3t
e
dt
=
e
4π
t
3
⋅[
−
1
4π
∫
t
0
e
−4π
t
3
d( −4π
t
3
)]
=
e
1
⋅[
− e
4π
1
⋅[
− e
4π
3 t
4π
t
−4π
t3
3
4
4 π t
− π t
= e
+
3
0
]
1
]
4π
=
1 4
3
(
4π
e π t
−
1).

. 12000,
0.1%
CO 2
, CO2
, 2000
0.03%
CO 2
,
, 6
, CO 2
?
t CO x(t)%
2
[ t , t + dt]
,
CO
2
CO
2
= 2000⋅
dt ⋅
0.03,
= 2000 ⋅ dt ⋅ x(
t),

CO2 =
CO −CO
2
2
12000dx = 2000⋅
dt ⋅ 0.03 − 2000⋅
dt ⋅ x(
t),
d
d
x
t
1
= − ( x −
6
0.03),
x
=
1
− t
0.03
+ Ce 6
Q x | 0 = 0.1, ∴C = 0.07,
x = 0.03
+ 0.07e
6 ,
t=
−1
x | t= 6=
0.03 + 0.07e
≈
0.056,
,
1
− t
6, %. 0.056
CO 2

. f ( x)
(1)
⎧
⎨
⎩
y′
y(0)
a
+
;
ay
=
=
0
f
(
x)
y(
x)
(2)
f
(
x)
≤
k(
k
)
x
≥
0
1
k − ax
y(
x)
≤ (1 − e ).
a
( 1)
y(
x)
=
e
− ∫
a d
x
[ ]
= e − ax F ( x)
+
C

F ( x)
f
( x)
e
ax
(2)
y(
y ( 0) = 0 C = − F (0),
ax
= −
[ ]
y( x)
e F ( x)
− F
− ax
x
at
(0)
= e ∫ f ( t ) e d t
0
− ax
x
= ∫
at
x)
e f ( t ) e d t
0
−
≤ ax
x
∫
at k −
[
ax
]
ke e d t ≤ e
ax e − 1
0 a
k
(
ax
= 1 − e ) ( x ≥ 0)
a

:
2
2
1sec
x tan ydx + sec y tan xdy = 0
x+
y x
x+
y y
2( e − e ) dx + ( e + e ) dy = 0
2 dy 3
3(
y + 1) + x = 0.
dx
:
1cos x sin ydy = cos y sin xdx ,
y
x= 0
π
=
4
π
=
4
− x
2cos ydx + (1 + e )sin ydy = 0,
y .
x= 0

1 ,
,.t
= 10
2
,50 / ,4 ⋅ / ,
?
0 ().
a ,,
h ,
(
k ).
.

x y
1tan x tan y = C 2( e + 1)( e − 1)
= C
3 4
34 ( y + 1) + 3x
= C .
1 2 cos y = cos x 2e x + 1 = 2 2 cos y.
v
≈ 269. 3/.
0 , x , y
k h 2 1 3
, x = ( y − y ).
a 2 3

.
d y = 3x
2 y
dx
.
dy
:
2
= 3x
dx
:
y
dy
∫
2
= ∫3
x dx
,
y
.
3
ln y = x + C
y
= ±
e x 3
+ C1
1
C
C = ± e 1
y = C e
x
3
1 x
= ± e C e
3
ln
y
=
x
( C )
( y = 0 )
3 +
ln
C

.
xydx
y( 0) =
2
+ ( x + 1) dy
1
=
0
:
ln y
dy
y
=
x
= −
1+
x
ln
x
1
2
2
+ 1
dx
+
ln
C
2
y x +1 =
C
( C )
C = 1,
y
x
2
+ 1
= 1

. :
y′ = sin 2 ( x − y + 1)
: u = x − y +1,
u ′ = 1−
y′
1 − u′
= sin
2
u
sec 2 u du
=
dx
tan u = x +
C
:
tan( x − y + 1)
= x +
C
( C )

y +
d x y
. = e .
dx
− y x
1 e d y = e dx
2
:
y
x
− e
− = e + C
x y
( e + C ) e + 1 = 0 ( C < 0 )
u = x +
y,
u ′ = 1+
y′
u
u ′ = 1+ e
d u
∫ = x
1 + e
u
+
u
u − ln (1 + e ) =
x
y
x
C
+
C
u
(1 + e
∫
1+
)
e
−
u
e
u
d u
ln (1+ e
+ ) = y − C ( C )

.
M , t = 0 M 0,
M(t) t .
dM
= −λ M ( λ > 0)
: , dt
M t = 0 = M 0
()
, :
ln
M = −λ
t +
,
lnC,
C = M 0
M
M
=
∫
=
dM
M
C e
M
0
= ∫( −λ )d
−λ t
e
−λ t
.
M 0
o
t
M
t

. 1m , ,
2
S = 1cm . ,
, h t
.
: ,
dV
Q = = 0.62 S 2 g h
d t
d V = 0.62
2gh
d t
h
o
h
h
r
+ d h
100cm
[ t , t + d t ] h h + d h ( d h < 0),

2
dV = −π r dh
dV
0.62
r
=
2
100 − (100 − h)
= −π
(200h
−
:
h
t
2gh
d t
=0 = 100
h
2
) dh
= −π
(200h
−
2
h
=
2
) dh
:
d π
1 3
t = − (200h
2 − h
2
0.62 2g
2
200h − h
h
o
) dh
h
r
h + d h
100cm

,
π
t = −
0.62 2g
π
= −
0.62 2g
,
∫
1
2
3
2
( 200h
− h
400 3
2
( h 2 5
− h
2 )
3 5
C
=
π
0.62
2
g
) dh
+
14
⋅ ⋅10
15
h t :
π
5 3
3
2
t = (7×
10 −10
h +
4.65 2 g
C
5
3h
h
5
2
h
o
)
t
h
r
h + d h
100cm
=0 = 100

.
:
(1)
( x
2
+ xy )d x − ( x y + y)dy
2
=
0
( 2) y′
+ sin( x + y)
= sin( x − y)
y x
: (1) dy
= dx
2
2
1+
y 1+
x
(2) y′
= −2cos xsin
y
y
ln tan = −2sin
2
x
+
C

.
,
y =
F ∈C
f
(x).
1
,
∫
L
F( x,
y) [ y sin xdx
− cos x dy]
F(0,1)
= 0,
F( x,
y)
=
: ,
∂
∂
[ −F(
x,
y)cos
x ] = [ F(
x,
y)
y sin x]
∂x
∂ y
− cos x + F sin x = y sin x + F sin x
F x
y′
−
y′
= y tan
y x=0 = 1
F
F
x
y
x
=
F y
y
tan
y
x
= 1
= sec
cos x
x
0

d y 2y
5
. − = ( x + 1)
2 .
dx
x + 1
dy
2y
dy
2dx
: − = 0 , =
dx
x + 1 y x + 1
ln y = 2ln x + 1 + ln C , y = C( x +1)
. y = u ( x)
⋅(
x
2
y′ = u′⋅(
x + 1) + 2u
⋅(
x + 1)
1
2
+ 1)
u′ = ( x +1)
2 3
u = ( x + 1)
2
+ C
3
3
2 ⎡ 2
= ( x + 1) ( x + 1) + C
⎢⎣ 3
y
2
2
,
⎤
⎥⎦
2

dx
⎡ 2 x ⎤
. + − dy
= 0 .
3
x y ⎢⎣ y y ⎥⎦
dx
: x, y , x > 0 ,
= 2d x ,
x
d x x 2
2 − = − x , y
dy
y y
,
x = e
∫ dy
dy
−
2y
1 ∫
[ ∫ − e
2y
d x + ln C
y
1 1
= y [ −∫ ⋅ dy + ln C ] =
y y
y e
x
y
= C ( C ≠ 0)
1
P( y)
= −
2y
] 1
Q( y)
= −
C y
y ln
y

.
,
E = Em sinω
t,
R
L , i (t) .
: . :
L
R
E
K
, 0
R R i
di
L L d t
di
E − L − Ri
= 0 ,
d t
: i t 0 = 0
=
di R Em sinω
t
+ i =
d t L L

d i R E t
i
m sin
+ =
ω
d t L L
i
:
t =
0 = 0
L
R
E
K
R
L
− ∫ dt
i (t) = ⎡ E
e ⎤
⎢⎣ ∫
m
∫ d t
sinω
t e d t + C
L
⎥⎦
Em
= ( Rsinω
t −ω
Lcosω
t ) + C
2 2 2
R + ω L
−
y = e ∫ P(
x)d
x ⎡ ∫ P(
x)d
x ω LE
: i
⎤
t=0 =
⎢⎣ ∫Q
0(
x
) e C = d2
x + m
C2
⎥⎦
2
R + ω L
R
L
e
−
R
L
t

i ( t)
= m
2 2 2
R
ω LE
+ ω
E m
L
e
−
R
t
L
+ ( Rsin
t Lcos
t)
2 2 2 ω − ω ω
R + ω L
ω L
: ϕ = arctan , R
L
R
E
K
i(
t)
=
R
ω LE
2
+ ω
m
2
L
2
e
−
R
L
t
E m
+ sin( ω t
2 2 2
R + ω L
−ϕ
)

.
dy
dx
−1
y
2
+ = a (ln x)
y .
x
: z = y ,
dz
z
− = −a
ln x
dx
x
z
−1
= y
z = e
∫
1
x dx
[ ∫ −a ln x)
e
a 2
= x C − (ln x)
2
, :
a 2
y x
2
[ ]
( − ∫ x
[ C − (ln x)
] = 1
1
x d
dx + C ]

.
:
:
x
y
xy
y
x
y
x
d
d
d
d
1)
( =
+
)
ln
(ln
d
d
2)
( x
y
y
x
y
x
−
=
x
x
y
y
y
d
d
1
=
−
x
y
x
y
x
y
ln
d
d = 2
2
1
d
d
2
x
y
x
x
y
= −
−
2
2
1
d
d
2
y
x
y
y
x
= −
−
2
sin
2
d
d
y
x
x
y
x
x
y
=
+
0
d
2
)d
(
(3)
3
=
−
−
y
x
x
x
y
0
)d
(
d
2
(4)
3
=
−
+ y
x
y
x
y
y
x
x
y
x
y
d
d
2)
ln
(
5)
( =
−

.
1. f (x)
f
( x)
x
= sin x − ∫ f ( x − t)dt
u = x − t
0
:
:
f ( x)
= sin x − ∫ f ( u)du
0
f ′( x)
+ f ( x)
= cos
f ( 0) = 0
f
( x)
=
1
2
x
(cos
x
+
sin
x
−
x
e
−x
)