文档内容
2024
专项刷题班第四章
常微分方程x y x y
例4.1 微分方程 y sin sin 的通解是 .
2 2
| =s
-
m
9
sm
xty
smt sm β= 2 10s
Ʃ+ , β
smπ
- B
' - -
2
州 xy xY xey
x
= 8 +Ʃ Ʃ ⼀
2 vs sm
2 z
= g
.
(
osƩ
δ
mi
=
±| dx sl ) == →>Ʃ
- 2 冲Ʃ →/ )d ) . d 三
smil
( t ) Ʃ
(cid:15482) |u |(s l - 10 = -2sm + C
例4.2 满足 f x xf x x的函数 f x ________.
x t
t - λ = -
=
,
→ fi - t) - t -
fit
) = - t (cid:15482)
f'
ix ) -
xfin
= - x
→fi
- x ) = xfix 1 - x
代⼊原⽅程
fix [xfix x (cid:15482) x fix xi
(cid:15482) 1 + x
1
-
)
= x (+ ) )
=
x+
x
fix x+
(cid:15482) 1 =
2
1Hx
ax
时
f =| 品
→ *
ax
=
Ʃ|nG x
= + ) + x - arltmx + c
例4.3 求微分方程 ydx x x2 y2 dy 0, (y 0)的通解
y
解 →
: Ydx = lo t xreyp ) ay x
=xt xeys
xtxep
= ⼆年 P
→ ++
y
d
(u y)
令 年 y - . y 代⼊
U y uty
= ,
, =
* Y . ut
→ u = ts
1
y
= s
f(cid:15482) ieuu
f
yy h ut ieur= lulMl + lhC= licly
(cid:15482) u+ (tu = =± CY
=
C
.
φ (C
.
=± C)U+ 1+ W = G . y
=
+ cy
P
可
y
xty== G
(cid:15482) x+
.
= 平⽅
→
xty=
=
C
. -
x
,
4
2 yP , Pye :
> x * = -2 C x+ x
c
. )
(cid:15482) 1 CP 2Gx
= _
(cid:15482) 9 1 + 2 C x
.
=
GE
例4.4 x , x 为已知的连续函数，则方程 y x y x x
通解是_____．
Seix
) dx
1 uindx cfe
①(+) .ω ix 1x tc )
φ= e ,
-e
e " . fee )
c * e*do tc
=
Jtdt
letedt ett . ettc
-
= = t t -
λ
↓
(
t ) C
= t -1 +
eel,
4 ( een! - φ(X)
: - (( ew 1 ) + c ) =φ x1-1 + ce例4.5 求方程 xy(4) y(3) 0的通解.
3) = pi
令 φ P 则Y 性 → x . P = 0
= p -
.
1 iax
/
P ⽚ P → P c o dx tc )
→ - =0 = e ,
lmxl
→ P e C C (λ =± Cx Cx
= - = =
"
叫 Gx
→ =
= x
" =|
→φ c. xx + Cz
→ Y = / ( Ʃ x + C 2 ) dx =Ʃ . x + 2 x + G
→φ=|③ x =☆ 4 +Ʃ=
+ Crx + ( 3 ) dx x x + ( 3 x + C 41
例4.6 已知 y cos2x xcos2x是二阶常系数非齐次微分方程一个解，
4
则该方程是( B )
(A) y 4 y sin2x (B) y 4 y sin2x
1 1
(C) y 4 y sin2x (D) y 4 y sin2x
4 4
E 代⼊ 1 1 (cid:49455)1 ①)
φ= 10s2x .
法 _ : 1032× - ③ .
* ⻬ Gerxtaerx
⾮⻬通 C 4 C 叫ztY : ① r .tr 2 φ=
法 = : = . . + 2
⻬通中⽆⼀式 形式 ② r = rzzr
acoszx .
∵
erxexrotG
* φ=
c
⼀书 是 y
xc
0
s2x ,
∴
是⼀个⻬次解 ③ r 1 =α+βi r 2 = 5 - βi
∴ (os2x
q = C et* C 0s β xtrexsm βx
, .各通
之 r 2i Gcos 2x 2 sm2x
r 2 2 = - +
∴ =
.
特征⽅程午
: 4 = 0
=ψ
⻬次 +4
" =
叫
" ⼀
原公程
ψ
4y f 代⼊ Y 1oszx
: + = ⼭ *
,
f Sm2x
(cid:15482) 1
× =
例4.7 设函数 y y x 满足微分方程 y 4 y 4 y 0和初始条件
y 0 2, y 0 4，求广义积分 y x dx.
0
'
γ+
特征⽅程 : 4 r+4 = 0 (cid:15482) ( r+2) =0 → r . =γ . = - 2
2x 2X
-
. - Cax
Yun + e
e .
= C
qi
: φ 0) = 2 , 0 ) = -4
1 . C 1 = 2 C 2 =
2X
φ ⼀
. *)= . e
2
|^∞ ex 9t e -^
: 1 [ q .* dx = 2 . dx = -
(cid:49478)
∞ x 2 d )
(cid:49478) ⼼
0
=* (
t"
-
1
e =
=
- (cid:49478)例4.8 求微分方程 y 4 y sin3xsin x cos2 x的通解.
γ=
特征⽅程 : - 4 =0 → 1 . 22 , γ 2 =- 2
: φ= C e 2x + 2e 2x sm 2 . sm β= - Ʃ [ cos(σ+β ) -10 (σ - β ]
.
Ʃ
\+ Cos2X
sm 3x.smX +c 0 x = - [os 4x - cos2x ] +
2
三
Ʃ
( 0s 4x +co> 2X +
= -
"
9 4Y
Ʃ
cos
4x ①
.
- = -
* * Fi
Hi
= acos4
xtbsm4x
,
M
,
代⼊ ① → a
=
b
, = 0
1 . ,
* Ficos
: 9 4x
=
i" ×
y -4 Y = ( 0s2
-
9 Ʃ
φ % * = 210s2 x + b =sm2 X * 代⼊ ②(cid:15482) G 2 = b →=
a
⼀⽚
上
φ* los2x
=
,
"
ψ 44 Ʃ ③
- =
* i ; Ʃ
:
φ = A (cid:15482) A = - ∴ φ * = -
;
-
-
:
*
= φ
.
*
+ φ
*
+φ ;
*
= icosφ
x
Ʃlos2
x
:
通解
:
φ= G +
e
2 x +
(
os
4
x-δ cos 2x-
Ʃ
,例4.9 设函数 f (x), g(x)满足 f (x) g(x), g(x) 2ex f (x)，且
g(x) f (x)
f (0) 0, g(0) 2，求 dx.
0 1 x (1 x)2
. -. a
=| qu xlvix
保 ux
xrdx
f
↑
↓
f^
t
x
f
itaixn
kax
-
xi
f iit
" . .
- .
fi feo
π ) )
=
-
(以
fix 1 = 94) (cid:15482) qix ) = f" ) , ⽽ qix 1 = 2e * - fi +)
,
"
f ex fi
1 (x) = 2 - *)
.f" fex *
: (*+ )
=
2
.
γ=+
特征 = 1 = 0 (cid:15482) γ= i r 2 =- i
T C C smx
: = cosx + 2
.
* fY fx
× *
y = ae ' 代⼊ 1 + 1 = 2 e 中 (cid:15482) a = 1
.
ex
性
.
fi
∴ *1 = C , cosxt ( 2 Smx t ex
⼜ fo 1 = 0 , q . 0) = fi 0 )= 2 ∴ . C . = - 1 ( 2= 1
f
f ex ltex
cosxtsmx ( )
∴ ×|= - + π =
=
原
∴例4.10 求微分方程 y
6
2 y
3
y 0的通解.
特征⽅程 γ
6
+ 2γ
3
+ 1 = 0 → ( r 3 + 1
]
= 0
1 ± 1 4 l± 给之
P ) ≥ -
[ 1 ( r = -r+ 1 == r r + 1 = 0 r = =
→
2
2
1+ o 3i 1 ⼀场之
(cid:15482) V = - | V 2 =- | r 3 =γ 4 = γ 5 =γ 0 =
.
2
2
起
esmBx
cos
* GxeX
q = C e + + ( 3 . ③ x + 4
.
10 eismiix
xt
t ( xe Ʃ ③ ( 0
5 s
例4.11 设 f (u)有二阶连续导数,z f x2 y2 满足
2 2
z z
x2
y2
，求
2 2
x y
z
x
的表达式. f
z = u ( xtyl
-
y
f
令 U= x 2 y 2 _ u
x
az fiu
)
= .
x 2 2
中 xty
北
xey -
X
= " 以 efiu = 2)0 ep
↑ )
f
2
eyz
x
2
" tfiu 时
f
ul )
= ( .
3
=+ }
(x y
收
y xx
= "( f'
f u) , t ial
. 3
2
X yp ( x =+ y ]xeyz
代⼊
(
'typ
f" fiul x
(cid:15482) ul =
( +
⿏
2
f" fl π × 缺型 ,
5 ult . = ,
+
令 P =
f '
u )
f"
(u) =
1
E =
E
( x
+
y
=|P
C
. 1
n
xeyz
+ Cz
,
2
aP
p u
→ + =
letau '
efaau
u
c ut al
dutcd
→ P . . +
=
=
'
aG
fiul u
→ +
=
fll 1 ( a π 4 ( lm 1 G
→ = + ) u = u + , +例4.12 （仅数一数二）已知 y (x) ex , y (x) u(x)ex 是二阶微分方程
1 2
(2x 1) y (2x 1) y 2 y 0的解，若u(1) e,u(0) 1，求 u ( x ) ，
并写出该微分方程的通解.
Ur
=
uxiet 4 i
=
Wix )
-
et
+
ux1 . eX W = N
+ux1
上 ex
,
φ " = ω× 1e* + 2ω ix 1. e * + u1x1 . e * ) = ω "(x +2 Wix ) tuix)] ex
i
代⼊原公程
ψ eX ' e e.
(cid:15482) ( 2x - 1 ) [ "(x) +2 Wix ) + U(x) ] . - [2x+1) [ω (x) + U1×1 ] . * 2φ(x ) *
.
Wix
+
(cid:15482) ( 2x-
1
) U"x ) (x - 3
)
) =0
p "
令 ix
P=ω ) =ω (x)
..
(cid:15482) ( 2x-1
)
P + (2x -3
)
.
P
=
☆ IP
→ (2x +)
. = -
- 2x
3
P 2x-
→|µ l dx
= -
2× -
e*
(cid:15482) P C (2x -1 .
= . )
ex
:
uix
) =
[
.
( 2x-1
)
.
(cid:49456) txdx -XeC
∴ u*1 = C . (2x - 1 ) . = - C . ( 2x+1 ) , e 2
⼜ UC1|= e φ ( 0) = - , : . C . = ( C 2 =
ex
∴ U(N) = - ( 2x+1) . ∴ φ 2 = - ( 2x+1 )
原通解 φ= c ex C [ 2xt1 )
∴ . - 2例4.13 设函数 f (x)在[1,)上连续.若由曲线 y f (x)，直线
x 1, x t(t 1)与 x 轴所围成的平面图形绕 x 轴旋转一周所成的旋转体
体积为V t t2 f t f 1 ，试求 y f x 所满足的微分方程，
3
2
φ=和
并求该微分方程满足条件 y 的解.
x2
9 ⼀
?
,ta f i ③
[
ife
*
f
↓
) 对球导 ! D
ut= ax -
f =
,
>
③
xfii 达 fit tfit] 七
2 ) + ) |
(cid:15482)
=
( tfe
f tfit
=(cid:15482) 3 t =2 1+ )
,
==
×φ+ xx
(cid:15482) 3 2= dy
+ .
→ 3 2 x φ x 2
dx
dl 3 _
作恐 i xy
(cid:15482)
3 2
= =
dx
y
dy
a
令 U = 元 =U +
,
dx ax
x . 2
u
(cid:15482) u+ x . = 3 = - 2 U → x = 3u - 3U = 3 u (u- 1)
= xaxx
fucu
3
_yu
f ui ildu = 31 ux1 hc
s +
( -
β
u 出 ucx
→
=β
u 出 =
→ ncx
⼗ " 3
a-
(cid:15482) =± ( x = C x
. .
u
9
1
- 3
九
(cid:15482) 4 x
C x - 43
= . → C
= .
Ψ
y
可
Y|
⼜ x=2 =
x
x
:
φ=
C =+ - :
= . 單 = 3
ltx
