第四節(jié)

一階線性微分方程一、一階線性微分方程二、伯努利方程三、小結dxdy

+

P(

x)

y

=

Q(

x)一、線性方程一階線性微分方程的標準形式:當Q(x)”0,上方程稱為齊次的.上方程稱為非齊次的.當Q(x)”0,【例如】dy

=y

+x2

,dxyy

-

2

xy

=

3,dx

=

x

sin

t

+

t

2

,dty

-

cos

y

=

1,線性的;非線性的.一階線性微分方程的解法d

xd

y+

P(

x)

y

=

01.解齊次方程分離變量兩邊積分得故通解為ln

y

=

-

P(

x)dx

+

ln

Cy

=

C

e-

P

(

x

)d

x2.解非齊次方程d

xd

y

+

P(

x)

y

=

Q(

x)對應齊次方程通解

y

=

C

e-

P(

x)dxu￠e-

P

(

x)

d

x

-

P(

x)

ue-

P

(

x

)

d

x

+

P(x)

u

e-

P(

x)

d

x

=

Q(x)即對應齊次方程通解非齊次方程特解y

=

Ce-

P

(

x)

d

x用常數變易法:作變換

y(x)

=

u(x)

e-

P(

x)

d

x

,

則故原方程的通解+

e-

P(

x)

d

x

Q(x)

e

P(

x)

d

x

dx

y

=

eQ(

x)

e

d

x

+

C-

P

(

x

)

d

xP

(

x

)

d

x即u

=

Q(x)

e

P(

x)

d

x

dx

+

C兩端積分得【常數變易法】把齊次方程通解中的常數變易為待定函數的方法.【實質】未知函數的變量代換.新未知函數u(x)

原未知函數y(x),作變換y

=

u(

x)e-

P

(

x

)dx求方程y￠+1

y

=sin

x

的通解.x

xxP(

x)

=

1

,xQ(

x)

=

sin

x

,e dx

+

C

y

=

exsin

xx

1

dxx-

1

dx

e dx

+

C=

e-ln

xxsin

xln

xxx=

1

(

sin

xdx

+

C

)

=

1

(-

cos

x

+

C

).【解】【例1】積，求曲線【例2】如圖所示，平行于y軸的動直線被曲線與

y

=

x

3

(

x截?下0的)

線段PQ之長數值上等于陰影部分的面y

=

f

(

x)0xf

(

x)f

(

x)dx

=

(

x3

-

y)2

,x03ydx

=

x

-

y,y

+

y

=

3

x2

,兩邊求導得解此微分方程【解】y

+

y

=

3

x2C

+y

=

e-

dxdx23

x

e

dx由y

|x=0

=0,所求曲線為得C

=-6,y

=

3(-2e

-x

+

x2

-

2x

+

2).yo

x

x=

Ce-

x

+

3

x2

-

6

x

+

6,PQy

=

x

3y

=

f

(

x)【例3】求方程x【解】注意

x,

y

同號,

當

x

>

0

時,

d

x

=

2d1P(

y)

=

-

2

yQ(

y)

=

-

1y由一階線性方程通解公式,得x

=

ey[

-

1

ex

,故方程可變形為y3x

d

y

=0

的通解.

y

x

yd

x

2+

-y[-

1d

y

+

ln

C所求通解為ye

=

C

(C

?

0)xy這是以

為x

因變量,y為自變量的一階線性方程伯努利(Bernoulli)方程的標準形式dxdy

+

P(

x)

y

=

Q(

x)

yn(n

?

0,1)當n

=0,1時，方程為線性微分方程.當n

?0,1時，方程為非線性微分方程.二、伯努利方程【解法】需經過變量代換化為一階線性微分方程.d

xy-n

d

y

+

P(

x)

y1-n

=

Q(

x)令

z

=

y1-n

,則

dz

=

(1

-

n)

y-n

d

yd

x

d

xd

xdz

+(1

-n)P(x)z

=(1

-n)Q(x)

(關于z,x的一階線性方程)求出此方程通解后,

換回原變量即得伯努利方程的通解.\

y1-n

=

z

=

e-(1-n)

P

(

x

)dx

(

Q(

x)(1

-

n)e(1-n)

P

(

x

)dxdx

+

C

).除方程兩邊,得y

的通解.【例3】

求方程

dy

-

4

y

=

x2dx

xy

=

x2

,1

dy

-

4y

dx

x令

z

=

y

,2

dz

-

4

z

=

x2

,dx

x2

2+

C

,

解得z

=xx

22+

C

.

x即y

=x4

【解】兩端除以

y，得【例4】用適當的變量代換解下列微分方程:1.

2

yy￠+

2

xy2

=

xe-

x2

;【解】2y￠+

xy

=

1

xe-

x2

y-1

,令z

=y1-(-1)=y2

,則dz

=2

y

dy

,dx

dxz

=

e-

2

xdx

[

xe-

x2

e

2

xdx

dx

+

C

]dx\

dz

+

2

xz

=

xe-

x2

,所求通解為222

xy2

=

e-

x

(

+

C

).dx

x

sin2

(

xy)

x-

y

;dy

=

12.【解】令z

=xy,則dz

=y

+x

dy

,dx

dx1x

sin2

(

xy)

x

sin2

z1dx-

y

)

=

,dz

=

y

+

x(2z

-

sin

2z

=

4

x

+

C

,分離變量法得將z

=xy

代回,所求通解為2

xy

-

sin(2

xy)

=

4

x

+

C

.可分離變量的方程;dy

=

13.dx x

+

y【解Ⅰ】令x

+y

=u,則dy

=du

-1,dx

dx代入原式dxdu

-

1

=

1

,uu

-

ln

|

u

+

1

|=

x

+

C

,分離變量法得將u

=x

+y

代回,y

-

ln

|

x

+

y

+

1

|=

C

,所求通解為y或

x

=

C1e

-

y

-

1dy【解Ⅱ】

方程變形為

dx

=

x

+

y.可分離變量的方程dydx

=

x

+

y.dy

dx

-

x

=

y一階線性非齊次微分方程則有x

=

e-

P

(

y

)dy

(

Q(

y)e

P

(

y

)dydy

+

C

)=

e-(-1)dy

(

ye(-1)dydy

+

C

)=

e

y

(

ye-

ydy

+

C

)=

e

y

(-

ye-

y

-

e-

y

+

C

)=

Ce

y

-

y

-

1三、小結1.齊次方程2.線性非齊次方程3.伯努利方程xy￠=

f

(

y

)令

y

=

xu

;令y

=u(x)e-

P

(x

)dx

;令

y1-n

=

z

;思考與練習判別下列方程類型:(1)

x

dy

+

y

=

xy

dydx

dxdx(2)

x

dy

=y

(ln

y

-

ln

x)3(3)

(

y

-

x

)

dx

-

2x

dy

=

03(4)

2

y

dx

+

(

y

-

x)

dy

=

0(5)

(

y

ln

x

-

2)

y

dx

=

x

dy提示:y

-

1d

y

=

dxy

x可分離變量方程d

y

=

y

ln

y齊次方程x2dx

x

xd

y

-

1

y

=

-線性方程2y2d

y

2

y

2dx

2

xdx

1-

x

=

-線性方程d

y

+

2

y

=

ln

x

y2dx

x

x伯努利方程【思考題】求微分方程y￠=的通解.cos

ycos

y