数学双专-常微分方程第四章

§4.2

内容回顾dtnnd

xn1d

x1

a

(t)dtn1

an

(t)x

f

(t)1n1d

xnd

x1

a

(t)dtn1n

a

(t)x

0dtn2x(t)

c1x1

(t)

c2

x2

(t)

cn

xn

(t)x(t)

c

x

(t)

c

x

(t)

c

x

(t)

~x

(t)1

1

2

2

n

n方程类型L[x]

f

(t)

L[x]

0F()

n

a

n1

a

01

n基本解组或通解常数变易法特解相加比较系数法拉

斯变换法求解方法本节内容/Contents/几类可降阶高阶方程幂级数解法（求特解）1)

方程不显含未知函数

x

及

x,

x

,,

x(k

1)F

(t,

x,

x,,

x(n)

)

0F

(t,

x(k

)

,

x(k

1)

,,

x(n)

)

01

k

n则方程可降为

n

k

阶的方程，即可降

k

阶§4.3

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Order

Method

and

Series

Method4.3.1

可降阶的方程的类型n

阶方程的一般形式

y

则令

x(k

)F

(t,

y,

y,,

y(nk

)

)

0(4.58)y

(t,

c1

,

c2

,,

cnk

)方法(4.58)

的通解若可求得x(k

)

y

(t,

c1

,

c2

,,

c

)nk逐次积分k

次，可得原方程的通解。§4.3

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Order

Method

and

Series

MethodF

(t,

x,

x

)

0x

y,

x

yF

(t,

y,

y)

0y

(t,

c1)x

(t,

c1

)积分，可得原方程的通解x

(t,

c1,

c2

)特别，对于二阶方程§4.3

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Method

and

Series

Method解

令t

dt4d

5

x

1

d

4

x求方程

dt5d

4

x

ydt4dt

0

的通解。y

1

y

0t11t

c

t1y

c

ex(4)1

c

t2x(3)

cc

1

t

2232x(2)6

c1

t

3

c t

c例143x

c1

t

4

c2

t

2

c

t

c24

2

ct

cx

ct

5

ct

3

ct

21

2

34

5§4.3

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Method

and

Series

Method2)不显含自变量

t

的方程F

(x,

x,,

x(n)

)

0(4.59)

可降低一阶方法

令x

yx

d

(x)

d

y

dy

dx

y

dy

dt

dt

dx

dt

dxx

d

(

ydy)

d

(

y

dy)

dxdx2

y

2d

2ydydx2

dxdxdy)2dx

dx

dtd

2

y)

y()2dt

dx

y((

y§4.3

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Order

Method

and

Series

Method假定)

fdx

dt

dxdxn2dt

dx

d

f

(

y,

dy

,,

d

n2

y

dx(n)将x,

x

,,

x(n)(4.59)(n2)

)x

xdx

y

d

(

f

(

y,

y

,,

y代入原方程x(n1)d

n2

ydxn2dxdy

f

(

y,

,,)

f

(

y,

y

,,

y

(n1)

)1

x

x§4.3

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Order

Method

and

Series

MethodG(x,

y,

dy

,,

d

n1

y

)

0dxdxn1降低一阶y

(x,

c1

,

c2

,,

cn1

)dt分离变量，可得原方程的解。1

2n1y

dx

x

(x,

c

,

c

,,

c

)§4.3

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Order

Method

and

Series

Method例2

求解方程§4.3

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Order

Method

and

Series

Methodxx

(x)2

0x

yx

y

dydxx

y

dy

y

2

0dxy

0或dxx

dy

y解令dy

dx

y

x1ln

y

ln

x

cxy

c1xx

c11xdx

c

dt211

x

22

c

t

c21x

2

2c

t

2cy

0

x

0x2

2c

t

2c1x

c2§4.3

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Order

Method

and

Series

Method已知(4.2)的k

个线性无关的特解，则(4.2)可降低k阶，即可得到n-k

阶的齐次线性方程。特别地，如果已知(4.2)的n-1

个线性无关的解，则(4.2)的基本解组可以求得。§4.3

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Order

Method

and

Series

Method3)

齐次线性方程n1d

x1

a

(t)dtn1n

a

(t)x

0dtn结论nd

x(4.2)§4.3

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Order

Method

and

Series

Method方法

设

x1

,

x2

,,

xk

是(4.2)的

k

个线性无关的解xi

0,

i

1,2,,

kan

x

xk

yan1

x

xk

y

xk

y(n1)

y

xk

kx(n1)

x

y(n1)an2

x

xk

y

2xk

y

xk

y令k

k

k

kx(n)

x

y(n)

nx

y(n1)

n(n

1)

x

y(n2)

x

(n)

ya12

an

(t)xk

]y

0k[x令§4.3

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Order

Method

and

Series

Methodx

y(n)

(nx

a

(t)x

)y(n1)

k

k

1

k

a

(t)x

(n1)

a

(t)x

(n2)(n)1

k

2

ky

zn1(t)z

0z(n1)

b

(t)z(n1)

b1(4.67)kxn-1阶线性方程z

y

(

x

)k

或

x

xzdt可将(4.2)化为n1阶线性方程同理，对于(4.67)

就知道了k

1个非零解ix

(

xi)i

1,2,,

k

1zz(n1)

b

(t)z(n1)

b

(t)z

01

n1(4.67)2

21

1

z

k

zk

1

k

1

2211)

)

(

(kk

1k

kx且其线性无关，z

0(

xk

1

)

0xxxx1

1

2

2kk

1

xk

1

)]

0x[

1

(

x

x

1x1

2

x2

k

1xk

1

k

xkxi

,i

1,2,,k

线性无关，1

2

k

0§4.3

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Order

Method

and

Series

Method类似地,令

u

z

k

1

z

或z

zk

1udt

c

(t)u(n3)

c

(t)u

01

n2u(n2)k

1izi

1,2,,

k

2u

(

zi

)

线性无关的解，继续下去，得到一个n-k阶的线性齐次方程若k=n-1，则可得到１阶线性齐次方程，则可求得通解。§4.3

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Order

Method

and

Series

Methodx

1令x

x1

ydtx

x1

ydt

x1

yx

x1

ydt

x1y

x1y

x1

y

x1

ydt

2x1y

x1

y若知其一非零解

x

x1

0

，则可求得通解。y

(

x

)§4.3

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Order

Method

and

Series

Method特别，对于二阶齐次线性方程d

2

x

dt2

dt

q(t)x

0x1

ydt

2x1y

x1

y

p(t)x1

ydt

p(t)x1

y

q(t)x1

ydt

0x1

y[2x1

p(t)x1]y

0yy

2x1

p(t)x1x112

x

p(t

)

x1dty

c

e

c

ex1[2

1x1(t

)dt

]11

c

eln

xx21

e

p

(t

)dt

c1

e

p

(t

)dt121§4.3

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Order

Method

and

Series

Methodx21y

c1

e

p(t

)dt基解组为1

e

p

(

t

)

dt

d

txx211x1

,通解1

e

p(t

)dtdt]x(t)

x1[c1

c2x21x

xydt1P.113§4.3

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Order

Method

and

Series

Method例4

已知

x

sin

ttt是方程

x

2

x

x

0解的解，试求方程的通解。p(t)

2sin2

tt

2tsin

ttx

(c1

c2

t

e

dt)12

dt

c2

sin2

t

sin

t

(cdt)t11tt12

12

sin

t

(c

c

ctgt)

1

(c

sin

t

c

cost)§4.3

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Order

Method

and

Series

Method§4.3

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Order

Method

and

Series

Method4.3.2

二阶线性方程的幂级数解法（求特解）解为方程的解设

y

a

a

x

a

x2

anxn012a0

0y(0)

0例5dx求方程dy

y

x

的满足初始条件y(0)

0

的解。y

a

x

a

x2

a

x3

anxn

1

2

3xn

y

a

2a

x

3a

x2

na

xn1

(n

1)a1

2

3

nn1

(

a x

a x

2

a

xn

)

x1

2

n

(a

1)x

a

x2

a

xn

1

2

n1a

0,22

a

1,22

(a

1)x

a

x2

a

xn

1

2

na

1§4.3

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Order

Method

and

Series

Methoda

2a

x

3a

x2

na

xn1

(n

1)a xn

1

2

3

n

n1nn1(n

1)a

aa

n

1ann

11(n

1)n3(n

1)!1112a

a

n1a

n

1

(n

1)na

n1n

1

n!nan

2

,

3

,

n!na

1

§4.3

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Order

Method

and

Series

Methodn

2

,

3

,2!

3!

xn

)y

(x2

x31!

2!

n!n!

xn

)

1

x

(1

x

x2

1

x

exy

edx

(edx

xdx

c)

ex

(ex

xdx

c)

ex

(e

x

x

e

x

c)

x

1

cexy

ex

x11

c

0,

c

1§4.3

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Order

Method

and

Series

Methody

2xy

4

y

0的满足初始条件y(0)

0解

设级数解为y(0)

1

的解。y

a

a

x

a

x2

a

xn

0

1

2na0

0,

a1

1y(0)

1

所以

na

xy

x

x

a

x

a x3

a

xn

n22

3

n由于y(0)

0n2

nn2n1y

1na

xy

n(n

1)an

xn2n2例8求方程

2x(1

nan

xn1

)

4(x

an

xn

)

0n2

n2n2n2nn(n

1)a

xn2n2n2n

4an

)xn

0

6x

(2nann(n

1)a

x0次项系数1次项系数2!a2

0

2

次项系数3!a3

6

0

a3

1(n

2)(n

1)an2

(2n

4)an

0a2

0§4.3

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Order

Method

and

Series

Methodn

为偶数时，即n

=2

ka2k

a2

02k

12k

12k

3k

112(k

1)2a

a

a2(k

1)12(k

1)1(k

1)1aa2k

12k

1a

1

aknna

a(2n

4)(n

2)(n

1)2(n

1)a

n2，由上述递推公式得n

为奇数时，即n

=2k+1§4.3

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Order

Method

and

Series

Methody

x

(1

x31!

1

x5

1

x7

1

x2k

1

)2!

3!

k!

x(1

x2

1

x4

1

x6

1

x2k

)2!

3!

k!2

xexk!a2k

1

12ka

0k!

k!k1k(k

1)2k

3

32k

12k

1a

1

a

a

1

a

1§4.3

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Order

Method

and

Series

Method例7

求初值问题dxx2

dy

y

x解

设n