書后習題答案第五章

5.6x,解：xx,2x解：x1 e2x,3e2e23e2

131ex,e

常數(shù)，所以線性無關cos2x,sincos2xsinex2,

sin2x,cosxsinsincosxsin

2excos2x,exsinexcosexsin

lnx,xlnlnxln

eax,ebxa

ex,

yexyex在y"yy"y 解：ex"exex"ex所以yexyex在上都是二階線性齊次微y"y0yCexCexy1y"y y1CexCe y1ysinxycosx在y"'y'y"'y'1"'01'0;sinx"'cosxsinx'cosxcosx"'sinxcosx'sinx，所y1ysinxycosx在y"'y'0x2xyC1C2sinxC3cosxy

"'

y

x2C1C2sinxC3cosx驗證函數(shù)y1,yx,yx2, ,yxn1在,上都是n階線性齊次微分方y(tǒng)nyn1n0;xn0;x2n

xn1n10.所以函數(shù)y1,yx,yx2, ,y在

n

yn

2C2CnyC1C2xC3

為方程 22Cnx為yn!C1C2xC3

cosx

sinxy"2y01解：cosx"2cosx;sinx"2sinx所以1

cosx

sinx y"2y0yCcosxC

ex2

xex2y"4xy'4x22y022ex2'2xex2ex2"2ex24x2ex2;xex2'ex22x2ex2xex2"6xex24x3ex2所以

及

都是方程y"4xy'4x22y0 yCex2Cxex2 （1）yCexCe2x

1e5x(C,

是任意常數(shù)）y"3y'2ye5x y'Cex2Ce2x5e5x,y"Cex4Ce2x25e5x yCexCe2x1e5x(C,

是任意常數(shù)）y"3y'2ye5x （2

yC1cos3x

sin3x

4xcosxsinx(C1,11

）y9yxcosx的通解。 y'3Csin3x y"9Ccos3x

cos3xsin3x

5cosx4xsinx,9sinx4xcosx, 所以yC1cos3x

sin3x

4xcosxsinx(C1,11

）y9yxcosx （3）yCx2Cx2lnx(CC是任意常數(shù)）x2y"3xy'4y y'2CCx2Cxlnx,y"2C3C

yCx2Cx2lnx(CC是任意常數(shù)）x2y"3xy'4y

2

2x

9lnx(C1,C2是任意常數(shù)x

3xy5y

lnx y'5Cx4C22xlnxx,y"20Cx32C22lnx1

2 y

2xxx

9lnx(C1,C2是任意常數(shù)）x

3xy5y

lnx1x（5）y1x

x2

(C1,C2是任意常數(shù)）是方程

"2y'xy

y'

CexC CeC 1x 1x y"

CexC 2CexC CeC 1x 1xx x1xy1x

x2

(C1,C2是任意常數(shù)）是方程

"2y'xy

（6）yCexCexCcosxCsinxx2(C,C,C

） y4yx2的通解。y'CexCexCsinx

cosx y"CexCexCcosx

sinx y"'CexCexCsinx

y4CexCexCcosx

sin 所以yCexCexCcosxCsinxx2C,C,C

） y4yx21yxex12x1y"2x1y'2y

yexu，則

y'exuexu',y"exu2exu'

2x1u2x3u'0。令u'pdp32xdxpC2x1ex 2x積分得uC2x1exCyC2x1C 1yxx是齊次線性方程x2y2xy'2y0的一個解，求非齊次線性方程1x2y"2xy'2y2x3 yxuy'uxuy"2u'xu，代入原方程得u"2ux2CxC yx3Cx2C yy0的通解為YxCcosx

yysecxyC1xcosxC2xsinx y'CxcosxCxsinxCxsin

xcosx令CxcosxCxsinx0y'CxsinxCxcosx y"C'xsinxC'xcosxCxcos 代入原方程得CxsinxCxcosxsecx

xsinC'xcosxC'xsinx C'xtan C'xsinxC'xcosxsec Cx

C1xlncosxC1,C2xx

yxsinxcosxlncosxC1cosxC2sin x2yxy'y0的通解為YxCxCxlnx x2y"xy'yx的通解。yC1xxC2xxlnx y'CxxC'xxlnxCx

x1lnx令CxxCxxlnx0y'Cx

x1lnx 2y"C'xC'x1lnx 2

xx 代入原方程得C'xxC'xx1lnx