常微分方程练习题及答案(复习题)

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常微分方程练习试卷

一、

23210dxxdt

()xdyfxyydx_______

3230dyyxdx(0)1,(0)2yy

xyyye*2()xxxyxeexe

()0Wt12(),(),,()nxtxtxtLaxb

22(2320)0xydxxydyy

()XAtX()t()At

20'05xx

251yyyy

20yyy

二、

13dyxydxxy

222()0dxdxxdtdt

sinyyx 22(cossin)(1)0xxxydxyxdy

3124A11XAdtdX)(tXAdtdX)0(x

2213dyxydx(1,0)

xAx(),t12(0),

(),()tt()XAtXC()()ttC

),()(0xxx

],[,,])([)(0200xxdyyxyxx

)}({xn],[)(x],[],[)()(xx

)(tAXdtdX)(0t)(exp)(0ttAt

uxy11(()1)dudxufux3,2,1 2114A32()480dydyxyydxdx3y1()()tt25 00tAtteee

13dyxydxxy

10,30xyxy1,2xy1,2,xy

.ddz2(1)1zdzdz21arctanln(1)ln||2zzC

222arctanln(1)(2)1yxyCx

222()0dxdxxdtdt

sinyyx

yyxyce()xycxe

()()()sinxxxcxecxecxex()sinxcxex

1()(sincos)2xcxexxc1(sincos)2xycexx

22(cossin)(1)0xxxydxyxdy

22(,)cossin,(,)(1)MxyxxxyNxyyx2MNxyyx

22cossin()0xxdxxydxyxdyydy

2222111(sin)()()0222dxdxydy2222sinxxyyC

3124A11XAdtdX)(tXAdtdX)0(x

31det()(2)(5)0,24AE

122,5122,51211,,(,0).12VV

2525().2tttteetee1211(0)113

)0()()(1tt2525211111132tttteeee25252134tttteeee

2213dyxydx(1,0) 0()0x

221001()[213()],xxyxxdxxx

223452011133()[213()],1025xxyxxdxxxxxx

3284dyydxxdyydxdypdx3284pyxyp

322322(4)(8)4dpypypypypdy

32(4)(2)0dppyypdy20dpypdy12pcy2()pyc

2224cpxc

22224()cpxcpyc

3240py123(4)py3427yx

xAx(),t12(0),

221()69014p1,2312n

12v111123322120()()(3)()!ititittteAEeti

10()!intiiteAEi

33310111exp(3)01111tttttAteEtAEetett 32()480dydyxyydxdx2114A

(),()tt()XAtXC()()ttC

()t1()t1()()()Xttt

()Xtdet()0Xt()()()ttXt

()()()()()ttXttXt()()()()()AttXttXt()()()()AtttXt()()()tAtt()()0tXt

()0,Xt()XtC()()ttC

),()(0xxx

],[,,])([)(0200xxdyyxyxx

)}({xn],[)(x],[],[)()(xx

xxdyx0,])([)(20

,)(00yxxxnnxxdyx0],[,,])([)(0120),2,1(n

xx00xx

),()|||)(|(|)()(|0200xxMdxxxx|}||)(|{max2],[xxxMx

00221000|()()|(|()()|)()(),2!xxxxMLxxdLMxdxx

}{max2],[xLxnnnnxxnMLxx)(!|)()(|011

021xnnx|(x)(x)|(|()()|)d,)(!)1()(!10010nxxnnnxxnMLdxnMLL

k

1110|()()|()()!!kkkkkMLMLxxxxkk

k0

)}({xnxx0)(x )()(xxxx0

)(tAXdtdX)(0t)(exp)(0ttAt

Attexp)(AXdtdX)(tC(t)expAtC

0ttCAt0exp10)(expAtC

1000(t)expAt(expAt)expAtexp(At)expA(tt)