常微分方程課后答案(第三版)

1、習(xí)題1.21=2xy,并滿足初始條件：x=0,y=1的特解。解：=2xdx 兩邊積分有：ln|y|=x+cy=e+e=cex另外y=0也是原方程的解，c=0時，y=0原方程的通解為y= cex,x=0 y=1時 c=1特解為y= e.2. ydx+(x+1)dy=0 并求滿足初始條件：x=0,y=1的特解。 解：ydx=-(x+1)dy dy=-dx兩邊積分: -=-ln|x+1|+ln|c| y=另外y=0,x=-1也是原方程的解 x=0,y=1時 c=e特解：y=3= 解：原方程為：=dy=dx 兩邊積分：x(1+x)(1+y)=cx4. (1+x)ydx+(1-y)xdy=0 解：原方程

2、為： dy=-dx兩邊積分：ln|xy|+x-y=c另外 x=0,y=0也是原方程的解。5（y+x）dy+(x-y)dx=0 解：原方程為： =-令=u 則=u+x 代入有：-du=dxln(u+1)x=c-2arctgu即 ln(y+x)=c-2arctg.6. x-y+=0 解：原方程為： =+-則令=u =u+ x du=sgnx dxarcsin=sgnx ln|x|+c7. tgydx-ctgxdy=0 解:原方程為：=兩邊積分：ln|siny|=-ln|cosx|-ln|c|siny= 另外y=0也是原方程的解，而c=0時，y=0.所以原方程的通解為sinycosx=c.8 +=0

3、 解：原方程為：=e2 e-3e=c.9.x(lnx-lny)dy-ydx=0 解：原方程為：=ln令=u ,則=u+ xu+ x=ulnuln(lnu-1)=-ln|cx|1+ln=cy.10. =e 解：原方程為：=eee=ce11 =(x+y) 解：令x+y=u,則=-1-1=udu=dxarctgu=x+carctg(x+y)=x+c12. =解：令x+y=u,則=-1 -1= u-arctgu=x+c y-arctg(x+y)=c.13. =解: 原方程為：（x-2y+1）dy=(2x-y+1)dx xdy+ydx-(2y-1)dy-(2x+1)dx=0 dxy-d(y-y)-dx+

4、x=c xy-y+y-x-x=c14: =解：原方程為：（x-y-2）dy=(x-y+5)dx xdy+ydx-(y+2)dy-(x+5)dx=0 dxy-d(y+2y)-d(x+5x)=0 y+4y+x+10x-2xy=c.15: =(x+1) +(4y+1) +8xy 解：原方程為：=（x+4y）+3令x+4y=u 則=-=u+3=4 u+13u=tg(6x+c)-1tg(6x+c)=(x+4y+1).16:證明方程=f(xy),經(jīng)變換xy=u可化為變量分離方程，并由此求下列方程：1） y(1+xy)dx=xdy2） = 證明： 令xy=u,則x+y= 則=-，有： =f(u)+1 du=

5、dx 所以原方程可化為變量分離方程。1） 令xy=u 則=- (1)原方程可化為：=1+（xy） (2)將1代入2式有：-=(1+u)u=+cx17.求一曲線，使它的切線坐標(biāo)軸間的部分初切點(diǎn)分成相等的部分。解：設(shè)（x +y ）為所求曲線上任意一點(diǎn)，則切線方程為：y=y(x- x )+ y 則與x軸，y軸交點(diǎn)分別為： x= x - y= y - x y 則 x=2 x = x - 所以 xy=c18.求曲線上任意一點(diǎn)切線與該點(diǎn)的向徑夾角為0的曲線方程，其中 = 。解：由題意得：y= dy= dx ln|y|=ln|xc| y=cx. = 則y=tgx 所以 c=1 y=x.19.證明曲線上的切線

6、的斜率與切點(diǎn)的橫坐標(biāo)成正比的曲線是拋物線。 證明：設(shè)(x,y)為所求曲線上的任意一點(diǎn)，則y=kx 則：y=kx +c 即為所求。 acknowledgements my deepest gratitude goes first and foremost to professor aaa , my supervisor, for her constant encouragement and guidance. she has walked me through all the stages of the writing of this thesis. without her consistent

7、 and illuminating instruction, this thesis could not havereached its present form. second, i would like to express my heartfelt gratitude to professor aaa, who led me into the world of translation. i am also greatly indebted to the professors and teachers at the department of english: professor dddd

8、, professor ssss, who have instructed and helped me a lot in the past two years. last my thanks would go to my beloved family for their loving considerations and great confidence in me all through these years. i also owe my sincere gratitude to my friends and my fellow classmates who gave me their h

9、elp and time in listening to me and helping me work out my problems during the difficult course of the thesis. my deepest gratitude goes first and foremost to professor aaa , my supervisor, for her constant encouragement and guidance. she has walked me through all the stages of the writing of this t

10、hesis. without her consistent and illuminating instruction, this thesis could not havereached its present form. second, i would like to express my heartfelt gratitude to professor aaa, who led me into the world of translation. i am also greatly indebted to the professors and teachers at the department of english: professor dddd, professor ssss, who have instructed and helped me a lot in the past two years. last my thanks would go to my beloved family for their loving considerations and great confidence in me all through these years. i also owe my sincere gratitu