(完整版)常微分方程基本概念习题及解答.doc

§=2xy,并满足初始条件:x=0,y=1的特解。dx解:dy=2xdx 两边积分有:ln|y|=x2+cyy=ex2+ec=cex2 另外y=0也是原方程的解,c=0时,y=0原方程的通解为y=cex2,x=0y=1时c=1特解为y=+(x+1)dy=0 并求满足初始条件:x=0,y=1的特解。解:y2dx=-(x+1)dy dyy2dy=-1 dxx 1两边积分:-1=-ln|x+1|+ln|c| y=y ln1|c(x1)|另外y=0,x=-1也是原方程的解 x=0,y=1时c=e特解:y=ln1|c(x1)|dy= 1dx xyy2x3ydy 1 y2 1解:原方程为: =dx yx x31 y2dy= 1 dx3y x x两边积分:x(1+x2)(1+y2)=cx24.(1+x)ydx+(1-y)xdy=0解:原方程为:1 ydy=-yx 1dxx两边积分: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:原方程为:dy=-x ydx x y令y=u 则dy=u+xdu 代入有:x dx dx-u 1du=1dxu2 1 xln(u2+1)x2=c-2arctgu即ln(y2+x2)=c-2arctg -y+dxx2 y2=0解:原方程为:dy=dxy+|xx|- 1x(y)2x则令y=udy=u+xdux11 u2ydxdu=sgnxdx1dxxarcsinx=sgnxln|x|+ctgydx-ctgxdy=0解:原方程为:dy=tgydxctgx两边积分:ln|siny|=-ln|cosx|-ln|c|siny=1 =osx另外y=0也是原方程的解,而 c=0时,y= sinycosx= dydxey23x+ =0y3xdy ey2解:原方程为:= edx y3x2e -3ey2=(lnx-lny)dy-ydx=0dy解:原方程为:dx=ylnyx x令y=u,则dy=u+xdux dx dxu+xdu=ulnudxln(lnu-1)=-ln|cx|10.1+lndy=exdxy=:原方程为:ey=cexdy=exeydx11 dy=(x+y)2dx解:令x+y=u,则du-1=u2dxdy=dxdu-1dx11 u2du=dxarctgu=x+carctg(x+y)=x+c12.dy=dx (x1y)2解:令x+y=u,则dy=dxdu-1dxdu-1= 1dx u2u-arctgu=x+cy-arctg(x+y)=.dy=2x y 1dx x 2y 1解:原方程为:(x-2y+1)dy=(2x-y+1)dxxdy+ydx-(2y-1)dy-(2x+1)dx=0dxy-d(y2-y)-dx2+x=cxy-y2+y-x2-x=c14:dy=x y 5dx x y 2解:原方程为:(x-y-2)dy=(x-y+5)dxxdy+ydx-(y+2)dy-(x+5)dx=0dxy-d(1y2+2y)-d(21x2+5x)=0215:y2+4y+x2+10x-2xy==(