求解方程dx分之dy=x-y 3分之x y-1

令u=y^(1-3)=y^(-2)du=-2y^(-3)dydy/dx-y=x*y^3dy/(y^3)dx-y^(-2)=x-0.5du/dx-u=xdu/dx+2u=-2x(e^(2x)u)'=-2

dy/dx=x/yydy=xdxxdx=∫ydyx^2/2=y^2/2+C1故y^2-x^2=C即为所求C为任意实数dy/dx=y/xdy/y=dx/x两边积分ln|y|=ln|x|+C1ln|y/x

x^2*dy/dx=xy-y^2dy/dx=y/x-y^2/x^2u=y/xy=xuy'=u+xu'代入：u+xu'=u+u^2xu'=u^2du/u^2=dx/x-1/u=lnx+lnCCx=e^(

(dy/dx)=e^(x+y)(dy/dx)=e^x*e^y分离变量dy/e^y=e^xdx两边积分-e^（-y）=e^x+C1则-y=ln（C-e^x)整理得y=-ln（C-e^x)

设u=x+ydy=du-dx原式可化为du/dx-1=(a/u)^21/(1+(a/u)^2)*du=dx两边积分得∫1/(1+(a/u)^2)du=x+c∫u^2/(u^2+a^2)du=x+c∫(

原式化为dy/dx=1/2-x/2y令u=y/x,y=ux则：dy/dx=xdu/dx+u代回有xdu/dx+u=1/2-1/(2u)du/dx=(1/2-u-1/(2u))/xdu/(1/2-u-1

令y=xu则dy/dx=u+xdu/dx代入原方程得：u+xdu/dx=x^2u^2/(x^2u-x^2)u+xdu/dx=u^2/(u-1)xdu/dx=1/(u-1)(u-1)du=dx/x积分：

令x+y=u,则y=u-x.两边求导得：y'=u'-1(y'=dy/dx,u'=du/dx)带入原方程得：u'-1=1/u所以u'=1+1/u=(u+1)/u对u'=(u+1)/u=du/dx进行分离

(dy/dx)^2-2/x*(dy/dx)+4=0(dy/dx-1/x)^2=1/x^2-4dy/dx=1/x+根号(1-4x^2)/x或dy/dx=1/x-根号(1-4x^2)/x①dy/dx=1/

两天同乘以e^(∫P(x)dx)则左边变成[ye^(∫P(x)dx)]',右边是Q(x)e^(∫P(x)dx)所以ye^(∫P(x)dx)=∫Q(x)e^(∫P(x)dx)dx+Cy=e^(-∫P(x

∵dx/(x+t)=dy/(-y+t)=dt==>dx/(x+t)=dt,dy/(-y+t)=dt==>dx-xdt=tdt,dy+ydt=tdt==>e^(-t)dx-xe^(-t)dt=te^(-

x+y+1=u求导得：1+y'=u'代入dy/dx=(x+y)/(x+y+1)u'-1=1-1/uu'=2-1/u=(2u-1)/uudu/(2u-1)=dx2udu/(2u-1)=2dx(2u-1+

先用分离变量法解dy/dx+P(x)y=0,得到的解y=...中含有一个任意常数C.再在y=...中令C是C(x),把含有C(x)的y=...代入原方程,解出C(x)=...,再把C(x)=...代回

这种题,如果题目没错的话,真要很高的水平才能解

令z=1/x,则dx=-x²dz代入原方程得(x²y³+xy)dy=-x²dz==>dz/dy+y/x=-y³==>dz/dy+yz=-y³

令y/x=uy=uxy'=u+xu'代入原式得:u+u'x=1/u+ku'x=1/u+k-udu/(1/u+k-u)=dxudu/(1+ku-u^2)=dx(2u-k+k)du/(1+ku-u^2)=

两边对x求导注意y为x的函数3x^2+3(y+x*dy/dx)+3y^2*dy/dx=0从而dy/dx=-(x^2+y)/(x+y^2)有不明白的留言

(dy/dx)=x(1-x)dy=x(1-x)dxdy=(x-x^2)dx两边都对x积分dy对x积分是y(x-x^2)dx对x积分是x^2/2-2x^3/3+c所以y=x^2/2-2x^3/3+c高等