求微分方程的特解(1+e^x)yy'=e^y,x=0时y'=0;
条件好像有错,∵若x=0时y'=0,代入原式则有e^y=0,这时y=-∞,那么特解不存在;故下面只
求通解,不求特解.要求特解,只需把条件代入求出C来就行了.)
(1+e^x)y(dy/dx)=e^y
分离变量得(y/e^y)dy=dx/(1+e^x)
积分之:-∫yd[e^(-y)]=∫{1-[(e^x)/(1+e^x)]}dx
于是得-[ye^(-y)-∫e^(-y)dy]=x-ln(1+e^x)+C
即有-[(y/e^y)+∫e^(-y)d(-y)]=x-ln(1+e^x)+C
-[(y/e^y)+(1/e^y)]=x-ln(1+e^x)+C
-(y+1)/e^y=x-ln(1+e^x)+C
这就是原方程的通解.