dy/dx=(x+y+1)/(x-y+1)的通解答案是e的arctg[y/(x+1)]次幂=c[(x+1)
问题描述：
dy/dx=(x+y+1)/(x-y+1)的通解答案是e的arctg[y/(x+1)]次幂=c[(x+1)
dy/dx=(x+y+1)/(x-y+1)的通解
答案是e的arctg[y/(x+1)]次幂=c[(x+1)
答
e的arctg[y/(x+1)]次幂=c[(x+1)
就是通解怎么得到e的arctg[y/(x+1)]次幂=c[(x+1)。。过程阿 大哥设X=x+1,Y=y代入：dY/dX=(X+Y)/(X-Y)=(1+Y/X)/(1-Y/X)设Y/X=U,Y=XUY'=U+XU'代入：U+XU'=（1+U)/(1-U)XU'=（1+U)/(1-U)-U=(1+U^2)/(1-U)(1-U)dU/(1+U^2)=dX/X积分得：arctanU-(1/2)ln(1+U^2)=lnX+lnC或：e^(arctanU)=(1/2)ln(1+U^2)(CX)通e^(arctany/(1+x))=(C/2)ln[(1+y^2/(1+x)^2)](1+x)亲哥哥，答案不一样阿答案是e的arctg[y/(x+1)]次幂=c[(x+1)呵呵，如果我没错，那就答案错