Question

（x²＋2xy－y²）dx＋（y²＋2xy-x²）dy=0 ， y|x=1=1特解

Answer 1

dy/dx=(y²-2xy-x²)/(y²+2xy-x²)
令u=y/x, y'=u'x+u
u'x+u=(u²-2u-1)/(u²+2u-1)
后面的自己做吧

Answer 2

dy/dx＝(u²-2u-1)/(u²+2u－1)＝xdu/dx+u有xdy/dx＝()/()－u＝(u²-2u-1－u(u²＋2u－1))/(u²+2u－1)＝－u²-u³-1-u/(u²+2u-1)＝-(u²+u³+1+u)/(u²+2u－1)＝－(u+1)(u²+1)/(u²+2u－1)分离变量，有(1-2u-u²)du/(u+1)(u²+1)＝dx/x又1/((u+1)(u²+1))＝1/u+1-2u/u²+1代入积分∫du/u+1－∫2udu/u²+1＝∫dx/x可得ln｜u+1｜-ln｜u²+1｜＝ln｜cx｜