[y+√（x²+y²）]dx=xdy满足y（1）=0的解
问题描述：
[y+√（x²+y²）]dx=xdy满足y（1）=0的解
答
[y+√(x²+y²)]dx=xdy满足y（1）=0的解 dy/dx=[y+√(x²+y²)]/x=y/x+√[1+(y/x)²].(1)令y/x=u,则y=ux,dy/dx=u+x(du/dx),代入（1）式得：u+x(du/dx)=u+√(1+u²),消去u即得：x(du/dx)=√(...