令x+y=u,则
dx+dy=du,代入换掉y,得
du/dx=tanu+1,分离变量,得
cosudu/(sinu+1)=dx,两边同时积分,得
ln(sinu+1)=x+lnc
所以
通解为
ln[sin(x+y)+1]=x+lnc化得
sin(x+y)+1=c*e^x
令x+y=u,则
dx+dy=du,代入换掉y,得
du/dx=tanu+1,分离变量,得
cosudu/(sinu+1)=dx,两边同时积分,得
ln(sinu+1)=x+lnc
所以
通解为
ln[sin(x+y)+1]=x+lnc化得
sin(x+y)+1=c*e^x