(x^2+2xy-y^2)dx+(y^2+2xy-x^2)dy=0 当x =1时 y =5 求特接

原式=d{[(x+y)(x-y)+2xy]x+[(x+y)(y-x)+2xy]
=d{[(x+y)(x-y)+2xy]x-[(x+y)(x-y)-2xy]}
=d*4xy
当x =1,y =5时
20d=0
d=0