1个回答
展开全部
方程两边求全微分:
(1/xyz)(yzdx+xzdy+xydz)+2xydx+x^2dy=e^(yz)(zdy+ydz)
移项整理得 (ye^(yz)- (1/z))dz =((1/x)+2xy)dx +((1/y)+x^2- ze^(yz))dy,
所以,dz =[((1/x)+2xy)dx +((1/y)+x^2- ze^(yz))dy]/(ye^(yz)- (1/z))
=[(z+2x^2yz)/(xyze^(yz)-x)]dx
+[(z+x^2yz- yz2e^(yz))/(y^2ze^(yz)-y)]dy.
(1/xyz)(yzdx+xzdy+xydz)+2xydx+x^2dy=e^(yz)(zdy+ydz)
移项整理得 (ye^(yz)- (1/z))dz =((1/x)+2xy)dx +((1/y)+x^2- ze^(yz))dy,
所以,dz =[((1/x)+2xy)dx +((1/y)+x^2- ze^(yz))dy]/(ye^(yz)- (1/z))
=[(z+2x^2yz)/(xyze^(yz)-x)]dx
+[(z+x^2yz- yz2e^(yz))/(y^2ze^(yz)-y)]dy.
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询