求由方程2y-x=(x-y)ln(x-y)所确定的函数y=y(x)的微分dy
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解决时间 2021-03-15 21:05
- 提问者网友:自食苦果
- 2021-03-15 01:10
求由方程2y-x=(x-y)ln(x-y)所确定的函数y=y(x)的微分dy.
最佳答案
- 五星知识达人网友:平生事
- 2021-03-15 01:34
2y-x=(x-y)ln(x-y)
2dy?dx=(dx?dy)ln(x?y)+(x?y)×
1
x?y (dx?dy)
2dy-dx=(dx-dy)ln(x-y)+dx-dy
[3+ln(x-y)]dy=[2+ln(x-y)]dx
(x-y)[3+ln(x-y)]dy=(x-y)[2+ln(x-y)]dx
[3(x-y)+(x-y)ln(x-y)]dy=[2(x-y)+(x-y)ln(x-y)]dx
因为2y-x=(x-y)ln(x-y),
所以,[3(x-y)+(2y-x)]dy=[2(x-y)+(2y-x)]dx
(2x-y)dy=xdx
①若2x-y=0,则dy=2dx
②若2x?y≠0,则dy=
x
2x?y dx;
2dy?dx=(dx?dy)ln(x?y)+(x?y)×
1
x?y (dx?dy)
2dy-dx=(dx-dy)ln(x-y)+dx-dy
[3+ln(x-y)]dy=[2+ln(x-y)]dx
(x-y)[3+ln(x-y)]dy=(x-y)[2+ln(x-y)]dx
[3(x-y)+(x-y)ln(x-y)]dy=[2(x-y)+(x-y)ln(x-y)]dx
因为2y-x=(x-y)ln(x-y),
所以,[3(x-y)+(2y-x)]dy=[2(x-y)+(2y-x)]dx
(2x-y)dy=xdx
①若2x-y=0,则dy=2dx
②若2x?y≠0,则dy=
x
2x?y dx;
全部回答
- 1楼网友:荒野風
- 2021-03-15 02:30
求方程2y-x=(x-y)ln(x-y)所确定的函数y=y(x)的微分dy…这题是让我们怎么求啊 最好能提供答案和过程啊,谢谢了
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