已知可微函数f(u,v)满足=2(u-v)e-(u+v),且f(u,0)=u2e-u. 记g(x,y)=f(x,y-x),求

admin2022-09-22  30

问题 已知可微函数f(u,v)满足=2(u-v)e-(u+v),且f(u,0)=u2e-u
记g(x,y)=f(x,y-x),求

选项

答案由g(x,y)=f(x,y-x)得[*]=f’1(x,y-x)-f’2(x,y-x),又[*]=2(u-v)e-(u+v),则2(x-y+x)e-(x+y-x)=2(2x-y)e-y,即[*]=2(2x-y)e-y

解析
转载请注明原文地址:https://kaotiyun.com/show/dPf4777K
0

最新回复(0)