Question

高数第七题

Answer 1

∂z/∂x=∂z/∂u+∂z/∂v
∂z/∂y=a *∂z/∂v-2∂z/∂u
于是得到∂^2z/∂x^2=∂(∂z/∂u+∂z/∂v)/∂x
=∂^2z/∂u^2+∂^2z/∂u∂v+∂^2z/∂v∂u+∂^2z/∂v^2
=∂^2z/∂u^2+2∂^2z/∂u∂v+∂^2z/∂v^2；
而∂^2z/∂x∂y=∂(∂z/∂u+∂z/∂v)/∂y
= -2 ∂^2z/∂u^2 +∂^2z/∂u∂v*a+∂^2z/∂v∂u*(-2)+∂^2z/∂v^2 *a
= -2∂^2z/∂u^2+(a-2)∂^2z/∂u∂v+a∂^2z/∂v^2；
∂^2z/∂y^2=∂(a *∂z/∂v-2∂z/∂u)/∂y
= -2a∂^2z/∂v∂u+a^2 *∂^2z/∂v^2- 2∂^2z/∂u^2 *(-2)-2a∂^2z/∂u∂v
=a^2∂^2z/∂v^2-4a∂^2z/∂v∂u+4∂^2z/∂u^2
所以代入得到
6*(∂^2z/∂u^2+2∂^2z/∂u∂v+∂^2z/∂v^2)
+[-2∂^2z/∂u^2+(a-2)∂^2z/∂u∂v+a∂^2z/∂v^2]
-(a^2∂^2z/∂v^2-4a∂^2z/∂v∂u+4∂^2z/∂u^2)=0
即(12+a-2+4a)∂^2z/∂u∂v+(6+a-a^2)∂^2z/d∂^2=0
能化简为∂^2z/∂v^2=0
于是 6+a-a^2=0且12+a-2+4a不等于0即a不等于 -2，
解得 a=3