Question

If u = y/z + z /x + x /y , then prove that x ∂u /∂x + y ∂u/ ∂y + z ∂u/ ∂z = 0

Answer 1

Answer:

We have:

u=yx+zx+xy

and we seek to validate that f satisfies the Partial differential Equation:

x∂u∂x+y∂u∂y+z∂u∂z

(In other words we are validating that a solution to the given PDE is u). We compute the partial derivative (by differentiating wrt to specified variable and treating all other variables as constants), and applying the chaiin rule:

ux=∂u∂x=−yx2−zx2+1y

uy=∂u∂y=1x−xy2

uz=∂u∂z=1x

Next we compute the LHS of the desired expression:

LHS=x∂u∂x+y∂u∂y+z∂u∂z

        =x(y(1−2x

Answer 2

Step-by-step explanation:

∂u /∂x =$(1/y,-z/x^2)$

∂u/ ∂y=$(1/z,-x/y^2)$

∂u/ ∂z=$(1/x,-y/z^2)$

x ∂u /∂x=(x/y,-z/x)

y ∂u/ ∂y =(y/z,-x/y)

z ∂u/ ∂z=(z/x,-y/z)

Answer:$x/y-z/x+y/z-x/y+z/x-y/z=0$