If u= y/x+ z/x + x/yu=yx+zx+xy, show that x(partial u)/(partial x)+y(partial u)/(partial y)+z(partial u)/(partial z)=0x∂u∂x+y∂u∂y+z∂u∂z=0 ?
1 Answer
We have:
u = y/x+z/x+x/y u=yx+zx+xy
and we seek to validate that
x(partial u)/(partial x) + y (partial u)/(partial y) + z (partial u)/(partial z)x∂u∂x+y∂u∂y+z∂u∂z
(In other words we are validating that a solution to the given PDE is
u_x = (partial u)/(partial x) =-y/x^2-z/x^2 + 1/y ux=∂u∂x=−yx2−zx2+1y
u_y = (partial u)/(partial y) = 1/x-x/y^2uy=∂u∂y=1x−xy2
u_z = (partial u)/(partial z) = 1/xuz=∂u∂z=1x
Next we compute the LHS of the desired expression:
LHS = x(partial u)/(partial x) + y (partial u)/(partial y) + z (partial u)/(partial z) LHS=x∂u∂x+y∂u∂y+z∂u∂z
\ \ \ \ \ \ \ \ = x((y)/(1-2xy+y^2)^(2)) - y ((x-y)/(1-2xy+y^2)^(2))
\ \ \ \ \ \ \ \ = x(-y/x^2-z/x^2 + 1/y) + y(1/x-x/y^2) + z(1/x)
\ \ \ \ \ \ \ \ = -y/x-z/x + x/y + y/x-x/y + z/x
Noting that all terms cancel, we then have the desired result:
LHS = 0
\ \ \ \ \ \ \ \ = RHS \ \ \ QED