Question

If `u= y/x+ z/x + x/y`, show that `x(partial u)/(partial x)+y(partial u)/(partial y)+z(partial u)/(partial z)=0` ?

Answer 1

We have:

`u = y/x+z/x+x/y`

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

`x(partial u)/(partial x) + y (partial u)/(partial y) + z (partial u)/(partial 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 chain rule:

`u_x = (partial u)/(partial x) =-y/x^2-z/x^2 + 1/y`

`u_y = (partial u)/(partial y) = 1/x-x/y^2`

`u_z = (partial u)/(partial z) = 1/x`

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)`

`\ \ \ \ \ \ \ \ = 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