Question

If f=x^3+y^3+z^3+3xy, then show that `x(partial f)/(partial x) + y (partial f)/(partial y) + z (partial f)/(partial z) = 3f`?

Answer 1

`f=x^3+y^3+z^3+3xyz`

`color(red)(delf/dx )= 3x^2 + 3yz`

`color(blue)(delf/dy) = 3y^2 +3xz`

`color(magenta)(delf/dz) = 3z^2 +3xy`

`color(white)(dd`

We need to show that `x color(red)(delf/dx)+ ycolor(blue)(delf/dy) + zcolor(magenta)(delf/dz)= 3f`

`LHS= x[color(red)(3x^2 + 3yz)] +y[color(blue)(3y^2 +3xz)] +z[color(magenta)(3z^2 +3xy)]`

`color(white)(dd`

`=> [3x^3 + 3xyz] + [3y^3 + 3xyz] + [3z^3 +3xyz]`

`=> 3x^3 +3y^3 +3z^3 +9xyz`

`=> 3(x^3 +y^3 +z^3 +3xyz) = 3f`

Answer 2

The function:

`f = x^3 + y^3 + z^3 + 3xy`

as given does not satisfy the Partial Differential Equation:

`x(partial f)/(partial x) + y (partial f)/(partial y) + z (partial f)/(partial z) = 3f`

and the question is in error.

If however, we modify the function then we can prove a modified result. Consider the modified function:

`f = x^3 + y^3 + z^3 + 3xyz`

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

`x(partial f)/(partial x) + y (partial f)/(partial y) + z (partial f)/(partial z) = 3f`

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

`f_x = (partial f)/(partial x) = 3x^2+3yz`

`f_y = (partial f)/(partial y) = 3y^2+3xz`

`f_z = (partial f)/(partial z) = 3z^2 + 3xy`

Next we compute the LHS of the desired expression:

`LHS = x(partial f)/(partial x) + y (partial f)/(partial y) + z (partial f)/(partial z)`

`\ \ \ \ \ \ \ \ = x(3x^2+3yz ) + y(3y^2+3xz) + z(3z^2+3xy)`

`\ \ \ \ \ \ \ \ = 3x^3+3xyz + 3y^3+3xyz + 3z^3+3xyz`

`\ \ \ \ \ \ \ \ = 3(x^3+xyz + y^3+xyz + z^3+xyz)`

`\ \ \ \ \ \ \ \ = 3(x^3+ y^3 + z^3+3xyz)`

`\ \ \ \ \ \ \ \ = 3f \ \ \` QED