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#?
2 Answers
We need to show that
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
# 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_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