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# ?

1 Answer
Steve M
Apr 5, 2018

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

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