If #u= (1-2xy+y^2)^-1/2# , then show that #x(partial u)/(partial x)-y(partial u)/(partial y)= y^2u^3# ?

1 Answer
Steve M · Shehroz B.
Apr 5, 2018

# x(partial u)/(partial x) - y (partial u)/(partial y) = 4y^2u^2 != y^2u^3#

Indicating an error in the PDE of the question.

Explanation:

We have:

# u = ((1-2xy+y^2)^(-1))/2 = 1/(2(1-2xy+y^2)) #

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

# x(partial u)/(partial x) - y (partial u)/(partial y) = y^2u^3#

(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) #

# \ \ \ \= ((-1)(1-2xy+y^2)^(-2)(-2y))/2 #

# \ \ \ \= (y)/(1-2xy+y^2)^(2) #

And

# u_y = (partial u)/(partial y) #

# \ \ \ \= ((-1)(1-2xy+y^2)^(-2)(-2x+2y))/2 #

# \ \ \ \= (x-y)/(1-2xy+y^2)^(2) #

Next we compute the LHS of the desired expression:

# LHS = x(partial u)/(partial x) - y (partial u)/(partial y) #

# \ \ \ \ \ \ \ \ = x((y)/(1-2xy+y^2)^(2)) - y ((x-y)/(1-2xy+y^2)^(2)) #

# \ \ \ \ \ \ \ \ = (xy-yx+y^2)/(1-2xy+y^2)^(2) #

# \ \ \ \ \ \ \ \ = (y^2)/(1-2xy+y^2)^(2) #

Using # u = ((1-2xy+y^2)^(-1))/2 => 1/((1-2xy+y^2)) =2u#

So that

# LHS = (y^2) * (2u)^2 #
# \ \ \ \ \ \ \ \ = 4y^2u^2 #
# \ \ \ \ \ \ \ \ != RHS #

Indicating an error in the PDE of the question.

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