Find a potential function for the vector field f(x, y)=2x/y i+(1-x^2)/y^2 j.

Potential function U(x,y) defines the vector field f(x,y) =(2x/y) i + [(1-x^{2})/y^{ 2}]j. The "potentiality" (curl = 0) of this vector field can be noticed from the fact that

∂U_{ y}/∂x = ∂U
_{x}/∂y = -2x/y^{2}

where

U_{y} = ∂U/∂y = (1-x^{2})/y^{2} and U_{x} = ∂U/∂x = 2x/y

Potential function is defined as the following integral

U(x,y) = ∫U_{x}dx + ∫U_{y}dy + const

Take a look at the first integral'

∫U_{x}dx = 2 ∫xdx/y = x^{2}/y + p(y)

For the second one

∫U_{y}dy = (1-x^{2}) ∫dy/y^{2} = -(1-x^{2})/y + q(x) = (x^{2}-1)/y + q(x)

p(y) is an arbitrary function of y, and q(x) is an arbitrary function of x. To make both integrals reflecting the same function we can put p(y) = -1/y and q(x) = constant (c). Thus, we have

U(x,y) = (x^{2}-1)/y + constant