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-x2)/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/y2
Uy = ∂U/∂y = (1-x2)/y2 and Ux = ∂U/∂x = 2x/y
Potential function is defined as the following integral
U(x,y) = ∫Uxdx + ∫Uydy + const
Take a look at the first integral'
∫Uxdx = 2 ∫xdx/y = x2/y + p(y)
For the second one
∫Uydy = (1-x2) ∫dy/y2 = -(1-x2)/y + q(x) = (x2-1)/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) = (x2-1)/y + constant