Find the ODE of the following

Question

                is not an exact ODE.

Select a function from below options so that if both sides of the equation is multiplied by the function, then the new ODE is exact.

a.) y^(-2)

b.) x

c.) y^(-1)

d.) x/y

Answer 1

This function is y^-1 = 1/y. If we multiplying equation by 1/y we get equation ydx + (x + 1/y)dy = 0 and this

equation is exact because ∂/∂y(y) = ∂/∂x(x + 1/y) = 1.