Chapter 1 - First-Order Differential Equations - 1.9 Exact Differential Equations - True-False Review - Page 91: g

True

Let $M=\frac{-2xy}{(x^2+y)^2}$ and $N=\frac{x^2}{(x^2+y)^2}$ Obtain: $\frac{\partial M}{\partial y}=\frac{\partial}{\partial y}(\frac{-2xy}{(x^2+y)^2})=\frac{-2x(x^2-y)}{(x^2+y)^3}\\ \frac{\partial N}{\partial y}=\frac{\partial}{\partial y}(\frac{x^2}{(x^2+y)^2})=\frac{-2x(x^2-y)}{(x^2+y)^3}$ The potential function $\Phi(x,y)=\frac{y}{x^2+y}$ Hence, the statement is true.