# Integrating Factor Counterexample

It has been my understanding that if there exist for an ODE in the form $$M(x,\ y)\ dx + N(x,\ y)\ dy = 0$$ integrating factors that depend only on $$x$$, then the formula $$u(x) = e^{-\int \frac{2D=curl\begin{bmatrix}M \\ N\end{bmatrix}}{N}\ dx}$$ will find at least some of them. Likewise, if there exist integrating factors that depend only on $$y$$, then the formula $$u(y) = e^{\int \frac{2D=curl\begin{bmatrix}M \\ N\end{bmatrix}}{M}\ dy}$$ should find at least some of them. For example, just before introducing the above formulas, my book notes

$$\frac{\partial}{\partial y}(uM) = \frac{\partial}{\partial x}(uN)$$ This is a partial differential equation for $$u$$. There is no procedure for solving this equation in general. However, sometimes we can make assumptions about $$u$$ that make this equation simpler...If there is an integrating factor that depends on only one variable, is it much easier to find.

variable names changed for consistency

However, I appear to have come across a counterexample; $$(\frac{2y + 8}{x})\ dx + (2 + \frac{7}{x})\ dy = 0$$ clearly has at least the integrating factor $$u(x) = x$$, but the above $$u(x)$$ formula leads to a mixed integral obstacle.

Is the above a correct principle? Is the mistake in my theory or calculations?

Note that you need to find $$u(x) = e^{\int \frac {M_y-N_x}{N}} .$$
$$(\frac{2y + 8}{x})\ dx + (2 + \frac{7}{x})\ dy = 0$$
$$M_y=2/x , N_x=-7/{x^2}$$
$$u(x) =e^{\int \frac {M_y-N_x}{N}}= e^{\int \frac {1}{x}dx } = e^{\ln x}=x$$