Showing that the differential equation $M(x, y)dx + N(x, y)dy = 0$ has an integrating factor of $exp\int Q(x)dx$ if $\frac{∂M/∂y − ∂N/∂x}{N} = Q$ Show that if $\frac{∂M/∂y − ∂N/∂x}{N} = Q$, where $Q$ is a function of $x$ only, i.e. $Q =Q(x)$, then the differential equation
$$M(x, y)dx + N(x, y)dy = 0$$
has an integrating factor of the form
$$I(x)=exp\int Q(x)dx$$
I understand that our differential equation is exact. But I can't work out the identities of M and N.
Can someone please explain this to me?
 A: Let $M(x,y)dx+N(x,y)dy=0$ is a differential equation of first order, and $I(x,y)$ is an integrating factor for this equation. With multiplying by $I$ we suppose $IMdx+INdy=0$ be exact. So we have
\begin{eqnarray*}‎
‎\frac{\partial(IM)}{\partial y}                      &=& \frac{\partial(IN)}{\partial x}                      \\‎
‎\frac{\partial I}{\partial y}M+I\frac{\partial M}{\partial y}  &=& \frac{\partial I}{\partial x}N+I\frac{\partial N}{\partial x}  \\‎
‎I(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}) &=&\frac{\partial I}{\partial x}N-\frac{\partial I}{\partial y}M‎     ~‎~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)‎
‎\end{eqnarray*}‎
Let $I$ be a function of $x$ only, then $\dfrac{\partial I}{\partial y}=0$ and $(1)$ concludes that
$$‎I(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}) =\frac{\partial I}{\partial x}N$$ 
which gives
$$‎\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N} =\frac{\frac{\partial I}{\partial x}}{I}$$ 
If $\dfrac{∂M/∂y − ∂N/∂x}{N} = Q$, so $Q$ is a function of $x$ only, then from the last
$$Q(x)=\frac{1}{I}\frac{\partial I}{\partial x}$$
or
$$Q(x)dx=\frac{\partial I}{I}$$
by integration of two sides, 
$$\int Q(x)dx=\int\frac{\partial I}{I}=\ln I$$
which obtains $I$ and we say that $I$ is an integrating factor of the form
$$I(x)=exp\int Q(x)dx$$
