The first order ordinary differential equation where one can separate the variables has the form where $\displaystyle\frac{dy}{dx}$ may be expressed as a product or a quotient of two functions, one of which depends only on $x$ and the other on $y$.  Such an equation may be written e.g. as

We notice first that if $Y(y)$ has real zeroes $y_{1},\,y_{2},\,\ldots$, then the equation (1) has the constant solutions  $y:=y_{1},\;y:=y_{2},\;\ldots$  and thus the lines  $y=y_{1},\;y=y_{2},\;\ldots$  are integral curves.  Similarly, if $X(x)$ has real zeroes  $x_{1},\,x_{2},\,\ldots$, one has to include the lines  $y=y_{1},\;y=y_{2},\;\ldots$  to the integral curves.  All those lines divide the $xy$-plane into the rectangular regions.  One can obtain other integral curves only inside such regions where the derivative $\displaystyle\frac{dy}{dx}$ attains real values.

Let $R$ be such a region, defined by

and let us assume that the $X(x)$ and $Y(y)$ are real, continuous and distinct from zero in $R$.  We will show that any integral curve of the differential equation (1) is accessible by two quadratures.

Let $\gamma$ be an integral curve passing through the point  $(x_{0},\,y_{0})$  of the region $R$.  By the above assumptions, the derivative $\displaystyle\frac{dy}{dx}$ maintains its sign on the curve $\gamma$ so long $\gamma$ is inside $R$, which is true on a neighbourhood $N$ of  $x_{0}$, contained in  $[a,\,b]$.  This implies that as $x$ runs the interval  $N$,  it defines the ordinate $y$ of $\gamma$ uniquely as a monotonic function  $y\mapsto y(x)$  which satisfies the equation (1):

The last equation may be written

Since $X$ and $Y$ don’t vanish in $R$, the denominators $Y(y(x))$ and $X(x)$ are distinct from 0 on the interval $N$.  Therefore one can integrate both sides of (2) from $x_{0}$ to an arbitrary value $x$ on $N$, getting

Because  $y=y(x)$  is continuous and monotonic on the interval $N$, it can be taken as new variable of integration in the left hand side of (3):  substitute  $y(x):=y$,  $y^{{\prime}}(x)\,dx:=dy$  and change the limits to  $y(x_{0})=y_{0}$  and  $y(x)=y$.

• Accordingly, the equality

  $\displaystyle\int_{{y_{0}}}^{y}\frac{dy}{Y(y)}\;=\;\int_{{x_{0}}}^{x}\frac{dx}{X(x)}$

  (4)

  is valid, meaning that if an integral curve of (1) passes through the point  $(x_{0},\,y_{0})$, the integral curve is represented by the equation (4) as long as the curve is inside the region $R$.
  

• Additionally, it is possible to justificate that if  $(x_{0},\,y_{0})$  is an interior point of a region $R$ where $X(x)$ and $Y(y)$ are real, continuous and $\neq 0$, then one and only one integral curve of (1) passes through this point, the curve is regular, and both $x$ and $y$ are monotonic on it.  N.B., the Lipschitz condition for the right hand side of (1) is not necessary for the justification.

• When the point  $(x_{0},\,y_{0})$  changes in the region $R$, (4) gives a family of integral curves which cover the region once.  The equations of these curves may be unified to the form

  $\displaystyle\int\frac{dy}{Y(y)}\;=\;\int\frac{dx}{X(x)},$

  (5)

  which thus represents the general solution of the differential equation (1) in $R$.  Hence one can speak of the separation of variables,

  $\displaystyle\frac{dy}{Y(y)}\;=\;\frac{dx}{X(x)},$

  (6)

  and integration of both sides.