Standard Form: \begin{equation} \frac{dy}{dx}=f(x,y) \label{standard} \end{equation}

Differential Form: Write $$f(x,y)=-\frac{M(x,y)}{N(x,y)}$$ (There are many ways to do this. Choose a way that is helpful for the problem at hand.) Then Eqn(\ref{standard}) becomes $$M(x,y)\, dx + N(x,y)\, dy =0$$

If $f$ and $\frac{\partial f}{\partial y}$ are continuous in a rectangle $\vert x-x_0\vert \le a$, $\vert y-y_0\vert\le b$, then there exists an interval about $x_0$ in which the initial value problem $y^{\prime}=f(x,y)$, $y(x_0)=y_0$ has a unique solution.